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+ Unit D: Equilibrium Textbook Reference: Chapters 15 and 16 + Equilibrium DAY 1 1.1K: Define equilibrium and state the criteria that apply to a chemical system in equilibrium; i.e., closed system, constancy of properties, equal rates of forward and reverse reactions 1.2K: Identify, write and interpret chemical equations for systems at equilibrium 1.4K: Define Kc to predict the extent of the reaction and write equilibrium-law expressions for given chemical equations, using lowest whole-number coefficients + Chemical Equilibrium 100% Not all reactions are quantitative (reactants products) Evidence: For many reactions reactants are present even after the reaction appears to have stopped Recall the conditions necessary for a chemical reaction: Particles must collide with the correct orientation and have sufficient energy If product particles can collide effectively also, a reaction is said to be reversible Rate of reaction depends on temperature, surface area and concentration + Chemical Equilibrium Consider the following reversible reaction: The final state of this chemical system can be explained as a competition between: The collisions of reactants to form products The collisions of products to re-form reactants We assume this system is closed (so the reactants and products cannot escape) and will eventually reach a: DYNAMIC EQUILBRIUM - Opposing changes occur simultaneously at the same rate + Chemical Equilibrium Consider the following hypothetical system: AB + CD AD + BC forward reaction, therefore AD + BC AB + CD reverse reaction • Initially, only AB and CD are present. The forward reaction is occurring exclusively at its highest rate. • As AB and CD react, their concentration decreases. This causes the reaction rate to decrease as well. • As AD and BC form, the reverse reaction begins to occur slowly. • As AD and BC’s concentration increases, the reverse reaction speeds up. • Eventually, both the forward and reverse reaction occur at the same rate = DYNAMIC EQUILIBRIUM + 4 Conditions of Dynamic Equilibrium* 1. Can be achieved in all reversible reactions when the rates of the forward and reverse reaction become equal Represented by rather than by 2. All observable properties appear constant (colour, pH, etc) 3. Can only be achieved in a closed system (no exchange of matter and must have a constant temperature) 4. Equilibrium can be approached from either direction. This means that the equilibrium concentrations will be the same regardless if you started with all reactants, all products, or a mixture of the two Types of Equilibrium 1. 2. Phase Equilibrium: a single substance existing in more than 1 phase Example: Liquid water in a sealed container with water vapour in the space above it Water evaporates until the concentration of water vapour rises to a maximum and then remains constant Solubility Equilibrium: a saturated solution Rate of dissolving = rate of recrystallization Types of Equilibrium 3. Chemical Equilibrium – reactants and products in a closed system Example: The Hydrogen-Iodine Equilibrium System The rate of reaction of the reactants decreases as the number of reactant molecules decrease. The rate at which the product turns back to reactants increases as the number of product molecules increases. These two rates become equal at some point, after which the quantity of each will not change. + Describing the Position of Equilibrium 1. Percent Yield- the yield of product measured at equilibrium compared with the maximum possible yield of product. % yield = product eq’m x 100 % product max The equilibrium concentration is determined experimentally, the maximum concentration is determined with stoichiometry + Describing the Position of Equilibrium 1. Percent Yield – Example If 2.50 mol of hydrogen gas reacts with 3.0 mol of iodine gas in a 1.00L vessel, what is the percent yield if 3.90 mol of hydrogen iodide is present at equilibrium % yield = product eq’m x 100 % product max H2(g) + I2(g) 2HI(g) 2.50 mol x (2 mol HI) = 5.0 mol HI 1 mol H2(g) % yield = 3.90 mol x 100 % 5.00 mol = 78% + Describing the Position of Equilibrium 2. Using an Equilibrium Constant, (Kc) Think “products over reactants” If the Kc > 1, the equilibrium favours products If the Kc < 1, the equilibrium favours reactants If the Kc > 100, the reaction is Quantitative If the Kc > 1/100, the reaction is considered non-spontaneous WHAT if Kc = 1 ?? + Describing the Position of Equilibrium Kc is a constant value at a given temperature. It only varies with changes in temperature which is why temp is always stated for Kc. The following do NOT affect Kc: concentration, catalyst, or time. Limitation- Kc tells you the position of a reaction it does not tell you the rate of a reaction. Eg. a large Kc does not mean the reaction is fast only that it favors products. + Describing the Position of Equilibrium 2. Using an Equilibrium Constant, (Kc) Example #1: Write the equilibrium law expression for the reaction of nitrogen monoxide gas with oxygen gas to form nitrogen dioxide gas. + Describing the Position of Equilibrium Using an Equilibrium Constant, (Kc) 2. Note: The Kc value describes the extent of the forward reaction. Kc reverse = 1 . = The reciprocal value Kc forward Example #2: The value of Kc for the formation of HI(g) from H2(g) and I2(g) is 40, at a given temperature. What is the value of Kc for the decomposition of HI(g) at the same temperature. Kc reverse = 1 . Kc forward = 1 40 = 0.025 + Describing the Position of Equilibrium Using an Equilibrium Constant, (Kc) 2. Note: For heterogeneous equilibrium systems, DO NOT include liquids and solids in the expression. (They are assumed to have fixed concentrations) Example #3: Write the equilibrium law expression for the decomposition of solid ammonium chloride to gaseous ammonia and gaseous hydrogen chloride The solid is omitted from the expression Example #4: Write the equilibrium law expression for the reaction of zinc in copper(II) chloride solution. The solids, as well as spectator ions (Cl-) are omitted from the expression + Describing the Position of Equilibrium + Describing the Position of Equilibrium GUIDED PRACTICE Determine the equilibrium law and constant for the following reactions. 2NO(g) + O2(g) <--> 2 NO2(g) 0.3 mol/L 0.5 mol/L 0.8 mol/L Mg(s) + Cu2+(aq) <--> Cu(s) + Mg2+ (aq) 2.0 M 4.0 M Hg(l) + Zn2+(aq) <--> Zn(s) + Hg2+(aq) 3.5 M 7.0 M C6H6(l) + Br2(l) <--> C6H5Br(l) + HBr(l) + PRACTICE Read Section 15.1 Equilibrium Concentrations + DAY 2 2.3K: Calculate equilibrium constants and concentrations for homogeneous systems when • concentrations at equilibrium are known • initial concentrations and one equilibrium concentration are known • the equilibrium constant and one equilibrium concentration are known. + Calculations in Equilibrium Systems Using the equilibrium law expression to determine whether a system is at equilibrium: Substitute in the given concentrations to the equilibrium expression. If the value is the equilibrium constant, the system is at equilibrium If the value is larger, this means there are more products that reactants. To reach equilibrium, the reaction must proceed to the left (towards the reactants) If the value is smaller, this means there are more reactants than products. To reach equilibrium, the reaction must proceed to the right (towards the products) + Calculations in Equilibrium Systems Example #1: In the following system: N2(g) + 3H2(g) ↔ 2NH3(g) 0.249 mol N2(g), 3.21 X 10-2 mol H2(g) and 6.42 X 10-4 mol NH3(g) are combined in a 1.00 L vessel at 375oC, Kc = 1.2 Is the system at equilibrium? Kc = NH3 2 N2(g) H2(g) 3 = (6.42 x 10 -4)2 =0.0500 (0.249)(3.21 x 10-2)3 If not, predict the direction in which the reaction must proceed. Value is below Keq = therefore more reactants than products, so reaction must proceed to the right + Calculations in Equilibrium Systems Example #2: Find the equilibrium concentration of the ions that are formed when solid silver chloride is dissolved in water. The equilibrium constant for this reaction is Kc = 5.4 X 10-4. AgCl(s) Ag+(aq) + Cl-(aq) Kc = Ag+(aq) Cl-(aq) 5.4 x 10 -4 = x2 0.023 mol/L = x = Ag+(aq) eq = Cl-(aq) eq + ICE Tables and Equilibrium Calculations STEPS: Always write out the equilibrium reaction and equilibrium law expression if not given. Draw an ICE Table (Initial, Change in and Equilibrium concentrations) (I + C = E) Substitute values where appropriate Solve for x Solve for equilibrium concentrations + ICE Charts and Equilibrium Calculations Example #3: Consider the following equilibrium at 100 oC: N2O4(g) ↔ 2 NO2(g) 2.0 mol of N2O4(g) was introduced into an empty 2.0 L bulb. After equilibrium was established, only 1.6 mol of N2O4(g) remained. What is the value of Kc? 2.0 mol = 1.0 mol/L (I) 2.0L 1.6 mol = 0.8 mol/L (E) 2.0L N2O4(g) I: 1.0 mol/L C: E: -x Solve for Kc = (0.40)2 (0.80) = 0.20 0 + 2x 1.0 – x = 0.80 E: 1.0 – x = 0.80 solve for x 2NO2(g) x = 0.20 2x = 0.40 2x + ICE Charts and Equilibrium Calculations Example #4 A 10 L bulb is filled with 4.0 mol of SO2(g), 2.2 mol of O2(g) and 5.6 mol of SO3(g). The gases then reach equilibrium according to the following equation: 2SO2(g) + O2(g) ↔ 2SO3(g) At equilibrium, the bulb was found to contain 2.6 mol of SO2(g). Calculate Kc for this reaction. 4.0 mol = 0.40 mol/L (I) 10.0L 2.6 mol = 0.26 mol/L (E) 10.0L 0.40 – 2x = 0.26 X = 0.07 2.2 mol = 0.22 mol/L (I) 10.0L 2SO2(g) O2(g) 2SO3(g) I: 0.40 0.22 0.56 C: - 2x - x + 2x 0.22 - x 0.56 + 2x 0.15 0.70 E: 0.40 – 2x = 0.26 E: Kc = 5.6 mol = 0.56 mol/L (I) 10.0L (0.70)2 (0.15)(0.26)2 0.26 = 48 + ICE Charts and Equilibrium Calculations Example #5 In a certain experiment 1.000 mol of NH3(g) was placed in an empty 1.000L flask at 500oC. After equilibrium was reached, 0.399 mol of N2(g) was found to be in the flask. Calculate the concentrations of NH3(g) andH2(g) at equilibrium by setting up an ICE table. 1.0 mol = 1.0 mol/L (I) 1.0L 0.399 mol = 0.399 mol/L (E) 1.0L 0.0 + x = 0.399 X = 0.399 0 mol = 0.0 mol/L (I) 1.0L 2NH3(g) 0 mol = 0.0 mol/L (I) 1.0L N2(g) 3H2(g) I: 1.0 0.0 0.0 C: - 2x + x + 3x 0.0 + x 0.0 + 3x 0.399 1.197 E: 1.0 – 2x = E: 0.202 + ICE Charts and Equilibrium Calculations Example #6 In a certain experiment, 3.00 mol of SO2(g), 2.00 mol of O2(g) and 4.00 mol of SO3(g) were placed in an empty 1.000L vessel at 500oC. After equilibrium was reached, 5.00 mol of SO3(g) was found to be in the flask. Calculate the concentrations each entity at equilibrium by setting up an ICE table. + PRACTICE Chemical Equilibrium WS + ICE Charts and Equilibrium Calculations Example #6 In a certain experiment, 3.00 mol of SO2(g), 2.00 mol of O2(g) and 4.00 mol of SO3(g) were placed in an empty 1.000L vessel at 500oC. After equilibrium was reached, 5.00 mol of SO3(g) was found to be in the flask. Calculate the concentrations each entity at equilibrium by setting up an ICE table. 3.00 mol = 3.00 mol/L (I) 1.0L 5.00 mol = 5.00 mol/L (E) 1.0L 4.00 + 2x = 5.00 X = 0.500 2.00 mol = 2.00 mol/L (I) 1.0L 4.00 mol = 4.00 mol/L (I) 1.0L 2SO2(g) O2(g) 2SO3(g) I: 3.00 2.00 4.00 C: - 2x - x + 2x 2.00 – x 4.00 + 2x 1.50 5.00 E: 3.0 – 2x = E: 2.00 + ICE Charts and Equilibrium Calculations Example #7: Using a perfect square Given the following reaction: N2(g) + O2(g) ↔ 2NO(g) Determine the equilibrium concentrations for all species present given that the initial concentration of each reactant is 0.200 mol/L. N2(g) O2(g) 2NO(g) I: 0.200 0.200 0 C: -x - x + 2x 0.200 - x 2x 0.195mol/L 0.00976mol/L E: E: Kc = 0.00250 0.00250 = 0.200 - x 0.195mol/L (2x)2 square root both sides (0.200-x)2 0.05 = 2x 0.200 – x = 0.01 – 0.05x = 2x = 0.01 = 2.05x HOW DOES THE Kc value relate to the amount of change ?? = 0.00488 + ICE Charts and Equilibrium Calculations Example #8: Using the Approximation Rule Calculate the concentration of gases produced when 0.100 mol/L COCl2(g) decomposes into carbon monoxide and chlorine gas. The Kc for this reaction is 2.2 X 10-10. 0.100 = >1000 -10 2.2x 10 2.2 x10-10 = COCl2(g) CO(g) Cl2(g) I: 0.100 0 0 C: -x + x +x x x E: 0.100 - x E: 0.100 mol/L (x)2 (0.100-x) 2.2 x 10-10 = x2 = 0.100 x 4.7 x 10-6 mol/L 4.7 x 10-6 mol/L Approximation Rule: When the reactants/Kc is larger than 1000, you can disregard the change in concentration for the initial entities. So 0.100 – x = 0.100 = 4.69 x 10-6 mol/L + ICE Charts and Equilibrium Calculations Example #9: Using Graphing to solve Gaseous NOCl decomposes to form NO(g) and Cl2(g). At 35oC the equilibrium constant Keq = 1.60 X 10-5. If 1.00 mol of NOCl(g) is placed in a 1.0 L flask, what are the equilibrium concentrations of each species 2NOCl(g) 1.00 = >1000 1.6x 10 -5 1.00 0 0 C: - 2x + 2x +x 0 + 2x 0+x 1.00 – 2x E: 1.6 x10-5 = Cl2(g) I: E: 2NO(g) (2x)2 (x) (1.00-2x)2 + PRACTICE Pg 682 Q 4, 6 Lab Excerise 15.B pg 686 Text Questions # 1-10 pg 688-689 Finish Equilibrium WS Quiz on ICE tables Friday Coming Up Tomorrow Le Chatelier’s Principle Chapter 15 Exam + Le Chatelier’s Principle 1.3K: Predict, qualitatively, using Le Chatelier’s principle, shifts in equilibrium caused by changes in temperature, pressure, volume, concentration or the addition of a catalyst and describe how these changes affect the equilibrium constant + Le Chatelier’s Principle Le Chatelier’s Principle is very useful in predicting how a system at equilibrium will respond to change. It states that when a system at equilibrium is disturbed, the equilibrium shifts in the direction that opposes the change, until a new equilibrium is reached. There are three common ways an equilibrium may be disturbed: Change in the concentration of one of the reactants or products Change in the temperature Change in the volume of a container Addition of a catalyst (less common) Changes in the temperature, will change the Kc value. No other changes will affect this value. + Le Chatelier’s Principle Effect of Changes in Concentration If a system at equilibrium is disturbed by the addition of a reactant (or the removal of a product), Le Chatelier’s principle predicts that the equilibrium will shift right. 2N2O(g) + 3O2(g) If the disturbance is the removal of a reactant (or the addition of a product), Le Chatelier’s principle predicts that the equilibrium will shift left. 2N2O(g) + 3O2(g) 4 NO2(g) 4 NO2(g) Since concentrations of solids are constants and do not appear in expressions for K, removing or adding some solid does not cause shifts. + Concentration Change If you see spike in either a reactant or product, which causes a gradual change in the other entities – a substance has been added or removed. Addition of reactant, HF(g) – will shift equilibrium to the right Removal of product, HCl(g), will shift the equilibrium to the right More products will be produced, and a new equilibrium is established More products will be produced, and a new equilibrium is established + Le Chatelier’s Principle Effect of Changes in Temperature In endothermic reaction, heat acts like a reactant. Increasing the temperature shifts the reaction right. Decreasing the temperature, shifts the reaction left Heat + 2N2O(g) + 3O2(g) In exothermic reactions, heat acts like a product. Increasing the temperature shifts the reaction left. Decreasing the temperature, shifts the reaction right. 4 NO2(g) 4 NO2(g) 2N2O(g) + 3O2(g) + Heat The equilibrium constant, Kc is temperature dependent Remember K gets bigger if there are more products being created (i.e. shifts to the right) Reaction Type Role of heat Effect of T Effect of T Endothermic Reactants + heat products K K Exothermic Reactants products + heat K K + Temperature Change The temperature decreases, at the time indicated by the dotted line. This will cause the equilibrium to shift to the right, creating more products, until a new equilibrium is established. + energy If you see a gradual change in the reactants with an opposite change in the products, you have a temperature change going on + Le Chatelier’s Principle Effect of Changes in the Volume of the Container If volume is decreased, pressure increases (Boyle’s Law – Chem 20) So the reaction will shift in the direction which contains the fewest moles of gas Pressure 2N2O(g) + 3O2(g) 4 NO2(g) 4 moles are fewer than 5 If volume is increased, pressure decreases (Boyle’s Law – Chem 20) So the reaction will shift in the direction which contains the most moles of gas Pressure 2N2O(g) + 3O2(g) 4 NO2(g) 5 moles are more than 4 If both sides of the equation have the same number of moles of gas, the change in volume of the container has no effect on the equilibrium. + Volume Change The volume of the container decreased, at the time indicated by the dotted line. This will cause a pressure increase. This will cause the equilibrium to shift to the right, the side with fewer moles of gas, creating more products, until a new equilibrium is established. If you see a spike in all of the entities = P increase, V decrease If you see a drop in all of the entities = P decrease, V increase + Le Chatelier’s Principle Effect of the Addition of a Catalyst Catalysts speed up the rate at which equilibrium is obtained, but have no effect on the magnitude of K. They increase both the forward and backward rate of reaction. + + Identify the nature of the changes imposed on the following equilibrium system at the four times indicated by coordinates A, B, C and D • At A, the concentration (or pressure) of every chemical in the system is decreased by increasing the container volume. Then the equilibrium shifts to the left (the side with more moles of gas) • At B, the temperature is increased. Then the equilibrium shifts to left. • At C, C2H6(g) is added to the system. Then the equilibrium shifts to the left. • At D, no shift in equilibrium position is apparent; the change imposed must be addition of a catalyst, or of a substance that is not involved in the equilibrium reaction. + Practice Change Direction of Shift (, , or no change) Effect on quantity a) Decrease in volume Kc b) Raise temperature Amount of CO(g) c) Addition of I2O5(s) Amount of CO(g) d) Addition of CO2(g) Amount of I2O5(s) e) Removal of I2(g) Amount of CO2(g) Effect (increase, decrease, or no change) + Practice Le Chat WS Text questions Pg 695 Q 1-3 Pg 699 Q 1-3, 5-7 Coming Up Quiz Tomorrow – ICE tables Review Le Chat Friday – Study day Monday Exam Chapter 15 + Acids and Bases - Kw 2.1K: Recall the concepts of pH and hydronium ion concentration and pOH and hydroxide ion concentration, in relation to acids and bases 2.2K: Define Kw to determine pH, pOH, [H3O+] and [OH–] of acidic and basic solutions + The Water Ionization Constant, Kw Even highly purified water has a very slight conductivity. This is due to the ionization of some water molecules into hydronium ions and hydroxide ions. The water ionization equilibrium relationship is so important, it gets its own special symbol and name: Ionization Constant for Water, Kw + The Water Ionization Constant, Kw Since the mathematical relationship is simple, we can easily use Kw to calculate either the hydronium or hydroxide ion concentration, if the other concentration is know. The presence of substances other than water decreases the certainty of the Kw value to two significant digits; 1.0 x 10 -14 + Kw Calculations Example: A 0.15 mol/L solution of hydrochloric acid at 25°C is found to have a hydronium ion concentration of 0.15 mol/L. Calculate the amount concentration of the hydroxide ions. + Kw Calculations Example #2: Calculate the amount concentration of the hydronium ion in a 0.25 mol/L solution of barium hydroxide. Ba2+ = 0.25 mol/L x 2 mol = 0.50 mol/L 1 mol + Kw Calculations Example #3: Determine the hydronium ion and hydroxide ion amount concentration in 500 mL of an aqueous solution for home soap-making containing 2.6 g of dissolved sodium hydroxide. + Do You Remember? pH and pOH + Acid Strength as an Equilibrium Position Do you remember the difference between strong and weak acids? Strong acids – ionize completely (quantitatively), even though this could be written with a double arrow, it is simpler to use a single arrow to show the the reaction is >99.9% Strong Acids: HClO4(aq), HI(aq), HBr(aq) , HCl(aq) , HSO4(aq), HNO3(aq) Weak acids – ionize (react with water) only partially (<50%) + % Ionization In a 0.10 mol/L solution of acetic acid, only 1.3% of the CH3COOH molecules have reacted at equilibrium to form hydronium ions. Calculate the hydronium ion amount concentration. + % Ionization The pH of 0.10 mol/L methanoic acid solution is 2.38. Calculate the percent reaction for ionization of methanoic acid. + Practice pH and pOH Review WS Kw, pH and pOH WS Pg. 716 #1,-6 Pg. 718 #7-8 + Bronsted-Lowry Acids and Bases 1.5K: Describe Brønsted–Lowry acids as proton donors and bases as proton acceptors 1.6K: Write Brønsted–Lowry equations, including indicators, and predict whether reactants or products are favoured for acid-base equilibrium reactions for monoprotic and polyprotic acids and bases 1.7K: Identify conjugate pairs and amphiprotic substances + Bronsted-Lowry Acid-Base Concept Focuses on the role of the chemical species in a reaction rather than on the acidic or basic properties of their aqueous solutions. Bronsted Lowry Definition for an Acid: proton donor Bronsted Lowry Definition for an Base: proton acceptor + Bronsted-Lowry Acid-Base Concept Bronsted-Lowry Reaction Equation: is an equation written to show an acid-base reaction involving the transfer of a proton from one entity (an acid) to another (a base) Bronsted-Lowry Neutralization: is a competition for protons that results in a proton transfer from the strongest acid present to the strongest base present. The Bronsted-Lowry concept does away with defining a substance as being an acid or base. Only an entity that is involved in a proton transfer in a reaction can be defined as an acid or base – and only for a particular reaction Remember this is just a theoretical definition, not a theory, because it does not explain why a proton is donated or accepted, and cannot predict which reaction will occur for a given entity in any new situation. + Bronsted-Lowry Acid-Base Concept Protons may be gained in a reaction with one entity, but lost in a reaction with another entity. The empirical term, amphoteric, refers to a chemical substance with the ability to react as either an acid or base. The theoretical term, amphiprotic, describes an entity (ion or molecule) having the ability to either accept or donate a proton. Example: When bicarbonate ions are in aqueous solution, some react with the water molecules by acting as an acid, and some react by acting as a base. Kc values given for these reactions, show that one predominates. The number of ions acting as a base is over 2000x more than the number reacting as an acid. = BASIC Solution + Bronsted-Lowry Acid Base Concept The Amphoteric Nature of Baking Soda It can partially neutralize a strong acid, but also a strong base Practice: pg. 724 #2-4 + Conjugate Acids and Bases In an acid-base reaction, there will always be two acids and two bases. The original acid on the left and the acid on the right created by adding a proton. The original base on the left and the base on the right created by removing a proton A pair of substances with formulas that differ only by a proton is called a conjugate acid-base pair This analogy shows why acetic acid is a weak acid. The proton is more strongly attracted to the acetic acid molecule than it is to the water. + Acids/Bases What about strong acids: When HCl reacts with water, the water wins the competition against the Cl- for the proton. This is why at equilibrium, essentially all of the HCl molecules have lost protons to water. >99% NOTE: The stronger the base, the more it attracts a proton (proton acceptor). The stronger the acid, the less it attracts its own proton (proton donor) + Conjugate Acids and Bases RULE: The stronger the base, the more it attracts a proton (proton acceptor). The stronger the acid, the less it attracts its own proton (proton donor) What does this mean about their conjugate pair?? The stronger an acid, the weaker is its conjugate base. If you are good at donating a proton, this means the conjugate base is not good at competing for it (weak attraction for protons) The stronger a base, the weaker is its conjugate acid. If you are good at accepting a proton, this means the conjugate acid is not good at giving it up (strong attraction for protons). + + Practice Bronsted-Lowry WS Pg 724 Q 2-4 Pg 726 Q 7 Predicting Acid and Base + Reactions 1.6K: Write Brønsted–Lowry equations, including indicators, and predict whether reactants or products are favoured for acid-base equilibrium reactions for monoprotic and polyprotic acids and bases 1.7K: Identify conjugate pairs and amphiprotic substances + Acid-Base Reactions What is happening? Collisions are constantly occurring. Each time, a proton is transferred to a stronger proton attractor. Theoretically, it could transfer several times (each time to a stronger proton attractor.) But once, it is transferred to the strongest base present, the proton will remain there as nothing outcompetes it. Likewise, once the strongest acid has given up its proton, its conjugate base cannot gain one back (as it is the weakest proton attractor in the whole system) So what does this mean? Proton transfer occurs between the strongest acid and strongest base. All other transfers are negligible so are ignored. + Predicting Acid-Base Reactions + Predicting Acid-Base Reactions 1) List all entities present initially (ions, atoms, molecules, H2O(l)) as they exist in aqueous solution. No entity can react as a base if it is weaker than water. For this reason, the conjugate bases of the strong acids are not considered bases in aqueous solutions H3O+(aq) is the SA that can exist. If a stronger acid is dissolved, it reacts instantly and completely with water to form H3O+(aq). So all strong acids are written as H3O+(aq) OH-(aq) is the SB that can exist. If a stronger base is dissolved, it reacts instantly and completely with water to form OH-(aq). The only example of this: soluble ionic oxides, write the cation and the oxide ion is written as OH-(aq) + Predicting Acid-Base Reactions 1) List all entities present initially (ions, atoms, molecules, H2O(l)) as they exist in aqueous solution. Example: What will be the predominant reaction if spilled drain cleaner (sodium hydroxide) solution is neutralized by vinegar? List entities present: Na+(aq) OH-(aq) CH3COOH(aq) H2O(l) + Predicting Acid-Base Reactions 2) Identify and list all possible aqueous acids and bases, using the Bronsted-Lowry definitions. Use the entity lists of the Relative Strengths of Acids and Bases (in your data booklet). Amphiprotic entities are labeled fro both possibilities. Conjugate bases on SA’s are not included. Metal ions are treated as spectators. Example: What will be the predominant reaction if spilled drain cleaner (sodium hydroxide) solution is neutralized by vinegar? List entities present: Na+(aq) OH-(aq) B A CH3COOH(aq) A H2O(l) B + Predicting Acid-Base Reactions 3) Identify the strongest acid and the strongest base present, using the table of Relative Strengths of Acids and Bases. Use the order of the entities in the Relative Strengths of Acids and Bases table to identify the SA (the highest one on the table) and the SB (the lowest one on the table) Example: What will be the predominant reaction if spilled drain cleaner (sodium hydroxide) solution is neutralized by vinegar? List entities present: SA Na+(aq) OH-(aq) SB CH3COOH(aq) H2O(l) + Predicting Acid-Base Reactions 4) Write an equation showing a transfer of one proton from the SA to the SB, and predict the conjugate base and the conjugate acid to be the products. Example: What will be the predominant reaction if spilled drain cleaner (sodium hydroxide) solution is neutralized by vinegar? List entities present: Na+(aq) OH-(aq) SB SA CH3COOH(aq) H2O(l) + Predicting Acid-Base Reactions 5) Predict the approximate position of equilibrium, using the following Learning Tip Example: What will be the predominant reaction if spilled drain cleaner (sodium hydroxide) solution is neutralized by vinegar? Na+(aq) OH-(aq) SA CH3COOH(aq) SB The reaction of H3O+(aq) and OH-(aq) is always quantitative (100%) so a single arrow is used H2O(l) + Practice Pg. 731 #9-17 Tomorrow: Table Building and Thought Lab Five-step method practice + Any SA will react quantitatively with any base from water down Any SB will react quantitatively with any acid from water up If WA above WB > 50% If WA below WB <50% + For which of these reactions would Kc > 1? Kc < 1 ? + Predict % reaction for these rxs + Table Building Lab Exercise 16.D If a reaction is >50%, it favours products and the SA is above the SB. If a reaction is <50%, it favours reactants and the SA is below the SB. + Table Building Lab Exercise 16.D + Practice - Work on 5 step method questions - Quiz Tuesday + Acids and Bases – Ka 2.2K: Define Ka , Kb and use these to determine pH, pOH, [H3O+] and [OH–] of acidic and basic solutions 2.3K: Calculate equilibrium constants and concentrations for Bronsted-Lowry acid base systems when: • concentrations at equilibrium are known • initial concentrations and one equilibrium concentration are known • the equilibrium constant and one equilibrium concentration are known. + The Acid Ionization Constant, K a One important way that chemists communicate the strength of any weak acid is by using the equilibrium constant expression for the ionization of weak acids. (Ka) Look at your Relative Strengths of Acids and Bases Table: All of the strong acids have a Ka value listed as very large; remember they ionize quantitatively, so the actual acid species present is H3O+(aq) All other acids are weak and vary greatly in the extent of reaction at equilibrium. We use the equilibrium law to write the formula for Ka Ka Values have only two sig digs because they are somewhat inaccurate because we calculate them based on assumptions. We omit liquid water because we assume its value is essentially constant for dilute solutions + Ka Calculations Two common Ka calculations: Calculating Ka from empirical data (Examples #1-3) Using Ka to predict hydronium ion concentration from an initial concentration of weak acid. (Example #4 and 5) + Ka Calculations Example #1: The pH of a 1.00 mol/L solution of acetic acid is carefully measured to be 2.38 at SATP. What is the value of Ka for acetic acid? 1.00mol/L – 0.0042 mol/L = 0.9958 (rounds to 1.00 – precision rule) Change in concentration is negligible in this case – but not always Regardless of size, Ka values are usually expressed in scientific notation = 1.7 x 10-5 + Ka Calculations Example #2: A student measures the pH of a 0.25 mol/L solution of carbonic acid to be 3.48. Calculate the Ka for carbonic acid from this evidence. + Ka Calculations Example #3: The pH of a 0.400 mol/L solution of sulfurous acid is measured to be 1.17. Calculate the Ka for sulfurous acid from this evidence. According to the equilibrium law, the Ka for sulfurous acid is1.4 x 10-2 + Ka Calculations Example #4: Predict the hydronium ion concentration and pH for a 0.200 mol/L aqueous solution of methanoic acid. Approximation Rule: 0.200 = >1000 -4 1.8 x 10 So (0.200-x) = 0.200 1.8 x 10-4 = x2 (0.200) x = 0.006 = H3O+(aq) concentration + Ka Calculations Example #5: Predict the hydronium ion concentration and pH for a 0.500 mol/L aqueous solution of hydrocyanic acid. Approximation Rule: 0.500 = >1000 -10 6.2 x 10 So (0.500-x) = 0.500 6.2 x 10-10 = x2 (0.500) x = 1.8 x 10-5 = H3O+(aq) concentration + Practice Ka Text pg 743 Q1-9 (any not finished from yesterday) Read pg 744-750 + Acids and Bases – Kb 2.2K: Define Ka , Kb and use these to determine pH, pOH, [H3O+] and [OH–] of acidic and basic solutions 2.3K: Calculate equilibrium constants and concentrations for Bronsted-Lowry acid base systems when: • concentrations at equilibrium are known • initial concentrations and one equilibrium concentration are known • the equilibrium constant and one equilibrium concentration are known. + Base Strength & Ionization Constant, K Ionic hydroxides, such as NaOH(aq) or Ca(OH)2(aq), are assumed to dissociate completely upon dissolving. b Finding the hydroxide ion concentration does not involve any reaction with water Example: Find the hydroxide ion concentration of a 0.064 mol/L solution of barium hydroxide + Base Strength & Ionization Constant, K b We can communicate the strength of weak bases, with the equilibrium constant for its reaction with water. The equilibrium constant is called the base ionization constant, Kb Two common Kb calculations: Calculating Kb from empirical data (Example #1) Using Kb to predict the concentration of hydroxide ions when the initial concentration is known (Example #2) + Kb Calculations We will use the same method as Ka calculations, but there is usually one extra step because pH values need to be converted to find hydroxide ion concentrations Example #1: A student measures the pH of a 0.250 mol/L solution of aqueous ammonia and finds it to 11.32. Calculate the Kb for ammonia 14 = pH + pOH pOH = 2.68 10-2.68 = 0.0021 = OH-(aq) Remember Kb has only 2 sig digs Kb for ammonia is 1.8 x 10-5 + Calculating OH- from Kb First problem, the data booklet has Ka values not Kb values. What do we do? Kw = KaKb So Kb = Kw/Ka Example #1: Solid sodium benzoate forms a basic solution. Determine the Kb for the weak base present. remember Kw = 1.00 x 10-14 + Calculating OH- from Kb Example #2: Find the hydroxide ion amount concentration, pOH, pH and the percent reaction (ionization) of a 1.20 mol/L solution of baking soda. Baking soda = NaHCO3(s) Na+(aq) + HCO3-(aq) For HCO3-(aq), the conjugate acid is H2CO3(aq) whose Ka is = 4.5 x 10-7 Approximation Rule: 1.20 = >1000 -8 2.2 x 10 So (1.20-x) = 01.20 2.2 x 10-8 = x2 1.20 . x = 1.6 x 10-4 = OH-(aq) + Calculating OH- from Kb Example #2: Find the hydroxide ion amount concentration, pOH, pH and the percent reaction (ionization) of a 1.20 mol/L solution of baking soda. 2.2 x 10-8 = x2 1.20 . x = 1.6 x 10-4 = OH-(aq) + Amphoteric Species If an entity can react as either a Bronsted-Lowry acid or base, how do you know which will be the predominant reaction? Find the Ka value, calculate the Kb value, which ever is larger wins! Example: Which reaction predominates when NaHSO3(s) is dissolved in water to produce HSO3-(aq) solution? Will the solution be acidic or basic? Ka = 6.3 x 10-8 Kb = Kw Ka = 1.0 x 10-14 = 7.1 x 10-13 1.4 x 10-2 The Ka value far exceeds the Kb value, so an aqueous solution of this substance will be acidic because the hydrogen sulfite ion will react predominately as a Bronsted-Lowry acid. + Practice Pg. 746 #11-13 Pg. 750 #1-6 + Acids and Bases - Buffers 1.8K: Define a buffer as relatively large amounts of a weak acid or base and its conjugate in equilibrium that maintain a relatively constant pH when small amounts of acid or base are added. + Chem 20 Review: A graph showing the continuous change of pH during an acid-base titration, which continues until the titrant is in great excess, is called a pH curve Endpoint refers to the point in a titration analysis where the addition of titrant is stopped. The endpoint is defined (empirically) by the observed colour change of an indicator. Equivalence Point refers to the point in any chemical reaction where chemically equivalent amounts of the reactants have combined. This point is determined by stoichiometry. + Interpreting pH Curves Buffering: is the property of some solutions to resist any significant change in pH when an acid or base is added. Buffering region (nearly level portions of the graph) Why is it buffering? Initially, the solution is mainly all water and OH- ions. Any additional acid added (H3O+) immediately reacts with OH- to become water – which does not change the pH significantly. This “leveling effect” finally fails near the equivalence point, when the OH- is almost completely consumed. Once excess acid has been added, the solution consists of water and H3O+ ions, so the pH has dropped to the acid range. Then any additional acid that is added, simple increase the H3O+ concentration slightly, but does not change the pH much. + Titration Analysis Chem 20 Review: Selecting proper indicators Alizarin yellow is not a suitable indicator because it will change colour long before the equivalence point of this strong acid-strong base reaction, which theoretically has a pH of exactly 7. Orange IV is also unsuitable; its colour change would occur too late. The pH at the middle of the colour change range for bromothymol blue is 6.8, which very closely matches the equivalence point pH; so bromothymol blue should give accurate results. + Acid Base Indicators Any acid base indicator is really two entities for which we use the same name: a Bronsted-Lowry conjugate acid-base pair. At lease one of the entities is visibly coloured, so you can tell simply by looking when it forms or is consumed. Examples include: Phenolphthalein: conjugate acid is colourless, conjugate base is bright pink Bromothymol Blue: conjugate acid is yellow, conjugate base is blue, and when they are in equal quantities – (appear green to the human eye) Litmus Paper – red (HIn) to blue (In-) We will use the designation HIn for the conjugate acid and In- for the conjugate base as their actual formulas can be very complex. Summary: An indicator is a conjugate weak acid-weak base pair formed when an indicator dye dissolves in water. + Practice: + Polyprotic Entities Chem 20 Review: Polyprotic acids – can lose more than one proton Polyprotic bases – can gain more than one proton If more than one proton transfer occurs in a titration, chemists believe the process occurs as a series of single-proton transfer reactions. On a graph, this means there will be more than one equivalence point First proton transfer = 100% Second proton transfer = 100% Carbonate ion is a diprotic base + Polyprotic Entities A pH curve for the addition of NaOH to a sample of H3PO4(aq) displays only two rapid changes in pH even though H3PO4(aq) is triprotic. This is because only two of the transfers are quantitative. The third reaction never goes to completion, but instead establishes an equilibrium. General Rule: Only quantitative reactions produce detectable equivalence points in an acid-base titration. + pH Curves + Practice + General Rule Strong Acid to Weak Base: pH at equivalence point is always lower than 7 Strong Base to Weak Acid: pH at equivalence point is always higher than 7 + + pH Curve Shape SA-SB: water is the only acid or base present = neutral solution SA-WB: a weak acid (NH4+)is present along with water, at the equivalence point, so the solution is acidic (pH < 7) WA-SB: a weak base (CH3COO-) is present along with water, at the equivalence point, so the solution is basic (pH > 7) WA-WB: do not have detectable equivalence points, because the reactions are usually not quantitative. + Practice: Pg. 762 #11-14 + Buffers A buffer is a relatively large amount of any weak acid and its conjugate base, in the same solution. In equilibrium, they maintain a relatively constant pH when small amounts of acid or base are added. I.e. The addition of a small amount of base produces more acetate ions. The very small change in the acid-base conjugate pair ration and the complete consumption of the OH- explains why the pH change is very slight The addition of a small amount of acid produces more acetic acid. The very small change in the acid-base conjugate pair ratio and the complete consumption of the H3O+ explains why the pH change is very slight + Buffer Example: Blood Plasma Blood plasma has a remarkable buffering ability, as shown by the following table. This is very useful, as a change of more than 0.4 pH units, can be lethal. If the blood were not buffered, the acid absorbed from a glass of orange juice would likely be fatal. + Buffering Capacity The limit of the ability of a buffer to maintain a pH level. When one of the entities of the conjugate acid-base pair reacts with an added reagent and is completely consumed, the buffering fails and the pH changes dramatically. All of the CH3COOH(aq) is used up, OHadditions will now cause the pH to drastically increase All of the CH3COO-(aq) is used up, H3O+ additions will now cause the pH to drastically decrease + Practice Pg. 766 #16-21 + Modelling Dynamic Equilibrium Mini Investigation pg. 678 Volume of Cylinder #1 Volume of Cylinder #2 25.0 0.0 20.0 5.0 17.0 8.0 14.0 11.0 11.0 14.0 8.0 17.0 5.0 20.0 2.0 23.0 2.0 23.0 2.0 23.0 2.0 23.0 Assume large straw transfers 5 mL each time and the smaller straw transfers 2 mL each time