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 Chemistry ‐ The Acidic Environment Chemistry Notes HSC Core Topic 2 The Acidic Environment 1. Indicators were first identified with the observation that the colour of some flowers depends on soil composition Recall: An acid is a substance which in solution produces hydrogen ions, H+ or more strictly H3O+, sometimes called the hydronium ion. A base is a substance which either contains the oxide O2‐ or hydroxide ion OH‐ or which in solution produces the hydroxide ion. A soluble base is called an alkali. Common acids are hydrochloric acid (HCl), sulfuric acid (H2SO4) and nitric acid (HNO3). Common bases are sodium hydroxide (NaOH), barium hydroxide (Ba(OH)2), potassium oxide (K2O), magnesium oxide (MgO), iron III oxide (Fe2O3), copper hydroxide (Cu(OH)2), and ammonia (NH3). Of these seven substances only the first three and ammonia are alkalis (soluble in water). Common properties of all acids are:  Acids have a sour taste  Acids sting or burn the skin  In solution, acids conduct electricity  Acids turn blue litmus (a vegetable dye) red Common properties of alkalis are:  Alkalis have a soapy feel  Alkalis have a bitter taste  In solution, alkalis are good conductors of electricity  Alkalies turn red litmus blue © (2012) All Rights Reserved 1 of 15 For more info, go to www.hscintheholidays.com.au  Classify common substances as acidic, basic or neutral All of the substances we use are acidic, basic or neutral, eg. vinegar, lemon juice, aspirin (acid); cleaning ammonia, sodium carbonate, oven cleaners (basic).  Identify that indicators such as litmus, phenolphthalein, methyl orange and bromothymol blue can be used to determine the acidic or basic nature of a material over a range, and that the range is identified by change in indicator colour  Identify data and choose resources to gather information about the colour changes of a range of indicators An indicator is a substance (usually a vegetable dye) which in solution changes colour depending on whether the solution is acidic or alkaline. Simple indicators can show the nature of substances over a range through colour change Indicator Highly acidic Slightly acidic Yellow Yellow Red Neutral Slightly alkaline colourless Highly alkaline red Methyl orange red Bromothymol blue blue Litmus blue Phenolphthalein  Identify and describe some everyday uses of indicators including the testing of soil acidity/basicity Indicators provide a cheap and convenient way of determining the acidity or alkalinity of substances. Some everyday uses of indicators are:  Testing the acidity or alkalinity of soils (because some plants need an acidic soil – azaleas and camellias – while others need an alkaline soil – most annual flowers and vegetables)  Testing home swimming pools (these need to be approximately neutral, though adding chemicals to sanitise the water can change its acid‐alkali balance)  Monitoring wastes from photographic processing (discharges to the sewerage system must be nearly neutral: photographic solutions are often highly alkaline).  Perform a first‐hand investigation to prepare and test a natural indicator Natural indicators such as the above can be created by taking a relevant plant, crushing it into a powder if needed, and adding water/methanol, etc to extract the dye. (To make indicator paper, soak the paper in the indicator.) 2. While we usually think of the air around us as neutral, the atmosphere naturally contains acidic oxides of carbon, nitrogen © (2012) All Rights Reserved 2 of 15 For more info, go to www.hscintheholidays.com.au and sulfur. The concentrations of these acidic oxides have been increasing since the Industrial Revolution 
Identify oxides of non‐metals which act as acids and describe the conditions under which they act as acids Oxides are a class of compound that often displays acidic or basic properties. An acidic oxide is one which either:  Reacts with water to form an acid or  Reacts with bases to form salts (or does both). Common acidic oxides are carbon dioxide, CO2, and diphosphorus pentoxide, P2O5, because CO2(g) + H2O(l) ‐ H2CO3(aq) (carbonic acid) Or CO2(g) + 2NaOH(aq) ‐ H2O(l) + 2Na+(aq) + CO32‐(aq) (sodium carbonate) P2O5(s) + 3H2O(l) ‐ 2H3PO4(aq) (phosphoric acid) Or P2O5(s) + 6NaOH(aq) ‐ 3H2O(l) + 6Na+(aq) + 2PO43‐(aq) (sodium phosphate) A basic oxide is one that:  Reacts with acids to form salts but  Does not react with alkali solutions (such as NaOH or KOH). Common basic oxides are copper oxide, CuO, and iron III oxide, Fe2O3, because CuO(s) + H2SO4(aq) ‐ CuSO4(aq) + H2O(l) Fe2O3 + 6HNO3 ‐ 2Fe(NO3)3(aq) + 3H2O(l)  Analyse the position of these non‐metals in the Periodic Table and generalise about the relationship between position of elements in the Periodic Table and acidity/basicity of oxides The acidity of the oxide of an element increases from left to right on the periodic table. Basic
Acidic
oxides
 Define Le Chatelier’s principle Chemical equilibrium is where a reaction does not go to completion, since the products are causing the back reaction at the rate of the forward reaction occurring. The relative amount of the products and reactants at equilibrium can be affected by shifting the equilibrium in order to favour one side. Le Chatelier’s principle states that if an equilibrium is disturbed, the system adjusts itself to minimise the disturbance. © (2012) All Rights Reserved 3 of 15 For more info, go to www.hscintheholidays.com.au  Identify factors which can affect the equilibrium in a reversible reaction Factors that can affect the equilibrium in a reversible reaction are changes in:  Concentrations of species involved in the reaction  Total pressure on a reaction involving gases  Temperature  Describe the solubility of carbon dioxide in water under various conditions as an equilibrium process and relate this to Le Chatelier’s principle Solubility of CO2 in water: CO2(g) + H2O(l) H2CO3(aq) H<0 Disturbance Effect on equilibrium (right means inc. solubility) Increased pressure Favours reaction with less moles of gas in product, reducing pressure, therefore shifts right, CO2 dissolves more. Increase in concentration of CO2 in System aims to reduce concentration of CO2 and thus system it reacts, ie, shifts equilibrium right Increase in temperature Favours endothermic reaction, ie back reaction, ie shifts left, and CO2 is less soluble  Identify natural and industrial sources of sulfur dioxide and oxides of nitrogen Natural and industrial sources of SO2 and nitrogen oxides  Sulfur dioxide (SO2) o Geothermal hot springs, volcanoes o Processing or burning fossil fuels (S(compounds) + O2(g) ‐‐> SO2(g)) o Extracting metals from sulfide ores (eg 2ZnS(g) + 3O2(g) ‐‐> 3ZnO(s) + 2SO2(g))  Oxides of nitrogen: nitrous oxide (N2O), nitric oxide (NO), nitrogen dioxide (NO2) o Lightning (O2(g) + N2(g) ‐‐> 2NO(g)) o Same in high temps of combustion chambers (cars, power stations) o Further reaction of 2NO(g) + O2(g) ‐‐> 2NO2(g) o N2O formed by bacteria (from nitrogenous fertiliser, causes greenhouse effect) o Mixture of NO and NO2 referred to as NOx  Describe, using equations, examples of chemical reactions which release sulfur dioxide and chemical reactions which release oxides of nitrogen These pollutant oxides can become dissolved in water droplets, within large volumes of air. SO2(g) + H2O(l) ‐‐> H2SO3(aq) 2H2SO3(aq) + O2(g) ‐‐> 2H2SO4(aq) (catalysed) 2NO2(g) + H2O(l) ‐‐> HNO2(aq) + HNO3(aq) 2HNO2(aq) + O2(g) ‐‐> 2HNO3(aq) (catalysed) © (2012) All Rights Reserved 4 of 15 For more info, go to www.hscintheholidays.com.au 
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Assess the evidence which indicates increases in atmospheric concentration of oxides of sulfur and nitrogen Choose resources, gather and analyse information from secondary sources to summarise the industrial origins of the above gases and evaluate reasons for concern about their release into the environment There has been an increase in industrial areas since the Industrial Revolution, in SO2 and NOx pollution. This led to regulations to control emissions from factories, power stations and motor cars. The annual average concentration of SO2 and NO2 in most large cities around the world is 0.01 ppm for each gas. This is about 10 times the value for clean air, though a concentration of 0.01 ppm for either gas is not harmful. Globally, because SO2 and NO2 are washed out of the atmosphere by rain, there appears not have been any significant build‐up of their concentrations over the last century or so (unlike CO2 which there has been a 30% increase and N2O increased as well). Sulfur dioxide irritates the respiratory system and causes breathing difficulties at concentrations as low at 1 ppm. People suffering from asthma and emphysema are particularly susceptible. The effects of sulfur dioxide are magnified if particulates are present also. Nitrogen dioxide irritates the respiratory tract and causes breathing discomfort at concentrations above about 3 to 5 ppm and at higher concentrations does extensive tissue damage. Concentrations above 3 ppm have rarely been reached even in heavily polluted cities. The main problem with NO2 is that it leads to the formation of ozone in what is called photochemical smog. This form of air pollution in which sunlight acts upon nitrogen dioxide in the presence of hydrocarbons and oxygen to form ozone and other pollutants called peroxyaclynitrates (PANS). Ozone has harmful effects at concentrations as low as 0.1 ppm. Hence releasing these oxides into the atmosphere at high concentrations can be harmful. 
Calculate volumes of gases given masses of some substances in reactions, and calculate masses of substances given gaseous volumes, in reactions involving gases at 0C and 101.3kPa or 25C and 101.3kPa At 0oC and 1atm pressure, molar volume of gases is 22.41 L. At 25oC and 1atm pressure, molar volume of gas is 24.47 L. See CCHSC for examples 
Explain the formation and effects of acid rain High SO2 and NOx emission cause acidic rain (as seen above): acid rain is rain with a higher H+ concentration than normal, > 10‐5 molL‐1. Regular rain unaffected by these pollutants contains some acidic carbonic acid (from CO2) and usually contains 10‐6 – 10‐5 molL‐1 H+. Acid rain causes:  Increasing lake acidity (affecting fish populations)  Damage to pine forests  Erosion of marble and limestone buildings/decorations (these Contain carbonates which react with acids)  Process and present information from secondary sources to describe the properties of sulfur dioxide and the oxides of nitrogen © (2012) All Rights Reserved 5 of 15 For more info, go to www.hscintheholidays.com.au Properties of SO2 and NOx N2O SO2 Colourless Colourless Pungent odour Sweet smell Soluble Insoluble  food  anaesthetic preservativ
 propellant e  bleaching  fumigant  Breathing difficulties at 1ppm NO Colourless No smell Insoluble  synthesizing nitric oxide NO2 Reddish‐brown Choking odour Soluble in water  nitric acid  fertilisers  explosives 
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Breathing difficulties at 5ppm Leads to forming O3 3. Acids occur in many foods, drinks and even within our stomachs  Define acids as proton donors and describe the ionisation of acids in water A previous (Arrhenius) definition was that acids form H+ in water (ionisation reaction). The H+ joins with water to make the hydronium ion (H3O+). According to the Lowry‐Brönsted definition (1923), acids are proton donors and bases are proton acceptors. 
Identify acids such as acetic acid (ethanoic acid), citric acid (2‐hydroxypropane‐1‐2‐
3‐tricarboxylic acid), vitamin C and hydrochloric acid as naturally occurring acids, and acids such as sulfuric acid and hydrobromic acid as manufactured acids Naturally occurring acids Acetic acid (ethanoic acid) – vinegar Citric acid (2‐hydroxypropane‐1,2,3‐
tricarboxylic acid, C6H8O7) Vitamin C (ascorbic acid, C6H8O6) Hydrochloric acid (in our stomachs) Manufactured acids sulfuric acid hydrobromic acid nitric acid phosphoric acid 
Describe the use of the pH scale in comparing the concentrations of acids and alkalis The pH scale is a useful scale for comparing the concentrations of acids and alkalis, since the concentrations of the ions themselves involved range over a very large linear scale. It is more convenient to use a smaller, closer scale, and thus pH is used. © (2012) All Rights Reserved 6 of 15 For more info, go to www.hscintheholidays.com.au 
Describe acids and their solutions with the appropriate use of the terms strong, weak, concentrated and dilute A strong acid is one where all its molecules ionise to form H+. A weak acid is one where only some of its molecules ionise to form H+. A concentrated solution is where the total concentration of the acid is high. A dilute solution is where the total concentration of the acid is low.  Identify pH as –log10[H+] and explain that a change in pH of 1 means a ten‐fold change in [H+] pH = ‐log [H3O+] A change by 1 in pH is a change tenfold in [H+]. Note: the number of decimal places for pH should equal the number of significant figures for [H3O+] Since pH = ‐log10[H+] [H+] = 10‐pH pOH = ‐log10[OH–] pH + pOH = 14 pH = 14 + log10[OH–]A change of 1 in pH relates to a tenfold change in [H+] A neutral substance is one where [H+] = [OH–], at 25oC, [H+] = 1x10–7molL‐1 
Compare the relative strengths of equal concentrations of citric, acetic and hydrochloric acids and relate this to the degree of ionisation of their molecules We say that HA is stronger than HB for weak acids, if: HA(aq) H+(aq) + A–(aq lies further to the right than HB(aq) H+(aq) + B–(aq) Thus the degree of ionisation of HA is greater than HB For equal concentrations of the following acids, different pHs were found, and it was deduced that HCl is strong, citric acid is weak and acetic acid is weaker.  Describe the difference between a strong and a weak acid in terms of an equilibrium between the intact molecule and its ions For strong acids, there is no equilibrium in the ionisation reaction (the back reaction does not occur), while for weak acids there is.  Gather and process information from secondary sources to write ionic equations to represent the ionisation of acids Hydrochloric acid is a strong acid. This means that when we dissolve hydrogen chloride gas in aqueous solution all the HCl molecules react with water to form hydrogen ions: HCl(g) + H2O(l) ‐ H3O+(aq) + Cl‐(aq) For a strong acid the ionisation reaction with water goes to completion. On the other hand acetic acid is a weak acid, because when pure covalent acetic acid is dissolved in aqueous solution, only some of the acetic acid molecules actually react with water to form hydrogen ions: © (2012) All Rights Reserved 7 of 15 For more info, go to www.hscintheholidays.com.au CH3COOH(aq) + H2O(l) ‐> H3O+(aq) + CH3COO‐(aq) This reaction is an equilibrium one: it does not go to completion.  Gather and process information from secondary sources to explain the use of acidic oxides such as sulfur dioxide as food additives For example: sulfur dioxide can be used as a food additive. 4. Because of the prevalence and importance of acids, they have been used and studied for hundreds of years. Over time, the definitions of acid and base have been refined 
Outline the historical development of ideas about acids including those of: Arrenius, Lavoisier and Davy Historical development of ideas about acids: (1) Antoine Lavoisier (1780)  acids contained oxygen (proved wrong) (2) Humphry Davy (1815)  acids contained replaceable hydrogen  reacted with metals & bases to form H2 (3) Svante Arrhenius (1884)  Acids ionise in solution to produce H+  Strong if ionises completely, weak if not  Base is a substance producing OH– (but this excludes metal oxides)  Did not give enough recognition to the solvent (eg, HCl in some solvents is weak) (4) Brönsted‐Lowry (1923)  An acid is a proton donor  A base is a proton acceptor (5) G.N Lewis (1923)  Base is a lone electron pair donor (has 2 electrons available to covalently bond)  Acid is a lone electron pair acceptor (for covalently bonding)  Outline the Bronsted‐Lowry theory of acids and bases According to the Brönsted‐Lowry theory, if a substance has greater tendency to give up protons than the solvent, then it is an acid in the solvent. If the substance has greater tendency to accept protons than the solvent, it is a base in the solvent.  Describe the relationship between an acid and its conjugate base and a base and its conjugate acid An acid gives up a proton to form its conjugate base. A base accepts a proton to form its conjugate acid. © (2012) All Rights Reserved 8 of 15 For more info, go to www.hscintheholidays.com.au The conjugate of a weak substance is weak, and the conjugate of a strong substance is too weak to react with the solvent. 
Identify a range of salts which form acidic, basic or neutral solutions and explain their acidic, neutral or basic nature Multi‐protic acids go through stepwise ionisation, from which it can be seen that complex ions can be made, eg hydrogen carbonate (HCO3–) When a strong acid and weak base react to form a salt, the anion conjugate acid of the weak base acts as an acid (while the acid’s conjugate is too weak) and thus the salt is acidic. Similarly, the salt formed by a weak acid and a strong base is basic. If both are strong, the resulting salt is neutral, and approximately neutral for both being strong.  Identify conjugate acid/base pairs Conjugate base Acid + water ‐ H3O+ + conjugate base HCl + H2O ‐ H3O+ + Cl‐ CH3COOH + H2O <‐ H3O+ + CH3COO‐ Conjugate acid Base + water ‐ conjugate acid + OH‐ NH3 + H2O ‐> NH4+ + OH‐ CO32‐ + H2O ‐> HCO3‐ + OH‐ 
Identify amphiprotic substances and write equations to describe their behaviour in acidic and basic solutions Amphiprotic substances can act as both a proton donor, and as a proton acceptor. (In basic conditions they tend to act as acids and vice‐versa.) HA–(aq or s) + OH– ‐‐> H2O(l) + A2–(aq) HA–(aq or s) + H+ ‐‐> H2A  Outline the Lewis definition of an acid According to the Lewis definition:  A base is a substance that has a lone pair of electrons which can be used to form a covalent bond with another substance.  An acid is a substance that is able to form a covalent bond with a lone pair of electrons from another substance. This definition helps explain reactions like CaO + CO2 ‐‐> CaCO3. © (2012) All Rights Reserved 9 of 15 For more info, go to www.hscintheholidays.com.au 
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Assess the importance of each definition in terms of understanding Gather and process information from secondary sources to trace and describe developments in understanding and describing acid/base reactions and use available evidence to explain the expanded view of an acid as developed by Lewis Our understanding from different acid definitions The Davy definition simple classified the acids by relating them to their properties. Arrhenius interpreted these properties and related them to their chemical actions, and showed the chemical difference between weak and string acids. The Lowry‐Brönsted definition then showed that acidity also depends on its properties relative to the solvent. It showed that reactions with water were nothing more than acid‐
base reactions. Provided a basis for the quantitative treatment of acid‐base equilibria. Lewis helps our understanding by focusing on electron pair movements, helping on a closer level, and interpreting various reactions. It uses lone pairs to explain how many reactions (mostly organic) proceed.  Identify neutralisation as a proton transfer reaction which is exothermic  Analyse information from secondary sources to predict combining volumes and masses of acids and bases and products in neutralisation reactions Neutralisation is a proton transfer reaction (protons are transferred from one species to another to form a salt). All neutralisation reactions are exothermic (ΔH ≈ ‐56 kJmol‐1). Acid + base ‐ water + salt In the neutralisation of ammonia by nitric acid: HNO3(aq) + NH3(aq) ‐ NH4+(aq) + NO3‐(aq) The acid transfers a proton to ammonia (to form the NH4+ ion) Re. CCHSC p. 146 
Describe the correct technique for conducting titrations and preparation of standard solutions Method for titration (1)
(2)
(3)
(4)
(5)
© Fill a burette with substance of known concentration and record level. Pipette an accurate amount of the unknown substance into the flask. Place the flask underneath. Add one or 2 drops of a suitable indicator to the flask. Slowly add the solution in the burette to the flask until the indicator just changes colour (equivalence point), swirling the flask. (2012) All Rights Reserved 10 of 15 For more info, go to www.hscintheholidays.com.au (6) Read the change in volume of the solution in the burette. (7) Calculate the unknown concentration. Method for preparing standard solutions Since most of our common acids and bases absorb water or are unstable, it is difficult to make precise concentrations of these solutions. We need to use a primary standard, a highly pure and stable substance, which can make an accurate concentration in solution by weighing it out and adding a desired amount of water (eg, NaCO3, Na2HCO3). By titrating a primary solution with an unknown concentration of a solution, it can be standardised.  Qualitatively describe the effect of buffers with relevance to a specific example in a natural system A buffer is a solution containing similar amounts of a weak acid and its conjugate base (eg. H2CO3 and NaHCO3). By Le Chatelier’s principle, when small amounts of an acid or base are added, it will minimise the change and thus maintains an approximately constant pH. This occurs naturally in lakes and rivers, where carbonic acid is dissolved from CO2 in the air and hydrogen carbonate dissolves from rocks around the river/lake, to make a buffer. This way, when small amounts of acidic & basic substances are added to it, it does not largely affect the acidity of the lake (which could be harmful to fish and wildlife). In our bloodstream this occurs with haemoglobin and its conjugate acid. This buffer stops the changing concentrations of CO2 affecting the acidity of human blood.  Process information from secondary sources to: visualise the rearrangement of particles, and change in electrical conductivity and pH, which occurs in the reaction vessel during titration, explain the purpose of accuracy during titration, and identify and describe modern analytical methods used to accurately measure concentrations Titration curves If pH during titration is measured throughout the titration, the following graphs can be produced. Weak & Strong
Strong & Strong
Strong & Weak
Weak & Weak
pH
Amount of base added in acid & base titration
Conductance titration can also be used – the equivalence point is determined as the point where the solution has least electrical conductance (or current flowing). A graph is made of two straight lines with the minimum at equivalence. A
mL
© (2012) All Rights Reserved 11 of 15 For more info, go to www.hscintheholidays.com.au Accuracy during titration Accuracy is required during titration in order to give an accurate analysis of the substance in the solution and its concentration. To achieve such accuracy, it is important that equipment is washed with the correct substance, and that “pure” substances are as pure as possible. (Flasks used may be washed and left with some water, while pipettes and burettes for the known substance must be washed with that substance.) Special volumetric equipment is used for accuracy in titration: Pipettes allow very high accuracy if used properly – draw solution in, until the solution is above the graduation mark, and then run it out until the meniscus sits on the mark. When run out, it is let out vertically, with the tip against the receiving flask. There must be no air bubbles when drawing in the solution. The liquid in the tip remains there. Burettes are overfilled and the excess run out for precision, as with pipettes. There must be no air bubbles, again. Portions of a drop on the end of the burette that are wanted in the titration can be washed in with a wash bottle. Modern Analytical Methods Volumetric analysis (as with titration) is convenient and cheap, but not very accurate for some requirements. Other analytical methods include: 
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Atomic absorption spectrometry (metal ions at ppm concentrations) Flame emission spectroscopy (metals in alloys) Neutron activation (explosives in airport luggage) 
Analyse information from secondary sources to assess the use of neutralisation reactions as a safety measure or to minimise damage in accidents or chemical spills Another industrial use of neutralisation reactions is in safety. Spills of corrosive acids and alkalis require urgent neutralisation for safety. Sewage authorities also require neutralisation of discharge from factories. – To neutralise acid, sodium carbonate is used, since it is stable, solid, cheap, and less corrosive than others. – To neutralise alkaline wastes, HCl or H2SO4, but for emergency spills, NaHCO3 is used since it is easy to handle and not problematic in excess (it can also be used for acid spills since it is amphiprotic).  Process information from secondary sources to calculate pH of strong acids given appropriate hydrogen ion concentrations Eg. [H3O+] = 10‐7 mol/L pH = ‐log10[H3O+] = ‐log10[10‐7] = 7.0 © (2012) All Rights Reserved 12 of 15 For more info, go to www.hscintheholidays.com.au 5. Esterification is a naturally occurring process which can be modelled in the laboratory 
Describe the difference between the alkanol and alkanoic acid functional groups in carbon compounds Alkanols CnH2n+1OH Alkanoic acids CnH2n+1COOH O
H
R
R
C O H
H
C
O H
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Explain the difference in melting and boiling point caused by the alkanoic acid and alkanol functional groups Both alkanols and alkanoic acids are polar and can hydrogen bond. This stronger intermolecular force means that they have higher melting and boiling points than alkanes and alkenes of similar molecular weight. 
Identify esterification as the reaction between an acid and an alkanol and describe, using equations, examples of esterification Esterification is the catalysed reaction between carboxylic acid and alkanol. Alkanol + alkanoic acid ‐‐> alkyl alkanoate + water CnH2n+1OH + CmH2m+1COOH ‐‐> CmH2m+1COOCnH2n+1 + H2O O
R
C
H
+
O H
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H O C R’
H
O
R
C
+
H
O C
H
H-O-H
R’
Describe the purpose of using concentrated sulfuric acid in esterification for catalysts and absorption of water The reaction is slow and comes to equilibrium. Concentrated H2SO4 is used as a catalyst. It also absorbs water, and thus shifts equilibrium right.  Identify the need for refluxing during esterification Refluxing is the process of heating a reaction mixture in a vessel with a cooling condenser attached in order to prevent loss of any volatile reactant or product. Reaction performed at high temperatures for speed, but volatile reactants escape at these temperatures. Thus, © (2012) All Rights Reserved 13 of 15 For more info, go to www.hscintheholidays.com.au refluxing is used. The reaction flask is fitted with a cooling condenser, so that any escaping vapours are condensed and return to the reaction flask. (Boiling chips are also used to prevent bubbling of mixture.) 
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Outline some examples of the occurrence, production and uses of esters Process information from secondary sources to identify and describe the uses of esters as flavours and perfumes in processed foods and cosmetics Esters have pleasant odours, occurring in perfumes and flavouring, naturally in fruits, plants. They are manufactured as synthetic flavours and perfumes, often to mimic the flavours of plants (by first identifying the main esters present in a substance). Produced, these are much cheaper than natural extraction. Ethyl acetate is used as a solvent, and some are used as plasticisers (eg. for PVC). Other Syllabus Points 
Solve problems by applying information about the colour changes of indicators to classify some household substances as acidic, neutral or basic 
Identify data, plan and perform a first‐hand investigation to decarbonate soft drink and gather data to measure the mass changes involved and calculate the volume of gas released at 25C and 101.3kPa 
Solve problems and perform a first‐hand investigation to use pH meters/probes and indicators to distinguish between acidic, basic and neutral chemicals 
Plan and perform a first‐hand investigation to measure the pH of identical concentrations of strong and weak acids 
Use available evidence to model the molecular nature of acids and simulate the ionisation of strong and weak acids 
Identify data, gather and process information from secondary sources to identify examples of naturally occurring acids and bases, their chemical composition and describe their pH in their naturally occurring form © (2012) All Rights Reserved 14 of 15 For more info, go to www.hscintheholidays.com.au Hydrochloric acid (HCl) is produced by glands in the lining of our stomachs to form an acidic environment for the efficient operation of the enzymes that break complex food molecules into easily transportable small molecules that are absorbed into the blood stream when they pass into the intestine Acetic acid (systematic name ethanoic acid), CH3‐COOH. Acetic acid is present in vinegar which is commonly made from wine by oxidation of ethanol. Citric acid (systematic name 2‐hydroxypropane‐1,2,3‐tricarboxylic acid with molecular formula C6H8O7). It occurs in citrus fruit. Vitamin C or ascorbic acid which has the molecular formula, C6H8O6. It occurs widely in fruits and vegetables and is an essential part of our diet. pH needed  Choose equipment and perform a first‐hand investigation to identify the pH of a range of salt solutions  Perform a first‐hand investigation and solve problems using titrations and including the preparation of standard solutions, and use available evidence to quantitatively and qualitatively describe the reaction between selected acids and bases  Perform a first‐hand investigation to determine the concentration of a commercial acidic or alkaline substance such as vinegar, orange juice or window cleaner  Perform a first hand investigation to measure temperature change during neutralisation and calculate the molar heat of reaction  Identify data, plan selected equipment and perform a first‐hand investigation to prepare an ester using reflux © (2012) All Rights Reserved 15 of 15 For more info, go to www.hscintheholidays.com.au ACIDS AND BASESINDICATORS-
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Methyl Orange- for identifying strong acid, red for acid, transition orange 3-4.5
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Bromothymol blue- for identifying neutrality, transition green 7-8.5
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Litmus- red 0-7, blue 7-14
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Phenolphthalein- for identifying strong bases, pink for strong base, lighter mink
transition 8-10
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Indicators are used as solutions or can be adsorbed onto the surface of paper such as
filter paper.
Acid-base indicators and pH testing kits are used in a range of everyday situations,
including in:
o the soil testing part of a nursery or hardware store
o a swimming pool store or the pool chemicals in a large supermarket
o the aquarium supplies section of a pet shops
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ACIDS REACTIONS:
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strong acid + strong alkali  water + salt
o H2SO4(aq) + 2KOH(aq)  2H2O(l) + K2SO4(aq)
acid + metal oxide  water + salt
o 2HCl(aq) + MgO(s)  H2O(l) + MgCl2(aq)
acidic non-metal oxide + water  acid
o CO2(g) + H2O(l)  H2CO3(aq)
acidic non-metal oxide + base  water + salt
o CO2(g) + 2NaOH(aq)  2H2O(l) + Na2CO3(aq)
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OXIDES- ACIDIC, BASIC OR AMPHOTERIC-
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Very alkali
o Produces base easily in water
o Easily neutralised wit acid
o Doesn’t react with other bases
Moderate alkali
o Insoluble with water
o Does not react with a base
o Easily neutralises with acid
Amphoteric
o Reacts with acid or base to neutralise
o Insoluble in water
Very acidic
o Does not react with acid
o Reacts with water to form acid
o Neutralises with base
LE CHATELIER’S PRINCIPLE AND EQUILIBRIUM-
EQUILIBRIUM





Products react with one another to re-from the starting materials, if left for long
enough, will finally reach a balance between opposing reactions.
Closed system- no matter or energy enters or leaves the system
Macroscopic properties remain constant
Concentrations of reactants and products stay constant (not necessarily equal)
Continual microscopic change occurs between reactants and products
Rate of forward reaction = rate of backwards reaction
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FACTORS THAT AFFECT EQUILIBRIUM:



change in concentration
change in temperature
Change in gas pressure.
LE CHATELIER’S PRINCIPLE


If a system is in equilibrium and it is disturbed or changed in some way, then the
system adjusts itself to minimise the amount of change
Think of a skateboarder regaining their balance after a bump
Changing the surface area of one or more of the solid components usually means
that the equilibrium is unchanged
ACIDIC OXIDES OF NON-METALS AND BASIC OXIDES OF METALS-
Reasons








Water consists of two main parts, a positive hydrogen ion (proton) and a negative
hydroxide ion
Alkalis have high numbers of hydroxide ions
Acids have high numbers of hydrogen ions
Metal have very low electronegativities
Oxygen and other non-metals have very high electronegativities
In ionic bonds, big difference in electronegativities which results in the complete
transfer of electrons (no polarity)
In covalent bonds the small difference in electronegativities results in electrons being
shared, most of this sharing is uneven; hence polar molecules
Dissolution of sodium oxide in watero Na2O(s) + H2O(l)  2NaOH(aq)
o 2NaOH(aq) + 2Na+(aq) + O2-(aq) + H+(aq) + OH-(aq)
Dissolution of sulfur trioxide in watero SO3(g) + H2O(l)  H2SO4(aq)
SOLUBILITY OF CO 2 IN WATER:

The solubility of carbon dioxide gas in water can be fully described using four
equilibrium equations:
1. CO2(g)
CO2(aq)
2. H2O(l) + CO2(aq)
3. H2CO3(aq)
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H2CO3(aq)
H+(aq) + HCO3-(aq)
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4. HCO3-(aq)
H+(aq) + CO32-(aq)

An equilibrium shift to the left or right in any of the equations above leads to a change in
the concentration of the reactants and products in that reaction. This influences the
concentrations in the other three reactions causing the equilibrium in those reactions to
also shift to the left or right respectively. For example, a decrease in the pressure above
the solution causes reaction 1 to shift to the left and results in a decrease in the pH of
the solution as the other equilibrium reactions also shift to the left. Likewise, an
equilibrium shift to the right in equations 2, 3 or 4 results in more carbon dioxide gas
being dissolved in equation 1.

In accordance with Le Chatelier's principle:
o
Addition of acid (increased concentration of H+) shifts equilibrium to the left in
equations 3 and 4 above and therefore results in a decrease in dissolved carbon
dioxide.
o
Addition of base (reacts with and reduces the concentration of H+) shifts
equilibrium to the right in equations 3 and 4 above and therefore leads to an
increase in the concentration of dissolved carbon dioxide.
o
Addition of a soluble carbonate (increased concentration of CO32-) shifts
equilibrium in equation 4 above to the left and therefore results in a decrease in
dissolved carbon dioxide.
TRENDS IN OXIDES-



Increasingly acidic character moving left to right across a period
Left hand sideo Undergo complete transfers of electrons
o No polarisation to attract ions
Right hand sideo Form increasing polarised molecules (due to electronegativities)
o When dissolved in water, partial positive charge accepts negative hydroxide ion,
results in production of H+ and HxOz- ions.
OXIDES IN THE ATMOSPHERE-
ACIDIC OXIDES IN THE ATMOSPHERE

Oxides of carbon
o Carbon monoxide
o Carbon dioxide
Oxides of sulfur
o Sulfur dioxide
o Sulfur trioxide
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
Oxides of nitrogen (NOx)
o Nitrogen monoxide
o Nitrogen dioxide
o Dinitrogen oxide
NATURAL SOURCES OF OXIDES


Oxides of carbono Naturally forms carbonic acid with atmospheric moisture, natural rainwater has
a pH or 5-6
o CO2(g) + H2O(g) <-> H2CO3(l)
o Carbon monoxide Removed by soil organisms or oxidised to carbon dioxide
Oxides of sulfuro Sulfur dioxide
 Bushfires
 Decaying vegetation and animals (proteins and nucleic acids breakdown
to form H2S(g)
 2H2S(g) + 3O2(g)  SO2(g) + 2H2O(g)
 plants and animals contain proteins and nucleic acid which have sulfur
 volcanoes
o S(s) + O2(g)  SO2(g)
Oxides of Nitrogeno Nitric oxide and nitrogen dioxide are more common
o Minor amounts of dinitrogen oxide released through action of soil bacteria
o Nitric oxide Soil bacteria
 Lightning
 Enough energy for atmospheric nitrogen to combine with
atmospheric oxygeno N2(g) + O2(g)  2NO(g)
 UV light on nitrogen dioxide
 Splits oxygen atomo 2NO2(g)  2NO(g) + O2(g)
 Hot bushfires
o Nitrogen dioxide Slowly in atmosphere by oxidation of nitric oxide 2NO(g) + O2(g)  2NO2(g)
 Hot bushfires
o Heat is enough to break very strong triple bond in N2,
allowing it to combine with oxygen
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MAN-MADE SOURCES OF OXIDES

Sulfur dioxideo Combustion of fossil fuels Fossil fuels once living organisms which have sulfur containing proteins
and nucleic acids
o Burning of sulfurous minerals From sulfur in fossil fuels
 4FeS2(s) + 11O2(g)  2Fe2O3(s) + 8SO2(g)
o Smelting 2ZnS(s) + 4O2(g)  2ZnO2(l) + 2SO2(g)
o Incineration of garbage
o Petrol refining
o Paper manufacturing
o Sewage treatment
NOxo Combustion of fossil fuels
o Internal combustion engine
o Power stations
o Oil refineries
o High temperature incinerators
ONCE IN THE ATMOSPHERE


Sulfur dioxideo SO2(g) forms a weak acid with water, sulfurous acid
 H2O(g) + SO2(g) <-> H2SO3(aq)
o SO2(g) easily oxidises with oxygen in the atmosphere to SO3(g)
 2SO2(g) + O2(g)  2SO3(g)
o SO3(g) combines with water to form H2SO4(aq)
 SO3(g) + H2O(g)  H2SO4(aq)
Nitrogen dioxideo Combine with atmospheric moisture
 Can produce weak nitrous acid
 Can produce strong corrosive nitric acid
 2NO2(g) + H2O(g)  HNO2(aq) + HNO3(aq)
 Nitrous oxide can also be oxidised into nitric acid 2HNO2(aq) + O2(g)  2HNO3(aq)
Creates rain with pH 3-4 i.e. acid rain
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NATURAL ACID RAIN



localised
a ‘one off’ event
diluted by atmospheric moisture
pH 5-6
AFFECTS OF ACID RAIN


Aquatic ecosystemso Rivers and lakes become acidic
o Can result in death of organisms
o Most fish eggs won’t survive lower than 5.5
o >5 many adult fish cannot extract calcium from water, also increases toxicity of
dissolved metals
o Increase in hydrogen ions removes dissolved carbon dioxide
 2H+(aq) + CO32-(aq) <-> H2CO3(aq) <-> H2O(l) + CO2(aq) <-> CO2(g)
 Stresses photosynthetic organisms
 Less oxygen released
Vegetationo Change pH of soil
o Difficult for plants to absorb calcium or potassium
o Makes trace elements more toxic
o Crop production reduced
o Affect important soil bacteria
 Less nitrogen- plant growth limited
o Deforestation
o Soil erosion
o Habitat loss
Man-made structureso Marble, limestone, cement, render, mortar- primarily calcium carbonate
 CaCO3(s) + 2H+(aq)  Ca2+(aq) + CO2(g) + H2O(l)
o Large impact in Asia- no pollution laws
o Metal structures
 Corrode and lose strength
 H2SO4(aq) + Fe(s)  H2(g) + FeSO4(aq)
 Paint to protect bridges
EVIDENCE
Quantativeo Ice cores Monitor level of atmospheric gases over time
 Water freezes, traps bubbles of air- capturing sample of atmosphere
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

Show levels of CO2 and NOx were much lower before Industrial
Revolution
 Do not give as much on SO2 levels- tends to react with other substances
over time
o Levels of NOx and Sulfur Dioxide present in much smaller concentrations than
CO2
 Haven’t been able to measure these low levels til ‘70’s
o Large difference in levels between industrial and non-industrial areas
o Since anti-pollution laws there has been a decrease
Qualitativeo Photos
o Oral histories
ISSUES OTHER THAN ACID RAIN
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
Photochemical smog
Brown colour of NO2
NO2 absorbs UV radiation to form nitric oxide and oxygen free radicals
o Oxygen radicals combine to form ozone O(g) + O2(g)  O3(g)
 Very unstable molecule
 Readily oxidises body tissue, disrupts biochemical reactions in body
 Irritates eyes
 Can cause breathing difficulties
 Also attacks rubber and plastics
 Toxic to plants
Acid smog
Links to longer term health problems- asthma, lung cancer, chronic bronchitis and
emphysema
Nitrous oxide- greenhouse gas
o Global warming
MAY cause biological mutation
WHAT ARE WE DOING ABOUT IT?



Anti-pollution laws
Smelters and coal burning power stations must have a scrubber installedo Usually contains a base to trap much of the sulfur
o SO2(g) + Mg(OH)2(aq)  MgSO3(aq) + H2O(l)
Cars must have a catalytic converter
o Considerably reduces NOx emitted
o Converts carbon monoxide to carbon dioxide
o 2CO(g) + 2NO(g)  N2(g) + 2CO2(g)
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GAS CALCULATIONS-


SLC- 24.79L at 25oC & 100kPa
STP- 22.71 at 0oC & 100kPa
CONVERTING MASS TO VOLUME1.
2.
3.
4.
5.
Balanced equation
Molar ratio
No. of moles (limiting reagent)
New molar ratio
Convert molar amount of gas into litres (check conditions)
CONVERTING VOLUME INTO MASS
1.
2.
3.
4.
5.
Balanced equation
Molar ratio (limiting reagent)
Known no. of moles using volume equation (check conditions)
New molar ratio
Covert moles into mass
ACIDS:






Acid- a hydrogen ion donor
Base- hydrogen ion acceptor
Hydrogen sulfateo Sulfuric acid
o H2SO4
o Industrial and analytical process
o Strong
Hydrogen acetateo Acetic or ethanoic acid
o CH3COOH
o Vinegar, biochemical process
o Weak
Hydrogen Chlorideo Hydrochloric acid
o HCl
o Industrial and analytical process
o Strong
Hydrogen phosphateo Phosphoric acid
o H3PO4
o Starting base for phosphate fertiliser production
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


o Weak
Hydrogen carbonateo Carbonic acid
o H2CO3
o Rain, beverage production
o Weak
2-hydroxypropane-1,2,3-tricarboxylic acid
o Citric acid
o (CO2H)CH2C(OH)(CO2H)CH2CO2H
o Found in citrus fruits, important acid in body
metabolism
o Weak
Hydrogen nitrateo Nitric acid
o HNO3
o Industrial and analytical process; oxidising agent
o strong
CREATION OF ACIDS





H3O+ (hydronium ion)- an H+ attached to a water molecule
Free H+ doesn’t exist in aqueous solution
When common acids dissolve in water they react in water to form hydronium ions
Hydrogen needs water to carry it
HCl(g) + H2O(l)  H3O+(aq) + Cl-(aq)
o OR HCl(aq)  H+(aq) + Cl-(aq)
FINDING THE PH



Concentration of hydronium ions
Cover such a wide range- hence pH
pH= -log10[H3O+]
ion concentration= 10-pH n/L
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CALCULATING PH FROM A MOLARITY1.
2.
3.
4.
5.
6.
7.
Weak or strong acid?
Strong acid, write disassociation equation
Theoretical molar ratio
Assume 1L, actual molar ratio
Log(mole)
Change sign
pH
CALCULATING MOLARITY FROM PH1.
2.
3.
4.
5.
Weak or strong acid?
Write disassociation reaction.
Theoretical mole ratio
Shift, log, pH
Assume 1L of acid, write actual mole ratio
EQUILIBRIUM CONSTANT



WATER -
Kw
1.0 x 10-14 = [H3O+] x [OH-]
In any solution at 25oC
CALCULATING PH USING HYDROXIDE ION CONCENTRATION1.
2.
3.
4.
Write out Kw substituting in hydroxide ions concentration
Re-arrange formula
Calculate [H3O+]
Perform –log
CALCULATING HYDROXIDE ION CONCENTRATION USING PH1. Calculate concentration of hydrogen ions using pH
a. Shift, log
2. Substitute hydrogen ion concentration into Kw equation
3. Re-arrange formula
4. Calculate
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CONCENTRATED/ DILUTE AND STRONG/WEAK:


Strong electrolytes completely dissolve or ionise in solution e.g. strong acids and bases
Weak electrolytes (acids and bases) only partially ionise in water solution
o For any weak acid [HA] that ionises to produce hydrogen ions, equilibrium can be
expressed as HA(aq) + H2O(l) <-> H3O+(aq)
o Degree of ionisation expressed by [H3O+] / [HA] x 100
o [HA] is any weak acid [ ] concentration in moles/litre
o For any weak acid [B-] that ionises to produce hydroxide ions B-(aq) + H2O(l) <-> HB(aq) + OH-(aq)
o Degree of ionisation [OH-] / [B-] x 100
o [B-] is any weak base and [ ] in moles/litre
FOOD ACIDS:

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







Methods for before refrigeration
Adding acids inhibited growth of both harmful and spoilage bacteria and mould, food
lasted longer
o Spoilage organisms spoil food, don’t make you sick
o Harmful (pathogenic) organisms may not necessarily spoil the food but can make
you sick e.g. salmonella
Mechanism for preventing cell decomposition
Bacteria and moulds have a set range of conditions in which they grow best
As food ages, bacteria and mould grow as there’s the right supply of nutrients, water
and correct pH
Adding WEAK acids, foods brought down to a pH in which bacteria find it difficult to
grow
Most common way is to add vinegar
Fermentationo Promotes growth of organisms which produce acids such as ethanoic and lactic.
Some foods have salt added to themo Once salts are in contact with moisture in the food hydrogen ions are free to
form
o Sulfur dioxide and sulfur bisulphate salts
Some food acids act as antioxidants
Leavening agents
o React with NaHCO3 to produce carbon dioxide, CO2 inhibits many microbes
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



Some enzymes speed up decay reactions in cells
o Only work in a narrow range of conditions
o Particularly sensitive to pH
o Break down if conditions are too acidic
Some food acids enhance flavour e.g. citric
Some acids added to increase nutritional value of the food e.g. ascorbic acid
Common food acidso Acetic acid
o Citric acid
o Propanoic acid
NATURALLY OCCURRING ACIDS AND BASES-



Hydrochloric acido Salt acid, muriatic acid
o Found in stomachs of mammals
 Stomach is lined with mucous to stop acid from attacking it.
 Role is to denature proteins in food
 Allows digestive enzyme to convert protein into individual amino acids
 Also protects body from bacteria in food
 pH 1-2
o Strong
o HCl
Citric Acido 3-carboxy-3-hydroxypentanedioic acid
o 2-hydroxypropane-1,2,3- tricarboxylic acid
o Fruits
 In many fruit and vegetables, high levels in citrus fruits
o Intermediate product in metabolism
 In every cell in the body
 Formed when glucose is oxidised to release energy, carbon dioxide and
water in mitochondria
o Weak
o C 6 H8 O 7
o COOHCH2COH(COOH)CH2COOH
o One of few acids which can exist in crystalline form
Calcium Carbonateo Limestone, chalk, marble
o Soils, rocks, shells, pearls, eggs
o Weak
o CaCO3
o Ionic lattice mineral found in many rocks
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
o Limestone is a rock type consisting of metamorphosed calcium carbonate
o Forms large part of shells, eggs and pearls
Magnesium Hydroxideo Milk of magnesia
o Rocks
o Weak
o Mg(OH)2
o Found in rocks, easily broken apart
o Often found in marble giving it the ‘glassy’ look
o Causes cement and concrete to crack
o Used to make antacids and laxatives
HISTORY OF DEVELOPMENT OF ACIDS:
WHAT THEY FIRST THOUGHT


Knew that acids:
o Tasted sour
o Gave a stinging sensation
o Turned litmus red
Knew that bases:
o Tasted bitter
o Felt slippery/slimy
o Turned litmus blue
Didn’t know chemical reasons why
ANTONIE LAVOISIER (1743-1794)





First to try and explain what caused acidity
Burnt non-metals to make oxides
Placed these in water
Resulting solutions were sour and stung, made acids
Proposed oxygen in compounds gave acidic properties
Did not explain why oxides of metals were not acidic
HUMPHREY DAVY (1778-1829)







Pioneer in electrolysis
Performed electrolysis on molten salts and muriatic acid (hydrochloric acid)
Able to split muriatic acid into hydrogen gas and chlorine gas
Successfully proved that an acidic substance did not have to contain oxygen
Proposed presence of hydrogen in acids gave them their properties
Did not explain why many compounds containing hydrogen were not acidic
Suggested bases were substances that reacted to with acids to form salts and water
Did not explain their properties
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JUSTUS VON LIEBEG (1803-1873)




Extended Davy’s theory
Stated acids were substances containing ‘replaceable hydrogen’
Reasoned that when acids attack metals, metals replace hydrogen in the acid
o Metal + acid  salt + hydrogen
Did not account for production of gases other than hydrogen in certain cases (e.g. nitric
acid attacks metals producing nitrogen dioxide)
Since Davy made link between hydrogen and acids ‘replaceable hydrogen’ theory is
attributed to his work
SVENTE ARRHENIUS (1859-1927)









Worked with electrolysis
First to propose idea of ‘ions in solution’
Performed electrolysis on many different acids
Proposed that hydrogen in acids was actually hydrogen ions
Suggested acids were substances that provided hydrogen ions in an aqueous solution
Said that non-acid part of the acid must be in form of negative ions
Proposed that a base was a substance that provided hydroxide ions in solution
First to suggest chemical reason for basicity
Proposed that when an acid reacts with a base, hydrogen ions in the acid neutralise
hydroxide ions in the base
o HA + XOH  H2O + XA
Problems with his theorieso Why are metallic oxides and carbonates basic? There is no hydroxide.
o Why are some solutions of various salts acidic or basic?
o Does not account for role of the solvent.
JOHANNES BRONSTED (1873-1947) AND THOMAS LOWRY (1874-1939)








Independently proposed the same theory
Proposed that an acid was a proton donor and a base a proton acceptor
Recognised the importance of protons
Substance cannot act as an acid without another acting as a base
o With acids the proton is donated to the water
Assigned a role to the solvent
o It is not just an inert liquid, acts as an ionising solvent
Water can be both a solvent and an acid/base
o Amphiprotic (gives or receives electrons)
Accounted for why ammonia and carbonates acted as a base
Est. hydroxide ions were responsible for basic properties
Accounted for why some salts are acidic or basic when dissolved
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ADVANCED LOWRY-BRONSTED THEORY:
CONJUGATES AND AMPHIPROTIC SUBSTANCES




Strong acid produce very weak bases and vice versa etc.
A base, after it has accepted a proton is a potential donor
To work out conjugate base (connected base) of an acid, remove a hydrogen ion
HCl(aq) + H2PO4-(aq)  H3PO4(aq) + Cl-(aq)
NaOH(aq) + H2PO4-(aq)  Na2+(aq) + HPO4-(aq) + H2O(l)
MONOPROTIC, DIPROTIC AND TRIPROTIC ACIDS





Monoprotic: able to give only one hydrogen ion per molecule of acid e.g. HCl, HNO3
Diprotic: able to give up two hydrogen ions per molecule of acid e.g. H2SO4
Triprotic: able to give three hydrogen ions per molecule of acid e.g. H3PO4
Ionisation of diprotic and triprotic acids occurs in a stepwise fashion
o From second is equilibrium if a strong acid
In Lowry-Bronsted terms, HSO4- is conjugate base of H2SO4. However, HSO4- is an acid as
it can donate its hydrogen ion.
In Di and Triprotic acids once one hydrogen ion disassociates from the rest of the acid, it
is much harder for the next to disassociate; each successive ionisation is smaller and
smaller.
SALTS AS ACIDS AND BASES:







Acidic salts:
o Upon dissolving in water the react as acids, forcing the water molecule to be a
base
o HSO4-(aq) + H2O(l)  SO42-(aq) + H3O+(aq)
Basic salts:
o Dissolve in water and react as bases, forcing water molecule to be an acid
o CO32-(aq) + H2O(l)  HCO3-(aq) + OH-(aq)
Reaction of salt with water to produce a change in pH is called hydrolysis (hydro-water,
lysis-split)
Consequence of stepwise ionisation of diprotic and triprotic acids is that such acids can
form two series of salts:
o Normal e.g. sulfates
o Hydrogen or acid salts e.g. hydrogen sulfate ion (HSO4-)
HSO4- and H2PO4- are always acidic
HCO3- and HPO42- are always basic
Some acidic salts:
o NaHSO4
o NH4Cl
o NH4NO3
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


Some neutral salts:
o NaCl
o KNO3
o Na2SO4
Some basic salts:
o Na2CO3
o CH3COONa (sodium ethanoate)
Consequence of stepwise ionisation of diprotic and triprotic bases is the formation of
two series of salts:
o Normal e.g. carbonates
o Base salts e.g. hydrogen carbonate ion (HCO3-)
NEUTRALISATION:
1. Heat of neutralisation is exothermic.
Strong Base
Weak Base
e.g. KOH, NaOH
e.g. NH3
Strong acid
Forms neutral salts
Forms acidic salts
e.g. HCl, HNO3, H2SO4
e. g. KCl, NaNO3, K2SO4
e.g. NH4Cl, NH4NO3, (NH4)2SO4
Weak acid
Forms basic salts
Forms neutral salts
e.g. CH3COOH, H2CO3
e.g. CH3COONa, Na2CO3
e.g. CH3COONH4, (NH4)2CO3
BUFFERS:








Resists a change in pH when small quantities of acid or base are added.
Mixture comprises a conjugate pair
Useful in keeping pH of substances nearly constant
Used in industry, laboratories and founding living things
Work using the common ion effect, special case of Le Chatelier’s principle
In general, addition of salt with ion common to solution of weak electrolyte reduces
ionisation of weak electrolyte. Common ion effects behaviour of the electrolyte.
Acetic acid/ acetate ion mixture is a very common laboratory buffer
Important buffer is hydrogen carbonate/ bicarbonate mixture in the blood
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BUFFERS IN THE BLOOD:
o pH is maintained 7.35-7.45
o presence of a weak acid (carbonic acid) and conjugate base (hydrogen carbonate
ion).
o HHb+ + 4O2  Hb(O2)4 + H+
 Absorption of oxygen shifts equilibrium to right
o H+ + HCO3-  CO2 + H2O
 Carbon dioxide released as equilibrium shifts to right
o Hb(O2)4 + H+  HHb+ + 4O2
 Oxygen released to tissues
o CO2 + H2O  H+ + HCO3 Carbon dioxide absorbed in blood
ACID-BASE TITRATIONS:

Adding one solution to another until reaction is complete
STEPS FOR MOST ARE:
o
o
o
o
APPARATUS:
Definite volume of one solutions is placed in a conical flask using a pipette
Few drops of appropriate indicator placed in a flask
Other solution is added to flask from a burette
Sufficient solution added to completely react
o Volumetric flask
o Pipette
o Burette
CLEANING GLASSWARE
o Glassware usually soaked in industrial strength detergent rinsed thoroughly with
copious amounts of tap water. Then copious amounts of distilled water. Wear
gloves.
o Burette and pipettes rinsed with distilled water, then with small quantities of
solution to be placed in them- prevents any dilution of solutions
o Conical flasks and volumetric flasks only rinsed with distilled water
PRIMARY SOLUTION
o Occurs when known mass of a primary standard is dissolved in a known volume
of solution. Most are bases, occasionally crystalline acids.
MAKING A STANDARD SOLUTION:
o weigh out app. amount of solid e.g. sodium carbonate, 5g (chemical grade= ARanalytical reagent)
o place in an oven set at 104oC for 48hrs
o allow to cool in a desiccator
o weigh
o place back at 104oC for 24hrs
o remove, allow to cool in a desiccator
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o re-weigh, should be same weight as step 4. If not, continue cycle until constant
weight is reached
o accurately weigh out amount of dried standard to use in the titration
o quantitatively transfer into A grade volumetric flask, using distilled water, rinse
out all particles
o Add more distilled water to the flask (about ¾ full), lid. Mix to dissolve contents.
PRIMARY STANDARDS:
must be obtainable in a very pure form
have a known formula
don’t alter during weighing by picking up or losing moisture or reacting with air.
should have a reasonably high relative formula mass to minimise errors
some are anhydrous
 without water of crystallisation
 e.g. sodium carbonate
DESICCATORS:
o
o
o
o
o
o allow heated items to cool down to room temp without absorbing moisture form
the air
o contains a drying agent e.g. silica beads
o standard sits on a perforated mesh or plate
o air tight lid
SECONDARY STANDARDS:
o solution whose exact concentration is not known, can be standardised by
titration against a primary standard
TITRATION CALCULATIONS:
FINDING THE CONCENTRATION OF AN UNKNOWN SOLUTION:
1.
2.
3.
4.
5.
balanced equation for reaction
calculate mole ratios
no. of moles of reactant in standard solution
Use equation for reaction to find no. of moles in unknown solution
calculate concentration
CHANGES IN PH DURING TITRATION:



End point- point in titration when indicator changes colour
Equivalence point- point where acid has stoichiometrically reacted with the base
o all H+ and OH- have reacted together
Strong acid- strong base titration:
o strong acids and strong bases completely ionised
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


o neutralisation reaction involves hydronium and hydroxide ion
o H3O+(aq) + OH-(aq)  2H2O(l)
o doesn’t hydrolyse to any appreciable extent
o pH at equivalence point is 7
Strong acid –weak base titration:
o pH at the equivalence point is less than 7 due to hydrolysis of the acidic salt
produced
o HCl(aq) + NH4OH(aq)  NH4Cl(aq) + H2O(l)
Weak acid- strong base titration:
o pH at the equivalence point >7 due to hydrolysis of the basic salt produced
o CH3COOH(aq) + NaOH(aq)  CH3COONa(aq) + H2O(l)
Weak acid- weak base titration:
o no rapid change in pH even at equivalence point
o pH of equivalence point is pH 7
ESTERS:
ALKANOLS AND ALKANOIC ACIDS:



Alkanols
o contain OH groups, hydrogen bonding occurs between neighbouring molecules
o relatively high melting points and boiling points compared with hydrocarbons of
similar molecules
Alkanoic acids
o form hydrogen bonds
o contain C=O group and OH group
o number of hydrogen bonds formed is greater
o possess higher melting and boiling points that most other organic compounds of
similar size and molecular mass, inc. alkanols
tend to dissolve readily in water due to hydrogen bonds
ESTERS:


acid catalysed reaction between an alkanoic acid and an alkanol
o called esterification
o condensation reaction
Naming esters:
o alkyl alkanoate
 alkyl- alkanol
 alkanoate- alkanoic acid
 acid is always the double bond in the molecule
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o general shorthand RCOOR- or RC(=O above)OR








generally liquid at room temp
much lower boiling points than alkanoic acids but with similar molecular mass
intermolecular forces are dispersion and dipole-dipole rather than hydrogen bonding
Butyl ethanoate- nail lacquer
Propyl ethanoate- pear fragrance
most aren’t very soluble in water
powerful solvents of a wide range of organic substances
volatile esters have pleasant odours
CREATING ESTERS:


because reactants can be volatile esterification is carried out using refluxing apparatus
o also allows esterification to be carried out at higher temperatures
Refluxing:
o the process of heating a reaction mixture in a vessel with a cooling condenser
attached in order to prevent loss of any volatile reactant or product
SAFETY AND NEUTRALISATION REACTIONS:

Neutralisation= reaction between acid and base to produce new compound neither
acidic nor basic or is reduced in its acidic or basic properties when compared to the
reactants.
o exothermic
 heat liberated per mole when a strong acid is neutralized by a strong
base is almost the same no matter what acid or base is used
 same reaction- neutralisation of hydronium and hydroxide ions
to form water
 those involving weak acids or bases produce slightly less heat
DILUTING STRONG ACIDS SAFELY:



acid  water
small quantities of acid to large volume of water with constant stirring
heat could potentially rise to boiling point- hazard
o safety glasses etc.
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ACIDS AND NEUTRALISATION REACTIONS:



neutralisation reactions can be used to clean up after spilled acids or bases
should never use conc. acids or bases when cleaning up spills
Sodium hydrogen carbonate commonly used
o weak base
o non-toxic
o amphiprotic substance
 can be used to neutralize basic spills
o fairly safe to environment
o dry, absorbent  effective bund
o cheap
o safe to handle
o heat of neutralisation is less
o stops fizzing when reaction is complete
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Chemistry ‐ ACIDIC TRIAL Qs SORTED BY DOTPOINT Chemistry Stage 6 Syllabus 9.3 The Acidic Environment Contextual Outline Acidic and basic environments exist everywhere. The human body has a slightly acidic skin surface to assist in disease control and digestion occurs in both acidic and basic environments to assist the breakdown of the biopolymers constituting food. Indeed, microorganisms found in the digestive system are well adapted to acidic or basic environments. Many industries use acidic and basic compounds for a wide range of purposes and these compounds are found in daily use within the home. Because of this, an awareness of the properties of acids and bases is important for safe handling of materials. Currently, concerns exist about the increased release of acidic and basic substances into the environment and the impact of these substances on the environment and the organisms within those environments. This module increases students' understanding of the history, nature and practice of chemistry, the applications and uses of chemistry and implications of chemistry for society and the environment. Students learns to: Students: 1. Indicators were identified with the observation that the colour of some flowers depends on soil composition © classify common substances as acidic, basic or neutral perform a first‐hand investigation to prepare and test a natural indicator identify that indicators such as litmus, phenolphthalein, methyl orange and bromothymol blue can be used to determine the acidic or basic nature of a material over a range, and that the identify data and choose resources to gather information about the colour changes of a range of indicators (2012) All Rights Reserved 1 of 65 For more info, go to www.hscintheholidays.com.au range is identified by change in indicator colour identify and describe some everyday uses of indicators including the testing of soil aciditylbasicity classify common substances as acidic, basic or neutral perform a first‐
hand investigation to prepare and test a natural indicator solve problems by applying information about the colour changes of indicators to classify some household substances as acidic, neutral or basic Question 20 (4 marks) You have carried out a first‐hand investigation to prepare and test a natural indicator. (a) Outline the procedure used to prepare and test the natural indicator. 2 …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... (b) Draw a table to show the results obtained in testing this indicator. 2 …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... 20. (a) The leaves were chopped into small pieces, and placed in hot water to extract the purple dye. The solution was decanted and allowed to cool. A range of solutions was tested by adding a few drops of the cabbage dye to each. 2 (b) © (2012) All Rights Reserved 2 of 65 For more info, go to www.hscintheholidays.com.au Subtance water ammonia soln. Vinegar drain cleaner Dye Colour blue green red yellow 2 Question 19 (4 marks) You have performed a firsthand investigation to prepare and test a natural acid/base indicator. (a) Recall the procedure you used to prepare this indicator. 2 …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... (b) Identify an everyday situation in which an indicator is used and explain why it is necessary to use the indicator. 2 …..................................................................................................................................... …..................................................................................................................................... …..................................................................................................................................... …..................................................................................................................................... © (2012) All Rights Reserved 3 of 65 For more info, go to www.hscintheholidays.com.au Question 19 19.3.1 H4, H13 (a) (i) Chop red cabbage into thin pieces.• Clearly presents procedure used to (ii) Grind the cabbage in a mortar and prepare a natural indicator.............................................1‐2 pestle. (iii)
Add some methylated spirits to the cabbage. (iv) Drain the coloured liquid. (b) Indicators can be used to determine the pH of an aquarium. The pH needs to be monitored because the marine life can only survive if the pH is within a specific range. • Identifies an everyday situation in which an indicator is used and explains why it is necessary to use the indicator.................................................2 • Identifies an everyday situation in which an indicator is used OR • Explains why it is necessary to use the indicator.................................................1 •identify that indicators such as • identify data and choose resources litmus, phenolphthalein, methyl to gather information about the orange and bromothymol blue can colour changes of a range of indicators be used to determine the acidic or basic nature of a material over a range, and that the range is identified by change in indicator colour Questions 6 and 7 refer to the table below which shows the colour ranges of three acid‐base indicators. 6. A solution is yellow in bromothymol and methyl orange, and colourless phenolphthalein. What is the pH range of the solution? (A) 7.5 to 8.5 (B) 6.0 to 7.5 (C) 4.5 to 6.0 (D) 8.5 to 10.0 © (2012) All Rights Reserved 4 of 65 For more info, go to www.hscintheholidays.com.au 7. 0.1 mol L‐1 citric acid (C6H807) solution is neutralised with a solution of 0.1 mol L‐1 sodium hydroxide (NaOH). The best indicator for this titration would be: (A) methyl orange (B) phenolphthalein (C) a mixture of methyl orange and bromothymol blue. (D) bromothymol blue 6
C
7 B 8. The table shows the colours of three indicators at different hydrogen ion concentrations. [HCL] mol L‐1 10‐2
10‐4
10‐6
Methyl Orange red orange yellow Bromothymol Blue yellow yellow green Phenol Red yellow red red What is the pH of a solution that showed the following indicator colours? Methyl Orange Yellow Bromothymol Blue Green Phenol Red Red (A) 2 (B) 4 (C) 6 (D) 8 © (2012) All Rights Reserved 5 of 65 For more info, go to www.hscintheholidays.com.au • identify and describe some everyday uses of indicators including the testing of soil acidity/basicity 6. • solve problems by applying information about the colour changes of indicators to classify some household substances as acidic, neutral or basic Some household cleaners contain strong bases such as sodium hydroxide. A student tested household cleaning solutions with litmus and recorded the results in the following table: Cleaning solution Colour of blue litmus Colour of red litmus X blue red Y blue blue Z red red Sodium hydroxide could be present in: (A) X and Y (B) X and Z (C) Y only (D) Z only © 6
C
(2012) All Rights Reserved 6 of 65 For more info, go to www.hscintheholidays.com.au Chemistry Stage 6 Syllabus Students learn to: Students: • identify oxides of non‐ • identify data, plan and 2. While we metals which act as acids perform a first‐hand usually think of investigation to and describe the the air around us decarbonate soft drink conditions under which as neutral, the and gather data to they act as acids atmosphere measure the mass naturally changes involved and contains acidic calculate the volume oxides of carbon, nitrogen and of gas released at 25°C sulfur. The and I00kPa concentrations of these acidic oxides have been increasing since the Industrial Revolution • analyse information from • analyse the position of secondary sources to these nonmetals in the summarise the industrial Periodic Table and origins of sulfur dioxide outline the relationship between position of and oxides of nitrogen and elements in the evaluate reasons for Periodic Table and concern about their acidity/basicity of oxides release into the environment • Define Le Chatelier's Principle • identify factors which can affect the equilibrium in a reversible reaction • describe the solubility of carbon dioxide in water under various conditions as an equilibrium process and explain in terms of Le Chatelier's principle • identify natural and industrial sources of sulfur dioxide and oxides of nitrogen • describe, using equations, examples of chemical reactions which release sulfur dioxide and chemical reactions which © (2012) All Rights Reserved 7 of 65 For more info, go to www.hscintheholidays.com.au release oxides of nitrogen • assess the evidence which indicates increases in atmospheric concentra‐ tion of oxides of sulfur and nitrogen • calculate volumes of gases given masses of some substances in reactions, and calculate masses of substances given gaseous volumes, in reactions involving gases at 0°C and 100kPa or 25°C and 100kPa • explain the formation and effects of acid rain identify oxides of non‐metals which act as acids and describe the conditions under which they act as acids analyse the position of these nonmetals in the Periodic Table and outline the relationship between position of elements in the Periodic Table and acidity/basicity of oxides define Le Chatelier's principle • identify factors vvhichcan affect the equilibrium in a reversible reaction • describe the solubility of carbon dioxide in water under various conditions as an equilibrium process and explain in terms of Le Chatelier's principle © (2012) All Rights Reserved 8 of 65 For more info, go to www.hscintheholidays.com.au 2. The volume of a gas formed during an equilibrium process was monitored under various conditions. The following table summarises the results of these experiments. Based on these results, what can be concluded about the reaction used to produce this gas? (A) It is exothermic and favours high pressures. (B) It is exothermic and favours low pressures. (C) It is endothermic and favours high pressures. (D) It is endothermic and favours low pressures. Question 2 A The greater amount of gas was produced at lower temperature (indicating that it is an exothermic process) and higher pressure. 4. The following reaction is allowed to establish an equilibrium. When added to this equilibrium, which of the following soluble chemicals will least affect its position? (A) Copper (II) fluoride (B) Hydrogen chloride (C) Sodium hydroxide (D) Copper (II) nitrate Question 4 D A. Copper (II) fluoride provides F‐ ions and so equilibrium shifts to the left. B. Hydrogen chloride will dissolve in the water increasing the H30+ concentration and so shift equilibrium to the left. C. Sodium hydroxide would provide OH‐ ions; these would react with the H30+; the equilibrium would shift to the right in an attempt to replace the H30+. D. Cu2+ and N03‐ ions do not affect any of the species present in this equilibrium. © (2012) All Rights Reserved 9 of 65 For more info, go to www.hscintheholidays.com.au Question 24 (4 marks) An equilibrium exists between gaseous and dissolved carbon dioxide in water as shown by the following equation: CO2(g) CO2(aq) With reference to Le Chatelier's principle explain the following: (a) fizzing occurs when a bottle of a carbonated drink is opened. 2 …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... (b) It is observed that the fizzing is less if the bottle is kept under refrigeration rather than at room temperature. Deduce whether the dissolving process is exothermic or endothermic; explaining your reasoning. 2 …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... 24. (a) Opening the bottle causes a decrease in pressure causing the equilibrium to shift to the side of greatest number of gaseous molecu!es, therefore formation of more gaseous CO2 (2 mk) (b) As lowering the temperature favours the formation of aqueous CO2 , this reaction is the reaction that produces heat, therefore the forward reaction as written is exothermic. (2 mk) 7. What effect would the addition of dilute hydrochloric acid have on the following equilibrium? CO2(aq) + H20(I) H+(aq) + HCO3‐(aq) (A) The equilibrium would shift to the left, decreasing the concentration of HC03‐ (B) The equilibrium would shift to the right, increasing the concentration of HC03‐ (C) The equilibrium would not change (D) The rate at which equilibrium is attained would be increased © (2012) All Rights Reserved 10 of 65 For more info, go to www.hscintheholidays.com.au 8. It is known that gases A and B reach equilibrium as they react together to form gas C. the variationin concentration of these gases was monitored and graphed as illustrated below. By applying Le Chatelier's principle. it can be predicted that at time t1 the yield of the forward reaction will (A) increase if pressure is increased. (B) decrease if pressure is increased. (C) decrease if pressure is decreased. (D) not be affected by a change in pressure. Question 8 D 9.3.2, 9.4.2 H10,H14 The equation for the reaction is A + B 2C. Since one mole of each A and B are consumed to form 2 moles of C, there are 2 moles of gas on each side of the equation. Therefore the ratio of moles is unaffected by the change in pressure. 13. Sulfuric acid reacts with pyrosulfuric acid according to the equation:‐ Identify a method of increasing the concentration of H3S04 + in the mixture at equilibrium. (A) increase the pressure on the system (B) add H2S04 (C) add a catalyst (D) add HS207 ‐ © (2012) All Rights Reserved 11 of 65 For more info, go to www.hscintheholidays.com.au 11. Nitrogen dioxide, N02, a brown gas and dinitrogen tetroxide, N204, a colourless gas are in equilibrium according to the equation:‐ If a sealed tube of the gases is placed in an ice‐water bath the colour fades from brown to almost colourless. Which conclusion is correct? (A) The forward reaction is exothermic (B) The reverse reaction is exothermic (C) The pressure has increased. (D) The pressure has not changed Marks © (2012) All Rights Reserved 12 of 65 For more info, go to www.hscintheholidays.com.au Question 25 (7 marks) As part of your practical work you decarbonated a beverage. A student decarbonated a sample of soda water by opening the bottle it was in and leaving it for a period of time, weighing it at regular intervals. She also used a non‐carbonated sample of water as a control, recording its mass at the same intervals. Mass (g) Initial After 12 hours 24 hours 36 hours 48 hours 60 hours Soda water 385.0 382.6 381.1 380.7 380.3 380.0 Plain water 385.0 384.7 384.2 383.7 383.4 383.0 (a)
Graph the information shown for each water sample on the same graph. 2 (b) Interpret the trends shown in the graph. 2 …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... © (2012) All Rights Reserved 13 of 65 For more info, go to www.hscintheholidays.com.au (c) Use the graph to determine the volume of CO2 gas at C25°C and 100kPa. Show your working. 2 …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... © (2012) All Rights Reserved 14 of 65 For more info, go to www.hscintheholidays.com.au • analyse information from • identify natural and industrial secondary sources to summarise sources of sulfur dioxide and the industrial origins of sulfur oxides of nitrogen dioxide and oxides of nitrogen and evaluate reasons for concern about • describe, using equations, examples of chemical reactions which release sulfur dioxide and chemical reactions which release oxides of nitrogen • assess the evidence which indicates increases in atmospheric concentration of oxides of sulfur and nitrogen Marks Question 21 (6 marks) A sample of lignite, a high sulfur content coal, was analysed and found to contain 4.32% sulphur. (a) Calculate the volume of sulfur dioxide, at 25°C and 100 kPa, that would be produced by burning 1.0 kg of lignite coal. 3 …..................................................................................................................................... …..................................................................................................................................... …..................................................................................................................................... …..................................................................................................................................... …..................................................................................................................................... …..................................................................................................................................... (b) Assess the impact, on the environment, of using lignite as a fuel, writing equations where appropriate. 3 …..................................................................................................................................... …..................................................................................................................................... …..................................................................................................................................... …..................................................................................................................................... …..................................................................................................................................... …..................................................................................................................................... © (2012) All Rights Reserved 15 of 65 For more info, go to www.hscintheholidays.com.au 21. (a) (b) Moles of sulfur = 43.5/32 = 1.35 mol Moles of S02 formed = 1.35 mol Volume of S02 formed = 1.35 x 24.79 = 33.47 L 3 Combustion of the lignite releases S02 into the atmosphere. This gas causes breathing difficulties for some people. It also undergoes oxidation to S03 which dissolves in rainwater to form sulfuric acid. Acid rain causes many problems including corrosion of metals and building materials, damage to plants and aquatic systems such as freshwater lakes and release of heavy metals by accelerated weathering of rocks. 3 © (2012) All Rights Reserved 16 of 65 For more info, go to www.hscintheholidays.com.au Question 22 (6 marks) Marks Assess the evidence which indicates that the atmospheric concentration of oxides of sulfur and nitrogen have been increasing. 6 …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... Sample answer Syllabus content, course outcomes and marking guide Question 22 Oxides of sulfur and nitrogen are found in the 9.3.2 H3, H4, H8, H13, H14 atmosphere as a result of a number of • Assesses evidence relating to increases natural processes, notably volcanic action in concentrations of the oxides of and decomposition of organic wastes in the nitrogen sulfur.....................................5‐6 case of SO2 and lightning in the case of oxides of nitrogen. Since the beginning of the • Describes evidence that the concentrations of oxides of nitrogen and industrial revolution, processes associated with industrialisation and increased human sulfur have increased due to human causes.................................................5‐6 population have added greatly to these natural sources of these These manmade sources include the processing of sulfide ores • Describes effects of the oxides of of oxides of nitrogen and sulfur have to extract metals and production of sulfuric acid, nitrogen and sulfur..............................1‐2 which increase concentration of S02 and © (2012) All Rights Reserved 17 of 65 For more info, go to www.hscintheholidays.com.au combustion of fossil fuels in power plants,car engines and jet engines, which have increased concentrations of the oxides of sulfur and nitrogen. The increased concentrations of these oxides has been evident by the formation of acid rain, which has resulted in damage to waterways, forests, crops and buildings over many years. Other evidence for the increased concentration of these oxides has come from atmospheric pollution, with nitrogen oxides contributing to photochemical smog. Higher levels of these oxides have also contributed to high rates of breathing difficulties often seen in large cities. Over the last few decades, anti‐pollution measures have been introduced to reduce industrial S02 output and catalytic converters have been a compulsory part of the pollution control equipment of new cars to reduce oxides of nitrogen. Together with standards requiring lower levels of sulfur in fuels, these measures have resulted in many industrial countries experiencing a decrease in the level of these oxides. • identify data, plan and perform a • calculate volumes of gases given first‐hand investigation to masses of some substances in decarbonate soft drink and gather reactions, and calculate masses of substances given gaseous volumes data to measure the mass changes in reactions involving gases at 0°C involved and calculate the volume and 100kPa or 25°C and 100kPa of gas released at 25°C and 100kPa © (2012) All Rights Reserved 18 of 65 For more info, go to www.hscintheholidays.com.au Question 17 (4 marks) A small amount of pure sodium metal is dropped into 1.2 L of water. The reaction is summarised in the following equation. The gas collected occupied a volume of 4.68 L at 25°C and 1 atm pressure. (a) Calculate the amount in moles of gaseous product. 1 …............................................................................................................................ …............................................................................................................................ …............................................................................................................................. …............................................................................................................................. …............................................................................................................................. (b) Calculate the final pH of the water. 3 …........................................................................................................................... …........................................................................................................................... …........................................................................................................................... …........................................................................................................................... …........................................................................................................................... …........................................................................................................................... © (2012) All Rights Reserved 19 of 65 For more info, go to www.hscintheholidays.com.au Question 21 (2 marks) When 1.5 L of HCI gas and 1.8 L of NH3 gas are mixed, a white solid of NH4C1 is formed. Calculate the mass of NH4Cl formed if the gas volumes were measured at 25°C and 101.3 kPa. 2 …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... 21 Moles HC] (limiting reactant) = 1.5/24.47 = 0.0613 mol (1mk) Mass NH4Cl formed = 0.0613 x mol. mass = 0.0613 x 53.5 = 3.28 g (1mk) Question 27 (3 marks) The data below gives the percentage composition of air by volume at sea level for a town on the far north coast of NSW. Constituent Symbol Volume % in air Molar mass Nitrogen N2
78.084 28.01 Oxygen 02 20.9476 32.00 Argon Ar 0.934 39.95 Carbon dioxide CO2
0.037 44.01 Neon Ne 0.001818 20.18 Helium He 0.000524 4.00 Methane CH4
0.00017 16.04 (a) Calculate the moles of oxygen present in 20 litres of this air at 25°C and 101.3 kPa. 1 ….......................................................................................................................... …......................................................................................................................... …......................................................................................................................... (b) Calculate the mass of argon which could be extracted from 200 litres of this air. 2 …........................................................................................................................ © (2012) All Rights Reserved 20 of 65 For more info, go to www.hscintheholidays.com.au …......................................................................................................................... …......................................................................................................................... …......................................................................................................................... 27. (a) moles of oxygen = 20 x 0.2095/24.47 = 0.171 mol (1mk) (b) moles of argon = 200 x 0.00934/24.47 = 0.0763 mol (1mk) mass of argon = moles x mol. mass = 0.0763 x 39.95 =3.05 g (1mk) © (2012) All Rights Reserved 21 of 65 For more info, go to www.hscintheholidays.com.au Question 21 (5 marks) Marks Low sulfur diesel fuels used in coal mining must have a sulfur content ofless than 0.05% sulfur by mass. (a) Calculate the volume of sulfur dioxide at 25°C and 100 kPa produced by burning 1.0 kg oflow (0.05%) sulfur diesel. 2 …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... (b) Discuss the impact on the environment of using high sulfur fuels. 3 …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... …...................................................................................................................................... 21. (a) S(s) + O2(g) SO2(g) Mass of sulfur in lignite = 1000 x 0.05/100 = 0.5g Moles of sulfur = mass/molar mass = 0.5/32.07 = 1.56 x10‐2 1 Moles of sulfur dioxide = moles of sulfur =1.5x10‐2 Volume of S02 = Moles x Molar volume © (2012) All Rights Reserved 22 of 65 For more info, go to www.hscintheholidays.com.au = 1.56 x 10‐2 x 24.79 =0.386 L 1 (b) The impact ofbuming large quantities of high sulfur fuels is detrimental to the environment where that fuel is being burned. ‐ high sulfur fuels release large amounts of sulfur dioxide into the atmosphere. 1 ‐ Sulfur dioxide is poisonous to living things 1 ‐ Sulfur dioxide dissolves in water in the atmosphere to produce sulphurous/ic acid ‐ sulphurous/ic acids fall as acid rain lowering the pH of many natural systems with the potential to change the environment and the organisms in it 1 • explain the formation and effects of acid rain © (2012) All Rights Reserved 23 of 65 For more info, go to www.hscintheholidays.com.au Chemistry Stage 6 Syllabus Students learn to: • define acids as proton donors and 3. Acids occur in many foods, drinks describe the ionisation of acids in and even within our water stomachs • identify acids including acetic(ethanoic), citric (2‐ hydroxypropane‐l,2,3‐ tricarboxylic), hydrochloric and sulfuric acid • describe the use of the pH scale in comparing acids and bases • describe acids and their solutions with the appropriate use of the terms strong, weak, concentrated and dilute • identify pH as ‐loglO [H+] and explain that a change in pH of 1 means a ten‐fold change in [H+] • compare the relative strengths of equal concentrations of citric, acetic and hydrochloric acids and explain in tenns of the degree of ionisation of their molecules • describe the difference between a strong and a weak acid in tenns of an equilibrium between the intact molecule and its ions • define acids as proton donors and describe the ionisation of acids in water Students: • solve problems and perform a first‐hand investigation to use pH meters/probes and indicators to meters/probes and indicators to distinguish between aciilic, basic and neutral chemicals • plan and perform a first‐hand investigation to measure the pH of identical concentrations of strong and weak acids • gather and process information from secondary sources to wri te ionic equations to represent the ionisation of acids • use available evidence to model the molecular nature of acids and simulate the ionisation of strong weak acids • gather and process information from secondary sources to explain the use of acids as food additives • identify data, gather and process information from secondary sources to identify examples of naturally occurring acids and bases and their chemical composition • process information from secondary sources to calculate pH of strong acids given appropriate hydrogen ion concentrations • gather and process information from secondary sources to write ionic equations to represent the ionisation of acids © (2012) All Rights Reserved 24 of 65 For more info, go to www.hscintheholidays.com.au 6. Consider each of the following equations: (i) HCI(aq) ‐H+(aq) + Cl‐(aq) (ii) HCI(aq)+ H20(l) ‐ H30+(aq) + Cl‐(aq) (iii) NH3(aq) + H20(l) ‐NH4 +(aq)+ OH‐(aq) (iv) NaOH(aq) ‐Na+(aq) + OH‐(aq) An acid can be defined as a proton donor. This can be seen in (A) equation (i) only. (B) equation (ii) only. (C) equations (ii) and (iii). (D) all of the equations. Question 6 C 9.3.3 H6 Equation (i) shows the Hcl ionising. Equation (iv) shows the NaOH dissociating. In equation (ii), HCI is acting as a proton donor. In equation (iii), H20 is acting as a proton donor. © (2012) All Rights Reserved 25 of 65 For more info, go to www.hscintheholidays.com.au identify acids including acetic (ethanoic), citric (2‐ hydroxypropane‐l,2,3‐ tricarboxylic), hydrochloric and sulfuric acid describe the use of the pH scale comparing acids and bases solve problem and perform a first‐hand investigation to use pH meters/probes and indicators to distinguish between acidic, basic and neutral chemicals. describe acids and their solutions with the appropriate use of the tenns strong, weak, concentrated and dilute 9. The pH of four acids and their concentrations are shown in the table below. Acid Conc. (mol L‐1) pH A 0.1 1.0 B 0.05 1.0 C 0.01 2.0 D 0.1 2.0 Which acid in the table is the weakest? (a) A (b) B (c) C (d) D 3. Consider the following reagent bottles of acids: In comparing these two solutions we can say that A. The [H+] is greater in the solution of acid A B. The [H+] is greater in the solution of acid B C. The acids are of equal strength D. A is the stronger acid © (2012) All Rights Reserved 26 of 65 For more info, go to www.hscintheholidays.com.au Question 3 D 3.3.3 H8, H10 Since both acids has the same pH, their respective [H+] must be the same. However, acid A is of a lower concentration so must be a stronger acid compare the relative strengths of equal concentrations of citric, acetic and hydrochloric acids and explain in terms of the degree of ionisation of their molecules describe the difference between strong and a weak acid in terms of an equilibrium between the intact molecule and its ions 1.
6. plan and perform a first‐hand investigation to measure the pH of identical concentrations of strong and weak acids use available evidence to model the molecular nature of acids and simulate the ionisation of strong and weak acids In an aqueous solution of the weak acid nitrous acid, HN02, which of the following species is present in the highest concentration? A. HN02 B. H30+ C. OH‐ D. N02‐ Answer and explanation Question 1 A Since this is a weak acid, only a small proportion of molecules have ionised. Therefore, the majority of molecules have remained intact. Citric acid (2‐hydroxypropane‐l,2,3‐tricarboxyiic acid) is a weaker acid than sulfuric acid, even though citric acid is triprotic. Which of the following best explains the above statement? A. Citric acid ionises more completely than sulfuric acid. B. Sulfuric acid will react completely with a base, but citric acid will only react partially with a base. C. Sulfuric acid is diprotic and therefore ionises more easily. D. Citric acid ionises less completely than sulfuric acid. Question 6 D 9.3.3 Acid strength is measured by the extent of ionisation of the acid. Weak acids ionise less than stronger acids. © (2012) All Rights Reserved 27 of 65 For more info, go to www.hscintheholidays.com.au H8. H14 8. A number of solutions were tested with a conductivity probe attached to a data logger. Which of the following solutions would record the highest conductivity reading? A. 0.01 mol L‐1 HCl B. 0.1 mol L1 HCI C. 0.01 mol L‐1 CH3COOH D. 0.1 mol L‐1 CH3COOH 8 B © (2012) All Rights Reserved 28 of 65 For more info, go to www.hscintheholidays.com.au Question 22 (3 marks) Citric acid and acedic acid are both weak acid. (a)
Write the structural formula of each acid. …………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………… (b) State why citric acid is stronger than acetic acid 2 1 …………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………… Question 22 9.3.3 H6, H13 (a) Critic acid acetic acid Draws two correct structural formulae ……..2 Draws one correct structural formulae………1 (b) Critic acid ionizes more than acetic acid, hence it is a stronger acid. Note: not acceptable to state that critic is triproic. States a correct reason…………………….. 1 © (2012) All Rights Reserved 29 of 65 For more info, go to www.hscintheholidays.com.au identify pH as ‐loglO [H+] and explain that a change in pH of 1 means a ten‐
fold change in [H+] process information from secondary sources to calculate pH of strong acids given appropriate hydrogen ion concentrations 3. A solution of pH 9 has enough water added to it so that it now has a pH of 8. What effect does this have on the solution? A.
It becomes 10 times more concentrated and less alkaline. B.
It becomes 10 times more dilute and more alkaline. C.
It becomes 10 times more concentrated and more alkaline. D.
It becomes 10 times more dilute and less alkaline. Question 3 D By adding more water, the solution must now be more dilute. Since the pH has decreased, the solution is also less alkaline. 10. The pH of a solution of magnesium hydroxide of concentration 4.5 x 10‐3 mol L‐I is closest to: A. 11.9 B. 11.6 C. 2.1 D. 2.4 10 A 5. A solution of pH 3 has the necessary changes made to it so that it is now a pH of 5. What change has been made to the concentration of H+? A. It has become more concentrated by a factor of 2. B. It has become more concentrated by a factor of 100. C. It has become less concentrated by a factor of 2. D. It has become less concentrated by a factor of 100. Question 5 D At a pH of 3, the concentration of H+ is equal to 10‐3mol L‐1
At a pH of 5, the concentration of H+ is equal to 10‐5mol L‐1
Therefore, the solution is now more dilute, because the concentration of H+ has decreased by a factor of 10‐3/10‐5 = 100 . © 9.3.3 H8, H10 (2012) All Rights Reserved 30 of 65 For more info, go to www.hscintheholidays.com.au 9. The pH of a 10 mL HCI solution was determined to be a value of 1. Water was added to this solution until its pH changed to a value of 2. The amount of water added was A. 10 mL. B. 20 mL. C. 90 mL. D. 100mL. Question 9 C 9.3.3 H10 An increase in a value of pH of 1 means that the dilution factor was 10. Therefore, the total new volume of the acid solution became 100mL. This would have been achieved by adding 90mL to the original. © (2012) All Rights Reserved 31 of 65 For more info, go to www.hscintheholidays.com.au Question 23 (5 marks) Marks A student used indicator paper to estimate the pH of three different acids, to the nearest integer value. Each acid was at a concentration of 0.1 0 mol L‐l in aqueous solution. The table below records these measurements: Acid
pH
acetic
3
citric
2
hydrochloric
1
(a) Compare the hydrogen ion concentrations in these three solutions. 2 ……………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………….. (b) Account for the differences in these values. ……………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………….. © (2012) All Rights Reserved 32 of 65 For more info, go to www.hscintheholidays.com.au 23. (a) (b) The hydrogen ion concentration in the Hel is 0.1 mol Cl. This is 10 times the concentration in critic acid and 100 times the concentration in the acetic acid. 2 Hydrochloric acid is 100% ionised in dilute solution, so that the hydrogen ion concentration is equal to the acid concentration. (1mk) Critic acid is only partly ionized (~5%) and acetic acid is a still weaker acid and has lower degree of ionization. (1mk) In both weak acids the hydrogen ion concentration is much less than the concentration of the dissolved acid. (1mk) 3 gather and process information from secondary sources to explain the use of acids as food additives identify data, gather and process information from secondary sources to identify examples of naturally occurring acids an bases and their chemical composition © (2012) All Rights Reserved 33 of 65 For more info, go to www.hscintheholidays.com.au Question 21 (4 marks) Vinegar is an aquenos solution of acetic (ethanoic) acid, a weak acid. (a) Apart from its taste, explain why acids such as vinegar are often used as food additives. 2 ………………………………………………………………………………………………………………….. ………………………………………………………………………………………………………………….. ………………………………………………………………………………………………………………….. ………………………………………………………………………………………………………………….. (b) Explain why such a solution would have a higher pH than a hydrochloric acid solution of the same concentration. 2 ……………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………… Question 21 (a) The addition of the acid to food is as a 9.3.3 H3, H4, H8 preservative. The acitic condition inhibits the Identifies and explains growth of microorganisms. use as preservatives…2 (b) Acetic acid is a weak acid. Therefore the acid molecules do not fully ionize, leaving the [H+] lower than expected, resulting in a higher pH. Hydrochloric acid is a strong acid. Therefore it completely ionizes, making the [H+] high and so the pH is lower. © Identifies use as preservatives…………..1 9.3.3 H8, H13, H14 Correctly relates strengths of the acids to [H+] and correctly relates [H+] to pH… 2 Correctly relates strengths of the acids to [H+] OR Correctly relates [H+] to pH…………………… 1 (2012) All Rights Reserved 34 of 65 For more info, go to www.hscintheholidays.com.au Chemistry Stage 6 Syllabus Students learn to: 4. Because of the prevalence and importance of acids, they have been used and studied for hundreds of years. Over time, the definitions of acid and base have been refined © outline the historical development of ideas about acids including those of: - Lavoisier - Davy - Arrhenius outline the Bronsted‐Lowry theory of acids and bases describe the relationship between an acid and its conjugate base and base and its conjugate acid identify a range of salts which form acidic, basic or neutral solutions and explain their acidic, neutral or basic nature identify conjugate acid/base pairs identify amphiprotic substances and construct equations to describe their behaviour in acidic and basic solutions identify neutralisation as a proton transfer reaction which is exothermic describe the correct technique for conducting titrations and preparation of standard solutions qualitatively describe the effect of buffers with reference to a specific example in a natural system outline the historical development of ideas about acids including those of - Lavoisier Students: gather and process infonnation from secondary sources 10 trace developments in understanding and describing acid/base reactions choose equipment and perform a first‐hand investigation to identifythe pH of a range of salt solutions perform a first‐hand investigation and solve problems using titrations and including the preparation of standard solutions, and use available evidence to quantitatively and qualitatively describe the reaction between selected acids and bases perform a first‐hand investigation to determine the concentration of a domestic acidic substance using computer‐based technologies analyse information from secondary sources to assess the use of neutralisation reactions as a safety measure or to minimise damage in accidents or chemical spills gather and process information from secondary sources to trace developments in understanding and describing acid/base (2012) All Rights Reserved 35 of 65 For more info, go to www.hscintheholidays.com.au -
Davy Arrhenius -
reactions 9. According to Lavoisier's ideas about acids, which of the following compounds would be acidic in water? A. N02 B. NH3 C. HC1 D. HBr Question 9 A According to Lavoisier, oxygen in compounds formed with non‐metals causes acidity. © 9.3.4 H1, H8 (2012) All Rights Reserved 36 of 65 For more info, go to www.hscintheholidays.com.au outline the Bronsted‐Lowry theory of acids and bases describe the relationship between an acid and its conjugate base and a base and its conjugate acid 7. Consider the reactions shown below. I H2S03 + H20 ~ H30+ + HS03‐ II Fe3+ + SCN‐ ~ FeSCN2+ Which of the following statements is correct? A. I and II are both acid‐base reactions. B. H2S03/ HS03‐ and Fe3+ / FeSCN2+ are conjugate pairs. C. Reaction I involves the formation of a co‐ordinate covalent bond. D. Reaction I shows HS03‐ acting as an acid. Question 7 C Reaction IT is not an Arrhenius or Lowry‐Bronsted acid base reaction; Fe3+IFeSCN2+ are not a conjugate pair; HS03‐ acting as a base. © 9.3.4 H8 (2012) All Rights Reserved 37 of 65 For more info, go to www.hscintheholidays.com.au identify a range of salts which form acidic, basic or neutral solutions and explain their acidic, neutral or basic nature choose equipment and perform a first‐hand investigation to identify the pH of a range of salt solutions Question 20 (5 marks) An acid, HX, is prepared by dissolving 0.1 moles of it in enough water to make 1 litre of solution. A pH meter shows that the solution pH is 3.5. (a) Calculate the [H3O+] for the solution. 1 ……………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………….. (b) Explain whether HX is a weak or strong acid. 2 ……………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………….. (c) The salt, NaK is dissolved in water. Predict whether the solution is acidic, neutral or basis, using an appropriate equation to justify your prediction. 2 ……………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………….. Question 20 (a)
[H30+] = 10‐3.5mol L‐1 (or 3.2 x 10‐4 mol L‐I) (b)
HX is a weak acid, as the concentration of H30+ is much lower than that of the acid, indicating that it has only partially ionised. H10 Correct calculation…………………. 1 H12 States HX is.a weak acid AND has partially ionised since its pH is greater than I .... ……………………. 2 Poor explanation ……………………. 1 (c)
© X‐ is a strong conjugate base compared to the weak acid HX. X‐ will then ionise water to produce OH‐ ions, so the solution will be basic.
X‐ + H2O HX + OH‐ States solution is basic AND shows production of OH‐ ions ………….. 2 States solution is basic OR (2012) All Rights Reserved 38 of 65 For more info, go to www.hscintheholidays.com.au Shows production of OH‐ ions….. 1 10. Identify the pH at the neutralisation point when sodium hydroxide is neutralised by hydrochloric acid. A.
B.
C.
D.
pH =0 pH =7 pH >7 pH <7 B
Question 18 (3marks) Certain salt dissolved in water to lower its pH. (a) Identify such a salt. 1 ………………………………………………………………………………………………………………………. (b) With the help of an equation, explain how the pH is lower. 2 ………………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………….. Sample Answer Question 18 (a) Ammonium chloride Syllabus content, course outcomes and marking guide 9.3.4 H6, H8 Identifies ammonium chloride as the salt…………………………………………………. 1 (b) NH4CI+ H20 ~ H30+ + CI‐ + NH3 The H30+ produced in this reaction lowers the pH. Identifies that the formation of H30+ lowers the pH and provides a suitable quation ……………………………………….. 2
Identifies that the fonnation of H30+ lowers the pH or provides a suitable quation …………………………………………. 1
© (2012) All Rights Reserved 39 of 65 For more info, go to www.hscintheholidays.com.au Identify conjugate acid/base pairs Identify amphiprotic substance and construct equations to describe their behavior in acidic and basic solutions Question 22 (3 marks) Choose an example of an amphiprotic substance and write equations to help explain its behaviour in acidic and basic solutions. 3 …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. 22. eg: sodium hydrogen carbonate (or the hydrogen carbonate ion) (1 mk) This species acts as a proton donor and acceptor, shown by:‐ 10. The conjugate base of the NH/ion has the formula: A. NH40H B. OH‐ C. NH3 D. NH42+ 10 C 7. Which of the following statements identifies the conjugate base of the acid HN03? A. NaOH is the conjugate base of the acid HN03 B. OH‐ is the conjugate base of the acid HN03 C. N03 is the conjugate base of the acid HN03 D. N03‐ is the conjugate base of the acid HN03 D © (2012) All Rights Reserved 40 of 65 For more info, go to www.hscintheholidays.com.au Question 20 (3 marks) A student dissolved some NaHC03 in a small amount of water. She knew that HC03‐(aq) could react in each of the following ways. (a) Name the type of behaviour being shown by HCO3 ‐1 (aq) 1 ……………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………. (b) Describe a simple test you could perform to determine whether reaction I or II is more likely to occur. Give the expected result for your test. 2 …………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………... ……………………………………………………………………………………………………………………… Question 20 (a) It is amphiprotic, acting as a base in I by accepting a proton and as an acid in II by donating a proton to water. 9.3.4 H8 Names amphiprotic, no description necessary………………………. 1 9.3.1 H8, HI1 (b) Dissolve a small quantity of NaHC03 in water and add some universal indicator or litmus. If I is more Describes a suitable test and likely then the indicator will be blue or violet or gives the appropriate result red litmus will turn blue. If II is more likely then the ……………………………………….. 2 indicator will be yellow or red, and blue litmus will Describes a test turn red. OR Gives a result………………….. 1 identify neutralisation as a proton transfer reaction which is exothermic © (2012) All Rights Reserved 41 of 65 For more info, go to www.hscintheholidays.com.au 10. 40 mL of 5 mol L‐1 HCl is added to 20 mL of 5 mol L‐1 NaOH. Which of the following correctly summarises the results? Temperature Change (A). increase =7 (B). increase <7 (C). decrease >7 (D). decrease <7 Question 10 B The reaction of an acid with a base is exothermic, and so the temperature would have increased. However, in this case, the acid is in excess and so the final pH is still acidic. © Temperature Change 9.3.3 H8, H10 (2012) All Rights Reserved 42 of 65 For more info, go to www.hscintheholidays.com.au describe the correct technique for conducting titrations and preparation of standard solutions perform a first‐hand investigation and solve problems using titrations and including the preparation of standard solutions, and use available evidence to quantitatively and qualitatively describe the reaction between selected acids and bases perform a first‐hand investigation to determine the concentration of a domestic acidic substance using computer‐based technologies Questions 13 and 14 are based on the following information. For a practical test, a student carried out a series of titrations usin¥ hydrochloric acid to determine the concentration of household ammonia. The student used the 0.1 mol L‐ HCl provided for the titration and the household ammonia solution which had been diluted by a factor of 10. Three titrations were performed and the results were 12.3 mL, 12.5 mL and 12.3 mL. The expected concentration of the ammonia solution was 3% but the student's value was calculated to be 4.5%. 13. What should the student do to assess the validity of her results? A. Check the concentration of the O.l mol L‐1HCl against a primary standard. B. Perform more titrations and average the results. C. Use the full strength ammonia solution to obtain a more accurate result. D. Dilute the ammonia by a factor of 20 to obtain a more accurate result. 14. Which of the following reasons, given by other students, best justifies the difference between the calculated results and the expected value? A. Titration is not an accurate method. B. The difference between 3% and 4.5% is not significant. C. Titration is an accurate method and the solutions provided should be checked. D. Spectrophotometry would have been a more suitable method to determine concentration. Question 13 A Of the alternatives provided, this would be the most useful because the titration values were so close. Question 14 C There must have been an en‐or in preparation of the solutions as titration is a suitable method. © (2012) All Rights Reserved 43 of 65 For more info, go to www.hscintheholidays.com.au Question 19 (8 marks) The accuracy of acid‐base titrations depends on several factors. These include the primary standard used, how the glassware is prepared and how the equivalence point is determined. (a) Explain why sodium hydroxide is not used as a primary standard. 1 ……………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………. (b) Anhydrous sodium carbonate can be used as a primary standard. How can we ensure that the sodium carbonate remains anhydrous? 1 ……………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………. (c) During a titration, a conical flask is prepared by rinsing it with distilled water. While this flask is still wet, a clean, dry pipette is used to transfer 20 mL of a standard solution into it. Will the accuracy of the titration be affected? Explain your answer. 2 ……………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………. (d) Although an indicator can be used to determine the equivalence point of an acid‐base titration, an alternative method is to monitor the electrical conductivity of the reaction mixture during the titration. The following graph shows the variation in electrical conductivity during such a titration. volume of solution added from burette Explain why the electrical conductivity: © (2012) All Rights Reserved 44 of 65 For more info, go to www.hscintheholidays.com.au (i)
Starts at a minimum but then decreases to a minimum values. 2 (ii)
…………………………………………………………………………………………………………….. …………………………………………………………………………………………………………….. Does not reach a zero level. 1 (iii)
…………………………………………………………………………………………………………….. …………………………………………………………………………………………………………….. Starts to increase again after the minimum value. 1 …………………………………………………………………………………………………………….. …………………………………………………………………………………………………………….. Sample answer Syllabus outcomes and marking guides Question 19 (a) Sodium hydroxide deliquesces when exposed to moisture. Therefore as it is being weighed its mass increases as it absorbs moisture. Since it is not possible to know how much moisture it has absorbed, the mass measurement is inaccurate. (b) The sodium carbonate can be stored in a desiccator. HII, HI2 Sodium hydroxide reacts with the atmosphere …………………………….. 1 HI2 Keep sodium carbonate in a dry environment. …………………………… 1 (c) The accuracy of the titration will not be HIO affected. This is because the moles of the States moles of reactant is reactant transferred from the pipette into the unaffected with explanation …….. 2
conical flask is unaffected by the volume of States moles of reactant is water already in that flask. unaffected with no explanation .... 1 (d) (i) The concentration of ions at the beginning HIO of the titration is at a maximum, hence there Relates maximum conductivity to is maximum electrical conductivity. As the maximum concentration of ions solution from the burette is added,' AND relates decrease to decreasing neutralisation begins to occur. This effectively concentration of ions …………........ 2 decreases the concentration of the ions and Relates maximum conductivity to therefore the electrical conductivity of the maximum concentration of ion solution will also decrease. OR Relates decrease to decreasing concentration of ions ................... 1 © (2012) All Rights Reserved 45 of 65 For more info, go to www.hscintheholidays.com.au (ii) At minimum electrical conductivity, the equivalence point has been achieved. However there are still ions present from the salt produced by the reaction, therefore there is still some conductivity possible (iii) As solution from the burette is still being added to the reaction mixture, but there are no further ions available for reaction. the concentration of the ions in the solution increases and hence so does its electrical conductivity. HlO Relates minimum conductivity to equivalence point and minimum concentration of ions…………………. 1 HIO Relates increasing conductivity to increasing concentration of ions … 1 8. Three pieces of apparatus used in titrations are a conical flask, a burette and a pipette. Which of these pieces of apparatus should be rinsed with distilled water immediately prior to use in a titration? A. all three B. conical flask only C. burette only 8 B D. pipette only © (2012) All Rights Reserved 46 of 65 For more info, go to www.hscintheholidays.com.au Question 24 (4 marks) In a titration it is found that 20.0 mL of 0.200 mol L‐1 sulfuric acid is required to neutralise 25.0 mL of a potassium hydroxide solution. (a) Write a balance equation for the neutralization. 1 …………………………………………………………………………………………………………………………. (b) Calculate the concentration of the potassium hydroxide in mol L‐1. 2 …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. (c) Calculate the mass of potassium hydroxide in 5 litres of the above solution. 1 …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. 25. (a) H2S04 + 2KOH ‐7 K2S04 + 2H20 (1 mk) (b) no. of moles of KOH = 2 x no. of moles of H2S04 25.0 x [KOH] = 2 x 0.200 x 0.0200 (1 mk) [KOH] = 0.320 mol L‐l (c) Mass KOH = molarity x volume x mol. mass = 0.32 x 5 x 56.1 (1 mk) = 89_8g 9. A student used a pipette to transfer 25.0 mL of a solution to a flask. After draining the solution into the flask a small amount of solution remained in the tip of the pipette. To deliver the correct volume of solution to the flask the student should: A.
blow the remaining solution into the flask B.
touch the inside of the flask with the tip of the pipette C.
shake the pipette to dislodge the remaining solution D.
rinse the pipette with a small quantity of distilled water into the flask © (2012) All Rights Reserved 47 of 65 For more info, go to www.hscintheholidays.com.au 8. A student performed a titration and presented her results in the following graph. 14 pH 7 0 mL of base added What does the pH of the equivalence point suggest? A. The solution was neutral at that point. B. There are more OH‐ ions than H+ions. C. There are free OH‐present in the original solution. D. The acid/base mixture was never neutral. Question 8 B Since the equivalence point is at a pH > 7 the mixture is alkaline. This means there must be more OH‐ than H+ present. © 9.3.4 H13 (2012) All Rights Reserved 48 of 65 For more info, go to www.hscintheholidays.com.au Question 22 (7 marks) To prepare a standard solution of sodium hydroxide a student first dissolved 1.0 g of solid sodium hydroxide in 250 mL of distilled water. By titration, 25.0 mL of this solution required 23.2 mL of standard 0.100 mol L‐j hydrochloric acid for neutralisation. (a) Why is titration necessary to standardise the sodium hydroxide solution? 1 ……………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………… (b) Calculate the concentration of the standardized sodium hydroxide solution. 2 ……………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………… (c) Describe the titration procedure for the standardisation. 4 ……………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………… © (2012) All Rights Reserved 49 of 65 For more info, go to www.hscintheholidays.com.au 22. (a) sodium hydroxide absorbs both water and carbon dioxide from the air, so that it cannot be used as a primary standard. Titration is needed to determine its concentration. 1 (b) [NaOH] = 23.2 x 0.100/25.0 = 0.0928 mol L‐j (mole ratio HCI:NaOH = 1:1) 2 (c) Titration procedure to include use of pipette for NaOH solution, burette for HCI, conical flask and suitable indicator such as phenolphthalein. All glassware to be rinsed with distilled water, followed by the solutions for the pipette and burette. 2 © Titration carried out with swirling of flask contents, with the end point recorded as the volume of HCl to decolorise the indicator. Rinse flask with distilled water after each titration. Minimum of three titration measurements, with two agreeing within 0.1 mL. 2 (2012) All Rights Reserved 50 of 65 For more info, go to www.hscintheholidays.com.au Question 23 (5 marks) A titration was carried out using 0.246 mol L‐1 HCL to standardise 25.0 mL aliquots of a solution of the weak base, sodium carbonate. An appropriate indicator was chosen to show the end point of the neutralisation. The results gained are shown in the table below. Run 1 2 3 4 5 Initial burette volume (mL) 0.5 23.6 0.7 23.5 0.2 Final burette volume (mL) 23.5 45.8 23.0 46.2 22.4 (a) Calculate the concentration of the sodium carbonate solution. Justify the steps in your calculation. 3 ………………………………………………………………………………………………………………………….. ………………………………………………………………………………………………………………………….. ………………………………………………………………………………………………………………………….. ………………………………………………………………………………………………………………………….. ………………………………………………………………………………………………………………………….. ………………………………………………………………………………………………………………………….. ………………………………………………………………………………………………………………………….. ………………………………………………………………………………………………………………………….. ………………………………………………………………………………………………………………………….. (b) The student has a choice of indicators: methyl orange; changes from red to orange from pH 3.0 to 4.5. phenolphthalein; changes from colourless to pink from pH 8.3 to 10.0. 2 Select the indicator that should be used for this titration, giving a reason for your choice. ………………………………………………………………………………………………………………………….. ………………………………………………………………………………………………………………………….. ………………………………………………………………………………………………………………………….. ………………………………………………………………………………………………………………………….. ………………………………………………………………………………………………………………………….. © (2012) All Rights Reserved 51 of 65 For more info, go to www.hscintheholidays.com.au Sample Answer Question 23 (a) Run 1 2 (rough) Initial burette 0.5 23.6 volume (mL) Final burette 23.5 45.8 volume (mL) Volume used 23.0 22.2 3 Syllabus content, course outcome
and marking guide 9.3.3 H10, H12 4 5 0.7 23.5 0.2 23.0 46.2 22.4
22.3 22.7 22.2
Ignore the first (rough) titration and the fourth as they are too far away from the other readings. Therefore, the average volume of the concordant readings is 22.23 Na2C03 + 2HCI 2 NaCI + CO2(g) + H20(l) Mol Na2C03 = 1/2mol(HC1) cV (Na2C03) = 1/2cV (HCI) 0.25 x C = 1/2 x 0.0246 x 0.02223 Na2C03 = 0.109 mol L‐t (b) Reaction between a strong acid and a weak base wi
produce an acidic solution, so methyl orange will be suitable. © Correct calculation justifying average titre used ………………… 3 Correct calculation using average o
all values ………………………………. 2 Correct equation. OR Some correct working…………… 1 Chooses correct indicator and provides appropriate reason ………………… 2 (2012) All Rights Reserved 52 of 65 For more info, go to www.hscintheholidays.com.au Question 24 (6 marks) A bottle of vinegar is labelled 4.0% w/v (4.0 g per 100 mL of solution) acetic acid (ethanoic acid). (a) Describe the laboratory procedure you would use to verify this concentration. 3 …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. (b) Calculate the volume of O.118 mol L‐1 NaOH required to neutralise the acid in 5.0 mL of this vinegar. 3 …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. 24. (a) (b) © (2012) All Rights Reserved 53 of 65 For more info, go to www.hscintheholidays.com.au The vinegar was diluted accurately using a pipette and volumetric flask. 1 The diluted vinegar was titrated against a standard NaOH solution. 1 The end point was found using phenolphthalein indicator, and the concentration of the undiluted vinegar calculated from the titration result. 1 CH3COOH + NaOH CH3COONa + H2O mass of acid in 5 mL = 5 x 4/100 = 0.20 g ; mole mass of acetic acid = 60 g 1 moles of acid = mass/molar mass = 0.20/60 = 0.0033 mole moles ofNaOH required = 0.0033 Volume ofNaOH = moles/molarity=0.0033/0.118 = 0.0280 L = 28.0 mL qualitatively describe the effect of buffers with reference to a specific example in a natural system 1 1 Question 18 (4 marks) The pH of human blood is maintained at about 7.4 by various buffers. One of the most important of these is the dihydrogen phosphatelhydrogen phosphate (H2P04‐/HPO42‐) equilibrium. (a) Write an equation for this equilibrium. 1 ……………………………………………………………………………………………………………………………. (b) With reference to this equation, explain how a solution containing this buffer could resist 3 a change in pH if a small amount of acid were added to it. ……………………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………………. Question 18 (a) H13 Correct equation………………… 1
(b) In a buffer solution, the concentration of H8 the weak acid and its conjugate base is Detailed explanation (cause and
considerably greater than the effect) with reference to the concentration of the H+. In this case if a equation in part (a)…………….. 3
small amount of acid were added, the extra H+ added would react with the HPO42‐ ion forcing the equilibrium to shift Brief explanation with reference
to the left. However, since the [HP42‐] is to equation ………………………….. 2
so much greater than the [H"1, then the original amount of H+ remains virtually Brief explanation with no unchanged. The volume change is very © (2012) All Rights Reserved 54 of 65 For more info, go to www.hscintheholidays.com.au small and so the new [H+] is almost identical to what it was before the acid was added. Therefore, the pH remains almost the same. 1.
reference to equation ….. 1 A small amount of acid is added to a buffer solution. As a result the pH of this solution will A.
not change. B.
decrease slightly. C.
increase slightly. D.
approach pH = 7. Answer and explanation Question 1 B A buffer solution will resist a change in pH, but will not prevent it. © (2012) All Rights Reserved 55 of 65 For more info, go to www.hscintheholidays.com.au Syllabus content and course outcomes 9.3.4 H8 Question 24 (3 marks) Define the term buffer in relation to acid‐base systems and describe ONE example of buffer action in a natural system 3 ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… 24. A buffer is a solution which maintains almost constant pH when small quantities of acidor base are added. The buffer consists of a weak acid and its conjugate base, at roughly equal concentrations. An example is our blood, which is buffered by the presence of the hydrogen carbonate ion, maintaining a stable pH as it circulates though the body. 3 © (2012) All Rights Reserved 56 of 65 For more info, go to www.hscintheholidays.com.au Question 22 (4 marks) The phosphate buffer system operates in the internal fluid of all cells. This buffer system is represented by the chemical equation below: (a) Define the term ‘buffer’ and identify the key components of any buffer system. 2 …………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… (b) Using relevant equations explain what happens if: (i)
H+ ions are added to this system. 2 …………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… (ii)
OH‐ ions are added to this system. 22. © …………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… (a) A buffer is a system which can maintain approximately the same pH even when significant amounts of strong acid or base are added. 1 Buffered solutions contain comparable amounts of a weak acid and its conjugate base. 1 (b) (i) According to Ie Chateliers principle addition of H+ ions will move the equilibrium to the left, restoring the original H+ion concentration. 1 (2012) All Rights Reserved 57 of 65 For more info, go to www.hscintheholidays.com.au (ii) addition of OH‐ions will move the equilibrium to the right, restoring the original H+ ion concentration. 1 analyse information from secondary sources to assess the use of neutralisation reactions as a safety measure or to minimise damage in accidents or chemical spills © (2012) All Rights Reserved 58 of 65 For more info, go to www.hscintheholidays.com.au Chemistry Stage 6 Syllabus Students learn to: 5. Esterification is a naturally occurring process which can be performed in the laboratory © describe the differences between the alkanol and alkanoic acid functional groups in carbon compounds identify the IUP AC nomenclature for describing the esters produced by reactions of straight‐chained alkanoic acids from C1 to C8 and straight‐chained primary alkanols from Cl to C8 explain the difference in melting point and boiling point caused by straight‐
chained alkanoic acid and straight‐chained primary alkanol structures identify esterification as the reaction between an acid and an alkanol and describe, using equations, examples esterification describe the purpose of using acid in esterification for catalysis explain the need for refluxing during esterification outline some examples of the occurrence, production and uses of esters describe the differences between the alkanol and alkanoic acid functional groups in carbon compound Students: identify data, plan, select equipment and perform a firsthand investigation to prepare an ester using reflux process information from secondary sources to identify and describe the uses of esters as flavours and perfumes in processed foods and cosmetics explain the difference in melting point and boiling point caused by straight‐
chained alkanoic acid and straight‐chained primary alkanol structures identify esterification as the reaction between an acid and an alkanol and describe, using equations, examples of esterification (2012) All Rights Reserved 59 of 65 For more info, go to www.hscintheholidays.com.au identify the IUPAC nomenclature for describing the esters produced by reactions of straight‐chained alkanoic acids from CI to C8 and straight‐chained primary alkanols from CI to C8 5. The following structure describes an ester. H H H O H C C C O C H H H H Which organic reactants were used to form this ester? A.
Methanol and propanoic acid B.
I‐propanol and methanoic acid C.
Ethanol and ethanoic acid D.
I‐butanol and butanoic acid Question 5 B The esters structure has double bonded oxygen on the single carbon; this C must be from the acid therefore the acid was methanoic acid. The other carbon chain in the ester contains 3 consecutive C atoms therefore tlus section was originally I‐propanol. 9. What is a correct name for the compound with the molecular formula A.
B.
C.
D.
© ethanoic acid ethano methanol methanoic acid D (2012) All Rights Reserved 60 of 65 For more info, go to www.hscintheholidays.com.au describe the purpose of using acid in esterification for catalysis explain the need for refluxing during esterification identify data, plan, select equipment and perform a firsthand investigation to prepare an ester using reflux Question 23 (8 marks) (a) To perform an esterification reaction in the laboratory a student was provided with methanol and propanoic acid, which she heated together under reflux with a catalyst. (i)
Name the ester which could be synthesised. 1 (ii)
…………………………………………………………………………………………………………….. Draw the structural formula for this ester. 1 (iii)
(iv)
Name a suitable catalyst for this reaction. 1 ……………………………………………………………………………………………………………… Justify the use of heating under reflux for this experiment. 2 ……………………………………………………………………………………………………………… ……………………………………………………………………………………………………………… ……………………………………………………………………………………………………………… ……………………………………………………………………………………………………………… ……………………………………………………………………………………………………………… (b) The ester formed in the above reaction has a molar mass of 88 g and boils at 78°C. Two other substances with the same molar mass are: 1‐pentanol BP 138°C butanoic acid BP 163°C Explain the difference in boiling points between these three substances. © (2012) All Rights Reserved 61 of 65 For more info, go to www.hscintheholidays.com.au ………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………… 23. (a) (i) (ii) (iii)
(iv)
© methyl propanoate (1 mrk) CH3CH2COOCH3 or expanded formula (1 mrk) strong acid, phosphorus pentoxide etc (1 mrk) Heating under reflux increases the reaction rate (higher temperature) (1 mrk) while preventing loss of reactants or products by vaporisation to outside. (1 mrk) (b) The ester has low polarity resulting in much weaker intermolecular forces than in pentanol and butanoic acid which both have polar OH groups. With an additional 0 atom butanoic acid is still more polar Pentanol and butanoic acid also form hydrogen bonds. The boiling points reflect the strengths of these interm?lecular forces. (1 mrk) (1 mrk) OR (1 mrk) (2012) All Rights Reserved 62 of 65 For more info, go to www.hscintheholidays.com.au Question 23 (5 marks) In your studies, you have investigated the production of esters, an endothermic process, and the Haber process, an exothermic process. 5 Compare the application of reaction rate and equilibrium principles in the production of esters and the Haber process ……………………………………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………………………………. Question 23 9.3.5.9.4.2 H3, H8, Hl3 Esterification is an equilibrium reaction with the equilibrium position well to the left. It is also a slow Compares the processes in terms reaction. The reaction to produce ammonia is also of how rate and position of slow, although its equilibrium position is well to the equilibrium are controlled … 4‐5 right. In each case reaction rate is increased through Compare rate factors the use of higher temperatures and a catalyst OR (concentrated acid for esterification, iron for Equilibrium factors only ammonia). Increasing the temperature also shifts the OR equilibrium position to the right in the case of esters, Superficial treatment of both .2‐3 increasing yield, but to the left for ammonia, An understanding 'of rate or decreasing yield. High pressure (up to 350 quilibrium……………………………… 1 atmospheres is used to increase rate and yield in the Haber process, which is a gas phase reaction, whereas esterification proceeds at atmospheric pressure. Each process removes a product, ammonia is liquefied so that the equilibrium shifts further to the right, and © (2012) All Rights Reserved 63 of 65 For more info, go to www.hscintheholidays.com.au water is removed in esterification so that the equilibrium also shifts to the right. Question 24 (1 mark) Internal combustion engines use petrol or diesel as fuels. State why computers in modern cars monitor the levels of carbon monoxide and nitrogen oxides produced. 1 ……………………………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………………………… Question 24 Carbon monoxide is highly toxic to humans, while both substances contribute to environmental pollution. Monitoring the level of these oxides helps the car's computer to adjust settings to ensure complete combustion of fuel takes place, minimising the level of these oxides. © 9.4.1 H4.H8 States a reason for monitoring… 1 (2012) All Rights Reserved 64 of 65 For more info, go to www.hscintheholidays.com.au Question 23 (4 marks) During your practical work you performed a first hand investigation to prepare an ester. (a) Identify an ester by name and draw a structural formula. 2 ………………………………………………………………………………………………………………………. ………………………………………………………………………………………………………………………. ………………………………………………………………………………………………………………………. (b) Explain the need for refluxing in this investigation. 2 ………………………………………………………………………………………………………………………. ………………………………………………………………………………………………………………………. ………………………………………………………………………………………………………………………. ………………………………………………………………………………………………………………………. ………………………………………………………………………………………………………………………. ………………………………………………………………………………………………………………………. 23. © (a) eg., ethyl acetate (ethanoate) CH3CH20COCH3 or expanded structural formula. 2 (b) The reaction is slow and requires and is carried at by boiling with a catalyst. As the reactants and product are volatile, and highly flammable, a reflux condenser is needed to continuously condense the escaping vapour and return the condensate to the reaction flask. 2 (2012) All Rights Reserved 65 of 65 For more info, go to www.hscintheholidays.com.au Chemistry Research
1.
Describe the composition and layered structure of the atmosphere. [Draw a clearly labelled diagram of the layered structure of the Earth’s
atmosphere]

The atmosphere is made up of two main layers:
the troposphere and the stratosphere. It is within these
layers that ozone exists. The troposphere extends from
the Earth’s surface to 15 kilometres above sea level. Over
90% of Earth's gases are in the troposphere. Temperatures
drop with altitude in the troposphere. At the top there is a
region where the temperature remains reasonably stable.
This is called the tropopause and it reduces the mixing of
gases. The stratosphere (upper atmosphere) is above the
tropopause. In the stratosphere temperatures rise with
increasing altitude.

The atmosphere is predominantly composed of nitrogen (78% by volume), oxygen (21%), and argon (0.93%), with the other gases in very small
concentrations. Despite these small concentrations, many of the other gases (e.g. ozone) cause concern because of the reactions they can
undergo.
2.
Describe ozone as a molecule able to act as an upper atmosphere UV radiation shield and a lower atmosphere pollutant. [State the sources of
ozone in the lower atmosphere and describe how the ozone molecule acts as a lower atmosphere pollutant. Describe the production of ozone in the upper
atmosphere (give chemical equations) and describe how the ozone molecule acts as a Ultra-Violet Radiation shield]
The action (and value) of ozone depends greatly on its location.

Ozone in the troposphere is a pollutant, even at the very low concentrations compared with the other gases. It can be found as a component of
photochemical smog [sunlight acts on NO2 → NO + O•, free radical joins with O2 to form O3]

Ozone is formed by the interaction of sunlight, particularly ultraviolet light, with hydrocarbons and nitrogen oxides, which are emitted by cars,
petrol/gasoline vapours, fossil fuel power plants, refineries, and industries.

Ozone is poisonous to humans, it is an irritant to the eyes, noes, throat and lungs. It is very reactive and is a strong oxidant – it can react with
body tissue, esp., with sensitive mucous membranes when breathed in. The ozone layer blocks harmful UVC, UVB rays passing through the
atmosphere – these can cause cancers & severe sunburn .

Every time oxygen/ozone reacts with UV light, it absorbs it. Protection is due to the ability of ozone to absorb UV radiation.
O2 + UV radiation → 2 O•
O• + O2 → O3
O3 + UV radiation → O• + O2
O• + O3 → 2
In both of these sets of reactions, the second reaction is exothermic, which accounts for the increasing temperature in the stratosphere.
3.
Demonstrate the formation of co-ordinate covalent bonds using Lewis electron dot structures [What is a co-ordinate covalent bond? How does a coordinate covalent bond differ from other ‘normal’ covalent bonds? Use the structure of ozone and oxygen to illustrate your answer]
Co-ordinate covalent bonds involve one atom donating a pair of electrons to the other to achieve a stable number of
electrons (octet rule). All the electrons in the bond come from one atom (in a covalent bond, each atom supplies one
electron).
4.
Compare the properties of the oxygen allotropes O2 and O3 and account for them on the basis of molecular bonding and structure [Compare
chemical and physical properties of the oxygen allotropes O2 and O3. Set your answer out in the form of a table with appropriate headings. Account for the
properties you have described in the table above on the basis of molecular bonding and the structure of the O2 and O3 molecules. Include a diagram in your
answer]
Properties
colour
Oxygen
colourless
Ozone
blue
Explanation
boiling point
–183°C
–111°C
The boiling point of diatomic oxygen is lower than that of the ozone as diatomic oxygen has a lower molecular mass requiring less energy in the boiling
process.
solubility in
water
sparingly soluble
more soluble than
oxygen
Non-polar O2 does not form strong intermolecular forces in the polar water. Ozone has a bent structure, which provides for some polarity of the molecule in its
interaction with water.
chemical
stability
far more stable than the
ozone molecule
far less stable than the
oxygen molecule
Ozone is easily decomposed into oxygen molecules:
[To decompose oxygen, its double bond must be broken; this requires considerable amounts of energy. However the sing bond (co-ordinate covalent bond) in
ozone requires much less energy to be broken, and hence ozone is less stable (readily decomposes to oxygen)]
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oxidation
ability
less powerful oxidant
5.
more powerful oxidant
The oxidising strength comes from the weakness of the single bond; it easily releases an oxygen, which can then oxidise a compound.
e.g. reaction with metals: oxygen forms the oxide as the only product whereas ozone reacts more readily producing the metallic oxide and an oxygen
molecule.
Identify the origin of chlorofluorocarbons (CFCs) and halons in the atmosphere. [What are CFC’s and halons, give an example of each. Identify the
origin of CFC’s and halons in the atmosphere]
Chlorofluorocarbons - compounds that contain only carbon, fluorine and chlorine (i.e. no H) eg CFC-12, CCl2F2
Introduced in 1930’s as replacement for ammonia in refrigeration [they had the required pressure-dependent properties that refrigerants needed, as well
as being odourless, non-flammable, non-toxic and inert, unlike toxic, foul-smelling ammonia]. Widely used as Refrigerants in fridges and air-conditioners,
Propellants in aerosol spray cans and many other uses. These many uses released CFCs directly into the lower atmosphere. CFCs are Inert and insoluble in
water- remained in the troposphere and were spread throughout the atmosphere.
Halons – compounds that contain only carbon, bromine and other halogen (i.e. no H, basically CFCs that also contain bromine) eg CBrF3
[Bromotrifluoromethane]. Dense, non-flammable liquids widely used as fire extinguishers – they were released directly into the atmosphere, and slowly
diffused into the stratosphere. Weren’t used as extensively as CFCs. Halon use has been drastically reduced because bromine atoms are even more
effective than chlorine atoms in the chain reactions that lead to the depletion of the ozone layer.
The concentration of the chlorine radicals that react with ozone is increased in springtime in the Antarctic with the return to longer periods
of sunlight. The rate is reduced by summer when the chlorine is basically used up after the accumulation over winter. This leads to a large
hole in the ozone layer developing over the Antarctic during the spring when sunlight is plentiful.
6.
Discuss the problems associated with the use of CFCs and assess the effectiveness of steps taken to alleviate these problems [include: Equations
to show the reactions involving CFCs and ozone to demonstrate the removal of ozone from the atmosphere. Identify alternative chemicals used to replace CFCs
and evaluate the effectiveness of their uses as a replacement for CFCs. National policies and international treaties (protocols)]
CFCs cause ozone layer depletion. They are very inert and not washed out by rain, they can remain in the troposphere for many years and eventually
diffuse into the stratosphere where they deplete the ozone layer. Leads to more UV radiation reaching Earth, greatly increases the chances of mutation
and damage (esp. cancer) in living things. Ozone depletion is more frequent in winter and spring due to more ice particles. These provide a surface to act
as a catalyst.
Short wavelength UV radiation (that has not been removed by the ozone layer) attacks the CFC molecule and breaks off a chlorine atom
CCl2F2 (g) + UV radiation
·Cl (g) + ·CClF2 (g)
This chlorine atom (radical) then reacts with ozone, forming oxygen gas and a chlorine monoxide radical (ClO);
·Cl (g) + O3 (g)
·ClO (g) + O2 (g)
The chlorine monoxide radical then reacts with an O radical and the chlorine radical is regenerated;
·ClO (g) + ·O (g)
·Cl (g) + O2 (g)
The result is that an ozone molecule and an oxygen radical have been converted into 2 oxygen molecules and the chlorine has not been used up. The chlorine radical can then
attack another ozone molecule and repeat- chain reaction. Very small amount of CFC’s can cause a significant damage.
Chlorine radicals are removed by their reaction with methane, which ends the chain reaction.
·Cl (g) + CH4 (g)
HCl (g) + ·CH3 (g)
·ClO species reacting with nitrogen dioxide, forming chlorine nitrate also stops the reaction.
Alternatives
Ammonia- industrial refrigeration reverted back. Use great care - ammonia is dangerous and toxic.
Hydrochlorofluorocarbons [HCFCs] - first replacements for CFC’s. Contain C-H bonds that are susceptible to attack by reactive radicals in the troposphere
and so are decomposed rapidly to a significant extent, means that only a small proportion ever reaches the stratosphere. Replaced CFC’s in domestic
refrigeration, as propellant in spray cans, as an industrial solvent and as a foaming agent. Effectiveness - small amounts of HCFCs that do reach the
stratosphere are ozone depleting (10% CFC’s potential). Only seen as a temporary solution. They also contribute to the greenhouse effect, and so their use
is being phased out (complete ban by 2030).
Hydrofluorocarbons [HFCs]- contain only carbon, hydrogen and fluorine (no chlorine or bromine) widely seen as a viable CFC and HCFC alternative as they
contain reactive C-H bonds (so they degrade in the troposphere) and they do not contain any chlorine. Their ozone depleting potential is zero. Used in
refrigeration and air conditioning applications. Effectiveness - good alternative in terms of atmospheric health. They do not contain chlorine therefore not
expected to produce undesirable radicals in the stratosphere - more research is required.
Protocols
Vienna Convention of 1985 established the framework for the protection of the ozone layer.
The Montreal Protocol - treaty designed to gain cooperation for the global reduction in the production of CFCs and halons signed in 1987 by 27 nations,
and involved: Freezing CFC production at 1986 levels immediately & Reducing CFC production by 50% by the year 2000.
Agreement made in London in 1990 involved: Eliminating the production and use of CFCs, halons, and carbon tetrachloride by 2000, with this date being
extended by ten years for developing nations. Eliminating the production and use of 1,1,1-trichloroethane (methyl chloroform) by 2005. Eliminating the
production and use of HCFCs by 2040 at the latest, but preferably by 2020.
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Agreement made in Copenhagen in 1992 involved: Eliminating the production and use of halons by the end of 1994. Eliminating the production and use of
CFCs and 1,1,1-trichloroethane (methyl chloroform) by 1996. Providing financial aid to developing nations for the implementation of the above measures.
7.
Analyse the information available that indicates changes in atmospheric ozone concentrations, describe the changes observed and explain
how this information was obtained.
Scientists identified that a dramatic decline in springtime ozone occurred from the late 1970s over the entire Antarctic. The decline reached approximately
30% by 1985. In some places, the ozone layer had been completely destroyed. The ozone decline over Antarctica during springtime now exceeds 50%. Due
to the decrease in the rate of release of CFCs and halons into the atmosphere that has come about through international agreements, a decrease in the
rate of ozone depletion has been observed in recent years. In 2002, an Australian study announced that levels of ozone-destroying chemicals in the
stratosphere were falling and that the ozone hole would begin to start closing in five years, with the process being complete by 2050.
Ground based: UV spectrophotometers - directed vertically upwards through the atmosphere to measure the intensity of ultraviolet radiation of
wavelengths that ozone absorbs, as well as the intensity of wavelengths on either side of this range. Provides a measure of the total ozone in the
atmosphere per unit area of Earth surface at the location of the spectrometer
Balloon based: placed at high altitude weather balloons that can rise above the stratosphere, pointed down and measure ozone from above.
Satellite: TOMS (total ozone mapping spectrophotometers) have been placed on several US satellites. Works similar to the UV method. As the satellites
orbit the entire globe TOMS is able to scan the entire globe and measure ozone concentration as a function of altitude and geographical location. TOMS
have been especially important since the 1980s in producing maps of the extent of the ozone holes.
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9.3 – The Acidic Environment:
Δ. Construct word and balanced formulae equations of all chemical reactions as they are encountered in this module:

–
Note: In chemistry, [x] means “concentration of x” in moles per litre (mol/L).
–
Eg. [H3O+] means “concentration of H3O+ ions” in mol/L.
Basic reactions to remember:
–
–
Acid reactions:

acid + base

acid + metal

acid + carbonate

acid + hydrogen carbonate
–
HCl(aq) + Mg(s)
salt + carbon dioxide gas + water
MgCl2(s) + H2 (g)
HCl(aq) + CaCO3(s)
salt + carbon dioxide gas + water
CaCl(s) + CO2(g) + H2O
(note: there is CO2 solid, its dry ice)
H+ + H2O
H3O+
Non-metal (acidic) oxides:

CO2 (g) + H2O (l)
H2CO3 (aq) (carbonic acid)

SO2 (g) + H2O (l)
H2SO3 (aq) (sulfurous acid)

2NO2 (g) + H2O (l)
HNO3 (aq) + HNO2 (aq) (nitric and nitrous acid)

P2O5 (g) + H2O (l)
2H3PO4 (aq) (phosphoric acid)
Metal (basic) oxides:

K2O (s) + H2O (l)
2KOH (aq) (potassium hydroxide)

Na2O (s) + H2O (l)
2NaOH (aq) (sodium hydroxide)

MgO (s) + H2O (l)
Mg(OH)2 (aq) (magnesium hydroxide)
Various equilibrium reactions:
–

salt + hydrogen gas
NaCl(s) + H2O(l)
Reactions of various oxides with water:
–

HCl(aq) + NaOH(aq)
Formation of hydronium:


salt + water
Formation of carbonic acid: CO2 (g) + H2O (l)
2+
H2CO3 (aq)
CuCl42‫(־‬aq) + 4H2O (l)
–
Copper complex-ions: Cu(H2O)4
–
Decomposition of dinitrogen tetroxide: N2O4 (g)
2NO2 (g)
–
Decomposition of calcium carbonate: CaCO3 (s)
CaO (s) + CO2 (g)
(aq)
+
4Cl‫(־‬aq)
Non-Arrhenius acid/base reaction (ie no water present and no free H+ ions):
–
Gaseous hydrogen chloride and ammonia react:

HCl
(g) +
NH3 (g)
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

Ionisation of strong and weak acids:
Hydrochloric: HCl (g) + H2O (l)
–
Nitric: HNO 3 (l) + H2O (l)
–
Sulfuric: H2SO4 (l) + 2H2O (l)
–
Ethanoic: CH3COOH (s) + H2O (l)
–

Organic decomposition: 2H2S (g) + 3O2 (g)

Burning high-sulfur coals: S (s) + O2 (g)

Smelting metal sulfides: 2PbS (s) + 3O2 (g)
2SO2 (g) + 2H2O (l)
SO2 (g)
2PbO (s) + 2SO2
Nitrogen Oxides:

Lightning: N2 (g) + O2 (g)

Further Catalysed by oxygen particles: 2NO (g) + O2 (g)
2NO (g)
NO2 (g)
Hydrogen carbonate (ie bicarbonate):

HCO3‫( ־‬aq) + H3O+ (aq)
HCO3
‫־‬
(aq)
+ OH
H2CO3 (aq) + H2O (l)
‫־‬
CO32‫( ־‬aq) + H2O (l)
(aq)
Natural Buffers:
–
The carbonic acid/hydrogen carbonate ion buffer in the mammalian blood system:

H3O+ (aq) + HCO3‫( ־‬aq)
H2CO3 (aq) + H2O (l)
Esterification:
–
–

H3O+ (aq) + CH3COO‫( ־‬aq)
Sulfur Oxides:


2H3O+ (aq) + SO42‫־‬
Amphiprotic substances:
–

H3O+ (aq) + NO3-
Sources of sulfur and nitrogen oxides in the atmosphere:
–

H3O+ (aq) + Cl‫( ־‬aq)
–
General word-formula:

acid + alcohol

alkanoic acid + alkanol
ester + water
ester + water
Example:

butanoic acid + pentanol

C3H7COOH (aq) + C5H11OH (l)
pentyl butanoate
C3H7COOCH2C4H9 (aq) + H2O (l)
Miscellaneous Terms:
–
In naming the following signify: Mon = 1 , Di = 2 of a certain element.
–
The prefix ‘Bi’ is used to indicate the addition of a single hydrogen ion, Not 2! (as in ‘Di’)
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1. Indicators were identified with the observation that the colour of some flowers depends on soil composition:

RECALL:
–
–
–
General Properties of Acids:

They taste sour.

They are corrosive (ie sting/burn skin)

When in a solution, they can conduct electricity (ie electrolytes).

Acids are neutralised by bases.

pH < 7

For LITMUS: blue  acid  red (ie turns blue litmus red)

Note: litmus is a dye made from lichens
General Properties of Bases:

They usually taste bitter.

May be corrosive.

Mainly insoluble in water (note: aqueous bases are called alkalis).

When in a solution, they can conduct electricity (ie electrolytes), not all bases are soluble.

Bases are neutralised by acids.

Bases are usually in the form of metal hydroxides (e.g. NaOH) OR metal oxides (e.g. MgO).

pH > 7

For LITMUS: red  base  blue (ie turns red litmus blue)
Note: by definition, and electrolyte is any solutions that has free ions that is able to conduct electricity. An acid
dissociates to form ions, so does alkali bases. Hence they are able to form electrolytes.
–
Naming:
Binary Acids


A binary compound consists of two elements. Binary acids have the prefix hydro in front of the full
name of the nonmetallic element. They have the ending -ic. Examples include hydrochloric and
hydrofluoric acid.
Hydrofluoric Acid - HF
Hydrochloric Acid - HCl
Hydrobromic Acid - HBr
Hydrocynide Acid - HCn
Hydrosulfuric Acid - H2S
Ternary Acids

Ternary acids commonly contain hydrogen, a nonmetal, and oxygen. The name of the most common
form of the acid consists of the nonmetal root name with the -ic ending, The acid containing one less
oxygen atom than the most common form is designated by the -ous ending. An acid containing one
less oxygen atom than the -ous acid has the prefix hypo- and the -ous ending. The acid containing one
more oxygen than the most common acid has the per- prefix and the -ic ending.
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
Nitric Acid - HNO3
Nitrous Acid - HNO2
Hypochlorous Acid - HClO
Chlorous Acid - HClO2
Chloric Acid - HClO3
Perchloric Acid - HClO4
Sulfuric Acid - H2SO4
Sulfurous Acid - H2SO3
Phosphoric Acid - H3PO4
Phosphorous Acid - H3PO3
Carbonic Acid - H2CO3
Acetic Acid - HC2H3O2
Oxalic Acid - H2C2O4
Boric Acid - H3BO3
Silicic Acid - H2SiO 3
Bases

Sodium Hydroxide - NaOH
Potassium Hydroxide - KOH
Ammonia – NH3
Ammonium Hydroxide - NH4OH
Calcium Hydroxide - Ca(OH)2
Magnesium Hydroxide - Mg(OH)2
Barium Hydroxide - Ba(OH)2
Aluminum Hydroxide - Al(OH)3
Ferrous Hydroxide or Iron (II) Hydroxide - Fe(OH)2
Ferric Hydroxide or Iron (III) Hydroxide - Fe(OH)3
Zinc Hydroxide - Zn(OH)2
Lithium Hydroxide - LiOH
Alkali Bases: is a basic, ionic SALT of an alkali metal or alkaline earth metal element.


Sodium Carbonate – Na 2CO3
Bicarbonate (or hydrogen carbonate) - NaHCO3
Calcium Carbonate - CaCO3
Classify common substances as acidic, basic or neutral:
–
Acids:

Vinegar (acetic acid), vitamin C (ascorbic acid), lemon juice (citric acid), aspirin (acetyl salicylic acid), ‘fizzy’
drinks (carbonic acid).
–
Bases:

Drain cleaners (sodium hydroxide), household cleaners (ammonia), antacid tablets (calcium carbonate), baking
powder (sodium bicarbonate/sodium hydrogen carbonate).
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–
Neutral:


Pure water, milk, table salt (ex sodium chloride; but NOT all salts are neutral),
Identify that indicators such as litmus, phenolphthalein, methyl orange and bromothymol blue can be used to
determine the acidic or basic nature of a material over a range, and that the range is identified by change in indicator
colour:
–
An indicator is a substance (usually a vegetable dye) that, in solution, changes colour depending on whether the
chemical enviroment is acidic or basic.

–
Most indicators produce 2 different colours; one for acidic and one for basic.
–
Litmus, phenolphthalein, methyl orange and bromothymol blue are common indicators.
Identify data and choose resources to gather information about the colour changes of a range of indicators:
–

The range of pH range of the common indicators is shown below:
Identify and describe some everyday uses of indicators including the testing of soil acidity/basicity:
–
Testing Soil pH:

Some plants only grow within narrow pH ranges, so the pH of the soil needs to be regularly tested. Examples
include azaleas/camellias need acidic soil, while vegetables (ex cucumbers) need alkaline.

A neutral white powder (such as talc or barium sulfate) is sprinkled over the damp soil; a few drops of indicator
are placed on top.

–
The white powder allows the colour change to be seen clearly.
Testing pH of Pools:

Pool water must be near neutral to avoid health problems, the pH required for safety is ~7.5 for less irritation to
eyes and skin.

A few drops of indicator are placed in a sample of the pool water; alternatively, pH paper, already soaked in
indicator can be used.
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–

Monitoring pH of Chemical Wastes:

Wastes that are produced from laboratories or photographic film centres tend to be highly acidic.

The pH of the wastes must be neutralised before they can be safely disposed.

Indicators are used to measure the pH, and substances added to neutralise it.
Solve problems by applying information about the colour changes of indicators to classify some household substances
as acidic, neutral or basic:
–

You can be given substance, and its colour indication in various indicators, to find its pH range.
Perform a first-hand investigation to prepare and test a natural indicator:
–
Aim: to prepare an indicator solution using naturally occurring substances, in this case obtained from red cabbage,
and to then test them.
–
–
Risk Assessment:

This practical involves boiling a solution gently, safety glasses and protective clothing should be worn.

If these safety precautions are taken, then the risk is acceptable.
Method:

One large red cabbage was peel and chopped; it was then blended thoroughly using a food processor with 200 mL
of distilled water.

Boil the mixture until a strongly coloured extract forms. After this cools, strain and transfer the reddish-purple
solution to another beaker, this is the indicator.

Add the indicator solution to different substances in a Petri dish, leaving one dish as the control. Observe the
colour changes as the indicator is added to substances of high, low, and neutral pH’s.

In 4 separate dishes, one as control, one having: 3 mL of distilled water, hydrochloric acid (HCl) and sodium
hydroxide (NaOH) solution was placed.
–
–
Result:

In the distilled water, it was DARK PURPLE.

In the HCl solution, it was PINK.

In the NaOH solution, it turned YELLOW.
Justification:

Beetroot was used as it is a very vividly coloured plant; and its pigmentation is very easily extracted.

A fresh beetroot was used instead of canned beetroot as the canned version may contain preservatives (many of
which are weak acids) that may affect the results.

HCl and NaOH was used as they are on opposite ends of the pH scale; this was to show the range of colours the
indicator could produce.
–
Limitations:

Beetroot come in many sizes; this was not controlled.

The exact pH at which the transition of colours occurred was not determined.
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2. While we usually think of the air around us as neutral, the atmosphere naturally contains acidic oxides of carbon,
nitrogen and sulfur. The concentrations of these acidic oxides have been increasing since the Industrial Revolution:

RECALL:
–

The hydronium ion:

When in aqueous solutions, ACIDS disassociate into anions and H+ ions (ex HCl  H+ + Cl- )

Then the hydrogen ion reacts with water, ie this reaction occurs: H+ + H2O

The H3O+ ion is called the hydronium ion, and is more stable than the H+ ion.

Thus, in water, acids form hydronium ions.
H3O+
Identify oxides of non-metals which act as acids and describe the conditions under which they act as acids:
–
Oxide: is a compound containing at least one oxygen atom as well as at least one other element.
–
Nonmetal oxides are compounds which contain a non-metal and oxygen (ex CO2), these are formed by the combustion
of nonmetals or if a compound contains the nonmetal. Examples:
–

C(s) + O2(g)  CO2(g)

CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
These non-metal oxides:

React with WATER to form acids. CO2 (g) + H2O (l) →H2CO3 (aq)

React with BASES to form salts. CO2(g) + 2NaOH(aq) → Na2CO3(aq) + H2O(l)

Some non-metal oxides are: CO2 (carbon dioxide), SO2 (sulfur dioxide), NO 2 (nitrogen dioxide) and P2O5
(phosphorus pentoxide).
–
Oxides of nonmetals, such as carbon, nitrogen etc, are acidic. They react with water to form acidic solutions, hence
they are known as acidic oxides.

Reactions:

–
CO2 (g) + H2O (l)
H2CO3 (aq) (carbonic acid)
 SO2 (g) + H2O (l)
H2SO3 (aq) (sulfurous acid)
 2NO2 (g) + H2O (l)
HNO3 (aq) + HNO2 (aq) (nitric and nitrous acid)
 P2O5 (g) + H2O (l)
2H3PO4 (aq) (phosphoric acid)
Note: the exceptions are the neutral oxides (preferably known as monoxides, are those oxides which show neither
basic nor acidic properties in their reaction with water), including N2O (dinitrogen oxide), CO (carbon monoxide) and
NO (nitric oxide).
–
Metal oxides are compounds which contain a metal and oxygen (ex MgO), these are formed by the combustion of the
metal itself. Examples:

2Mg(s) + O2(g)  2MgO(s)

2Ca(s) + O2(g)  2CaO(s)
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–
Metal oxides:

React with WATER to form bases. MgO (s) + H2O(l)  Mg(OH)2(aq)

React with ACIDS to form salts. MgO(s) + 2HCl(l)  MgCl2(s) + H2O(l)

E.G. Some metal oxides that act as bases are: K2O (potassium oxide), Na 2O (sodium oxide), CaO (calcium oxide)
and MgO (magnesium oxide).
–
Oxides of metals are basic, such as magnesium, calcium etc. They react with water to form basic solutions, hence they
are known as basic oxides.

Reactions:
 K2O (s) + H2O (l)
2KOH (aq) (potassium hydroxide)
 Na2O (s) + H2O (l)
2NaOH (aq) (sodium hydroxide)
 CaO(s) + H2O(l)
 MgO (s) + H2O (l)
Ca(OH)2(aq) (calcium hydroxide)
Mg(OH)2 (aq) (magnesium hydroxide)
Note: Come to this section, after the topic the rest of topic has been finished
–
Amphoteric oxides are oxides that can act as BOTH acids and bases depending on the reaction they are put in.
–
Amphoteric oxides act like bases/acids, but don’t have to donate/accept protons. While amphiprotic must
accept/donate proton. Amphiprotic substances are a subset of amphoteric substances, every amphiprotic substance is
amphoteric, but not the other way around.
–
The only elements that combine to form amphiprotic oxides are BeO (beryllium oxide), Al2O3 (aluminium oxide),
ZnO (zinc oxide), SnO (tin oxide) and PbO (lead oxide).
–
Amphoteric oxides:

React with ACID to form salts.

React with BASE to form salts.

Lead oxide
 with acid: PbO + 2HCl
 with base: PbO + NaOH

Na2PbO2 + H2O
Aluminium Oxide
 with acid: Al2O3 + 6HCl
 with base: Al2O3 + 2NaOH

PbCl2 + H2O
2AlCl3 + 3H2O
Na2Al2O4 + H2O
Analyse the position of these non-metals in the Periodic Table and outline the relationship between position of elements
in the Periodic Table and acidity/basicity of oxides:
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
–
The acidic oxides are on the RIGHT side of the periodic table (non-metals).
–
The basic oxides are on the LEFT side of the periodic table (metals).
–
The amphoteric oxides are in-between.
–
Note: Nobel gases have no oxides.
Define Le Chatelier’s principle:
–
The Concept of Chemical Equilibrium:

Most chemical reactions go to completion. Mg (s) + 2HCl (aq)

However not all chemical reactions do NOT go to completion.

Many reactions are in fact reversible and two-directional; they can go from left to right, or right to left; this is
MgCl2 (aq) + H2 (g)
represented using a 2-directional arrow.

This reaction from LEFT-to-RIGHT is called the forward reaction, and the reaction from RIGHT-to-LEFT is
called the backward reaction. These are known as reversible reactions.

Reversible reactions do not go to completion, but they reach a position of STABILITY. This is called the point of
chemical EQUILIBRIUM.

At this point there is no longer any change in the concentrations of the reactants or products; the reaction has
come to a stop (macroscopically).

This does NOT mean that all the reactants have become products or vice-versa. At the point of equilibrium, there
will be reactants AND products present reacting to form each other (microscopically).

For example, the reaction of carbon dioxide & water is a reversible reaction.
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 Eg: CO2 (g) + H2O (l)

H2CO3 (aq)
The ‘double-arrow’ symbolises the 2-directional nature of the reaction; water and CO2 can react to form
carbonic acid, and carbonic acid can decompose to form water and CO2.

At the equilibrium the concentration of the reactants and products are NOT equal. The system is in a balance.
This does not mean that there are 2 moles of reactant and 2 moles of product, for instance. Just that both the
forward and reverse reaction is moving at the same rate.

The chemical equilibrium is does NOT have to be fixed. One side of the reaction can be favoured to another
depending on the conditions of the reaction; this is the basis of Le Chatelier’s Principle.
–
Le Chatelier’s Principle:

Le Chatelier’s Principle states that:
“If a chemical system at equilibrium is subjected to a change in conditions, the system will readjust itself to
counteract that change.”

A system will always seek the point of equilibrium.

So when a system at equilibrium has a change imposed on in, it will “readjust” to OPPOSE the change to return
to a point of equilibrium.

Identify factors which can affect the equilibrium in a reversible reaction:
–
Factors that affect (or shift) the equilibrium position are a change in concentration of a species, a change in pressure
in the system and change in temperature of the system.
–
1) Change in Concentration:

If the concentration of a particular substance (species) is:
 INCREASED, the equilibrium point will shift towards the opposite side of the equation; this opposes the
change, as it reduces the concentration of the species by producing more products on the opposite side.
 DECREASED, the equilibrium point will shift towards the same side of the equation the species is on; this
opposes the change, as it increases the concentration of the species by the opposite reaction.

Take for example, the copper complex ion equilibrium:
 Eg: Cu(H2O)42+ (aq) + 4Cl‫(־‬aq)
CuCl42‫(־‬aq) + 4H2O (l)
(BLUE)
(GREEN)
‫־‬
1. If Cl ions are increased the equilibrium will shift to the RIGHT, as more CuCl42‫־‬is formed hence the
system will become more green.
2. If more water is added the equilibrium will shift to the LEFT, as more Cu(H2O)42+ is formed, and the
system will become more blue.
3. If water is removed the equilibrium will shift to the RIGHT, so that the water lost will be replaced by
the forward reaction.
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–
2) Change in Gas Pressure (or change in Volume):

If the TOTAL pressure on the system is:
 INCREASED, the equilibrium will favour the side that reduces pressure, that is, has less moles; thus
opposing the change.
 DECREASED, the equilibrium will favour the side that increases pressure, that is, produces more moles;
thus opposing the change.
 Pressure is increased is by reducing the volume, or decrease by increasing the volume.

Take, for example, the equilibrium between dinitrogen tetroxide and nitrogen dioxide:
 Eg: N2O4 (g)
(1 mole)
2NO2 (g)
(2 moles)
1. If the TOTAL pressure is increased, the equilibrium will shift to the left, to decrease pressure, as there
are less moles produce on the left.
2. If the TOTAL pressure is decreased, the equilibrium will shift to the right, to increase the pressure, as
more moles are produced on the right.
–
3) Change in Temperature:

Recall:
 An exothermic reaction is one that produces heat energy. For exothermic reactions, the sign of the change in
heat (ΔH) is NEGATIVE.

In terms of reversible reactions, exothermic reactions produce heat through the forward reaction, but
absorb heat in the reverse reaction.
 An endothermic reaction is one that absorbs heat energy. For endothermic reactions, the sign of the change
in heat (ΔH) is POSITIVE.

In terms of reversible reactions, endothermic reactions absorb heat through the forward reaction, but
release heat in the reverse reaction.

If a system is exothermic, and the temperature is:
 INCREASED, the equilibrium will shift to the left, as the reverse reaction is endothermic, and will cool the
system to oppose the heating.
 DECREASED, the equilibrium will shift to the right, as the forward reaction is exothermic, and will heat the
system to oppose the cooling.

Take for example, the decomposition of calcium carbonate (within a closed system; that is, nothing is allowed to
escape). It is an endothermic reaction; the change in heat is positive:
 Eg: CaCO3 (s)
CaO (s) + CO2 (g)
ΔH = 178 kJ/mol
1. If the temperature is INCREASED, the forward reaction will increase, with equilibrium lying more on
the right, as the endothermic forward cooling opposes the imposed heating.
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2. If the temperature is DECREASED, the reverse reaction will increase, with equilibrium lying more on
the left, as the exothermic reverse heating opposes the imposed cooling.
–
There are many other factors that can affect equilibrium, but these are the most common; simply remember that the
system will OPPOSE any change, as stated by Le Chatelier’s principle of chemical equilibrium.
–
Note: CATALYSTS increase the rate of reaction, such that equilibrium is reached faster, but do not favour sides of
equilibrium.

Describe the solubility of carbon dioxide in water under various conditions as an equilibrium process and explain in
terms of Le Chatelier’s principle:

–
CO2 (g) + H2O (l)
H2CO3 (aq) + 50kj
ΔH = -50kj (ie exothermic)
The equilibrium between carbon dioxide, water & carbonic acid can be examined under different conditions, such as:

Effect of species concentration, according to Le Chatelier’s Principle:
 If the concentration of CO2 is INCREASED, such as by pumping more CO2 into the system, the system will
dissolve more CO2 to counteract the change; hence equilibrium shifts to the opposite site; the right.
 If the concentration of CO2 is DECREASED, such as by removing CO2 from the system, the system will
release more CO2 to oppose this loss; hence equilibrium shifts to the same side; the left.

Effect of pressure, according to Le Chatelier’s Principle:
 If pressure is INCREASED, the equilibrium will favour the side that reduces the pressure, and that is the
forward reaction; it reduces pressure by dissolving the CO2 (solutions take up less volume than gases).
 If pressure is DECREASED, the equilibrium will favour the side that increases the pressure, and that is the
reverse reaction; it increases pressure as the carbonic acid changes back into CO2 and water.

Effect of temperature, according to Le Chatelier’s Principle:
 If it is INCREASED, the equilibrium will favour the reverse reaction, as it is endothermic and will oppose
the change by cooling the system.
 If it is DECREASED, the equilibrium will favour the forward reaction, as it is exothermic and will oppose
the change by heating the system.

Effect of acids (H+) and bases (OH-), according to Le Chatelier’s Principle:
 CO2 (g) + H2O (l)
2H+ (aq) + CO32‫(־‬aq)
 If an ACID is added, the concentration of H+ will increase, and hence the equilibrium will shift to the left,
producing more gas.
 If a BASE is added, the OH‫ ־‬will react with the H+ (creating water) greatly shifting the equilibrium to the
right, as more H+ is produced to counteract the change. If enough basic substance is added, the reaction could
go to completion, and no longer have an equilibrium point.
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
Identify natural and industrial sources of sulfur dioxide and oxides of nitrogen:
–
Sulfur Dioxide (SO2):

NATURAL sources: Volcanic gases, bushfires, decomposition of organic matter (by bacteria)

INDUSTRIAL sources: Processing and burning of coal with sulfur impurities and extracting metals from sulfurrich ores, such as galena (PbS).
–
–
Nitrogen Monoxide (or Nitric oxide, NO):

NATURAL sources: The reaction of nitrogen and oxygen in the atmosphere due to high temperatures of lightning.

INDUSTRIAL sources: The nitrogen in the air reacts with oxygen in the hot engines and power stations.
Nitrogen Dioxide (NO2):

NATURAL sources: After nitric oxide is produced by lightning, it slowly reacts with oxygen to produce nitrogen
dioxide.


INDUSTRIAL sources: Nitrogen reacts with oxygen in engines and power stations.
Describe, using equations, examples of chemical reactions which release sulfur dioxide and chemical reactions which
release oxides of nitrogen:
–
Sulfur Dioxide:

Natural: When organic matter decomposes it produces hydrogen sulfide (H2S), which then reacts with oxygen to
produce sulfur dioxide:


2H2S (g) + 3O2 (g)
Industrial: The burning of sulfur-rich coal theses directly combine with oxygen:
 S (s) + O2 (g)

2SO2 (g) + 2H2O (l)
SO2 (g)
Industrial: The extraction of metals from metal sulfides also releases sulfur dioxide. E.g. smelting of galena for
lead:
 2PbS (s) + 3O2 (g)
–
2PbO (s) + 2SO2
Oxides of Nitrogen:

Natural and Industrial: Nitric oxide is produced either when lightning, with its high temperatures combines
nitrogen and oxygen:
 N2 (g) + O2 (g)

2NO (g)
The same reaction occurs in the high temperatures of engines or power plants, also combining nitrogen and
oxygen.

Nitrogen dioxide is formed when nitric oxide reacts with oxygen in the air:

2NO (g) + O2 (g)
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NO2 (g)
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
Assess the evidence which indicates increases in atmospheric concentration of oxides of sulfur and nitrogen:
–
It is difficult to quantitatively state that oxides of sulfur and nitrogen have been increasing in the atmosphere because
these oxides occur in relatively low concentrations, ie 0.01 ppm (parts per million) and chemical instruments able to
measure very low concentrations (like those for SO2), have only been available since the 1970s, so there is no reliable
data for these gases before this time.
–
HOWEVER, analysis of gas found in ice-core samples excavated from Antarctica shows that levels of N2O in the
atmosphere has increased by about 10%.
–
Also NO2 leads to the formation of photochemical smog, a direct indicator of excessive levels of nitrogen oxides in
the atmosphere. In 1952, the so called “Great Smog of December” killed some 4000 people, mainly the frail and
elderly.
–
The increase of the incidence of acid rain also points to the increase in atmospheric concentrations of SO2 and NOx
compounds. Acid rain forms when atmospheric water reacts with these compounds; hence an increase in acid rain
points to an increase in these compounds.
–
It can also be stated that the increased burning of fossil fuels after the Industrial Revolution did indeed lead to a rise
in oxides of sulfur; the effect of acid rain is that it ate away the carbonate statues.

Analyse information from secondary sources to summarise the industrial origins of sulfur dioxide and oxides of
nitrogen and evaluate reasons for concern about their release into the environment:
–
See ABOVE for industrial sources of sulfur dioxide and oxides of nitrogen.
–
Reasons for concern about these Oxides:

Acid Rain: Sulfur dioxide and nitrogen dioxide are acidic oxides that react with water in the atmosphere to form
acids:
 SO2 (g) + H2O (l)
 2NO2 (g) + H2O (l)
H2SO3 (aq) (sulfurous acid)
HNO3 (aq) + HNO2 (aq) (nitric and nitrous acid)
These acids then combine with rain droplets, forming acid rain. Acid rain is very destructive; it can decimate
entire forests, corrode limestone structures and disrupt natural ecosystems by altering natural pH levels.

Health Problems: Sulfur dioxide is a respiratory irritant and can cause breathing difficulties at concentrations as
low as 1 ppm. It triggers asthma attacks and aggravates emphysema. Nitrogen dioxide is also a respiratory
irritant. At concentrations above 3 ppm, it can begin to destroy tissue.

Explain the formation and effects of acid rain:
–
Pure water has a pH of 7.
–
Acid rain is any rain that has a pH of less than 5.
–
FORMATION of acid rain:

Sulfur dioxide reacts with rain in the atmosphere forming sulfurous acid:
 SO2 (g) + H2O (l)
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H2SO3 (aq)
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Page 14 of 49

Sulfurous acid then reacts with oxygen; this is catalysed by air particles:
 2H2SO3 (aq) + O2 (g)

Nitrogen dioxide also reacts with rain, making nitrous and nitric acids:
 2NO2 (g) + H2O (l)

HNO2 (aq) + HNO3 (aq)
Nitrous acid then reacts with oxygen, this is catalysed by air particles:
 2HNO2 (aq) + O2 (g)

2H2SO4 (aq)
2HNO 3 (aq)
Thus, in industrialised areas, rain can contain relatively high levels of very strong acids; that is nitric and sulfuric
acids.
–
Effects of acid rain:

Plants destruction:
 Acid rain settles into soil and causes chemical reactions there which affect plants, causing defoliation and a
decrease ability to withstand frost.
 Sulfate ions in acid rain react with calcium and magnesium in soil so that these essential minerals become
unavailable for plants, hence defoliating and reducing growth aswell, example spruce trees. More than half
the trees in Germany’s Black Forest were damaged by acid rain.

Animal life:
 Acid rain lowers the ph of lakes and streams e.g. in Scandinavia, due to emissions from Central Europe,
killing fish eggs and other aquatic life. In some pH-sensitive fish species, such as brook trout, many
populations have been completely wiped out.
 In other areas the acid has been able to react with aluminium compounds and releasing toxic aluminium that
causes the fish gills to clog with mucus.

Erosion of marble and limestone buildings: These contain carbonate (esp. calcium carbonate) which reacts
readily with acid forming salt.

 CaCO3 (s) + H2SO4 (aq)
CaSO4 (s) + H2O (l) + CO2 (g)
 CaCO3 (s) + HNO3 (aq)
Ca(NO3)2 (s) + H2O (l) + CO2 (g)
Calculate volumes of gases given masses of some substances in reactions, and calculate masses of substances given
gaseous volumes, in reactions involving gases at 0˚C and 100kPa or 25˚C and 100kPa:
–
Thus, the list of Chemistry equations so far are:
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–
Note: The concentration (ie molarity/molar) of a solution is defined as moles per litre or mol/L, ie 1 molar = 1 M = 1
mole/litre.
–
Basic Stoichiometry:

Stoichiometry is the study of the calculation of quantitative (measurable) relationships of the reactants and
products in chemical reactions.

One of the basic laws of stoichiometry is the ‘Law of Definite Proportions’:
 In chemical reactions, reactants combine in fixed, definite proportions (or RATIOS) to form products. The
ratios are determined by the coefficients of the species in a BALANCED chemical reaction. The coefficient
of a species is the number before the species:

Eg: S (s) + O2 (g)
SO2 (g)
 Looking at the coefficients, you can see that in this reaction, S will react with O2 in the ratio of 1:1, forming 1
SO2; e.g. if 3 mol of sulfur is burnt, then 3 mol of oxygen is also burnt, and 3 mol of SO2 is produced.

Eg: 2H2S (g) + 3O2 (g)
2SO2 (g) + 2H2O (l)
 In this reaction the ratio of reactants to products is 2:32:2. RATIOS are extremely important in
determining the masses of reactants and products.
 EG: Calculate the moles, and the mass, of sulfur dioxide produced when 50g of hydrogen sulfide is oxidised:

First, we work how many moles of H2S is in 50g:
n
= m/M (mass/molar mass)
= 50/(2 × 1 + 32.1); molar mass of H = 1, S = 32.1
= 1.466 mol

Since we are only concerned with H2S and SO2, we look at the ratio of H2S:SO2; the ratio is 2:2, which is
the same as 1:1 (simplified). Hence, there were also 1.466 mol of SO2 produced.

To calculate the mass:
n
= m/M
1.466 = m/(32.1 + 2 × 16); molar mass of S = 32.1, O = 16
1.466 = m/64.1
Hence, m

= 93.97 grams
Therefore, even though there were the same number of moles of both compounds, 50g of H2S burned to
produce 93.97g of SO2
–
Working with Volumes:

It was discovered that ALL gaseous substances, with the same temperature and pressure, occupied the same
VOLUME, no matter how massive the molecules that made up the gas were.

This means that a fixed amount of molecules of hydrogen gas (with an atomic weight of 2) will occupy the same
volume as a fixed amount of molecules of radon gas (which has an atomic weight of 444!). Logically, you can
have 10 small , 2 atomic weight units OR 1 large 20 atomic unit, still both same volume.
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
Thus, it was discovered that:

This is a simple concept that is to apply:
 EG: Calculate the volume of sulfur dioxide produced when 50g of hydrogen sulfide is oxidised at 25°C and
100kPa.

From the working produced above, we know that 1.466 mol of sulfur dioxide are produced. Since one
mole of gas at 25°C and 100kPa occupies 24.79 L, 1.466 moles of gas will occupy:
V = n × MV (moles × molar volume)
= 1.466 × 24.79
= 36.34 L

Identify data, plan and perform a first-hand investigation to decarbonate soft drink and gather data to measure the
mass changes involved and calculate the volume of gas released at 25˚C and 100kPa.
–
Aim: To decarbonate a soft drink bottle and to use the information gathered to determine the mass, and therefore the
volume of gas released.
–
Risk Assessment:

–
A very low level of risk is associated with this practical.
Method:

An unopened 600ml can of soft drink was weighed.

The cap of the bottle was then unscrewed, so that gas could be released from it.

The cap was then retightened, and the bottle inverted gently several times, before being reopened to release the
gas. This procedure was then repeated several times.

The process of decarbonation was stopped when the bottle maintained a constant weight over 3 consecutive
weightings after being shaken/opened.
–
Results:

2.4 g of CO2 was released. To calculate the volume of gas released at 25˚C:
n
= m/M (mass/molar mass)
= 2.4/44; molar mass of CO2 = 44
= 0.0545 mol
V = n × MV (moles × molar volume)
= 0.0545 × 24.79
= 1.35L
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–
–
Justification:

No liquid was allowed to spill to try to keep a high level of accuracy.

Electronic scales were also used for accuracy.
Limitations:

The accuracy of this method is limited, as there may be loss of water as gas, OR there may still be CO2 dissolved.

The assumption that all gases lost was from the carbon dioxide pumped into the bottle – some oxygen (which has
a low solubility) may have been lost.
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3. Acids occur in many foods, drinks and even within our stomachs:

Define acids as proton donors and describe the ionisation of acids in water:
–
An acid is a substance that releases H+ ions.
–
Another name for the H+ ion is a ‘proton’; this is because a positive hydrogen atom (H+) is just a hydrogen nucleus
(no electrons) and hence is just a proton.
–
When acids react with other substances, the H+ ion is transferred to another species; that is why acids can be defined
as proton-donors.
–
In water, acids IONISE (separate into its ions).
–
Eg: Pure hydrogen chloride added to water:

–

H+ (aq) + Cl‫(־‬aq)
However, H+ ions are very reactive, they immediately react with water molecules close by.

–
HCl (g)
HCl (g) + H2O (l)
H3O+ (aq) + Cl‫(־‬aq)
Acids ionise to form hydronium ions (H3O+)
Identify acids including acetic (ethanoic), citric (2-hydroxypropane-1,2,3-tricarboxylic), hydrochloric and sulfuric acid:
–
–
–
–
Acetic Acid:

Systematic name: Ethanoic acid

Molecular formula: CH3COOH

It is the acid present in vinegar.

It is classified as a weak acid.
Citric Acid:

2-hydroxypropane-1,2,3-tricarboxylic acid OR tri-ethanoic instead of tricarboxylic

Molecular formula: C6H8O7

It occurs in citrus fruit.

Widely used as a preservative.

It is classified as a weak acid.
Hydrochloric Acid:

Molecular formula: HCl

It is a very strong acid.

It is produced in the stomach to aid in digestion.

It is also industrially made in large quantities, with many uses.
Sulfuric Acid:

Molecular formula: H2SO4

It is also a strong acid.

Most industrially produced chemical. Used to make batteries, fertilizers etc.
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
Describe the use of the pH scale in comparing acids and bases:
–
pH: ‘power of hydrogen”, is a measure of the acidity or basicity of a solution depending on the concentration of
hydrogen ions.
–
The pH scale is used to determine the acidity or basicity of a substance.


It is numbered from 0 to 14.
–
A pH of 7 is attributed to neutral substances.
–
A pH < 7 refers to acidic substances; a strong acid is close to 0.
–
A pH > 7 refers to basic substances; a strong base is close to 14.
Describe acids and their solutions with the appropriate use of the terms strong, weak, concentrated and dilute:
–
A strong acid is an acid that releases ALL its H+ ions when in solution.
–
Its molecules are completely disassociated in solution.
–
Its ionisation reaction goes to completion:

–
Eg: HCl (g) + H2O (l)
H3O+ (aq) + Cl‫(־‬aq)
A weak acid is an acid that does NOT completely release all its H+ ions; hence it is ‘weaker’ than a strong acid of the
same concentration. Some of its molecules remain intact in solution.
–
Its ionisation reaction with water is a reversible reaction that reaches equilibrium when a certain number of H+ ions
are released:


Eg: CH3COOH (s) + H2O (l)
H3O+ (aq) + CH3COO‫( ־‬aq)
–
The terms concentrated and dilute refer ONLY to the amount of molecules present in the solution.
–
A CONCENTRATED acid has approximately more than 5 mol/L of solute.
–
A DILUTE acid has less that 2 mol/L of solute.
Identify pH as -log10[H+] and explain that a change in pH of 1 means a ten-fold change in [H+]:
–
pH is calculated through an equation related to H+ concentration.
–
This equation is:

–
pH = -log10[H+] OR pH = -log10[H3O+]
A change in pH of 1 means a 10-fold change in [H+]:

Eg: pH of 4; [H+] = 10-4 while a pH of 3; [H+] = 10-3. Therfore an increase in 1 ph, is a 10 fold , ie 10-3 / 10-4 = 10.
–
SIMILARLY, there is another scale of acidity and basicity: it is the pOH scale.
–
It is almost the same as pH, except that is uses OH‫ ־‬concentration as a measure, instead of H+ concentration:

–
pOH = -log10[OH-]
The RELATIONSHIPS between pH and pOH are:

pH + pOH = 14

[H+][OH-] = 10-14
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
Process information from secondary sources to calculate pH of strong acids given appropriate hydrogen ion
concentrations:
–
The pH of a strong acid can be calculated, given the molar concentration, as ALL its protons are released into solution.
–
However, an acid may be able to release MORE than one proton. These acids are called polyprotic acids.
–
CALCULATING pH of a strong, monoprotic acid:

Monoprotic acids release only one proton, e.g. HCl:
 HCl

H+ + Cl-
Eg: Calculate the pH of 0.01 M hydrochloric acid.
 M means mol/L; hence this is 0.01 mol/L, or 10-2 mol/L
 Then pH = -log10(10-2) = 2
–
CALCULATING pH of a strong, diprotic acid:

Diprotic acids release 2 protons, e.g. H2SO4.
 H2SO4
2H+ + SO42-

Hence, for every mole of acid, it releases 2 protons.

Eg: Calculate the pH of 0.1 M sulfuric acid:
 It has 0.1 mol/L; therefore its [H+] = 2 × 0.1 = 0.2 mol/L
 So, pH = -log10(2 × 0.1 ) = 0.7
–
Extra: CALCULATING pH of a WEAK acid:

Weak acids do not fully ionise; if the degree of ionisation is given, usually as a percentage.

Eg: Calculate the pH of 0.1 mol/L ethanoic acid if only 1.3% ionises:
 [H+] = 1.3% of 0.1 = 0.0013
 So, pH = -log10(1.3% of 0.1) = -log10(0.0013) = 2.9
–
Extra: CALCULATING pH of a BASE:

Just as acids can be polyprotic, bases can release more than one OH- ion.

This must be taken into consideration when calculating the pH.

EG: Calculate the pH 0.1 M of Ca(OH)2 if it fully ionises:
 Ca(OH)2
2OH‫ ־‬+ Ca2+
 [OH-] = 2 × 0.1 = 0.2
 pOH = -log10(0.2) = 0.69
 pH = 14 – pOH = 14 – 0.69 = 13.3
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Note: Come to this section, after the topic the rest of topic has been finished
–
Extra: Calculating the pH of a solution after DILUTION:

The dilution formula is used; c1v1 = c2v2.

Eg: Find the pH of the solution formed when 10 mL of 0.1 M sulfuric acid is diluted to 250 mL (it is diprotic):
 H2SO4
2H+ + SO42-
 From above the concentration of sulfuric acid is 0.1mol/L , hence concentration of hydrogen ion is
c1 = [H+] = 2 × 0.1 = 0.2, the volume of the solution remains the same, hence v1 = 0.01 L and v2 = 0.25 L
 Hence: c2 = (c1 × v1) ÷ v 2 = (0.2 × 0.01) ÷ 0.25 = 0.008 M
 Therefore the original concentration of hydrogen ions ie [H+] was 0.1 mol/L, after dilution this has dropped
down to 0.008.
 The original pH would be –log10(0.1 × 2) = 0.69
 The new pH is –log10(0.008) = 2.1
–
Extra: Calculating the pH of a solution after NEUTRALISATION:

The neutralisation of a acid and a base can leave a solution that is slightly acidic or basic; the volumes and
concentrations of solutions used determines the pH of the final product.

Note: The solution will only be neutral when there are exactly the same number of moles of H+ and OH- reacted.

Eg: 10 mL of 0.5 M hydrochloric acid is mixed with 25 mL of 0.35 M sodium hydroxide. Find the pH of the final
product, assuming salt is neutral:
 Firstly, a balanced chemical equation must be written:

HCl (aq) + NaOH (aq)
NaCl (aq) + H2O (l)
 As can be seen, the molar ratio of reaction for the acid and base is 1:1.
 Calculating the number of moles of acid and base, using n = c × v:

Acid: n = 0.5 × 0.01 = 0.005

Base: n = 0.35 × 0.025 = 0.00875
 There are more moles of base than there are acid; hence there will be an excess of base after neutralisation
completes.
 Moles of base remaining = 0.00875 – 0.005 = 0.00375
 As NaOH only releases one OH- ion, [OH-] = 0.00375

pOH = -log10(0.00375) = 2.426

Hence pH = 14 – pOH = 11.6
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
Compare the relative strengths of equal concentrations of citric, acetic and hydrochloric acids and explain in terms of
the degree of ionisation of their molecules:
–
Equal concentrations (1 mol/L) of citric, acetic and hydrochloric acid were made, and their pH was measured using a
pH probe.
–
The pH’s were: HCl = 1, Acetic (CH3COOH) = 2.9 and Citric (C6H8O7) = 2.1
–
Using the formula, [H+] = 10-pH (derived from the pH formula), concentrations of hydrogen ions released by each acid:
–

HCl: [H+] = 10-1 = 0.1 mol/L

CH3COOH: [H+] = 10-2.9 = 0.00126 mol/L

C 6H8O7: [H+] = 10-2.1 = 0.0079 mol/L
Degree of Ionisation: is the percentage of H+ ions that have been released by an acid; it is calculated by expressing
the actual [H+] over the concentration (ie number of molecules) of the acid.
–
–
Noting that brackets are “concentration of”

HCl: 0.1/0.1 × 100 = 100% ionisation (a strong acid)

CH3COOH: 0.00126/0.1 × 100% = 1.26% ionisation (a very weak acid)

C 6H8O7: 0.0079/0.1 × 100% = 7.9% ionisation (a moderately weak acid)
Hence, looking at the degrees of ionisation, the strongest acid is hydrochloric acid, followed by citric acid, and then
ethanoic acid.


Gather and process information from secondary sources to write ionic equations to represent the ionisation of acids:
–
Hydrochloric: HCl (g) + H2O (l)
–
Nitric: HNO 3 (l) + H2O (l)
–
Sulfuric: H2SO4 (l) + 2H2O (l)
–
Ethanoic: CH3COOH (s) + H2O (l)
H3O+ (aq) + Cl‫( ־‬aq)
H3O+ (aq) + NO32H3O+ (aq) + SO42‫־‬
H3O+ (aq) + CH3COO‫( ־‬aq)
Describe the difference between a strong and a weak acid in terms of an equilibrium between the intact molecule and
its ions:
–
A strong acid is an acid that releases ALL its H+ ions when in solution. Its molecules are completely disassociated in
solution.
–
Its ionisation reaction goes to completion:

–
Eg: HCl (g) + H2O (l)
H3O+ (aq) + Cl‫( ־‬aq)
Conversely, a weak acid is an acid that does NOT completely release all its H+ ions; hence it is ‘weaker’ than a strong
acid of the same concentration.
–
Some of its molecules remain intact in solution.
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Page 23 of 49
–
Its ionisation reaction with water is a reversible reaction that reaches equilibrium when a certain number of H+ ions are
released:


EG: CH3COOH (s) + H2O (l)
H3O+ (aq) + CH3COO‫( ־‬aq)
Use available evidence to model the molecular nature of acids and simulate the ionisation of strong and weak acids:
–
Take these two beakers as a model of the molecular nature of acids, and the level of ionisation of different acids.
–
In the LEFT beaker, there is a solution of sulfuric acid. All the molecules have completely ionised, as it is a strong
acids. Each H2SO 4 molecule has released 2 protons, as it is a diprotic acid.
–
However, in the RIGHT beaker, there is a solution of ethanoic acid, of equal concentration. Only half of the acid
molecules have ionised (in reality the percentage is much lower, 1.26% ionisation). The unionised molecules remain
as intact molecules, holding on to their hydrogens.

Gather and process information from secondary sources to explain the use of acids as food additives:
–
Acids are added to food for 2 reasons: as preservatives, and to add flavour.
–
Preservatives:
–

Ethanoic acid (in the form of vinegar) is used as a preservative in ‘pickling’.

Citric acid is a natural preservative, often added to jams and conserves.
Flavourings:

Carbonic acid is added to soft drinks to add ‘fizz’.

Ethanoic acid, as vinegar, is also used as a flavouring on salads etc.
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
Identify data, gather and process information from secondary sources to identify examples of naturally occurring acids
and bases and their chemical composition:
Name
Formic acid
Acetic acid
Citric acid
Ascorbic acid
Carbonic acid
Hydrochloric acid
Hydrogen
carbonate ion
Calcium carbonate
Ammonia

Acid or base
Acid
Acid
Acid
Acid
Acid
Acid
Base
Chemical formula
HCOOH
CH3COOH
HOC(CH2COOH)2COOH
C6H8O6
H2CO3
HCl
HCO3-
Base
Base
CaCO3
NH3
Source
Ant and bee stings
Vinegar
Fruits
Vitamin C
Soda water
Stomach Acid
Biochemical reactions in
the body
Limestone, marble
Urine
Solve problems and perform a first-hand investigation to use pH meters/probes and indicators to distinguish between
acidic, basic and neutral chemicals:
–
–
Part A: Testing Substances with Methyl Orange (3.1-4.4):

10 test-tubes were set up with 5 mL of acidified solution (0.1 M HCl).

2 drops of methyl orange were placed in each test-tube.
Results:

It is known that methyl orange is red in strongly acidic solutions, and yellow in slightly acidic to highly alkaline
solutions.
–
Part B: Testing Common Substances Using a pH Meter:

A small amount of each solution was placed in a separate beaker, and a few drops of universal indicator solution
added. The colour change of the solution was matched with a provided colour chart to determine the solution’s
approximate pH.

A pH probe connected to a data logger was then calibrated using a solution of known pH, before being used to
measure the pH of the solution more exactly.
–
Results:

–
pH of solutions was found.
Justification:

Methyl orange is only red in strongly acidic solutions, so even a very weak base would have been able to elicit a
colour change.

A wide range of substances was used to portray the wide range of possible bases. This showed that bases are not
only limited to metal hydroxides.

A pH meter gave instantaneous accurate results.
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–
Limitations:

The methyl orange test was not able to distinguish basic and neutral substances, as both caused no colour change
from the red.

Using a pH meter or probe is a non-destructive way of testing whether a chemical solution is acidic, basic or
neutral. Provided the pH meter electrode or probe is washed well with distilled water between measurements, the
solutions tested should be unaffected, but the indicator will contaminate the portion of solution tested.

Plan and perform a first-hand investigation to measure the pH of identical concentrations of strong and weak acids:
–
A pH meter was used to measure the pH’s of the solutions.
–
RESULTS:

Sulfuric acid; pH = 0.7

Hydrochloric acid; pH = 1.2

Citric acid; pH = 2.9

Ethanoic acid; pH = 3.3
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4. Because of the prevalence and importance of acids, they have been used and studied for hundreds of years. Over
time, the definitions of acid and base have been refined:

Outline the historical development of ideas about acids including those of Lavoisier, Davy and Arrhenius:
–
Lavoisier: ACIDS contain OXYGEN:

He hypothesised that since many of the common acids contain oxygen (such as acetic acid, C2H3COOH, carbonic
acid, H2CO3, sulfuric acid, H2SO4).

–
He thought that oxygen was the source of the acidity.
Davy: ACIDS contain HYDROGEN:

He showed that hydrochloric acid (HCl) did not contain oxygen, disproving Lavoisier’s hypothesis.

Many other non-oxygen containing acids had been discovered, such as hydrofluoric acid (HF), hydrobromic acid
(HBr) and hydrocyanic acid (HCN).

–

Thus he hypothesised that all acids contain hydrogen.
Arrhenius: ACIDS produce H+ IONS in WATER, BASES produce OH- IONS in WATER:

Things like NaOH, contained hydrogen, but were not acids, thus the theory was neglected.

Arrhenius proposed the idea that acids disassociate into their ions when they are dissolved in water.

Thus, he hypothesised that acids release a H+ when in an aqueous solution.

He also said that bases release OH‫ ־‬ions in aqueous solutions.
Gather and process information from secondary sources to trace developments in understanding and describing
acid/base reactions:
–
In modern times, the theories of Lavoisier and Davy have been deemed redundant, or insufficient, while Arrhenius
acid/base theory is still used as a simple model.
–
The 2 most recent theories are the Brönsted-Lowry and the Lewis acid/base model:
Scientist(s)
Acid Definition
Base Definition
History
Lavoisier
Corrosive
substances that
contain oxygen
Corrosive
substances with
hydrogen
Substances
disassociate into H+
ions in solution
No definition
Worked with metal oxides that form the common
oxyacids with water.
No definition
Worked with hydrohalic acids such as HCl and HBr,
disproving Lavoisier.
Substances that
disassociate into
OH- ions in
solution
Proton (H+)
acceptor
Electron pair
donator
Made theories of acid/base ionisation. Only aqueous
(water) solutions were considered.
Davy
Arrhenius
Brönsted-Lowry
Proton (H+) donor
Lewis
Electron pair
acceptor
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Extended acid/base reactions to those without water.
Acids must contain hydrogen; no solution required
No hydrogen required. Not required for HSC course.
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
Outline the Brönsted-Lowry theory of acids and bases:
–
–
–
The Arrhenius definition of acids and bases is LIMITED in that it:

Only applies to aqueous (water) solutions.

Only accounts for substances that already have hydrogen or hydroxide ions in their structure.

Cannot explain how some substances can act as BOTH an acid and a base (ie amphiprotic).
The Bronsted-Lowry theory states:

A acid is a PROTON donor (ie gives protons to a base)

A base is a PROTON acceptor (ie accepts protons from an acid)
Thus is different from Arrhenius theory as acids and bases no longer have to be aqueous solutions; by BrönstedLowry definition they can be solids, gases, cations, anions, non-water solutions.
–
Eg:

–
–

HCl (g) and NH3 (g)
According to Arrhenius theory, this is not an acid OR base reaction, as there is:

No water is present (they are all gases), hence no free H+ ions.

HCl and NH3 are gases in this reaction, so they are not Arrhenius acids and bases.

NH3 is a base, but doesn’t have OH- in its structure.
Hence, by Brönsted-Lowry definition, HCl gas is an acid, and NH3 gas is a base.
Describe the relationship between an acid and its conjugate base and a base and its conjugate acid:
–
Through the acceptance of the Bronsted-Lowry theory, an acid is a proton-donor, and a base is a proton-acceptor, the
concept of conjugate acids and bases can be introduced:
–

A conjugate base is the acid with a proton removed

A conjugate acid is the base with a proton added
Eg: Reaction between hydrochloric acid and water:

HCl (aq) + H2O (l)
H3O+ (aq) + Cl‫( ־‬aq)
–
The original acid is HCl; its conjugate base is Cl‫( ־‬acid minus proton).
–
The original base is H2O; its conjugate acid is H3O+ (base plus proton).
–
Eg: Reaction between ammonia and phosphine:

NH3(g) + PH3(g)
PH4 (aq) + NH2- (aq)
–
The original acid is NH3; its conjugate base is NH2-
–
The original base is PH3; its conjugate acid is PH4+
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–
Speacial case: Reacting hydrogen chloride gas and ammonia:

HCl (g) + NH3 (g)
NH4+ + Cl-
NH4Cl(s)
-
–
The original acid is HCl; its conjugate base is Cl
–
The original base is NH3; its conjugate acid is NH4+
–
BUT Ammonium Chloride is a salt compound.
–
Note 1: Unlike the previous definitions, the Brønsted-Lowry definition does not refer to the formation of salt and
solvent, but instead to the formation of conjugate acids and conjugate bases, produced by the transfer of a proton
from the acid to the base. An acid and a base react not to produce a salt and a solvent, but to form a new acid and a
new base.
–
Note 2: Ammonia is a weak base, and a unique thing about it, is NEARLY EVERY reaction with an acid, it does not
produce gas, every other base produces gas (ie CO2).

Identify conjugate acid/base pairs:
–
Conjugate base is acid without a proton; conjugate acid is base with a proton.
–
It can get confusing during exams when under stress to think which side is the base and to minus or plus hydrogen.
–
A simple formula:
–
Eg: HCl (aq) + H2O (l)

–
–
H3O (aq) + Cl‫( ־‬aq)
Acid/base pairs are HCl/Cl‫ ־‬and H3O+/H2O
Eg: NH3 (aq) + H2O (l)

BASE + H+ = ACID
NH4+ (aq) + OH‫( ־‬aq)
Acid/base pairs are H2O/OH‫ ־‬and NH4+/NH3
Eg: What is the conjugate acid of NH2‫?־‬

When it asks for conjugate acid , it assumes automatically the given is a base , hence to get to acid , it is the base
plus a proton; hence it is NH3.

Identify a range of salts which form acidic, basic or neutral solutions and explain their acidic, neutral or basic nature:
–
A salt is as a compound that forms when an acid reacts with a base (neutralisation). They are composed of cations
(positively charged ions) and anions (negatively charged ions)
–
Salts are NOT all neutral; there are many acidic and basic salts.
–
The reason why many salts solutions are NOT neutral (pH ≠ 7.0) is that the ions making that salt, can act as BrönstedLowry acids/bases, that is they are called conjugate acids + conjugate bases.
–
From above we acknowledged the fact that B-L theory does not recognise salts/solvents etc , there is only conjugate
pairs. Hence when these ions bring along their properties to the salt formed.
–
The acidity or alkalinity of a salt depends on the reactants of the neutralisation reaction:

A strong acid and a strong base will make a neutral salt

A weak acid and a weak base will make a neutral salt

A strong acid and a weak base will make a slightly ACIDIC salt
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
A weak acid and a strong base will make a slightly BASIC salt
–
These are the general rules for neutralisation reactions, but the acidity/alkalinity of a salt must be proved by hydrolysis.
–
Hydrolysis is simply reacting the salt’s ions with water or more specifically The reaction of an anion with water to
produce OH- or the reaction of a cation to produce H3O+ (its is the opposite of neutralisation).

Just as neutralisation is reaction of acid and base to form salt and water, salt can react with water to go the other
way around.

Reaction between a STRONG acid and a WEAK base:
 Theoretically the salt will be slightly acidic; pH < 7.

Eg: HNO3 (aq) + NH3 (aq)
NH4NO3 (aq)
 Nitric acid is a strong acid, and ammonia is weak base; hence ammonium nitrate will be a slightly acidic salt.
 HOWEVER you must prove it. React the salt’s ions with water:

NH4+ + H2O

NO3- + H2O
NH3 + H3O+
No reaction
 The presence of hydronium ions proves it is an acidic solution.

Reaction between a WEAK acid and a STRONG base:
 The salt will be slightly basic; pH > 7.

Eg: NaOH (aq) + CH3COOH (aq)
CH3COONa (aq) + H2O (l)
 Sodium hydroxide is a strong base, and ethanoic acid is a weak acid; hence sodium ethanoate will be a
slightly basic salt.
 To prove it:

Na+ + H2O

CH3COO‫ ־‬+ H2O
No Reaction
CH3COOH + OH‫־‬
 The presence of hydroxide ions proves it is a basic solution.

Reaction between a STRONG acid and a STRONG base:
 This salt will be neutral; pH close to 7.

Eg: KOH (aq) + HCl (aq)
KCl (aq) + H2O (l)
 Neither of the ions in the product reacts with water; hence it is neutral.

Choose equipment and perform a first-hand investigation to identify the pH of a range of salt solutions:
–
Aim: To test the pH of a range of salt solutions.
–
Method:

Each salt was placed in a beaker and their respective pHs were first estimated by looking at what acid and base
constitutes the salt, The acidity of salt solutions can be explained by the Bronsted-Lowry acid-base behaviour of
their constituent ions.

They were then measured with a pH meter. Then several drops of universal indicator were added to each.
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
–

Results:

Na2CO3: Approximately 11

NH4CH3COO: Just over 7

KNO3: Approximately 9

NaHCO3: Approximately 9

NH4NO3: Approximately 6

MgCl2: Approximately 6

NaNO3: Approximately 7

CH3COONa: Approximately 8.9
Identify amphiprotic substances and construct equations to describe their behaviour in acidic and basic solutions:
–
An amphiprotic substance is one that can act as BOTH an acid and a base.
–
Their behaviour depends on the environment they are placed in.
–
EG: The hydrogen carbonate ion (bicarbonate ion) HCO3‫ ־‬is amphiprotic:


–

Note: never learn these; just remember combinations of weak/strong acid-bases.
HCO3‫( ־‬aq) + H3O+ (aq)
HCO3
‫־‬
(aq)
+ OH
‫־‬
H2CO3 (aq) + H2O (l)
CO32‫( ־‬aq) + H2O (l)
(aq)
Other amphiprotic species include the hydrogen sulfite ion (HSO 3‫ ) ־‬and water.
Identify neutralisation as a proton transfer reaction which is exothermic:
–
Neutralisation reactions are reactions between acids and bases. As acids are proton-donors & bases are protonacceptors, neutralisation reactions between acids and bases are proton-transfer reaction reactions. Ie protons (H+) are
transferred from the acid to the base.
–
The first definition of neutralisation was “acid + base = salt and water”, but not all neutralisation reactions release
water, but all neutralisation donate/accept H + hence it needed to be changed.
–
Note: nearly every hsc question uses the defintion acid + base = salt + water, keep it in hand, and know about proton
transfer as a second defintion.
–
–
Eg: HCl (aq) + KOH (aq)
KCl (aq) + H2O (l)

H + Cl + K + OH

If we remove all the spectator ions, the basic underlying proton-transfer reaction is easily seen:

H+ (aq) + OH ‫( ־‬aq)

Hence, the proton is transferred to the hydroxide ion, forming water.
+
-
+
-
Eg: H+ (aq) + NH3 (aq)

K+ + Cl- + H2O
H2O (l)
NH4+ (aq)
This reaction is taken under ‘net ionic equation’ the original reaction is (HBr + NH3
NH4Br), and it can
be done rightfully so.

The proton is transferred to the ammonia, forming ammonium.
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
–
As can be seen, no water is produced; it is still considered neutralisation.
Note: Most neutralisation reactions are exothermic; they liberate heat energy, (example of endothermic reaction is
vinegar and baking powder)
–
The ΔH ≈ -56 kJ/mol, depending on the strength of the reactants.

Recall: Exothermic reactions release energy as more energy is released in bond formation than is absorbed in
bond breaking.

Analyse information from secondary sources to assess the use of neutralisation reactions as a safety measure or to
minimise damage in accidents or chemical spills:
–
It is important to immediately neutralise any chemical spills involving strong acids and bases, as they are corrosive
and dangerous.
–
Neutralisation reactions (converting an acid/base into salt) are used as safety measures in cleaning up after such
incidents.
–
When neutralising an acid or a base the following procedure is followed:

The most preferred agents of neutralisation has the properties of being stable, easily transported, solid
(powdered), cheap and weakly amphiprotic (so it can act as a WEAK acid or a WEAK base). This is the safest
material, as it can neutralise both acids and bases; even if an excess is used, it is very weak, and so does not pose
any safety risks.
–
The most common substance used to neutralise spills in laboratories is powdered sodium hydrogen carbonate, known
as sodium bicarbonate (NaHCO3); this is because the hydrogen carbonate ion (HCO3-) is an amphiprotic species.
–

NaHCO3 + HCl

NaHCO3 + NaOH
NaCl + H2O + CO2 (acting as base)
Na2CO3 + H2O (acting as an acid)
Strong acids and bases must never be used to neutralise spills; if an excess is used, the spill will become dangerous
again.

Describe the correct technique for conducting titrations and preparation of standard solutions:
–
Titration is a chemical analysis technique in which neutralisation is used to experimentally determine the unknown
concentration of a solution through the reaction of that solution and a solution of a known until the reaction between
them is complete.
–
The Chemical Theory of Titration:

The point of acid/base titration is to determine the concentration of an unknown solution (secondary solution) by
slowly reacting a certain volume of this solution with another solution of known concentration (standard
solution), until the equivalence point/endpoint is reached.
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
The equivalence point of a chemical reaction occurs when all available molecules have reacted, and the reaction
comes to an end. The endpoint is when the indicator (which is used to give a visual indication of the molecules
have reacted), changes colour.

The volumes of the reactants at this end/equiv points are carefully measured; using the knowledge of these
volumes, and the original concentration of the standard solution, the concentration of the unknown solution can
be calculated.

–
Note: because volume measurements play a key role in titration, it is also known as volumetric analysis.
Primary Standards:

Primary standards are the substances which can be used to prepare standard solutions (ie they are the solutes of
the solvent solution).

There are certain criteria chemicals need to satisfy before they can be used to create standard solutions; these
must have the following properties:
 HIGH PURITY:

This is to produce accurate results, untainted by chemical impurities (ex HCl can vary from batch to
batch).
 CHEMICAL STABILITY (low reactivity):

Standards must be chemically stable so they do not react with the water/gases in the air (e.g. CO2).
 NON-HYGROSOPIC AND NON-EFFLORESCENT:

Hygroscopic substances absorb water from their surroundings, while efflorescent substances release
water into their surroundings.

Both these processes change the concentration of solutions, resulting in imprecise standard solutions.
 HIGH SOLUBILITY:

Primary standards need to dissolve completely into their solutions.
 HIGH MOLECULAR WEIGHT:


The high molecular weight of primary standards minimizes many errors in measurement.
Unsuitable chemicals to use as primary standards include sodium hydroxide, which in solution
will react with gases in the air, hydrochloric acid, which is efflorescent, and sulfuric acid, which
is hygroscopic.
–
Preparing Standard Solutions:
–
Calculating the amount of primary standard needed:

Before making standard solutions, the amount of primary standard to be used, needs to be
calculated. The typical laboratory glassware in which standard solutions are made are called
volumetric flasks; 250 mL flasks are most commonly used.

We will determine the amount of primary standard needed to create 250 mL of the 0.05 M
standard solution of sodium carbonate.
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
CALCULATIONS:
 Molar Mass (Na2CO3) = 2 (22.990) + (12.011) + 3 (15.999) = 105.987
 n = c × v = 0.05 x 0.25 = 0.0125 mol
 mass = 0.125 x 105.987 = 1.325 g

–
Hence, to make up 250 mL of 0.5 M solution, 1.325 g of primary standard Na2CO3 is needed.
Method:
1.
Firstly, the primary standard must be as pure as possible. The primary standard must be placed in an oven, and
cooled in a dessicator to remove all traces of moisture.
2.
Thoroughly rinse a 250 mL volumetric flask, a small beaker and a glass funnel with distilled water. Place the
funnel in the neck of the volumetric flask, and place the beaker on an electronic scale.
3.
Zero the scale, and using a very clean spatula, transfer as accurately as possible, 1.325 g of sodium carbonate
into the beaker.
4.
Using a wash-bottle of distilled water, transfer the powder into the flask by ‘washing’ it into the funnel. Ensure
that the entire beaker is washed, and all water that touches the beaker flows into the funnel, to ensure all solute is
transferred. Wash the funnel, allowing the water to flow into the flask, and then remove it.
5.
Fill the flask half-way up to the 250 mL graduation mark, and gently swirl the flask until all the solute has
dissolved. Set the flask on the bench.
6.
Using the wash-bottle, fill the flask with water until it is just under the graduation mark. Using a Pasteur pipette,
add distilled water until the meniscus sits exactly on the 250 mL mark.
7.
Cover the flask with its glass stopper and invert the entire flask three times. If the meniscus has lowered, add a
few more drops. If not, the standard solution is complete.
–
The Correct Technique for Conducting Titrations:

There is a very precise and specific technique to titration, which uses a variety of calibrated glassware. These
include:
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 Volumetric Flask: This is used to prepare and hold standard solutions.
 Conical Flask (ie Erlenmeyer flask): This is used to hold the reactants during titration. Its shape prevents the
reactants from spilling as they are swirled together.
 Burette: The burette is a piece of cylindrical glassware, held vertically, with volumetric divisions on its full
length and a precision tap (stopcock), on the bottom. It is used to dispense precise amounts of a liquid reagent
in titration. Burettes are extremely precise and accurate to ±0.05 mL.
 Pipette: The pipette is a glass tube used to transfer precise volumes of liquid reagents. Pipettes are usually
designed to transfer one measurement of volume, such as only 25 mL. The reagent is drawn up the pipette
using a pipette filler (e.g. a rubber bulb).

Before titration, all glassware must be RINSED appropriately. The technique for the different glassware is:
 Volumetric flasks (including glass stoppers) are rinsed thoroughly with distilled water, preferably multiple
times. Close the flask with the stopper until it is going to be used; it is left wet.
 Conical flasks are rinsed thoroughly with distilled water and left wet.
 Burettes: the distilled water is filled into the burette and the tap opened. Water is allowed to flow out and
thoroughly rinse the tip. More water is then added, and the entire glass tube is swirled in your hands to wash
the sides of the burette. Burettes are rinsed THREE TIMES with distilled water and then ONCE more with
the solution it is going to contain.
 Pipettes are also rinsed THREE times with distilled water, and then ONCE with the solution it is going to
contain.

After everything is rinsed appropriately, the glassware is filled.

By convention, the acid is placed in the burette (unknown), and the base in the conical flask
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
Using the above example again:
 Using a funnel, the nitric acid is poured into the burette until ABOVE the zero mark. Hold a white card
behind the zero mark, and open the tap slowly until the meniscus sits JUST on top of the mark. The white
card makes the meniscus clearer; the volume delivered by the burette is called a titre.
 Using a pipette, a fixed volume (aliquot) of sodium carbonate (say 25 mL) is drawn from the volumetric flask
and deposited into the conical flask; the volume measured out by a pipette is called an aliquot.
–
Indicators:

To determine the endpoint (ie to also determine the pH), we use indicators. A suitable indicator should be chosen,
preferably one that will experience a change in color (ie end point) close to the equivalence point of the reaction.

However, a suitable indicator must be used.

For strong-acid/weak-base titrations, as above, methyl orange is the most suitable indicator. This is because it
changes from yellow to pale orange/pink within the slightly acidic range, corresponding with the endpoint of pH
= 2.5.

–
For the following titrations, these indicators are suitable.
The Titration (ie where location of the endpoint is determined):

Finally, the titration can be performed. A few drops of methyl orange indicator are added to the conical flask, and
the solution turns a clear yellow colour (ie yellow because of weak base).

The conical flask is placed on a white tile (to make the solution’s colour clear) under the burette, which is held in
a retort stand.

The tap is slowly opened, and the conical flask is continuously swirled.

When the first colour change is noticed, the tap is closed. A swirl of the conical flask will likely return the
solution back to its original colour.

Slowly open the tap so that solution flows out in drops, and stop when the endpoint is reached, as shown by the
colour of the indicator.

The endpoint of a chemical reaction is the point where the reaction STOPS, because all the species have reacted.
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
In acid/base titration, the reaction is a neutralisation reaction and hence pH can determine when the endpoint has
been reached, however, this does NOT mean that the pH at the endpoint is 7.

The pH at the endpoint depends on the acid and base being used.

Eg: For the titration above, nitric acid is titrated into the sodium carbonate:
 Before titration occurs, for sodium carbonate, pH is much greater than 7.
 As nitric acid is slowly added, the pH decreases steadily, until the endpoint is reached, at 22.2 mL of acid.
 BUT the salt formed between a strong acid and a weak base is slightly acidic; hence at the endpoint, the
solution in the flask is slightly acidic.
–

The first titration performed is a rough draft and often overshoots the endpoint. This first titration is rejected.

Titration is performed multiple times to achieve the accurate results.
Eg: Nitric Acid & Sodium Carbonate:

The titration of a solution of nitric acid (HNO3) of UNKNOWN concentration with a known solution of basic,
0.05 M sodium carbonate (Na2CO3). Note: you can have the concentrations other way around.

The solution of known concentration is called the standard solution.

The first step in any acid/base titration is to identify the chemical reaction that is going to occur, by writing a
balanced chemical equation. In this case:
 2HNO3 (aq) + Na2CO3 (aq)

2NaNO3 (aq) + CO2 (g) + H2O (l)
20 mL of the 0.05 M sodium carbonate was titrated with the nitric acid and the endpoint was reached after 22.2
mL of nitric was used.

Looking at the chemical equation, the molar ratio of reaction is 2:1

Hence, 22.2 mL of nitric acid contains TWICE as many moles as 20 mL of sodium carbonate. Calculating:
 n 1 = c1 × v1 = 0.05 x 0.02 = 0.001 mol of sodium carbonate.
 n 2 = 2 × n1 = 0.002 mol of nitric acid.
 c2 = n 2 / v2 = 0.002 / 0.0222 = 0.09 M

From this titration, the concentration of nitric acid was calculated to be 0.09 M
–
Many rules must be adhered to for an accurate titration.
–
Most important is the preparation of an extremely precise standard solution.

This is the titration curve of the above reaction:
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–
The Calculations Involved With Titration:

All the calculations required are detailed above; all titration concentrations can be calculation using the equation
n = c × v.

But another way to calculate the concentration:
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
Qualitatively describe the effect of buffers with reference to a specific example in a natural system:
–
A buffer is a solution that is able to maintain a constant pH; even the addition of a strong acid or base does not
change its pH. They can do this by accepting or providing H+ ions as necessary to respond to the change in pH,
keeping the concentration of H+ in solution relatively constant, and thus resisted a change in pH.
–
The buffer solution contains approximately equal amounts of a weak acid and its conjugate base at equilibrium.
–
The equilibrium involved can be represented as:

HA + H2O
H3O+ + A-
–
‘HA’ is the weak acid, which gives its proton to water, forming its conjugate base, ‘A-’
–
An example of a NATURAL buffer system is the carbonic acid system:

CO2 from respiration dissolves in the blood, forming the HCO3- ion.
 H2CO3 (aq) + H2O (l)
–
H3O+ (aq) + HCO3‫( ־‬aq)
Using Le Chatelier’s principle, we can deduce why the pH remains constant:

Addition of any acid (regardless of its strength) increases [H3O+]. However, this does not decrease the pH; the
additional acid simply reacts with the conjugate base and forces the equilibrium to the left, forming more of the
weak acid to even out the added concentration, so the concentrations cancels each other out and the pH returns to
its original value.

Addition of any base (that is OH- ions) does not increase the pH; as the base reacts with the hydronium, [H3O+],
thus this decreases. This shifts the equilibrium to the right, to replenish the lost [H3O] +. Or another reason is the
fact that water is created, hence increasing the concentration of water, this wants to reduce this, hence equilibrium
pushed to the right to minimise this. Thus the pH returns to its original value.
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
Perform a first-hand investigation and solve problems using titrations and including the preparation of standard
solutions, and use available evidence to quantitatively and qualitatively describe the reaction between selected acids and
bases:
–
Another titration performed was between potassium hydrogen phthalate (a weak acid) and sodium hydroxide (a strong
base), using phenolphthalein indicator.

Perform a first-hand investigation to determine the concentration of a domestic acidic substance using computer-based
technologies:
–
50 mL of household vinegar was placed in a small beaker.
–
A pH probe was attached to the laboratory computer; the probe was rinsed with distilled water, and then rinsed with
left-over solution.
–
The probe was then placed in the solution, and the pH measured.
–
Given the fact that food-grade ethanoic acid only has about 0.4% ionisation, the concentration of ethanoic acid was
then calculated.
–
Results:

The average pH measured was 2.5

[H+] = 10-pH = 10-2.5 = 0.00316 mol/L

But [H+] = 0.4% × c ; (where c = concentration of ethanoic acid)

Hence c = 0.00316 ÷ 0.004 = 0.79 M

Therefore vinegar is 0.79 M ethanoic acid.
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5. Esterification is a naturally occurring process which can be performed in the laboratory:

RECALL:
–
A functional group is a specific group of atoms within a molecule that is responsible for the chemical properties of
that molecule.
–
Polarity occurs within a bond when one of the atoms is more electronegative than the other; the bond has a slight
charge.
–
Alkyl Group: An alkyl is a functional group of an organic chemical that contains only carbon and hydrogen atoms,
which are arranged in a chain. They have general formula C nH2n+1.

Describe the differences between the alkanol and alkanoic acid functional groups in carbon compounds:
–
Alkanols: Organic compounds containing the hydroxyl group (-OH), attach to an alkyl group. Their general formula
is ROH, where R stands for a saturated alkyl group.
–
Alkanoic Acids: Organic compounds containing the carboxyl group (-COOH), attached to an alkyl group. Their
general formula is RCOOH, where R stands for a saturated alkyl group.
–
–
The functional group of alkanols is the hydroxyl group (–OH):

Structure: An oxygen and a hydrogen molecule covalently bonded.

Note: the ‘H’ ion is very hard to remove in alkanols.
The functional group of alkanoic acids is the carboxyl group (–COOH).

Structure: An oxygen is double-bonded to a central carbon, and an –OH group is single-bonded to the same
carbon.

Note: The –OH is not called a hydroxyl group when within a carboxyl group. And they can release there
‘H’ ions, just the equilbirum lies well to the reacatant side.
–
The differences between the functional groups, also differentiates the carbon compounds containing each respectively.
–
Polar Bonding: Both the hydroxyl and carboxyl group contain polar bonds (ie C-O & O-H bonds), allowing the
molecules to form hydrogen bonds. However, the hydroxyl group contains two polar covalent bond, whilst the
carboxyl group contains three polar covalent bonds (C-O, O-H and C=O bonds) this is significant in influencing the
melting points and boiling points of these compounds.
–
Extra molecules: The carboxyl COOH has C=O which adds to the polarity of alkanoic acids. This means that they
have stronger dipole-dipole interacts as well as bonding and dispersion forces and added molecular weight. Which
means they require more energy to overcome these forces than their respective alkanols.
–
Nature of atoms: The H atom in the hydroxyl functional group is hard to remove, whilst the H group in the carboxyl
functional group can be reversibly lost COOH
COO- + H+. This enables the carboxyl group to act as a weak
acid, giving the name “alkanoic acid”. Alkanols, however, cannot act as acids or bases, they are nuetral.
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–
Note: a dimer is a molecule consisting of two identical simpler molecules, if 2 molecules share hydrogen
bonds/dipole-dipole interactions.

Explain the difference in melting point and boiling point caused by straight-chained alkanoic acid and straight-chained
primary alkanol structures:
–
For the same number of carbons in a straight carbon-chain, the highest boiling points and melting points are in the
following order: alkanes  alkanols  alkanoic acids
–
The intermolecular forces between molecules determines what physical state they will exist in (solid, liquid or gas)
for a given pressure and temperature. The intramolecular (ie covalent bonds) holding the molecules together are very
strong, but these are largely irrelevant to the physical properties of the substance. Physical properties are governed by
the intermolecular forces. For example when you heat water, you don’t decompose it to oxygen and hydrogen gas,
you change the state in which the molecules exist in.
–
The stronger the intermolecular forces, the more ‘tightly bound’ the molecules are to each other, and hence more
energy needs to be forced into the system to overcome these forces (i.e. higher melting or boiling point).
–
ALKANES: the only intermolecular forces between molecules are dispersion forces. These forces are caused by the
movement of electrons around the molecule, which create instantaneous moments of charge. These charges attract
each other, creating these dispersion forces. They are very weak, and hence the boiling and melting points of alkanes
are low. Note: ALL alkanes are non-polar.
–
ALKANOLS: they have dispersion forces, as well as dipole-dipole forces (ie interactions between the polar bonds of
molecules). Alkanols contain the polar C-O and O-H bonds, as these have a partial charge, they attract nearby
molecules this is dipole-dipole interactions, the O-H molecule can also undergo hydrogen bonding (form of dipole
dipole interaction where hydrogen is involved) which occurs between molecules. Their intermolecular interactions are
stronger than those of alkanes, and hence they have higher melting and boiling points.
–
ALKANOIC ACIDS: have the strongest intermolecular interactions, as they have three polar bonds in each molecule:
C-O, C=O and O-H bonds hence more dipole dipole interactions. In addition, hydrogen bonding (between –OH
molecules) and dispersion forces occur. This greatly increases the melting and boiling points of alkanoic acids.
–
Also the longer the carbon chains, the higher the boiling and melting points are, but less soluble they become.
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–
Note: However, as the alkanes and corresponding alkanoic acids/alkanols start to get longer, then the effect of the
polar functional groups become less important in comparison to the large non-polar carbon-carbon section, which has
only dispersion forces. Thus, as alkanoic acids and alkanols begin to become longer, their melting and boiling points
start to become closer to the corresponding alkanes.
–
Thus, if we look at a homologous series of alkanes, alkanols and acids, which is a sequence of compounds with
increasing numbers of carbons, we see that the alkanes boiling points start to curve towards the other two.
Boiling Points of Alkanes, Alkanols and Alkanoic Acids
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
Identify esterification as the reaction between an acid and an alkanol and describe, using equations, examples of
esterification:
–
Esters are organic compounds, volatile (readily evaporating) containing the ester functional group: “–COO–”. They
are made through the condensation reaction (where H2O condensers out) between an alkanol and alkanoic acids.
–
Their general formula is RCOOR’
–
In the diagram, R and R' represent a hydrogen, or carbon chains (mainly). Hence the simplest ester is HCOOCH3; this
is called methyl methanoate.
–
Other general forms of esters can be written as:
–
Esterification is the process which forms esters. In the most general sense it is the reaction between an acid and an
alcohol.

For the HSC course, we limit our knowledge of esterification as the reaction between straight-chained primary
ALKANOIC (ie carboxylic) acids, and straight-chained primary ALKANOLs. This forms carboxylate esters.
–
Carboxylate esters are formed when a carboxyl functional group (–COOH) of an alkanoic acid reacts with the
hydroxyl functional group (–OH) of an alkanol.
–
Esterification is a condensation reaction; a water molecule is generated.
–
An interesting point is that in eliminating the water molecule, it is the carboxylic acid (-COOH) that supplies the OH,
and the alkanol (-OH) that supplies just an H.
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–
Esterification is a reversible reaction in which equilibrium lies much to the left at room temperature. It is a moderately
slow endothermic reaction and the reaction comes to an equilibrium rather than to completion.

–

alkanoic acid + alkanol + ~50kj
ester + water
ΔH > 0 (ie +ve, endothermic)
Thus to alter the system, understand of Le Chatelier's principle is needed.
Describe the purpose of using acid in esterification for catalysis:
–
Concentrated sulfuric acid is often added during the process of esterification; this serves 2 purposes:

Sulfuric acid acts as a catalyst, ie it speeds up the rate of reaction by lowering the activation energy, and thus the
rates of the forward and reverse reactions will be increased equally allowing the point of equilibrium to be
reached faster.

Sulfuric acid increases the yield of the reaction. It does this by acting as a dehydrating agent; it absorbs the water,
encouraging the forward reaction, shifting equilibrium to the right according to Le Chatelier’s principle.

Explain the need for refluxing during esterification:
–
Refluxing: a technique involving the condensation of vapors and the return of this condensate to the system from
which it originated, this is done by a refluxing apparatus which is basically a condenser placed
vertically onto a boiling flask; it cools any vapours that boil off so that they drip back into the flask.
–
The need for it includes the fact that the natural production of ester is a slow process (esp at normal
SLC) and a high un-yeilding process and dangerous.
–
Heating the reaction flask has 2 main benefits:

The higher the temperature (more kinetic energy), the faster the rate of reaction; equilibrium
can be reached much faster than if it was left at room temperature.

Also, esterification is an endothermic reaction; increasing the heat of the flask encourages the
forward reaction, creating more ester.
–
Hence through refluxing the reaction can be reached faster , with more yields.
–
Also the products and reactants of an esterification reaction (acids, alcohols, esters) are volatile substances, and thus
would escape from the container as they were heated, presenting a hazard to students and loss of reactant and
produces. Reflux enables safe technique and more viable reaction.
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
Identify the IUPAC nomenclature for describing the esters produced by reactions of straight-chained alkanoic acids
from C1 to C8 and straight-chained primary alkanols from C1 to C8:
–
Straight-chained alkanoic acids only have one carboxyl functional group, located on one of the end-carbons.
–
Naming alkanols:

Count the number of carbons, take the parent alkane name, drop the ‘e’ and add on an ‘-ol’.

There are no exceptions. The IUPAC names coincide with all the systematic names; methanol and ethanol are
considered correct.
–
Naming alkanoic acids:

Count the number of carbons; taking the name of the parent alkane with the same number of carbons, drop the ‘e’
and add on ‘-oic acid’.
 Eg: The 3-C alkane is propane. Hence 3-C alkanoic acid is propanoic acid.

There are 2 exceptions to the rule; the IUPAC-preferred name for the alkanoic acids for 1-C and 2-C are not
methanoic or ethanoic acids, but rather formic acid and acetic acid respectively.

–
Methanoic = Formic, Ethanoic = Acetic
Naming esters:

An ester is always in the following order: alkanol then alkanoic acid.

There are two different situations when esters would have to be named.
1. Given the alkanol and the alkanoic acid, name the ester:
 Firstly, take the alkanol and replace ‘-anol’ with ‘-yl’. Secondly, take the alkanoic acid and replace ‘-oic
acid’ with ‘-oate’. Finally, place the 2 words together, alkanol then alkanoic acid, and you have named the
ester.
 Eg: Name the ester formed by reacting propanol and acetic acid.

Propanol  Propyl; Acetic acid  Acetate.

Therefore the ester is propyl acetate.
 Or another method would be to remember that esters are alkyl alkanoates, where alk is the alkanol(alcohol)
and akano is the alkanoic acid.
2. Given the ester structural formula, name the ester:
 Firstly, you have to identify how many carbons were in the alkanol and the alkanoic acid; split the ester along
the C-O-C bond. The side with the C=O bond is the acid, as only the carboxyl group has a double bond.
 Now that the acid and alkanol has been identified, the steps above are easily followed.
 EG: Name the following ester:
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
Using an imaginary shade, cover the part which includes the C=O bond, and anything before it. Ie

To the left, there is the C=O bond, and 4 carbons (counting the C=O); hence, the alkanoic acid used had
4 carbons. It was butanoic acid.


Note: The C=O bond belongs to alkanoic acids; hence it’s given to it in the ester naming process.

It follows then that the right side is the alkanol, which has 3 carbons. Therefore propanol was used.

Therefore the ester is propyl butanoate.
Note: In most of the ester examples, the ester bond will be attached to the ends of both the carboxylic acid (which
must be true, as the carboxyl group is always on the end of a molecule) and the alcohol. But, as a hydroxyl group
can be anywhere on a molecule, it is possible to get esters where the bond is not attached to the end of the alcohol.

In this case, numbering is used to indicate which carbon of the alcohol the ester bond is attached to. So for
example, you would name 2-propyl acetate to indicate (2-propyl ethenate is wrong, from above):

Outline some examples of the occurrence, production and uses of esters:
Ester

Production
Uses
Methyl Salicylate
(Oil of Wintergreen)
Esterification of salicylic acid with
methanol.



Ethyl Acetate
Esterification of acetic acid, ethyl
alcohol and sulfuric acid.


As an essence in perfumes
As a flavouring in sweets
Can be used medicinally as a counterirritant.
Artificial fruit essence
Solvent for varnishes, lacquers and glues
(both polar and non-polar compounds).
Process information from secondary sources to identify and describe the uses of esters as flavours and perfumes in
processed foods and cosmetics:
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–
Esters are industrially produced to mimic flavours and scents found in nature
–
These are for use in processed foods, or mixed to produce unique perfumes.
–
Many processed foods are flavoured artificially, e.g. banana-flavoured milk is flavoured with the ester iso-pentyl
acetate.
–
Cosmetics contain esters as scents, such as perfumes, with are comprised almost exclusively of a mixture of esters in a
solvent, or to give soaps, hand-lotions or other cosmetics a pleasant smell.

Identify data, plan, select equipment and perform a firsthand investigation to prepare an ester using reflux:
–
Aim: This practical aimed to prepare the ester butyl ethanoate by reflux from 1-butanol and ethanoic acid.
–
Risk Assessment:

This practical involves the use of corrosive concentrated ethanoic and sulfuric acids. Skin contact should be
avoided and safety glasses worn. It also involves the prolonged heating of volatile, flammable substances. Hence,
the heating should take place using reflux, and a hotplate should be used rather than a Bunsen burning for heating.
Boiling chips should be placed inside the reaction flask to avoid spitting of the hot reaction mixture.

–
If these safety precautions are taken, then the risk is acceptable.
Method:

The ester prepared was ethyl ethanoate.

Ethanol and ethanoic acid were used as the reactants.
 ethanol + ethanoic acid

ethyl ethanoate + water
In a round-bottom flask, 40 mL of ethanol, 30 mL of ethanoic acid and 10 drops of concentrated sulfuric acid
were placed.

Boiling chips were also dropped into the flask.
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
The condenser attachment of the reflux apparatus was attached to the flask, and secured onto a retort stand; water
was passed into the condenser.

The flask was placed in a water-bath in a large beaker.

A Bunsen-burner was used to bring the water to a gentle boil, which brought the reflux apparatus to a temperature
of 100ºC; it was left to reflux for 30 minutes.

Heating was stopped and the entire apparatus was left to cool, with the water still flowing.

Disassemble the apparatus, carefully holding on to the flask.

The refluxed contents were poured into a separating funnel; the mixture was washed with 50 mL of distilled water
and shaken. The lower aqueous layer was drained off, and this washing was repeated twice more.

Finally 15 mL of saturated NaHCO3 solution and 35 mL of distilled water was added, the mixture shaken, and the
aqueous layer drained off.

–
The final product is relatively pure ethyl ethanoate.
Results:

It was observed that the ester, ethyl ethanoate had a varnish-like odour. The refluxed mixture was initially
colourless, but upon reaction with sodium hydrogen carbonate formed two layers, one which appeared yellowish
and oily (the ester).
–
Justification:

See justification for refluxing & justification of the addition of sulfuric acid.

Boiling chips were added to facilitate a slow and gentle heating.

A water bath was also used to allow for gentle heating

The use of the separating funnel for extracting the ester was justified because all the reactants and products except
the ester are water soluble; ethanol, ethanoic acid, and of course water, are all water soluble. Hence they are all
immiscible (not blending) with the ester.

The ester also is less dense and floated on top, allowing the aqueous reactants to be drained off.

Washing with distilled water was used to remove as much ethanol, ethanoic acid and sulfuric acid from the ester
as possible; multiple washings increased the purity of the ester.

Finally, the sodium hydrogen carbonate was used to neutralise any remaining sulfuric acid in the mixture.
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Chemistry_The Acidic Environment
Anthony Liang
©
Indicator
Colour in lower
pH
Colour in
transition pH
Colour in
higher pH
Transition
pH range
Methyl Orange
[acidic]
Red
Orange
Yellow
3.1 → 4.4
Bromothymol
blue
[neutral]
Yellow
Green
Blue
6.0 → 7.6
Litmus
[neutral]
Red
Purple
Blue
5.5 → 8.0
Phenolphthalein
[basic]
Colourless
Colourless
Pink
8.3 → 10.0
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Acidic
Neutral
Basic
Citric juices
Salt
Oven cleaner
Carbonated drinks
Sugar
Ammonia
Milk
Pure Water
Detergent
Aspirin
Alcohol-water solutions
Blood
Vinegar
Lactose solutions
Soda (Bicarb, washing)
Battery acid
Drain cleaner
Stomach acid
Soap
Lactic acid
Lime water
Vitamin C tablets
Pool Chlorine
Rust convertors
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#
1.
The Acidic Environment
•
An acid is a substance which in solution
produces hydrogen ions, H+, or more strictly
hydronium ions, H3O+.
o
pH below 7
o
Sour taste
o
Sting or burn the skin
o
Conduct electricity in solution
o
Turns blue litmus red
•
classify common substances as
acidic, basic or neutral
[NB: Arrhenius definitions]
•
A base is a substance which either contains
the oxide O2- or the hydroxide ion, OH-. That, or a
substance that when in solution produces the
hydroxide ion.
o
pH above 7
o
bitter taste
o
Slippery or soapy feel
o
Conduct electricity in solution
o
Turns red litmus blue
•
A neutral solution is one in which the
hydrogen ion concentration and hydroxide ion
concentration is equal.
o
pH of 7
Examples of common substances and their acidity:
•
Indicators are strongly coloured
substances that change colour over a specific pH
range in solution depending on whether the
solution is acidic or basic [solution’s acidity]
•
identify that indicators such as
litmus, phenolphthalein, methyl orange
and bromothymol blue can be used to
determine the acidic or basic nature of a
material over a range, and that the range
is identified by change in indicator
colour
•
Examples such as litmus, phenolphthalein,
methyl orange and bromothymol blue can be used
to determine the acidic or basic nature of a material
over a range, and that range is identified by change
in indicator colour
•
They are particularly useful as their
indicator ranges over the common titration ranges
and real- life acidities.
•
identify data and choose resources to
gather information about the colour
changes of a range of indicators
•
©
identify and describe some everyday
uses of indicators including the testing
of soil acidity/basicity
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• Soil acidity/basicity is tested to ensure that
the plants intended for the area have suitable
pH in which to live
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o Some plants such as azaleas and
camellias need acidic soils whilst most
annual flowers and vegetables need
alkaline soil.
o Small sample of soil (1/3 of test tube)
is moistened with distilled water and
mixed with universal indicator solution
(or measured electronically).
o White neutral, insoluble powder,
BaSO4 is sprinkled on the surface,
providing a white background against
to observe the colour of the universal
indicator
o The colour change observed is
compared with a standard pH chart
o Addition of calcium oxide (lime) to
increase pH or ammonium sulphate or
compost to decrease it
• Swimming pools are also monitored for their
pH levels for their suitability for human use
o The addition of pool cleaning
chemicals, NaOCl, produces HOCl
which is used to kill microbes, but
increases pH.
o Sample of pool water is collected in a
vial, phenol red is added and compared
against a colour chart, where low pH
(<6.8) is given by yellow, high (>8.4)
given by red-purple, and satisfactory
levels given by pinkish orange.
o HCl or NaOCl is added as necessary to
control pool pH levels, as a pH under
6.5 causes metals to be attacked and
pH over 8 can cause irritation to lungs,
skin and eyes.
o An ideal pH is 7.2 to 7.6
• Testing aquarium pH – many fish and other
marine life can only survive at very narrow pH
ranges so it is important to constantly monitor
the pH of an aquarium
• Monitoring wastes from photographic
processing– the pH of wastes discharged from
laboratories that process photographic film
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need to be monitored to ensure that pH is
within an environmentally safe range.
• Domestic waste water and waste water from
light industries is often test to ensure that
waste water is not acidic so that it will not
corrode sinks, drains and sewerage pipes
• In chemical research they are used to
determine the acidity and basicity of a
solution, and to monitor changes in acidity
during accurate volumetric analysis (titrations)
2.
•
perform a first-hand investigation to
prepare and test a natural indicator
•
solve problems by applying
information about the colour changes of
indicators to classify some household
substances as acidic, neutral or basic
•
identify oxides of non-metals which act
as acids and describe the conditions
under which they act as acids
[memorise definition + specific
examples + equations]
• Skill
• Indicators can be used to differentiate between
the possible pH ranges of substances. For
example, litmus is a common indicator and its
presence shows whether the substance has pH
less than 4.8 or higher than 8.1.
• The colour change of indicators occurs over a
range of pH values. One indicator by itself
cannot be exact.
• However, if we use more indicators, we can
narrow the range of the pH. Thus indicators
can be used to find the pH range of a
substance, classifying it as acidic, neutral or
basic.
• To get an exact pH, we use a pH meter: a
digital device which uses a glass electrode.
• NB: safety goggles must be worn and warning
labels read
Acidic oxides
• When non-metal oxides react with water,
acids are produced, creating H3O+ ions.
Acidic oxides
o React with water to form an acid
[and/or]
o React with bases to form salts (+water)
• Examples: Most non-metal oxides (except for
CO, NO and N2O which are neutral)
Carbon dioxide (CO2), Nitrogen Dioxide
(NO2), Diphosphorus Trioxide (P2O3), Sulfur
dioxide (SO2)
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• Equation:
In water (solution) carbon dioxide reacts to form
carbonic acid:
CO
2(g)
+ H O → H CO
2
(l)
2
3(aq)
When carbon dioxide reacts with the base sodium
hydroxide it forms the salt sodium carbonate:
CO
2(g)
+ H O → H CO
2
(l)
2
3(aq)
• Notably, the oxides of more electronegative
elements make strong acids.
• To detect that a non-metal oxide gas is acidic
with indicator paper, the paper must be moist.
Moisture enables the gas to dissolve and form
the acid that produces hydrogen ions. Reaction
of a hydrogen ion with an indicator causes the
colour change.
Basic oxides
• Basic oxides
o React with acids to form a salt
o Do not react with alkali solutions
• Examples: Copper Oxide (CuO), Iron (II)
Oxide (Fe2O3), Magnesium Oxide (MgO),
Calcium Oxide (CaO), Sodium Oxide (Na2O)
• Equation:
An example of a basic oxide is copper oxide
which reacts with sulfuric acid to produce copper
sulphate salt:
CuO + H SO
(s)
2
4(aq)
→ CuSO
4(aq)
+H O
2
(l)
Amphoteric oxides
• Amphoteric oxides
o React with acids and bases
• Examples: Zinc Oxide (ZnO), Aluminium
Oxide (Al2O3), Lead Oxide (PbO), Tin Oxide
(SnO) [ZAPS]
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Neutral oxides
• Neutral oxides
o Do not react with acids or bases
• Examples: Carbon Monoxide (CO), Nitrous
Oxide (N2O)
•
analyse the position of these nonmetals in the Periodic Table and outline
the relationship between position of
elements in the Periodic Table and
acidity/basicity of oxides
• In general, the more electronegative an
element, the more acidic its oxide is when
reacting with water. For example, the highly
electronegative S, N and Cl form very strong
(~100% ionisation) acids in water.
• Hence, Oxides of elements are increasingly
acidic going from left to right across a period
in the periodic table
• That is why non-metals form more acidic
oxides.
o They are located towards the righthand side of the period table and are
concentrated in the top right-hand
corner (upper-right hand corner)
o They are highly electronegative and
thus mostly covalent compounds
o Exceptions are the Noble Gases which
do not form oxides
o More weak semi-metals form weaker
acids, notably HSiO3, analogous to
silicon dioxide dissolved in water – but
note that SiO2 doesn’t dissolve in water
readily. This is because the weaker
acid, with their weaker A- group does
not release the hydrogen as easily.
o For example:
SO3(g) + 2NaOH(aq) → Na2SO4(aq) + H2O(l)
• On the other hand, metals form basic oxides.
o They are located towards the left-hand
side of the periodic table an increase in
character towards the bottom-left
hand side (lower-left hand corner) as
the metallic character of metals
increases.
o These have very little electronegativity
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o
o
o
o
and as such form ionic solids with
oxygen
These ionise completely in water to
form oxide ions, which react with
water to form hydroxide
The more ‘electropositive’ a metal is,
the more it ionises in water to form a
strong base.
Some metal oxides cannot be dissolved
in water fully but that part that does
completely ionises if the metal is
sufficiently ‘electropositive,’ forming
hydroxide ions.
For example:
O2- + H2O → 2OHCuO(s) + 2HCl(aq) → CuCl2(aq) + H2O(l)
• There are some elements which form
amphoteric oxides, which can act as either an
acid or base, depending on the conditions of
reaction. For example:
Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O(l)
Where aluminium oxide acts as a base, or
here, where it acts as an acid:
Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2NaAl(OH)4
o Generally the oxides of semi-metals.
They are located near the borderline
between metals and non-metals.
o Many transition metals form
amphoteric oxides, such as zinc,
vanadium, chromium, and manganese.
For example:
ZnO(s) + 2KOH(aq) → K2[Zn(OH)2](aq)
ZnO(s) + 2HCl(aq) → ZnCl2(aq) + H2O(l)
Here, zinc oxide can be seen to be able
to neutralise strong acids and bases.
• Also, metals with higher oxidation states
tend to form oxides which are more acidic.
For example, Cr oxides can be basic
(+2),amphoteric (+3), or acidic (+6)
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•
describe, using equations, examples
of chemical reactions which release
sulfur dioxide and chemical reactions
which release oxides of nitrogen
•
identify natural and industrial sources of
sulfur dioxide and oxides of nitrogen
Natural Sources of Sulfur dioxide
• Sulfur is present in organic matter as part of
proteins. Bacteria decompose organic matter
to produce hydrogen sulphide (H2S), which
oxidises to form sulfur dioxide
[learn the relevant equations and
always include them in your exam
answers]
2H S +3O
2 (g)
2(g)
→ 2 SO
Alternatively, S + O
(s)
2(g)
+2H O
2(g)
→ SO
2
SO is readily oxidised to SO , 2SO
2
3
(g)
2(g)
2(g)
+O
2(g)
→
2SO
3(g)
SO reacts with water in the atmosphere to form
3
acid rain
• Volcanic gases released during eruptions and
from geysers
• Geothermal hot springs
• Combustion of organic matter, e.g. bushfires
Industrial sources of sulfur dioxide
• Processing or combustion of low grade fossil
fuels with sulfur impurities (brown coal and
petroleum) in power plants and motor vehicles
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e.g. When coal is burnt, iron sulphide (FeS )
2
impurities oxidise to form sulfur dioxide
4 FeS
2(s)
+ 11 O
2(g)
→ 2 Fe O
2
3(s)
Alternatively, S + O
(s)
2(g)
+ 8 SO
2(g)
→ SO
2(g)
• Extracting metals from their sulfide ores:
The smelting of sulfide ores in industrial
plants such Cu2S and ZnS involves roasting the
ore in air which release SO2 into atmosphere
e.g. roasting of zinc sulphide in extraction
of zinc
2 ZnS + 3 O
(s)
2(g)
→ 2 ZnO + 2 SO
(s)
2(g)
• Incineration of garbage
• Petroleum refineries
• Industries using sulfur dioxide for production
of sulfuric acid, or other products
Natural sources of nitrogen oxides (N2O, NO, NO2)
• NO is primarily produced in high-temperature
combustion environments (particularly high
voltages), such as during lightning strikes, as
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these high localised temperatures cause
nitrogen and oxygen in air to react.
N
•
2(g)
+O
2(g)
2 NO
(g)
NO2 (nitrogen dioxide) produced by action of
sunlight on NO
2 NO + O
(g)
2(g)
→ 2 NO
2(g)
•
NO (nitric oxide) and N2O (nitrous oxide)
produced by the bacterial action on
nitrogenous material in soils
Industrial sources of nitrogen oxides
• NO and NO2 produced in high temperature
combustion of fuel in motor vehicles and
power stations (stationary sources)
N
2(g)
+O
2(g)
2 NO + O
(g)
•
•
•
©
analyse information from secondary
sources to summarise the industrial
origins of sulfur dioxide and oxides of
nitrogen and evaluate reasons for
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2(g)
2 NO
(g)
→ 2 NO
2(g)
o Occurs at oil refineries and electrical
power production sites, where arcing is
present.
o In Sydney, around 85% of total NOx
emissions comes from vehicles, which
combust impure fossil fuels.
o Processes that use a high ratio of air to
fuel release more NOx (e.g. diesel
engines, power stations)
NO produced from burning of biomass
N2O manufactured as a fuel for racing cars
and for use as sedative/analgesic
Concentrations of oxides of sulfur and nitrogen has
increased as a result of human activity, particularly
since onset of Industrial Revolution
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concern about their release into the
environment
[A good way to structure such as
response is to address three key areas ]
1. The formation and effects of
acid rain
2. The concern with pollution
3. The concern with respiratory
disorders and the effect on
human health
Sulfur and nitrogen oxides are detrimental to the
environment for several reasons.
[NB: take note of ways to reduce
emissions]
• Sulfur dioxide (SO2) is a respiratory irritant.
At concentrations of as low as 1ppm it causes
breathing difficulties, particularly affecting
asthmatics
• Nitrogen oxides can irritate or damage the
respiratory system.
• NO2 is harmful to vegetation and can cause
damage to foliage.
• NO2 causes irritation of the eyes, particularly
in young children and older people. At high
concentrations it causes extensive tissue
damage.
• NOx leads to the formation of photochemical
smog when sunlight acts upon NO2 to form
ozone. This is both visually unattractive and a
health hazard, such as irritation of respiratory
systems.
• In addition, they can cause acid rain (rain
with high [H+] than normal) through
dissolution in water, forming strong acids
which can adversely affect the environment
and humans.
o Sulfur dioxide (and its further oxidised
state SO3), react with water vapour to
form sulphurous and sulfuric acids ,
which are weak and strong
respectively.
o NO2 reacts to form nitrous and nitric
acid, which are of similar strength.
o These chemicals are highly corrosive.
• Evaluation: The release of sulfur and nitrogen
oxides into the environment is detrimental not
only to the environment and its complex
interactions, but also to humans and society.
Thus, the emission of these acidic oxides into
the atmosphere is undesirable, and steps
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should be further taken to reduce their effect
on the atmosphere.
•
assess the evidence which indicates
increases in atmospheric concentration of
oxides of sulfur and nitrogen
[remember at the end or beginning of
answer to provide own judgement as it
is worth one mark]
• There is both direct and indirect evidence
suggesting the atmospheric concentrations
of oxides of sulfur and nitrogen have been
increasing in localised areas (but not
globally) since the Industrial Revolution
• Direct evidence for increasing concentrations
of SO2 and NOx in localised areas:
o Quantitative analysis of Antarctic ice
core samples by the CSIRO (also,
measurement of carbon isotopes in old
trees) have shown steady increases in
concentrations of these oxides in recent
times
o While the annual averages in cities
haven’t increase significantly, the
number of days where concentrations
have been above safe levels are
increasing.
[Use dot-points]
• However, this evidence is not every reliable
for the following reasons:
o Levels of these oxides are below
0.01ppm and are difficult to detect and
measure
o There is no reliable data on levels
before the 1950s and it wasn’t until the
1970s that sensitive gas analysis
techniques developed that allowed
accurate and close monitoring of
levels.
o SO2 and NO2 both form sulphate and
nitrate ions which are hard to detect
and measure. [washed out in acid rain]
• Indirect evidence is also useful in suggesting
increased concentrations of SO2 and NOx in
localised areas. However there are a range of
other variable which need to be taken into
account, meaning that indirect evidence is not
particularly valid.
o Increases in atmospheric
concentrations of these chemicals have
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led to increasing incidence of acid rain,
and resulting damage to ecosystems
and structures
o The incidence of pollution and
photochemical smog is on the rise
•
explain the formation and effects of
acid rain
[Scaffold + assessment for holistic
questions]
1. Industrial processes such as …
(equation) and natural
processes (equation) produce
…. (appropriate gas/acidic
oxide)
2. This acidic oxide then dissolves
in water in the atmosphere …
(gas’s reaction with water). The
(appropriate acid) ionises,
producing hydrogen ions which
lower the pH of the water
(ionisation equation) making
the resulting rain acidic.
3. This acid rain has a negative
impact as it (2-4 effects
depending on the marks
awarded. Make sure you include
the effect on marble and
concrete buildings as you can
write out another equation
here]
[Qualitatively: Sulphur dioxide
released into the atmosphere oxidises
and dissolves in water to form
sulphuric acid]
[Dangers of sulphur:
a. Identify: Burning of Sulphur
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• Assessment – Overall, although we do not
have accurate measurements for the
concentrations of sulfur dioxide and nitrogen
dioxide, there is still evidence, both direct and
indirect that the concentrations of oxides of
sulfur and nitrogen in localised areas have
been increasing since the Industrial
Revolution.
• Acid rain has a higher hydrogen ion
concentration than normal: higher than 105
mol/l (i.e. pH<5)
• Unpolluted rain is slightly acidic (~5.6 pH)
due to dissolved carbon dioxide in water to
form carbonic acid
CO
2(aq)
H CO
2
+ H O ⇌ H CO
3(aq)
2
(l)
+
⇌H
(aq)
2
3(aq)
+ HCO
3(aq)
• The formation of acid rain. Acid rain is
mainly the result of rain dissolving non-metal
oxides present in the atmosphere, particular
sulfur and nitrogen oxides. Sulfurous and
nitrous acids are weak but nitric and sulfuric
acids are strong. They are formed upon
oxidation of the weaker acidic oxides
dissolved water.
1. Oxides enter the atmosphere either
through natural or industrial
processes
2. These gases then react with water in
rain droplets to produce acidic
solutions (though dilute).
Oxides of Sulfur
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b.
c.
d.
e.
fuels (and other processes)
result in large amounts of SO2
released into the atmosphere
<Equation>
Sulphur dioxide is poisonous to
living things
Dissolves in water in
atmosphere to produce
sulphurous/ic acid <equations>
These acids form acid rain,
potentially changing the
environment and organisms in
it.
Always link to examples]
o Sulfur dioxide dissolves in water to
form sulfurous acid (H2SO3)
SO
2(g)
+ H O → H SO
2
(l)
2
3(aq)
o Substances in the upper atmosphere
then catalyse the reaction between
sulfurous acid and oxygen to form
sulfuric acid (H2SO4)
2H SO
2
3(aq)
+O
2H SO
2(g)
2
4(aq)
o Sulfur dioxide is readily oxidised in air
to form sulfur trioxide.
2 SO
2(g)
+O
2(g)
→ 2 SO
3(g)
o Sulfuric acid is water-soluble and so is
removed from the atmosphere
Oxides of Nitrogen
o Nitrogen dioxide reacts with water to
form a mixture of nitric acid (HNO3)
an nitrous acid (HNO2)
2 NO
2(g)
+ H O  HNO
2
(l)
HNO
3(aq)
+
2(aq)
o Substances catalyse the reaction
between nitrous acid and oxygen to
form nitric acid
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2HNO
2(aq)
+O
2HNO
2(g)
3(aq)
o Alternatively formation of nitric acid is
as follows:
4 NO2(g) + 2 H2O(l) + O2(g)  4 HNO3(aq)
o Nitric acid is water soluble so is
removed from the atmosphere.
3. These acids ionise to form hydrogen
ions, which lower the pH of the
water and making the rain acidic.
+
H2SO4(aq) → H
HNO3(aq) → H
(aq)
+
(aq)
+ HSO
+ NO
4 (aq)
-
3 (aq)
4. The rain then falls on the Earth’s
surface causing many harmful
effects on plants and animals. This
often occurs a considerable distance
from the pollution source.
Acid rain has varied and adverse effects on the
environment and human society.
• Causes surface waters and lakes,
particularly in industrialised regions to
become too acidic to support fish and other
aquatic life, which have a limited pH range
for survival, often with drastic effects on
aquatic ecology.
o At lower pHs, fish eggs can’t hatch and
adult fish may die.
o Acid rain disrupts the CO2
gaseous/aqueous equilibrium and
stresses fish life.
o Lowered pH irritates external contacts,
such as skin and gills.
o Aluminium which has been leached
from soils by the acidic rain is toxic to
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marine life.
o Biological magnification of acid rain
through the food chain can seriously
damage predators
o Acidity leads to reduced calcium
uptake in fish, causing reduced skeletal
growth
• Causes damage to plant foliage, including
crops and forests, via their corrosive nature
o Pine needles, for example, lose their
waxy coating, or acid rain can cause
defoliation.
o The adverse effect nature is also
amplified due to the changes in soil
pH, particularly as seedlings can
become damaged and resulting in
reduced productivity.
o Minerals such as K, Ca, and Mg can be
removed from the ground by
dissolution via acid rain, removing
plant nutrients important for plant
growth. E.g. sulfuric acid ionises in
water, resultant sulfate SO4- ions
precipitate ions out of the soil.
2 H+(aq) + Mg2+(clay) → 2 H+(clay) + Mg2+(aq)
o Acids dissolve minerals, releasing
toxic heavy metals
• Acid rain damages structures made of stone
or metal, corroding them
o In particular, the historical heritage in
the world is slowly being destroyed by
the erosional effects of acid rain on
marble and limestone buildings.
 The hydrogen ion reacts with
carbonate (a base)
H
+
(aq)
+ CO
3 (s)

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2-
→ HCO
3 (aq)
Sulfuric acid in rain causes
destruction of limestone and
marble buildings/statues
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CaCO3(s) + H2SO4(aq) → Ca2+(aq) + SO42-(aq) + CO2(g) +
H2O(l)
As water evaporates in rock
crevices, calcium sulfate
(gypsum) can crystallise out,
causing rocks to crumble
Gypsum forms a surface to
which dirt/soot particles can
readily bind
 Iron in buildings can react with
hydrogen ions to produce
hydrogen gas
+
Fe(s) + 2H (aq) → Fe2+(aq) + H2(g)
• Bronchitis and asthma in humans are
worsened by acid rain
•
define Le Chatelier’s principle
•
identify factors which can affect the
equilibrium in a reversible reaction
• In 1885, the French chemist, Le Chatelier,
proposed a principle for predicting the effect
of change on reversible reaction:
Le Chatelier’s Principle: If a system at
equilibrium is disturbed (subjected to
changes in conditions), then the system
adjusts itself so as to minimise the
disturbance (imposed change)
[NB: favour one side of equation]
In answering equilibrium problems,
the following format should be used as
to avoid any confusion or assumptions:
The system acts to minimise the
disturbance/oppose the change (Le
Chatelier’s principle) – that is, [insert].
The reaction that does this is [insert];
thus this direction of reaction is
preferred. The equilibrium position thus
shifts [insert]. Subsequently, answer
any extra parts, such as qualitative
concentrations.
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• Dynamic process occurring in a closed system
at constant temperature
• Products will initially be formed by means of
forward reaction, and as the concentration of
reactants decrease, the rate of forward reaction
decreases because collisions between reactant
molecules become less frequent.
• While the reverse reaction, initially zero, will
gradually increase as concentration of
products increase because collision of
products become more frequent.
• Systems at equilibrium (reversible reactions)
have constant concentrations of reactants and
products, and equal/opposing reaction rates.
• Hence, macroscopic properties do not change,
but at a molecular level, the forward and
reverse reactions occur at equal rates.
1. A change in concentration
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A + B ⇌ C +D
a. Removing A or B or adding C or D will
increase the backward (reverse) reaction as
there is relatively less reactants than before the
change.
b. Removing C or D or adding A or B will
increase the forward reaction rate as there is
relatively less products than before the change.
NB: However, in an equilibrium, changing the
amount of liquids or solids will not affect the
equilibrium as they always have a fixed
concentration (pure substances
Changing the concentration or amount of
aqueous or gaseous reactant systems will cause
the equilibrium to shift in such a way that all the
ratio of all species is as close as it can be to
before the change. However, they can never reach
levels before the change.
2. A change in pressure (volume) for gaseous
systems:
mA ⇌ nB
If we assume n > m, and A, B are gases. External
pressure and volume have an inverse relationship.
a. Increasing pressure decreases volume,
therefore, reverse reaction will be promoted,
lowering the total number of moles occupying
the lesser volume (promoting less gas
particles/volume)
b. Decreasing pressure increases volume,
therefore, forward reaction is promoted,
increasing number of moles of gas to expand
into the greater volume (promoting more gas
particle/volume)
• Volume changes can also be thought of
pressure changes but note that concentration
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changes as well. (c = n/v)
• Adding an inert gas will not change
equilibrium position
• If m > n, the direction of preferred reactions
are the opposite.
• If m=n, equilibrium will remain the same
despite pressure change. [total number of
moles on both sides equal]
• Changing the pressure (or volume) of a
gaseous system will cause the system to shift
in such a way that the number of particles per
unit volume is as close as it can be to before
the change, but it cannot achieve previous
levels, as with any other equilibrium change.
• In addition, increased pressure increases
chance of collision, and hence increases rate of
reaction, therefore systems achieve
equilibrium faster. While converse is also true.
•
3. A change in temperature
A ⇌B
ΔH
If we assume the forward reaction has a positive ΔH,
heat is absorbed in the forward reaction
[endothermic].
a. Increasing temperature will favour the forward
reaction, which increases heat input
b. Decreasing temperature will favour the reverse
reaction, which increases heat output
If ΔH is negative, heat is released in the forward
reaction. [exothermic]
a. Decreasing temperature will favour the
forward reaction, which increases heat output
b. Increasing temperature will favour the reverse
reaction, which increases heat input
• If the temperature of an equilibrium system is
lowered, Le Chatelier’s principle predicts that
the equilibrium will be re-established so that
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the change is partially counteracted by the
increase in temperature, favouring exothermic
reactions.
• An increase in temperature, according to Le
Chatelier’s principle, will result in a reestablished equilibrium which partially
counteracts the change, decreasing
temperature and favouring endothermic
reactions.
• Changing the temperature of an equilibrium
will result in the temperature change being
partially counteracted by increasing or
decreasing heat output as necessary, so that the
heat of surrounds is as close as it can be to
before the change.
•
4. Addition of a Catalyst
• When a catalyst is added to an equilibrium
system, activation energy is lowered for both
forward and reverse reactions and increases
the rate of reaction for reactions in both
directions equally.
• Does not change the relative proportions of
reactants and products, hence not affecting the
equilibrium, however allows the system to
reach equilibrium in less time.
• In a closed system such as a bottle of soft
drink, gaseous carbon dioxide establishes an
equilibrium with dissolved CO2
(1)
• Dissolved CO2 reacts with water to form
weakly acidic solution of carbonic acid
(2)
describe the solubility of carbon
dioxide in water under various
conditions as an equilibrium process
and explain in terms of Le Chatelier’s
principle
Guideline:
• Define Le Chatelier’s principle
• Identify the disturbance
• Identify how it will minimize the
disturbance
• Identify which way it will shift
Scaffold:
If temperature/pressure/concentration/
volume of the system is
increased/decreased, by Le Chatelier’s
Principle the equilibrium will shift to
minimize the disturbance and (insert
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General net reaction*: ΔH < 0
Concentration
a. If CO2(g) concentration is increased, by Le
Chatelier’s Principle the equilibrium will shift
to minimise the disturbance and consume the
extra CO2(g). The forward reaction uses up
CO2(g) and so the equilibrium will shift to the
right, increasing [H2CO3] and thus increasing
the amount of CO2(g) dissolved in solution.
b. Conversely, if [H2CO3] is increased, the
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system will shift to the left to consume the
extra carbonic acid.
change). This change occurs because
(explain change) and causes the
equilibrium to shift to the left/right.
Pressure
a. If pressure (4-5x atmospheric pressure) was
increased (increases concentration), by Le
Chatelier’s Principle the equilibrium will shift
to minimise the disturbance and decrease
(reduce/minimise) the pressure. The forward
reaction produces less moles of gas (left side
has much more volume relative to the right
side) so reduces the pressure. Therefore, the
equilibrium shifts to the right and more CO2(g)
is dissolved (also increasing acidity).
o This is done in bottling plants where
pressurised conditions are used to
supersaturate the solution with carbon
dioxide.
b. Conversely, decrease pressure, system will
shift to increase (create more) pressure, hence
shift to the left with more moles of gas.
o When a bottle of soft is opened, the
pressure of CO2(g) is decreased and the
reaction favours the evolution of gas
(equilibrium shifting to left), seen as
bubbles/fizzing coming out of solution,
and eventually the drink becomes flat.
Temperature
a. If temperature is increased, by Le Chatelier’s
Principle the equilibrium will shift to minimise
this disturbance and consume the excess
(extra) heat. The forward reaction is
exothermic, whilst the reverse reaction is
endothermic which decreases the temperature.
Therefore, the equilibrium shifts to the left
(favours the reverse reaction) and less CO2(g)
is dissolved (evolves gas and eventually
results in flatness)
b. Conversely, if temperature is decreased, the
system will looks to create heat and favour the
exothermic (forward) reaction and shift to the
right, increasing solubility.
• Therefore, for gases, solubility decreases as
temperature increases. In contrast, for solids,
solubility increases as temperature increases.
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Acidity (pH)
• Carbonic acid establishes equilibria involving
hydrogencarbonate and carbonate ions
(3)
(4)
• This explains rain water’s slight acidity and
soda pop’s tangy taste.
a. If an additional source of hydrogen ions (H+(aq))
from an acid such as vinegar, it drives both
(3) and (4) equilibria to the left and increase
the amount of carbonic acid. As a
consequence, increased H2CO3(aq) shifts the
equilibria in (2) to the left, and creates more
CO2(aq). This then drives equilibrium (1) to
produce CO2(g). As such, an increase in [H+]
will lead to less CO2 being dissolved.
b. Conversely, in basic conditions, CO2 dissolves
more readily as more H2CO3 is consumed in
reaction.
3.
©
•
identify data, plan and perform a
first-hand investigation to
decarbonate soft drink and gather
data to measure the mass changes
involved and calculate the volume
of gas released at 25˚C and 100kPa
•
calculate volumes of gases given
masses of some substances in
reactions, and calculate masses of
substances given gaseous volumes,
in reactions involving gases at 0˚C
and 100kPa or 25˚C and 100kPa
•
define acids as proton donors and
describe the ionisation of acids in water
(2012) All Rights Reserved
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• Avogadro’s hypothesis states that: Equal
volumes of all gases, measured at the same
temperature and pressure, contain equal
numbers of particles (molecules).
Alternatively, equal numbers of molecules of
different gases occupy the same volume.
• At 0°C (273.15K) and 100kPa, molar volume
is 22.71L/mol
At STP:
• At 25°C (298.15K) and 100kPa, molar volume
is 24.79L/mol
At SLC:
• Ionise: To dissociate atoms or molecules
into electrically charged species (ions)
• By Brønsted-Lowry’s definition of acids and
bases, acids are proton (H+) donors and
bases are proton acceptors.
• In water, an acid HA donates a proton to
water, forming H3O+ and A-, and this process is
called ionisation.
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+
HA + H O → H O + A
2
-
3
It can be alternatively written as
+
HA → H + A
•
-
• Skill
• All monoprotic acids (1 ionisable hydrogen
atom per molecule) ionise via
HA → H+ + A- for strong acids, and
HA H+ + A- for weak acids
Examples are:
gather and process information from
secondary sources to write ionic
equations to represent the ionisation of
acids
HCl
+
(aq)
→H
CH COOH
3
-
(aq)
(aq)
+ Cl
H
+
(aq)
and
(aq)
+ CH COO
3
(aq)
Polyprotic acids ionise in step, though with each
step the readiness of ionisation decreases
drastically e.g.
• A diprotic acid (2 ionisable hydrogen per
molecule)
+
H SO
2
4(aq)
HSO
H SO
2
+H O → H O
2
4 (aq)
4(aq)
(l)
+H O
2
(l)
+ 2H O
2
3
HO
3
(aq)
+
(aq)
+ HSO
+ SO
+
(l)
2H O
3
(aq)
4 (aq)
2-
4 (aq)
+ SO
2-
4 (aq)
When reacting with a strong base (e.g. NaOH), one
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mole of sulfuric acid will react with 2 moles of
hydroxide ions
2OH
(aq)
+ H SO
2
→ 2H O + SO
4(aq)
2
(l)
2-
4 (aq)
• A triprotic acid (3 ionisable hydrogen per
molecule)
H PO
3
+H O
4(aq)
H PO
2
2
4 (aq)
HPO
3
2
2-
•
identify acids including acetic
(ethanoic), citric (2hydroxypropane-1,2,3tricarboxylic), hydrochloric and
sulfuric acid
4(aq)
(l)
+H O
4 (aq)
H PO
+
(l)
+HO
2
+ 3H O
2
(l)
HO
3
HO
(aq)
+
3
HO
(aq)
+ H PO
2
+ HPO
+
3
(aq)
+ PO
+
(l)
3H O
3
(aq)
4 (aq)
2-
4 (aq)
3-
4 (aq)
+ PO
34 (aq)
Some common acids are:
• Acetic (ethanoic) acid – CH3COOH
(monoprotic), the main component of vinegar
(4%). Acetic acid occurs naturally in the
decomposition of biological material, such as
oxidation of wine alcohol, but most acetic acid
used by humanity is manufactured industrially.
It is a weak acid, ionising to a small degree at
standard conditions only. [pH of 0.1M solution
= 2.9; degree of ionisation = 1.3%]
CH COOH
3
(aq)
H
+
(aq)
+ CH COO
3
(aq)
• Citric (2-hydroxypropane-1,2,3tricarboxylic) acid. Citric acid occurs
naturally in citrus fruits such as oranges, limes
and lemons, but it is also manufactured to be
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added to food as preservatives. It is formed
during cellular respiration in body cells. It is a
weak triprotic acid, ionising in 3 steps, all of
which proceed little and are progressively
weaker. [pH of 0.1M solution = 2.1; degree of
ionisation = 8%]
C H O(COOH)
3
5
+
3(aq)
3H
(aq)
+ C H O(COO)
3
5
33
(aq)
• Hydrochloric acid (HCl) is an important acid.
It is found in human stomach acid and aids
digestion, by activating enzymes. It is a strong
acid due to its ease of ionisation, caused by its
highly polar covalent bonding. It ionises, to all
intents and purposes, 100% at standard
conditions.[pH of 0.1M solution = 1]
HCl → H+ + Cl• Sulfuric (H2SO4) acid is produced industrially
on a large scale. Most sulfuric acid is
manufactured, but it can also occur naturally.
For example, most sulfur dioxide released into
the earth’s atmosphere is oxidised and
dissolved in water to form the sulfuric acid in
acid rain. If the acid rain results from volcanic
eruption it could be regarded as natural, but if
acid rain results from smelting of sulfide ores,
it could be regarded as manufactured. It is
industrially important, being involved in
fertiliser, explosive, and petroleum production.
It is a strong acid, due to the sulfate group
being very electronegative and thus allowing
easy ionisation. [pH of 0.1M solution = 0.69]
H SO
2
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+
4(aq)
+H O → H O
2
(l)
3
(aq)
+ HSO
4 (aq)
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HSO
H SO
2
•
identify data, gather and process
information from secondary sources to
identify examples of naturally
occurring acids and bases and their
chemical composition
4 (aq)
4(aq)
+H O
2
(l)
+ 2H O
2
HO
3
+
(aq)
+ SO
+
(l)
Acid
Stomach acid/Hydrochloric
Acid (HCl)
Vinegar/Acetic Acid (ethanoic
acid) (CH3COOH)
Citric acid (C6H8O7)
2H O
3
(aq)
2-
4 (aq)
+ SO
2-
4 (aq)
Base
Ammonia (NH3)
Nicotine
(C8H14N2)
Limestone
(CaCO3)
Acidic rain (H2CO3 and others)
Some examples of naturally occurring bases and acids
are:
•
identify pH as -log
+
[H ] and explain
10
that a change in pH of 1 means a ten+
fold change in [H ]
[Remember to use log10, not ln or loge]
•
©
pH = - log [H ]
10
• From this equation we can see that a change in
one unit of pH means a tenfold change in the
concentration of H+, as the scale is
logarithmic.
• pH of solution refers to the negative base ten
logarithm of the hydrogen ion concentration
• The pH (potential Hydrogen) scale offers a
logarithmic scale for the relative strengths of
acids and bases (their degree of ionisation).
This is used rather than the molarity of
hydrogen ions as the use of indices is
cumbersome.
• In pure water without any dissolved gas, [H+]
= [OH-] =10-7 mol L-1 and so pH = 7.
• In an acidic solution,[H+] > 10-7 mol L-1 and
pH < 7
• In a basic solution, ,[H+] < 10-7 mol L-1 and pH
>7
• The following table relates pH to the hydrogen
describe the use of the pH scale in
comparing acids and bases
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+
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ion concentration, [H+], and provides examples
of common aqueous solutions for each pH
value.
• Similarly, a pOH scale can be used to
represent the [OH-] of a solution.
• Note that, at standard conditions, pH + pOH =
14.
Auto-ionisation of water
+
2H O
2
(l)
HO
3
(aq)
+ OH
(aq)
Ionisation constant at 25⁰C for water is given by:
+
-
K = [H ] [OH ] = 1.0 x 10
w
•
process information from secondary
sources to calculate pH of strong
acids given appropriate hydrogen
ion concentrations
[to find the pH of a base, you can
calculate the pOH and then convert to
pH]
•
-14
Skill
+
pH = - log [H ]
10
+
[H ] = 10
-pH
-
pOH = - log [OH ]
10
-
[OH ] = 10
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-pOH
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+
•
•
-
pH + pOH =14 and [H ][OH ]=10
-14
All strong acids ionise 100% in water for
their first proton. Thus, the pH is usually
the same as their concentration.
The number of decimal places in the value
of pH should equal the number of
+
•
•
significant figures in the value for [H ]
For weaker acids, the degree of ionisation
must be given, as use of ka is not covered
in the syllabus.
Find [H+] and use the same above formula
for pH. The concentration of hydrogen ions
and hence pH will depend on whether the
acid is monoprotic, diprotic or triprotic.
+
H SO → 2H + SO
2
•
•
describe acids and their solutions
with the appropriate use of the terms
strong, weak, concentrated and
dilute
[for the purposes of the HSC you can
consider
©
•
hydrochloric
•
Sulfuric
•
Nitric acid
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4
2–
4
[H+] = 2 x concentration of H SO
2
4
For calculations involving a hydroxide
concentration rather than H+, use the fact
that pH + pOH = pKw = 14
Strong, concentrated, weak and dilute have
different meanings to their everyday use when
referring to acidic solutions. They refer to degree
of ionisation and concentration, rather than
strength.
• A strong solution is one in which a high
degree (~100%) of the acid in the solution
is ionised (or termed completely
dissociates) to produce hydrogen and
hydronium ions, but not necessarily
concentrated.
• A concentrated solution is one in which
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As the only strong acids]
•
describe the difference between a
strong and a weak acid in terms of an
equilibrium between the intact
molecule and its ions
there is a high amount of acid per unit
volume (total concentration of a solute is
high), but not necessarily ionised. e.g. any
solution with high molarity.
• A weak solution is one in which a low
degree of acid in solution is ionised e.g.
vinegar
• A dilute solution is one in which there are
few acid molecules per unit volume (total
concentration of solute is low), even
though they might be all ionised.
• From this, it can be seen that a dilute
strong solution can still have a lower pH
than a concentrated weak solution.
• Note that there is no arbitrary cut-off
between weak and strong or concentrated
and dilute. When asked to give quantitative
values, one should use extreme,
indisputable concentrations or acid
strengths, such as 0.001 or 5M, or 100%
dissociation versus 1% dissociation.
• A strong and weak acid are differentiated by
the degree of ionisation they undergo in
solution. The more H+ per molecule, the
stronger the acid is.
For any given acid, an equilibrium is set with
the surroundings
HA
(aq)
H
+
(aq)
+A
(aq)
• The strength of an acid is dependant on this
degree of ionisation, or, in other words, the
position between the intact and its ions. The
degree of ionisation is calculated as the
percentage of acid molecules that have
dissociated into ions.
+
Degree of ionisation = [H ]/[HA] x 100%
• A strong acid’s equilibrium is (almost, and for
all intents and purposes is) shifted all to the
right, giving a very high ratio, while in a weak
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acid the forward reaction proceeds very little,
leading to a very low ratio.
For example:
HCl
+
(aq)
→H
CH COOH
3
(aq)
-
+ Cl
(aq)
H
+
(aq)
(aq)
+ CH COO
3
(aq)
o Hydrochloric acid thus produces much
more H+ than acetic acid and thus is
stronger, having a lower pH. This is
due to the fact that HCl has a very high
equilibrium ratio of [H+]/[HA]
compared to acetic acid, which has a
very low position.
Applying Bronstod-Lowry theory/Extension
Strong Acids
• Strong acids are completely ionised to
produce hydrogen ions in aqueous solution
• E.g. hydrochloric acid

HCl(g) → H+(aq) + Cl-(aq)

HCl(g) + H2O(l) → H3O+(aq) + Cl(aq)
o Bronsted-Lowry: HCl is a stronger
acid (proton donor) than H3O+, and
H2O is a stronger base than Cl-, so
forward reaction is favoured
• Hydrochloric, sulfuric and nitric acid are all
better proton donors than H3O+ ion, and hence
are strong acids
Weak Acids
• Weak acids are only partially ionised in
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water; an equilibrium is established
between intact acid molecules and its ions
• E.g. acetic acid
 CH3COOH(aq) + H2O(l) H3O+(aq)
+ CH3COO-(aq)
o Equilibrium established between acetic
acid molecules, and hydronium and
ethanoate ions
o Bronsted-Lowry: CH3COOH Is a
weaker acid (proton donor) than H3O+,
and H2O Is a weaker base than
CH3COO-,
• Aquated metal ions can donate a proton from
one of their surrounding water molecules (e.g.
Al3+ ion)
Strong Bases
• Strong bases completely dissociate to produce
hydroxide ions in aqueous solution
• E.g. potassium hydroxide

KOH(s) → K+(aq) + OH-(aq)
o Group I/II metal hydroxides are strong
bases
o Group II metal hydroxides produce
two moles of hydroxide ion for every
mole of metal hydroxide, but have
limited solubility
• Metallic oxides are also strong bases; oxide
ion is stronger base than hydroxide ion

Na2O(s) + H2O(l) → 2Na+(aq) +
2OH-(aq)
Weak Bases
• Only a small proportion of molecules or ions
react with water to form hydroxide ions in
aqueous solution
• E.g. ammonia ion
 NH3(aq) + H2O(l) NH4+(aq + OH(aq)
o Bronsted-Lowry: NH4+ is a stronger
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•
compare the relative strengths of equal
concentrations of citric, acetic and
hydrochloric acids and explain in terms
of the degree of ionisation of their
molecules
acid than H2O and OH- is a stronger
base than NH3, so reaction occurs to a
small extent
• Many weak bases are anions such as
carbonate, acetate, fluoride and phosphate,
which can accept protons from water to a
small extent, producing hydroxide ions
Generally, strong acids will react with strong bases to
form weaker conjugate acids and weaker conjugate
bases. i.e. acids on the ‘strength of acids/bases’ table
will react with bases below them.
For the equation
Account for differences in pH of acetic,
citric and hydrochloric acid:
1. HCl acid is 100% ionised in
solution so [H+] is equal to acid
concentration
2. Citric acid is partly ionised
(about 5%) and acetic acid is
weaker still and has an even
lower degree of ionisation.
3. In both weak acids, [H+] is
much less than concentration of
dissolved acid.
Note: always compare [H+] with the
concentration of acid molecules (i.e.
degree of ionisation)
HA
H+(aq) + A-(aq)
(aq)
• The magnitude of Ka shows how much the
reaction shifts to the right hand side, and
depicts the relative strengths of acids.
Ka of acetic acid = 1.8 x 10-5
Ka of citric acid = 7.4 x 10-4; 1.7 x 10-5; 4.0 x
10-7 (citric acid is triprotic)
• From this we see that acetic acid is the
weakest of the three, then citric acid, and
hydrochloric acid is by far the strongest.
• This is due to the degree of ionisation of the
molecules in water; less strong acids ionise
less readily in water and require a much high
concentration to equal that of a stronger acid.
• Acetic and citric acid usually ionise to less
than 5% (depending on their concentration)
but hydrochloric acid ionises, for all intents
and purposes, 100% at standard conditions.
• Note that the 2nd and 3rd ionisations of citric
acid do not affect the pH significantly, as they
are orders of magnitude less than the first.
However contributes to a higher [H+],
resulting in lower pH than acetic acid
-
+
C H O +H O C H O +H O
6
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8
7
2
6
7
7
3
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C H O +H O C H O
6
7
7
2
6
6
C H O +H O C H O
6
6
7
2
6
5
27
37
+H O
+
3
+H O
3
+
• Thus it can be seen that HCl is stronger
than citric acid, which in turn stronger than
acetic acid, due to their degrees of
ionisation.
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•
plan and perform a first-hand
investigation to measure the pH
of identical concentrations of
strong and weak acids
•
use available evidence to
model the molecular nature of
acids and simulate the
ionisation of strong and weak
acids
• A molymod kit can be used to model
strong/weak/concentrated/dilute acids.
• The hydrohalic acids HCl and HF (or citric acid)
were modelled with a molymod kit.
• HF is a weak acid and ionises little in solution,
while HCl has essentially complete ionisation.
• H is a black circle, F is a red circle and Cl blue.
Possible method
1. Make 3 HCl molecules, 3 citric acid molecules
and eight water molecules.
2. First remove the hydrogen from HCl and attach it
to the water molecule to model the hydronium ion.
Do this 3 times. This will simulate complete
ionisation.
3. Remove one hydrogen from HF (or citric acid)
and attach to water. Do this only once and it will
model incomplete ionisation.
In their non-solution state they exits as molecular gases,
as below.
In solution, however, they ionise to form H+ ions and
their conjugate base.
• Note the differences between their degrees of
ionisation – HCl, being a strong acid, completely
ionises, while HF, as a weak acid, ionises to a
small extent only.
• The evidence for this small extent of ionisation
comes from testing pH using a pH meter – HF’s
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pH is higher than HCl’s at the same concentration
meaning that the HF solution has less H+ ions,
meaning it ionises less.
©
•
solve problems and perform a
first-hand investigation to use
pH meters/probes and
indicators to distinguish
between acidic, basic and
neutral chemicals
•
gather and process information
from secondary sources to
explain the use of acids as food
additives
(2012) All Rights Reserved
Many acids, biological and otherwise, can be used as food
additives for a wide range of functions.
• They can be added to improve the taste of foods
and drinks by adding a certain flavour, especially
malic, acetic, citric and tartaric acids.
• They inhibit growth of microbes such as bacteria
and mould, due to the low pH preventing excess
enzyme action. This preserves the food, increasing
its shelf life and making it safe to eat. Propanoic
acid is used for bread, potato crisps, and cake
mixes.
• Acting as anti-oxidants, they prevent spoilage of
foods in use as preservatives by slowing the
oxidation of oils. Citric acid is used as a
preservative in soft drinks, and vitamin C is also
used.
• Acids can also be added to improve nutrient
content (e.g. ascorbic acid or vitamin C)
• They act as leavening agents (substances which
react with NaHCO3 to produce CO2 gas). Tartaric
acid acts as one in desserts.
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•
outline the historical
development of ideas about
acids including those of:
- Lavoisier
- Davy
- Arrhenius
•
gather and process information
from secondary sources to trace
developments in understanding
and describing acid/base
reactions
• Antoine Lavoisier (1780s) proposed that acids
contained oxygen.
o As a result of his studies, Lavoisier shower
that many non-metal oxides (e.g. SO2 and
CO2), when dissolved in water, formed
acids (e.g. H2SO4 and H2CO3.
o Therefore, Lavoisier concluded that
oxygen was the element in a compound
that was responsible for general acidic
properties, as his knowledge was mostly
restricted to oxyacids.
o Whilst, his Oxygen Theory of Acids was
erroneous, as metal oxides in fact form
bases, it was important as the first attempt
to classify acids and bases based on their
chemical composition, rather than
property.
• Sir Humphry Davy (1815) proposed that acids
were substances that contained replaceable
hydrogen*.
o He continued Lavoisier’s investigation,
and this definition of acids based on
chemical composition was extended by
decomposing hydrochloric acid (among
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others), showing that it contained no
oxygen.
o Further, it was shown that many oxygen
containing substances were basic (e.g.
sodium oxide), effectively disproving
Lavoisier’s definition. Therefore he
proposed that all acids contained hydrogen
rather than oxygen.
o However, there was no proposition for the
mechanism of acidity, and this theory did
not explain why some compounds
containing hydrogen, such as CH4, are not
acidic, but could be basic e.g. NH4+
o In 1838, German chemist Justus von
Liebig extended Davy’s theory by
proposing that acids had ‘replaceable
hydrogen’
 He provided a more refined
definition than Davys’s describing
some precise behaviour. He
reasoned that when acids attacked
metals, the metals replace the
hydrogen to form a salt and
hydrogen gas.
 However, this still failed to account
for some other properties of acids,
such as production of NO2 and
hydrogen from nitric acid on metal.
• Svante Arrhenius proposed that acid is a
substance that produces hydrogen ions (H+) in
aqueous solution, and that a base is a substance
that produces hydroxide ions (OH-) in aqueous
solution. [H+/OH- in solution]
o Proposed that an acid was a substance that
ionised in solution to produce hydrogen
ions
+
HA → H + A
-
o The hydrogen ions thus cause the acidic
properties. For example:
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HCl
+
(aq)
→H
+ Cl
-
(aq)
(aq)
o This explained the fact that acids can
conduct electricity (electrolytes), acids
react with metals for form hydrogen gas
and the differing strengths of acids, which
was explained by degree to which the
forward reaction occurs (degree of
ionisation). For example, CH3COOH only
produces a small amount of hydrogen ions,
and is thus is weaker than HCl.
CH COOH
3
+
(aq)
H
-
(aq)
+ CH COO
3
(aq)
o Arrhenius also suggested that reactions
between acids and bases, called
neutralisation, produced water when H+
and OH- react.
+
-
HA + XOH → X + A + H O
2
H
+
+ OH
(aq)
(aq)
→H O
2
(l)
o In water, hydrogen ions, due to their small
size and high charge (and status as a single
proton), react with water molecules to
form H3O+, a coordinate covalent
compound. They are represented thus as
H+(aq) or H3O+(aq) even though the latter is
more correct.
o Arrhenius also proposed that a base was a
substance that, upon dissolution in water,
produced OH- ions. For example:
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+
XOH → X + OH
NaOH
(aq)
Ba(OH)
2(s)
→ Na
+
→ Ba
(aq)
+
(aq)
-
+ OH
(aq)
+ 2OH
(aq)
o Although the Arrhenius definition is
suitable for many common acids and
bases, interpreted the properties of acids
and provided a chemical explanation for
their nature, it has its limitations.
 Many substances which behave as
acids or bases, such as NH3 and
Na2CO3, do not contain an OH
group, and their reactions with
acids could not be explained.
 Some substances which react with
acids and have OH in their
structure, but are insoluble in water
(e.g. some group two hydroxides).
These couldn’t be classified as
bases, as Arrhenius’ theory only
applied to aqueous media.
 It also does not fully explain or
account for the relative strengths of
acids and bases (i.e. why they
ionise to different levels.)
 Also it could not explain why
certain salts, when dissolved in
water, created acidic or basic
solutions, such as NaS or ZnCl
while NaCl is neutral. [assumed
salts were neutral]
• The Brönsted-Lowry theory of acids and bases
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addressed these shortcomings. Further
development led to the Lewis (based on
electrons) theory which is more generalised, and
incorporates the solvent-system definition.
•
outline the Brönsted-Lowry
theory of acids and bases
• In 1923, Brönsted and Lowry independently
developed a more general theory of acids and
bases, which involves proton transfer and
acceptance in acid-base reactions, and addressed
shortcomings of previous theories.
• An acid is a proton donor (H+) while a base is a
proton acceptor, for the ionisation of HCl:
HCl + H O → H O
(g)
2
(l)
3
+
(aq)
+ Cl
(aq)
HCl is donating a proton and is therefore acting as
an acid. Cl- is the conjugate base of HCl. H2O is
accepting proton and therefore acting as a base.
H3O+ is the conjugate acid of H2O
• Brönsted-Lowry concept effectively relates acidity
and basicity to the structure and properties of the
substance relative to those of the solvent. This
theory added a role to the solvent, and mainly
focuses on water as an ionising solvent.
+
H O +H O H O
2
•
(l)
2
(l)
3
(aq)
+ OH
(aq)
It addressed the limitation of Arrhenius’ theory:
o Explains basic behaviour of many
substances which do not contain an OH
group
NH
3(aq)
+ H O NH
2
(l)
+
4 (aq)
+ OH
(aq)
Here ammonia acts as a base by accepting
H+ while water, which donates the proton,
acts as an acid. Here, we can see that while
ammonia contains no OH- in its structure,
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it creates it upon reaction.
o Extends acid-base reactions and
acidic/basic properties to solvents other
than water, and reactions without a
solvent. Allowed chemists to venture out
of aqueous chemistry to the realm of
non-aqueous or gas-phase reactions.
NH
3(g)
+ HCl → NH Cl
(g)
4
(s)
o Explains amphoterism; why some
substances can act as both acids and
bases
HCO
-
3 (aq)
+H O
2
(l)
H CO
2
3(aq)
+ OH
-
(aq)
HCO
-
3 (aq)
+H O
2
(l)
CO
2-
3 (aq)
+
+H O
3
(aq)
o Proposed that neutralisation could
proceed by direct proton transfer
o Showed that the hydrolysis of salts were
simple acid-base reactions.
o Explained how water could participate
in acid-base reactions.
•
©
describe the relationship
between an acid and its
conjugate base and a base and
its conjugate acid
(2012) All Rights Reserved
• In the Brönsted-Lowry theory, a protonated base
has the potential to act as an acid, and, similarly, a
deprotonated acid as a potential base.
• When an acid donates a proton, it forms its
conjugate base. Therefore the conjugate base of
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an acid is the acid minus one H+ ion.
• When a base accepts a proton, it forms its
conjugate acid. The conjugate acid of base is
the base plus one H+ ion.
For example, ammonia’s reaction can be written
backwards:
NH
3(g)
+H O
2
(l)
NH
+
4 (aq)
+ OH
-
(aq)
NH
+
4 (aq)
+ OH
(aq)
NH
3(g)
+H O
2
(l)
In this reaction NH4+ is now donating a proton to
water and is thus an acid, while OH- is the
accepting base. Also,
CH COOH
3
(aq)
+ OH
(aq)
-
CH COO
3
(aq)
+
HO
2
(l)
Acid + Base Conjugate base +
Conjugate acid
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CH COO
3
-
+
(aq)
+H O
3
(aq)
CH COOH
3
(aq)
+
HO
2
(l)
Base + Acid Conjugate acid + Conjugate
base
These reactions show that ethanoic acid acts as a
weak acid while the ethanoate ion acts as a weak
base.
• Thus, CH3COOH/ CH3 COO- and NH4+/ NH3 are
conjugate acid/base pairs.
• In general, with acid HA,
HA
(aq)
H
+
(aq)
+A
(aq)
HA is the acid, and A- is its conjugate base.
• The strength of an acid and its conjugate base are
related. (explained by position of equilibrium)
o The conjugate base of a strong acid is an
extremely weak (negligible) base
o The conjugate base of a weak acid is a
weak base
o The conjugate acid of a weak base is a
weak acid
o The conjugate acid of a strong base (such
as hydroxide ion i.e. OH-) is an extremely
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weak (negligible) acid
•
identify conjugate acid/base
pairs
[Ensure that you clearly identify
both pairs involved and which
species is the conjugate or base
in each pair]
•
identify a range of salts which
form acidic, basic or neutral
solutions and explain their
acidic, neutral or basic nature
1. Consider ions
- A ionises to form B and
C ions
2. Consider conjugates
3. Consider effects
- the X does not hydrolyse,
- However Y is a weak
acid/base in aq solution
so reacts by
accepting/donating a
proton. State whether
water acts as base or
acid
- produce OH- or H3O+
©
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• Skill
• Conjugate acid-base pairs are defined by the
Brönsted-Lowry theory acid-base theory.
• An acid is a substance which readily donates a
proton (H+ ion) in order to become its conjugate
base.
• Similarly, bases accept protons to become the
conjugate acid. To change one to the other, simply
add H+ (for base to acid) or remove one (for acid
to base). Some examples:
• The reaction of a salt with water to produce a
change in pH is called hydrolysis.
• Any salt consists of two ions, a cation and an
anion. Each of these, by the B-L definition of
acids and bases, has the capacity to act as an acid
or base, reacting with water.
• The pH of the solution depends on the degree to
which these ions act as B-L acids or bases.
• In a salt formed from a weak acid and a strong
base, the salt in aqueous solution is basic (pH >
7)as the anion is a weak base [form OH-]
e.g. CH COOH
3
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(aq)
+ NaOH
(aq)
→ CH COONa
3
(aq)
+
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and therefore raise or
lower pH
4. Take basic part and
react with water OR take
acidic part and react
with water
[hydrolysis reaction –
note in equilibrium]
Ion rules (consider ions step)
1. Neutral ion (do not react
with water)
- Anions (conjugate
bases) from strong
acids: Cl-, NO3-, Br-, I- Group I/II cations
from strong bases: Li+,
Mg2+, Na+, Ca2+, K+,
Ba2+
2. Basic ion (react with
water to form hydroxide
ions
- Anions (conjugate
bases) from weak acids:
F-, S2-, SO42-, ClO-,
CH3COO-, CO32-,
HCO3-, PO43-, HPO42- Some anions from
polyprotic acids:
HCO3-, HPO423. Acidic ion (react with
water to form hydronium
ions)
- Cations from weak
bases: NH4+
- Some anions from
polyprotic acids: HSO4-,
H2PO4- Cations from aquated
metal ions: Al3+, Fe3+
HO
2
1. CH COONa
3
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-
(aq)
→ CH COO
3
(aq)
+ Na
+
(aq)
2. Na+ is essentially the conjugate acid of NaOH
(strong base), therefore Na+ has negligible
acidity (does not hydrolyse).
CH3COO- is the conjugate base of CH3COOH
(weak acid), therefore CH3COO- is a weak
base
3. Overall salt is basic
4. CH COO
-
3
(aq)
+ H O CH COOH
2
(l)
3
(aq)
+ OH
(aq)
• In a salt formed from a weak base and a strong
acid, the salt in aqueous solution is acidic (pH <
7), as the cation is a weak acid [form H3O+]
e.g. NH
1. NH Cl
4
•
©
(l)
(aq)
3(aq)
+ HCl
→ NH
(aq)
+
4 (aq)
→ NH Cl
+ Cl
4
(aq)
(aq)
2. NH4+ is the conjugate acid of NH3 (weak base)
and will therefore be a weak acid
Cl- is the conjugate base of HCl (strong acid)
and will therefore be a negligible base
3. Overall, acidic salt
4. NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq)
In a salt formed from a strong base and a
strong acid neither ion hydrolyses and so the
salt in aqueous solution has a pH close to 7
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e.g. HCl
1. NaCl
+ NaOH
(aq)
(aq)
(aq)
→ Na
+
(aq)
→ H O + NaCl
+ Cl
2
(l)
(aq)
(aq)
+
2. Na is a negligible acid
-
Cl is a negligible base
3. Therefore, negligible + negligible = neutral
•
In a Salt formed from a weak base and a weak
acid both the anion and cation approximately
cancel each other out, and the alt in aqueous
solution has a pH close to 7
e.g. NH
3(aq)
CH COO
3
+ CH COOH
3
+
(aq)
NH +
4
(aq)
2. Conjugate acid of NH3 (weak base), therefore
weak acid
Conjugate base of CH3COOH (weak acid),
therefore weak base
3. The two parts of the salt roughly cancel each
other and pH is close to 7
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4. NH
+
4 (aq)
+H O
2
CH COO
3
•
choose equipment and perform
a first-hand investigation to
identify the pH of a range of
salt solutions
•
identify amphiprotic substances
and construct equations to
describe their behaviour in
acidic and basic solutions
[when writing out examples of
how amphiprotic substance act
as both acids and bases, always
show it reacting with either
hydroxide ions or hydronium
ions as the HSC markers will
not accept reactions with just
water]
[NB: most other amphiprotic
substances are anions produced
as a result of the ionisation of
diprotic and triprotic acids e.g.
HSO3-, HSO4- HPO4-, HPO42-]
(aq)
+
NH
(l)
3(aq)
+H O
3
+ H O CH COOH
2
(l)
3
(aq)
(aq)
+ OH
(aq)
•
An amphiprotic substance is one which can act as
both a proton donor (acid) and a proton acceptor
(base), thus acting both as an acid and a base
according to Bronsted-Lowry theory
o An amphiprotic substance must contain a
hydrogen atom, as it can donate a proton
•
Also, since amphiprotic substances can react both as
an acid and as a base, all amphiprotic substances are
amphoteric, but not all amphoteric substances are
amphiprotic
o Some metal oxides are amphoteric but not
amphiprotic; they don’t donate a proton,
but can participate in neutralisation
reactions, e.g. zinc oxide
 As a base: ZnO(s) + 2H+(aq) → Zn2+
(aq) + H2O(l)
 As an acid: ZnO(s) + H2O(l) + 2OH2(aq) → [Zn(OH)4] (aq)
• In the auto ionisation of water, one molecule
donates a proton and ther other accepts it.
H O + H O → OH
2
(l)
2
(l)
(aq)
+
+H O
3
(aq)
• Water (H2O) can react as a base in reaction with
HCl and as an acid in reaction with NH3. Thus,
water is acting primarily as either a proton donor
or acceptor, depending on its conditions.
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Acid: NH
3(aq)
+ H O NH
2
(l)
+
4 (aq)
+ OH
-
+
(aq)
[H O → H +
2
-
OH ]
+
Base: HCl + H O
(g)
2
(l)
HO
3
-
(aq)
+ Cl
(aq)
+
[H O + H →
2
+
HO ]
3
• In similar ways the hydrogencarbonate ion
(HCO3-) can react with water or hydroxy or
hydronium ions, as a base or acid, donating or
accepting protons.
Acid: HCO3-(aq) + H2O(l) CO32-(aq) + H3O+(aq)
Base: HCO3-(aq) + H2O(l) H2CO3(aq) + OH-(aq)
Acid: HCO
Base: : HCO
•
©
identify neutralisation as a
proton transfer reaction which
is exothermic
(2012) All Rights Reserved
-
+OH-(aq) → CO32-(aq) + H2O(l)
3 (aq)
-
+ H3O+(aq) → H2CO3(aq) + H2O(l)
3 (aq)
• Neutralisation is a proton transfer reaction
which is exothermic.
• When an acid and a base meet (Acid-base
reactions), they undergo neutralisation.
• The meaning of acid-base reactions have changed
along with the definitions of acid and base, but in
the Arrhenius sense, neutralisation occurs when
the H+ ion form the acid and the OH- ion from the
base combined to form H2O
These general equations should be memorised (Arrhenius
definition):
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Acid + metal hydroxide → salt + water
Acid + metal oxide → salt + water
Acid + carbonate/hydrogencarbonate → salt +
water + carbon dioxide gas
HCl
•
(aq)
+ NaOH
(aq)
→ NaCl
(aq)
+H O
2
(l)
For a net ionic equation (holds true for any
monoprotic neutralisation reaction):
+
H
(aq)
+ OH
(aq)
→H O
2
(l) rxn
H° = -55 to 60kJ/mol
• It was observed that all neutralisation between
strong acids and base had very similar heats of
reaction.
• The theory explained that the only reaction
occurring was that of neutralisation, producing the
same amount of heat each time.
• In the new Brönsted-Lowry definition,
neutralisation is now a proton-transfer reaction,
explaining its exothermic nature. The reaction
results in the formation of a new covalent bond
between hydrogen and oxygen as water molecules
are formed E.g. reaction of hydrochloric acid and
ammonia solution can be written as:
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NH3(aq) + H+(aq) →NH4+(aq)
•
analyse information from
secondary sources to assess the
use of neutralisation reactions
as a safety measure or to
minimise damage in accidents
or chemical spills
[when assessing the use of a
potential chemical, look at both
Benefits (e.g. ease of storage,
cost, amphoteric property, no
damage when used in excess,
safe to handle/store, quickacting neutraliser, non-toxic)
And also the disadvantages
(e.g. large amount required,
heat of neutralisation and by
products etc)
o Strong acids and bases are a liability to
safety in the laboratory and the
commercial world.
o Many acids/bases are corrosive, and
chemical spills of these substances must be
cleaned quickly with neutralisation
reactions.
o To neutralise acidic spills, Na2CO3 could
be used because:
o Adding excess is not particularly
dangerous since Na2CO3 is a weak base.
However, this means a large amount is
required to neutralise strong acids.
o It is a stable solid so is easy to store and
handle.
HCl
(aq)
+ NaHCO
Net reaction:
H
3(aq)
+
(aq)
→ CO
2(g)
+ HCO
+ H O + NaCl
3 (aq)
2
(l)
→ CO
2(g)
(aq)
+H O
2
(l)
o To neutralise a basic spill, weak acids
such as benzoic acid, boric acid or acetic
acid could be applied. However a safe
option would be to neutralise it with
NaHCO3 because:
o Due to the presence of the hydrogen
carbonate ion, it is amphiprotic, which
makes it useful if the nature of the spill is
unknown
o It acts as a weak acid or base, so it causes
no problems even if excess is added.
o Both Na2CO3 and NaHCO3 release CO2,
and the fizzing allows us to monitor
reaction progress.
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OH
•
qualitatively describe the effect
of buffers with reference to a
specific example in a natural
system
(aq)
+ HCO
3 (aq)
→ CO
3
2-
+H O
2
Factors of consideration
• Small spills should be diluted with water.
However large spills, neutralisation is preferred.
• Location must also be considered. If it is on a
table bench, all options are possible. However, it
is on any person’s body, the exothermic nature of
neutralisation will generate a lot of heat causing
discomfort, and so washing with lots of running
water may be more appropriate.
• If the spill is in the environment, e.g. leakage from
a truck transporting acid, then the impacts on the
environment must also be considered before
neutralisation, e.g. the heat produced could cause
thermal pollution and there would be salt waste
product.
Assessment: Neutralisation is an exothermic reaction
which can be dangerous, due to the amount of heat
evolved upon large-scale neutralisations. However it is
otherwise a safe and elegant technique to neutralise acid
and base spills, and as such is a suitable technique to an
otherwise difficult problem.
(or generic) Neutralisation reactions are effective as a
safety measure to minimise the damage in accidents or
chemical spills involving acids and bases.
• A buffer solution contains approximately equal
amounts of weak acid and its conjugate base
(or vice versa). It is therefore able to maintain
an approximately constant pH when small
quantities of another acid or base are added.
(resist changes to pH)
• The main buffer system used to control the pH of
blood is the carbonic acid / hydrogencarbonate
ion buffer system. The presence of buffers in the
blood maintains the pH between 7.35 and 7.45
H CO
2
3(aq)
+ H O HCO
2
(l)
3 (aq)
+H O
3
+
(aq)
• The diffusion of oxygen into cells and the lungs
and carbon dioxide out affects the pH of the
blood, lowering the pH with every inhalation and
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•
describe the correct technique
for conducting titrations and
preparation of standard
solutions
[always write the reaction
equation]
[It is very important whenever
describing the measurement of
volumes in titration procedures
to say that the bottom of the
meniscus was touching the
gradation line]
[rinsing usually a MC q]
[when doing calculations using
the titres, never use the rough
titre]
©
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increasing it with exhalation, which can cause
problems due to fluctuations in pH on a regular
basis. Thus the buffer is needed to regulate these
changes to a small degree.
• Excess changes are regulated by the buffer system
by Le Chatelier’s Principle.
• Excess (addition of) H3O+ shifts the equilibrium to
the left and consume the extra hydronium ions,
increasing pH. In addition, it is neutralised by
HCO3-(aq)
• Excess OH- removes H3O+ from the system by
neutralisation, causing the equilibrium to shift to
the right to create extra hydronium ions and
decrease pH. Either way, the changes in pH are
minimised. In addition, it can be neutralised by
carbonic acid.
• The buffer system also works to control excess
addition from external sources of either acid or
base, effectively controlling the pH of the blood to
manageable values.
• Metabolic processes need constant pH because
they depend on enzymes, and enzymes only work
efficiently within a narrow pH range. However,
many waste products in the body are acids or
bases
• Also important in maintaining fairly constant pH
of 8.5 in ocean water
Preparation of Standard Solution
• A primary standard is a substance of such high
purity and stability that a standard solution of
it, of accurately known concentration, can be
prepared by weighting out the desired mass,
dissolving it in water and making the volume up
to an accurately known value.
• Being the first step, they are very important and
must meet a list of strict criteria:
1. Extremely high level of purity [usually
Analytical Reagent (A.R.) grade] (easy to
purify and store in pure form)
2. Extremely accurately known composition
(formula)
3. Moisture-free
4. Stable, and unaffected by air in weighing
(does not lose or gain water, or react with
oxygen or carbon dioxide in air)
5. Readily soluble in water (distilled/pure)
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6. High molecular mass in order to minimise
weighing errors
7. React instantaneously and completely
• Excellent starting products which display these
properties are:
o Basic primary standard: anhydrous sodium
carbonate [Na2CO3]
Note*: Na2CO3 for primary standard, NaHCO3 for
neutralisation
o Acidic primary standard: hydrated oxalic acid
[H2C2O4 . 2 H2O]
Other primary standards used are:
• If a particular solution must be used as the titrant,
but it is not composed of a primary standard, then
its concentration can be found (standardised) by
titration against a primary standard, known as a
secondary standard solution.
o Sodium hydroxide NaOH is difficult to
obtain in very pure form, and hygroscopic
(absorbs moisture from the air) and also
reacts with CO2 in the air
o Hydrochloric acid HCl is also not a
primary standard; its concentration varies
slightly among batches; concentrated HCl
fumes and releases HCl fumes
o Concentrated sulfuric acid absorbs water
from the air
o Hydrated sodium carbonate loses water to
the air as it is being weighed
o Possible sequence for preparing standard
sodium hydroxide solution (or titrate with
oxalic acid)
 Use a standard Na2CO3 solution to
standardise a HCl solution
 Use standard HCl solution to
standardise a NaOH solution
• To prepare a primary standard solution, high
degrees of care, precision, and accuracy need to be
taken to ensure good results. The equipment used
to create a primary standard solution are:
o primary standard, mass balance (3 or 4
decimal places), volumetric flask of
desired size, funnel, wash bottle with
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distilled/deionised/pure water, clean
beakers, glass rod, spatula
 Note all glassware should be
impeccably clean. All glassware
can be rinsed with supplied water
and cleaned, except the weighing
container.
For example, this is a method to prepare a primary
solution with high precision:
1. Weigh on analytical balance exact amount of solid
on watch glass
2. Transfer to beaker, dissolve in distilled water
3. Using a funnel, transfer to volumetric flask
4. Wash out beaker with distilled water and tip into
flask
5. Using distilled water, make up the volume in the
flask until the bottom of the meniscus is level with
etched mark on the flask.
6. Stopper the flask and shake to ensure thorough
mixing
7. Label solution, giving name and concentration
Conducting Titrations
• Titration involves reacting a solution of unknown
concentration with a solution of known
concentration (called a standard). By measuring
how much standard solution is required to
neutralise a predetermined volume of our
unknown solution, we can calculate the
concentration of the unknown solution
Titration Equipment
• Volumetric flask: holds an accurately known
volume of solution (usually 250 mL) indicated by
a line etched into the neck of the flask
o Used for preparation of standard solutions
• Pipette: used to accurately deliver a specific
volume of solution (usually 25 mL) into the
conical flask
o Do not shake out the last drop; the pipette
is calibrated to account for this
• Burette: used to accurately deliver variable
volumes of solution into the conical flask
o Reading is usually estimated to ± 0.05 mL
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Preparation of Glassware:
• Rinse the burette/ pipette with distilled water, and
then rinse (3 times) with the solution that will be
placed in there. The rinses are then discarded.
o If burette/pipette is rinsed with only
distilled water then water droplets
remaining on inside of glassware will
dilute the solution, and the number of
moles of solution added can no longer be
accurately determined
o Any solution left over after rinsing will not
affect the final answer as it is the
concentration of the solution in each
piece of equipment that is important
• Wash the conical flask with distilled water only
o Any water remaining in the flask will not
change the number of moles of solution
o Amount or moles of reactants that are
important not concentration
Titration Procedure
1. Wash burette with distilled water, and then with
some of the titrant solution
2. Set up burette with retort stand
3. Carefully pour titrant solution into burette, with a
funnel, until bottom of meniscus is level with the
zero mark, known as titrant.
4. Wash pipette with distilled water, and then with
some of the analyte solution to be titrated
5. Fill a 25mL pipette up to etched mark with analyte
solution
a. Note that solution should be poured from
original container (e.g. volumetric flask)
into a clean, dry beaker, and drawn into
pipette from the beaker. This is so that the
whole stock of analyte solution is not
contaminated by any foreign material in/on
the pipette.
6. Wash a conical flask with distilled water
7. Empty the analyte solution from the pipette into a
conical flask, and add 3 drops of indicator
8. Position conical flask under burette
9. Using burette, slowly add titrant solution into
the conical flask one drop at a time, swirling the
flask all the while, until the required colour
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change occurs
10. Record amount of titrant solution used up in
burette
11. Repeat all previous steps 3 more times, for a total
of 1 rough titration, and 3 precise titrations.
Definitions
• Titration: Procedure used to experimentally find
the concentration of a solution.
• Indicator: Substance which signals the
equivalence point of a titration by changing
colours
• Equivalence point: Point when reaction is
complete.
The equivalence point of a chemical reaction is
the point of the neutralisation reaction where the
amounts of the two reactants are just sufficient to
cause complete consumption of both reactants.
• End point: Point when indicator changes colour.
• Titrant: Standard solution of known
concentration and composition.
• Titre: Volume of solution delivered from the
burette.
• Analyte: any chemical to be analysed
• Aliquot: Fixed volume of solution delivered by
pipette.
Indicator choice
• The indicator we choose depends on the nature of
the acid-base reaction.
• Choice of indicator that undergoes a sharp colour
change at the exact equivalence point of the
titration.
• It is important to distinguish between the
equivalence (equal amounts of acid/base) and end
points (indicator colour changes).
Titration Curves
These indicators are used for specific titrations:
• Equivalence Point is at pH 7
• Neutral pH of salt produced, therefore use
Bromothyl Blue or Litmus
• Equivalence point is below pH 7
• Acidic pH of salt produced, therefore use Methyl
Orange or Methyl Red
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• Equivalent point is above pH 7
• Basic pH of salt produced, therefore use
Phenolphthalein or Thymol Blue
• No sudden change in pH as both reactants are
weak
• *NB learn back-titration
•
•
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Note that point of inflexion is at pH of the salt
solution produced
In a titration curve, independent variable (y-axis)
is volume of titrant, and dependent variable (xaxis) is pH of solution
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