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ACIDS, BASES and SALTS For thousands of years people have known that vinegar, lemon juice and many other foods taste sour. However, it was not until a few hundred years ago that it was discovered why these things taste sour - because they are all acids. The term acid, in fact, comes from the Latin term acere, which means "sour". Acids taste sour, are corrosive to metals, change litmus (a dye extracted from lichens) red, and become less acidic when mixed with bases. Bases feel slippery, change litmus blue, and become less basic when mixed with acids. Acids react with bases to form salts. Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Arrhenius Acids: Substances that when placed in water, will dissociate to produce H+ ions: HCl(aq) → H+(aq) + Cl-(aq) 1884 Arrhenius Bases: Substances that when placed in water will dissociate to yield OH- ions: NaOH(aq) → Na+(aq) + OH-(aq) Nitric Acid - HNO3 Chloric Acid - HClO3 Perchloric Acid - HClO4 Sulfuric Acid - H2SO4 Phosphoric Acid - H3PO4 Acetic Acid - HC2H3O2 Chemistry 21A How these acids and bases dissociate? Swedish chemist Svante Arrhenius, received the Nobel Prize in Chemistry in 1903 One of the founders of the science of Physical Chemistry Potassium Hydroxide – KOH Calcium Hydroxide - Ca(OH)2 Barium Hydroxide - Ba(OH)2 Dr. Dragan Marinkovic ACIDS, BASES and SALTS the Brønsted-Lowry theory is an acid-base theory, proposed independently by Danish Johannes Nicolaus Brønsted and English Thomas Martin Lowry in 1923. In this system, an acid is defined as any chemical species (molecule or ion) that is able to lose, or "donate" a hydrogen ion (proton), and a base is a species with the ability to gain or "accept" a hydrogen ion (proton). It follows that if a compound is to behave as an acid, donating a proton, there must be a base to accept the proton. So the Brønsted–Lowry concept can be defined by the reaction: acid + base conjugate base + conjugate acid CH3CO2H + H2O CH3CO2- + H3O+ H2O + NH3 OH- + NH4+ There is strong evidence that the hydrogen ion is never found free as H+. The bare proton is so strongly attracted by the electrons of surrounding water molecules that H30+ forms immediately. Brønsted Acids: Any substance that can transfer a proton (H+) to another substance Brønsted Bases: Any substance that can accept a proton (H+) from another substance Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Hydrochloric Acid - HCl Chloric Acid - HClO3 Perchloric Acid - HClO4 Sulfuric Acid - H2SO4 Phosphoric Acid - H3PO4 Acetic Acid - HC2H3O2 Chemistry 21A Show acid/base conjugate pairs. Potassium Hydroxide – KOH Calcium Hydroxide - Ca(OH)2 Barium Hydroxide - Ba(OH)2 Dr. Dragan Marinkovic ACIDS, BASES and SALTS AMPHIPROTIC a compound or ion that can either donate or accept H+ ions, i.e. can act both as an acid and as a base H2O, HSO4- , HPO42-, HSO3- etc. HSO4- + H3O+ H2SO4 + H2O HSO4- + OH- SO42- + H2O CH3CO2H + H2 O H2O + NH3 Chemistry 21A CH3CO2OH- + + H 3O+ NH4+ Dr. Dragan Marinkovic ACIDS, BASES and SALTS Classical acid (and salt) naming system: Anion (Salt) Prefix Anion (Salt) Suffix Acid Prefix Acid Suffix Example per ate per hypo ic acid perchloric acid (HClO4) ate ic acid chloric acid (HClO3) ite ous acid chlorous acid (HClO2) ite hypo ous acid hypochlorous acid (HClO) ide hydro ic acid hydrochloric acid (HCl) sulf nitr phosph carbon brom iod Chemistry 21A sulfur nitr phosphor carbon brom iod Dr. Dragan Marinkovic ACIDS, BASES and SALTS fertilizers, explosives Nitric Acid - HNO3 Strong Acids Nitrous Acid - HNO2 Hydrohalic acids: HCl, HBr, HI Hypochlorous Acid - HClO bleach Nitric acid: HNO3 Chlorous Acid - HClO2 Sulfuric acid: H2SO4 Chloric Acid - HClO3 Perchloric acid: HClO4 Perchloric Acid - HClO4 Sulfuric Acid - H2SO4 car batteries Sulfurous Acid - H2SO3 involved metabolism such as adenosine diphosphate Phosphoric Acid - H3PO4 (ADP) and triphosphate (ATP); DNA, RNA Phosphorous Acid - H3PO3 soft drinks, seltzer water Carbonic Acid - H2CO3 Acetic Acid - HC2H3O2 vinegar, pickles Oxalic Acid - H2C2O4 kidney and bladder stones Boric Acid - H3BO3 treatment of skin irritations; insecticide Silicic Acid - H2SiO3 orthosilicic acid H4SiO4, is the form predominantly absorbed by humans and is found in numerous tissues including bone, tendons, aorta, liver and kidney. Chemistry 21A Hydrofluoric Acid - HF glass etching Hydrochloric Acid - HCl stomach acid Hydrobromic Acid - HBr Hydroiodic Acid - HI Hydrosulfuric Acid - H2S rotten eggs smell Dr. Dragan Marinkovic ACIDS, BASES and SALTS SULFURIC ACID World production in 2001 was 165 million tonnes, Hot concentrated sulfuric acid is an oxidizing agent with an approximate value of US$8 billion. Fe(s) + 2H2SO4(conc) FeSO4(aq) + 2H2O(l) + SO2(g) Concentrated sulfuric acid is very good at removing the water from sugars. C12H22O11(s) + nH2SO4 (l) 12C(s) + 11H2O (l) + nH2SO4(l) Making hydrogen peroxide (H2O2) by reacting barium peroxide with sulfuric acid. BaO2(s) + H2SO4(aq) Chemistry 21A BaSO4(s) + H2O2(aq) Dr. Dragan Marinkovic ACIDS, BASES and SALTS HO Some important organic acids: HO O C H C C H H2 HO O OH ascorbic acid, vitamin C Scurvy is a disease resulting from a deficiency of vitamin C, which is required for the synthesis of collagen in humans CH3COOH acetic acid OH O CH3CHOHCOOH HC citric acid O lactic acid HO (milk acid) O OH O C H2 C H2 HO Chemistry 21A CH3 O HC C H CH O acetylsalicylic acid OH non-steroidal anti-inflammatory drug (NSAID) Dr. Dragan Marinkovic ACIDS, BASES and SALTS glycine an α-amino acid, with the amino group on the left and the carboxyl group on the right L- and D-alanine MONOSODIUM GLUTAMATE (MSG) as a food ingredient has been the subject of health studies. A report from the Federation of American Societies for Experimental Biology (FASEB) compiled in 1995 on behalf of the FDA concluded that MSG was safe for most people when "eaten at customary levels. Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS glycylglycine glycine glycine When two amino acids react (“head-to-tail”) they form a peptide bond (in reaction between the acid of one molecule and amine of another molecule). Thus, a PEPTIDE (bond) is formed, A polypeptide is a chain of amino acids. Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Bases lye, oven and drain cleaner Sodium Hydroxide - NaOH Potassium Hydroxide - KOH glass cleaner Ammonium Hydroxide - NH4OH caustic lime, mortar, plaster Calcium Hydroxide - Ca(OH) 2 laxatives, antacids Magnesium Hydroxide - Mg(OH)2 Barium Hydroxide - Ba(OH)2 antacids, deodorants Aluminum Hydroxide - Al(OH)3 Ferrous Hydroxide or Iron (II) Hydroxide - Fe(OH)2 Ferric Hydroxide or Iron (III) Hydroxide - Fe(OH)3 an absorbent in surgical dressings Zinc Hydroxide - Zn(OH)2 Lithium Hydroxide - LiOH Strong bases Sodium Hydroxide - NaOH Potassium Hydroxide – KOH Calcium Hydroxide - Ca(OH)2 Barium Hydroxide - Ba(OH)2 Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS CH3NH2 methylamine CH3CH2NH2 ethylamine atropine Injections of ATROPINE are used in the treatment of bradychardia (an extremely the first effective treatment for malaria low heart rate) Atropine occurs in the deadly All organic bases (like nightshade plant (Atropa belladonna) inorganic ones) react with acids to form salts. VINCRISTINE, one of the most potent ANTILEUKEMIC DRUGS in use today, was isolated in a search for diabetes treatments from Vinca rosea (now Catharanthus roseus) in the 1950's Chemistry 21A Morphine, C17H19NO3, is the most abundant of opium’s 24 alkaloids, accounting for 9 to 14% morphine of opium-extract by mass. Named after the Roman god of dreams, Morpheus. Dr. Dragan Marinkovic ACIDS, BASES and SALTS 2H2O Chemistry 21A H3O+ + OH- Dr. Dragan Marinkovic ACIDS, BASES and SALTS 2H2O H3O+ + OH- Keqx[H2O]2 = [H3O+][OH-] [H2O] = const. = 55.5 M Keqx[H2O]2 = Kw KW = const. at 25oC Kw = [H3O+][OH-] = 10-14 Chemistry 21A The product of water Dr. Dragan Marinkovic ACIDS, BASES and SALTS 2H2O H3O+ + OH- Keqx[H2O]2 = [H3O+][OH-] The product of water if [H3O+] = [OH-] = 10-7 THE SOLUTION IS NEUTRAL Chemistry 21A T (°C) 0 5 20 70 100 at 25oC Kw = [H3O+][OH-] = 10-14 Kw 0.114 × 10-14 0.186 × 10-14 0.681 × 10-14 15.85 × 10-14 51.3 × 10-14 12.29 pKw 14.94 14.73 14.17 12.80 6.14 neutral pH 7.47 7.37 7.08 6.40 Dr. Dragan Marinkovic ACIDS, BASES and SALTS H3O+ + OH- 2H2O Keqx[H2 O]2 = [H3 O+][OH-] [H2O+] = const. = 55.5 M Keqx[H2O]2 = Kw at 25oC Kw = [H3 [H3 O+] = = [OH-] 10-14 = For convenience instead of exponential numbers, negative logarithms of these numbers are used. pH = -log[H+] or -log[H3O+] IN THE NEUTRAL SOLUTION at 25oC pH = -log [1 x 10-7] = -(-7.00) = 7.00 KW = const. O+][OH-] THE pH CONCEPT 10-7 Chemistry 21A [H3O+] = 10-pH pOH = -log[OH-] Dr. Dragan Marinkovic [H+] pH Example 1 x 100 0 HCl 1 x 10-1 1 Stomach acid 1 x 10-2 Acids (acidic 1 x 10-3 Solutions) 1 x 10-4 2 Lemon juice 3 Vinegar 4 Soda (Coca-Cola) 1 x 10-5 5 Rainwater 1 x 10-6 6 Milk 1 x 10-7 7 Pure water 1 x 10-8 8 Egg whites 1 x 10-9 9 Baking soda 1 x 10-10 10 Tums® antacid Neutral Bases (basic 1 x 10-11 Solutions) 1 x 10-12 11 Ammonia 12 Mineral lime - Ca(OH)2 1 x 10-13 13 Drano® 1 x 10-14 14 NaOH Chemistry 21A ACIDS, BASES and SALTS pH = -log [H+] Note: concentration is commonly abbreviated by using square brackets, thus [H+] = hydrogen ion concentration. When measuring pH, [H+] is in units of moles of H+ per liter of solution. The pH of blood is maintained within the narrow range of 7.35 to 7.45. Normal urine pH averages about 6.0. Saliva has a pH between 6.0 and 7.4. Tear pH was measured in 44 normal subjects. The normal pH range was 6.5 to 7.6; the mean value was 7.0. Dr. Dragan Marinkovic ACIDS, BASES and SALTS pH in living systems Chemistry 21A Compartment pH Gastric acid 0.7 Lisosomes 4.5 Granules of chromaffin cells 5.5 Urine 6.0 Neutral H2O at 37 °C 6.81 Cytosol 7.2 Cerebrospinal fluid (CSF) 7.3 Blood 7.34 – 7.45 Mitochondrial matrix 7.5 Pancreas secretions 8.1 Dr. Dragan Marinkovic ACIDS, BASES and SALTS Calculate the molar concentration of H3O+ in water solutions with the following OH- molar concentrations: a) 6.9x10-5 b) 0.074 Calculate the molar concentration of OH- in water solutions with the following H3O+ molar concentrations: a) 0.0087 b) 9.9x10-10 Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Calculate the molar concentration of H3O+ in water solutions with the following OH- molar concentrations: a) 6.9x10-5 b) 0.074 Kw = [H3O+][OH-] = 10-14 [H3O+] = 10-14/[OH-] = 10-14/[6.9x10-5] = 1.449x10-10 M [H3O+] = 10-14/[OH-] = 10-14/[7.4x10-2] = 1.35x10-13 M Calculate the molar concentration of OH- in water solutions with the following H3O+ molar concentrations: a) 0.0087 b) 9.9x10-10 Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Calculate the molar concentration of H3O+ in water solutions with the following OH- molar concentrations: a) 6.9x10-5 b) 0.074 Kw = [H3O+][OH-] = 10-14 [H3O+] = 10-14/[OH-] = 10-14/[6.9x10-5] = 1.449x10-10 M [H3O+] = 10-14/[OH-] = 10-14/[7.4x10-2] = 1.35x10-13 M Calculate the molar concentration of OH- in water solutions with the following OH- H3O+ molar concentrations: a) 0.0087 b) 9.9x10-10 [OH-] = 10-14 /[H3O+] = 10-14/[8.7x10-3] = 1.15x10-12 M [OH-] = 10-14 /[H3O+] = 10-14/[9.9x10-10] = 1x10-5 M Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Determine the pH of water solutions with the following characteristics. Classify each solution as acidic, basic or neutral. Convert the following pH values in both [H+] and [OH-] values. Chemistry 21A a) [H+] = 7.5x10-6 b) [OH-] = 2.5x10-4 c) [OH-] = 8.6x10-10 a) pH = 3.95 b) pH = 4.00 c) pH = 11.86 Dr. Dragan Marinkovic ACIDS, BASES and SALTS Determine the pH of water solutions with the following characteristics. Classify each solution as acidic, basic or neutral. a) [H+] = 7.5x10-6 b) [OH-] = 2.5x10-4 c) [OH-] = 8.6x10-10 pH = -log [H+] pOH = -log [OH-] pH + pOH = 14 a) pH = -log [H+] = -log(7.5x10-6) = 5.12 b) pOH = -log [OH-] = -log(2.5x10-4) = 3.6 Convert the following pH values in both [H+] and [OH-] values. Chemistry 21A pH = 14 - pOH = 14 - 3.6 = 10.4 a) pH = 3.95 b) pH = 4.00 c) pH = 11.86 Dr. Dragan Marinkovic ACIDS, BASES and SALTS Determine the pH of water solutions with the following characteristics. Classify each solution as acidic, basic or neutral. a) [H+] = 7.5x10-6 b) [OH-] = 2.5x10-4 c) [OH-] = 8.6x10-10 pH = -log [H+] pOH = -log [OH-] pH + pOH = 14 pH = -log [H+] = -log(7.5x10-6) = 5.12 pOH = -log [OH-] = -log(2.5x10-4) = 3.6 Convert the following pH values in both [H+] and [OH-] values. a) pH = 3.95 b) pH = 4.00 c) pH = 11.86 [H3O+] = 10-pH = 10-3.95 = 1.12x10-4 Chemistry 21A pH = 14 - pOH = 14 - 3.6 = 10.4 [H3O+] = 10-pH [OH-] = 10-pOH [OH-] = 10-pOH = 10-10.05 = 8.91x10-11 Dr. Dragan Marinkovic ACIDS, BASES and SALTS pH meter Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Properties of Acids Types of Acids Monoprotic - a solution that produces one mole of H+ ions per mole of acid HCl , HNO3 Diprotic - a solution that produces two moles of H+ ions per mole of acid H2SO4 Triprotic - a solution that produces three moles of H+ ions per mole of acid H3PO4 Polyprotic - two ore more H+ per mole of acid STRONG ACIDS Acids that are essentially 100% Ionized in aqueous solutions Hydrohalic acids: HCl, HBr, HI Nitric acid: HNO3 Sulfuric acid: H2SO4 Perchloric acid: HClO4 Chemistry 21A WEAK ACIDS Acids that are partially ionized (usually less than 5%) in equilibrium. Nitrous Acid - HNO2 Sulfurous Acid - H2SO3 Phosphorous Acid - H3PO3 Carbonic Acid - H2CO3 Acetic Acid - HC2H3O2 Boric Acid - H3BO3 Silicic Acid - H2SiO3 Dr. Dragan Marinkovic ACIDS, BASES and SALTS Properties of Acids Acids react with: Metals Metal oxides Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) FeO(s) + 2HCl(aq) → FeCl2(aq) + H2O(l) Hydroxides/bases Al(OH)3(s) + 3CH3COOH(aq) → Al(CH3COO)3(aq) + 3H2O(l) Carbonates 3CaCO3(s) + 2H3PO4(aq) → Ca3(PO4)2(aq) + 3CO2(g) + 3H2O(l) Bicarbonates KHCO3(aq) + HNO3(aq) → KNO3(aq) + CO2(g) + H2O(l) Strong acids react with salts of weak acids Na3BO3(aq) + 3HBr(aq) → H3BO3(aq) + 3NaBr(aq) The major products of al these reactions (metal compounds with acids) are SALTS. Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Properties of Acids Making dilutions from stock solutions: If we have 1 L of 3 M HCl, what is M if we dilute acid to 6 L? M1 = 3 mol/L, V1 = 1 L, V2 = 6 L M1V1 = M2V2 M1V1/V2 = M2 M2 = (3 mol/L x 1 L) / (6 L) = 0.5 M Why does the formula work? Because we are equating mol to mol: M1V1 = 3 mol M2V2 = 3 mol Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Properties of Acids What volume of 0.5 M HCl can be prepared from 1 L of 12 M HCl? M1 = 12 mol/L, V1 = 1 L, M2 = 0.5 L M1V1 = M2V2 M1V1/M2 = V2 V2 = (12 mol/L x 1 L) / (0.5 L) = 24 L Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Properties of Acids 1. How many mL of a 14 M stock solution must be used to make 250 mL of a 1.75 M solution? 2. You have 200 mL of 6.0 M HF. What concentration results if this is diluted to a total volume of 1 L? 3. 100 mL of 6.0 M CuSO4 must be diluted to what final volume so that the resulting solution is 1.5 M? 4. What concentration results from mixing 400 mL of 2.0 M HCl with 600 mL of 3.0 M HCl? 5. What is the concentration of NaCl when 3 L of 0.5 M NaCl are mixed with 2 L of 0.2 M NaCl? 6. What is the concentration of NaCl when 3 L of 0.5 M NaCl are mixed with 2 L of water? 7. Water is added to 4 L of 6 M antifreeze until it is 1.5 M. What is the total volume of the new solution? 8. There are 3 L of 0.2 M HF. 1.7 L of this is poured out, what is the concentration of the remaining HF? Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Properties of Bases Slimy or soapy feel on fingers, due to saponification of the lipids in human skin. Concentrated or strong bases are caustic (corrosive) on organic matter and react violently with acidic substances Strong Bases NaOH - Sodium Hydroxide KOH - Potassium Hydroxide Ca(OH)2 - Calcium Hydroxide Ba(OH)2 - Barium Hydroxide NaOH(s) + H2O(l) → Na+(aq) + OH-(aq) Weak Bases Aluminum Hydroxide - Al(OH)3 Iron (II) Hydroxide - Fe(OH)2 Iron (III) Hydroxide - Fe(OH)3 Zinc Hydroxide - Zn(OH)2 Bases react with: Acids Al(OH)3(s) + 3CH3COOH(aq) → Al(CH3COO)3(aq) + 3H2O(l) “Acidic oxides” 2NaOH(aq) + CO2(g) → Na2CO3(aq) + H2O(l) Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Salts In medicine, saline (also saline solution) is a general term referring to a sterile solution of sodium chloride (table salt) in water. It is used for intravenous infusion, rinsing contact lenses, and nasal irrigation. In medicine, normal saline (NS) is the commonly-used term for a solution of 0.91% w/v of NaCl, about 300 mOsm/L. Less commonly, this solution is referred to as physiological saline or isotonic saline, Saline solution for intravenous infusion. The white port at the base of the bag is where additives can be injected with a hypodermic needle. The port with the blue cover is where the bag is spiked with an infusion set. Chemistry 21A NaCl(s) + H2O(l) → Na+(aq) + OH-(aq) NaOH HCl Dr. Dragan Marinkovic ACIDS, BASES and SALTS Salts According to chemistry, the term "salt" is used for ionic compounds that is composed of positively charged cations (usually metal or ammonium ions) and the negatively charged anions, so that the product remains neutral and without a net charge. The anions may be inorganic (Cl-) as well as organic (CH3COO-) and monoatomic (F-) as well as polyatomic ions (SO42-). Salt's solution in water is called electrolytes. Both, the electrolytes and molten salts conduct electricity. Salts with OH- are basic salts (CaOHCl, BaOHNO3) and with H+ are acidic salts (NaHSO4). Usually salts are solid crystals having high melting point. Taste - It differes from salt to salt. It can elicit all the five basic tastes, like salty (sodium chloride), sweet (lead diacetate very toxic!), sour (potassium bitartrate), bitter (magnesium sulfate), and umami or savory (monosodium glutamate). Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Salts Silver electroplating. The anode is a silver bar and the cathode is an iron spoon. anode cathode + Anode: The positive terminal of an electrical current flow. Cathode: The negative terminal of an electric current system. - Diagram of a Hoffman voltameter used for the electrolysis of water to produce hydrogen and oxygen gases 2H2O → 2H2 + O2 Cathode (reduction): 2H+(aq) + 2e− → H2(g) Anode (oxidation): 2H2O(l) → O2(g) + 4H+(aq) + 4e Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Salts Anode: The positive terminal of an electrical current flow. Anions – negative ions – in aqueous solutions move towards ANODE, e.g. Cl-, NO3-, SO42CATIONS – positive (usually metal) ions, in aqueous solutions move towards CATHODE, K+, Al3+, Ba2+ Cathode: The negative terminal of an electric current system. EQUIVALENT OF SALT is the amount that will produce 1 mol of positive electrical charge when dissolved and dissociated. Number of salt equivalents in 1 L of 1 M of MgCl2 is 2(+) x 1 M = 2 eq Chemistry 21A Number of salt equivalents in 3.12x10-2 mol of Fe(NO3)3 is 3(+) x 3.12x10-2 = 9.36x10-2 eq = 93.6 meq Dr. Dragan Marinkovic ACIDS, BASES and SALTS Salts When crysralluized from aqueous solutions many salts crystallise as hydrates: CuSO4•5H2O - copper (II) sulfate pentahydrate CoCI2•6H2O - cobalt (II) chloride hexahydrate SnCl2•2H2O - stannous (tin II) chloride dihydrate When heated, these salts lose their crystalline water and become “anhydrous salts”. HYDRATE is a salt containing specific numbers of water molcules as part of solid crystalline structure. CuSO4•5H2O CoCI2•6H2O CuSO4 WATER OF HYDRATION is water retained as part of the solid crystalline structure of some salts. CoCI2 Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Salts Metal + nonmetal Syntheses of salts – reactions yielding salts Metals Metal oxides Fe(s) + S(s) → FeS(s) Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) FeO(s) + 2HCl(aq) → FeCl2(aq) + H2O(l) Hydroxides/bases Al(OH)3(s) + 3CH3COOH(aq) → Al(CH3COO)3(aq) + 3H2O(l) Carbonates 3CaCO3(s) + 2H3PO4(aq) → Ca3(PO4)2(aq) + 3CO2(g) + 3H2O(l) Bicarbonates KHCO3(aq) + HNO3(aq) → KNO3(aq) + CO2(g) + H2O(l) Strong acids react with salts of weak acids Na3BO3(aq) + 3HBr(aq) → H3BO3(aq) + 3NaBr(aq) Two soluble salts when mixed forming a new insoluble salt BaCl2(aq) + K2SO4(aq) → BaSO4(s) + 2KCl(aq) “Acidic oxides” 2NaOH(aq) + CO2(g) → Na2CO3(aq) + H2O(l) Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS The strength of Acids and Bases Types of Acids Monoprotic - a solution that produces one mole of H+ ions per mole of acid HCl , HNO3 Diprotic - a solution that produces two moles of H+ ions per mole of acid H2SO4 Triprotic - a solution that produces three moles of H+ ions per mole of acid H3PO4 Polyprotic - two ore more H+ per mole of acid WEAK ACIDS Acids that are partially ionized (usually less than 5%) in equilibrium. STRONG ACIDS Acids that are essentially 100% Ionized in aqueous solutions Hydrohalic acids: HCl, HBr, HI Nitric acid: HNO3 Sulfuric acid: H2SO4 Perchloric acid: HClO4 Chemistry 21A % dissociation moderately weak 6.7 Nitrous Acid - HNO 2 34 Sulfurous Acid - H2SO3 Phosphorous Acid - H3PO3 0.2 Carbonic Acid - H2CO3 1.3 Acetic Acid - HC2H3O2 0.01 Boric Acid - H3BO3 Silicic Acid - H2SiO3 Dr. Dragan Marinkovic ACIDS, BASES and SALTS The strength of Acids and Bases ACID DISSOCIATION CONSTANT The equilibrium constant for the dissociation of an acid. HA + H2O A− + H3O+ CH3COOH + H2O CH3COO- + H3O+ n all but the most concentrated solutions it can be assumed that the concentration of water, [H2O], is constant, approximately 55 mol·dm−3. On dividing K by the constant terms and writing [H+] for the concentration of the hydronium ion the expression Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS The strength of Acids and Bases Polyprotic acids are acids that can lose more than one proton. The constant for dissociation of the first proton may be denoted as Ka1 and the constants for dissociation of successive protons as Ka2, etc. Phosphoric acid, H3PO4, is an example of a polyprotic acid as it can lose three protons. equilibrium H3PO4 H2PO4− + H+ H2PO4− HPO42− + H+ HPO42− Chemistry 21A PO43− + H+ Ka1 Ka2 Ka3 Dr. Dragan Marinkovic ACIDS, BASES and SALTS The strength of Acids and Bases Slimy or soapy feel on fingers, due to saponification of the lipids in human skin. Concentrated or strong bases are caustic (corrosive) on organic matter and react violently with acidic substances NaOH(s) + H2O(l) → Na+(aq) + OH-(aq) Strong Bases NaOH - Sodium Hydroxide KOH - Potassium Hydroxide Ca(OH)2 - Calcium Hydroxide Ba(OH)2 - Barium Hydroxide Chemistry 21A Weak Bases Aluminum Hydroxide - Al(OH)3 Iron (II) Hydroxide - Fe(OH)2 Iron (III) Hydroxide - Fe(OH)3 Zinc Hydroxide - Zn(OH)2 Dr. Dragan Marinkovic ACIDS, BASES and SALTS The strength of Acids and Bases NH3(aq) + H2O BASE DISSOCIATION CONSTANT The equilibrium constant for the dissociation of a base. NaOH(aq) NH4+(aq) + OH-(aq) Na+(aq) + OH-(aq) B + H 2O HB+ + OH− Using similar reasoning to that used before In water, the concentration of the hydroxide ion, [OH−], is related to the concentration of the hydrogen ion by Kw = [H+][OH−], therefore Substitution of the expression for [OH−] into the expression for Kb give Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Analyzing Acids and Bases Titration is a common laboratory method of quantitative chemical analysis that is used to determine the unknown concentration of a known reactant. Because volume measurements play a key role in titration, it is also known as volumetric analysis The EQUIVALENCE POINT, or STOICHIOMETRIC POINT, of a chemical reaction occurs during a chemical titration when the amount of titrant added is equivalent, or equal, to the amount of analyte present in the sample. known volume of unknown concentration (of acid) Titration curve of a strong base titrating a strong acid Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Analyzing Acids and Bases The volume of titrant added from the buret is measured. For our example, lets assume that 18.3 mL of 0.115 M NaOH has been added to 25.00 mL of nitric acid solution. The following setup shows how the molarity of the nitric acid solution can be calculated from this data. = or 0.0842 M HNO3 ENDPOINT - The volume or amount of acid or base added to a solution to neutralize the unknown solution during a titration. When using an indicator, the endpoint occurs when enough titrant has been added to change the color of the indicator. Titration curve of a strong base titrating a strong acid Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Analyzing Acids and Bases pH meter ENDPOINT - The volume or amount of acid or base added to a To determine endpoint indicators can be used solution to neutralize the unknown (paper or soluble indicator dyes) or a pH meter solution during a titration. Indicator Low pH color Transition pH range High pH color Gentian violet (Methyl violet) yellow 0.0–2.0 blue-violet Thymol blue (first transition) red 1.2–2.8 yellow Thymol blue (second transition) yellow 8.0–9.6 blue Methyl orange red 3.1–4.4 orange Bromocresol purple yellow 5.2–6.8 purple Bromothymol blue yellow 6.0–7.6 blue Phenol red yellow 6.8–8.4 red Cresol Red yellow 7.2–8.8 reddish-purple Phenolphthalein colorless 8.3–10.0 fuchsia Alizarine Yellow R yellow 10.2–12.0 red Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Titration Calculations HNO3(aq) + NaOH(aq) NaNO3(aq) + H2O(l) acid/base molar ratio 1 : 1 H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l) 1 : 2 H3PO4(aq) + 3NaOH(aq) Na3PO4(aq) + 3H2O(l) 1 : 3 M1V1 = M2V2 30 mL of 0.10 M NaOH neutralized 25.0 mL of hydrochloric acid. Determine the concentration of the acid NaOH(aq) + HCl(aq) -----> NaCl(aq) + H2O(l) 0.03 L x 0.1 mol/L = 0.025 L x M2 M2 = 0.003 mol/0.035 L = 0.12 mol/L Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Titration Calculations 50 mL of 0.2 mol L-1 NaOH neutralized 20 mL of sulfuric acid. Determine the concentration of the acid 1.Write the balanced chemical equation for the reaction NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2H2O(l) 2.Extract the relevant information from the question: NaOH V = 50mL, M = 0.2M H2SO4 V = 20mL, M = ? 3.Check the data for consistency NaOH V = 50 x 10-3L, M = 0.2M H2SO4 V = 20 x 10-3L, M = ? 4.Calculate moles NaOH n(NaOH) = M x V = 0.2 x 50 x 10-3 = 0.01 mol 5.From the balanced chemical equation find the mole ratio NaOH:H2SO4 2:1 6.Find moles H2SO4 NaOH: H2SO4 is 2:1 So n(H2SO4) = ½ x n(NaOH) = ½ x 0.01 = 5 x 10-3 moles H2SO4 at the equivalence point 7.Calculate concentration of H2SO4: M = n ÷ V n = 5 x 10-3 mol, V = 20 x 10-3L M(H2SO4) = 5 x 10-3 ÷ 20 x 10-3 = 0.25M or 0.25 mol L-1 Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Hydrolysis Reactions of Salts Hydrolysis is a chemical reaction with water. CH3COONa(aq) + H2O(l) Salts of weak acids and/or bases hydrolyze in aqueous solutions. CH3COOH(aq) + Na+(aq) + OH-(aq) basic pH NH4Cl(aq) + H2O(l) NH4OH(aq) + H3O+(aq) + Cl-(aq) acidic pH Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Buffers BUFFER A solution with the ability to resist changing pH when acids (H+) or bases (OH-) are added. BUFFERS usually consist of a pair of compounds one of which has the ability to react with H+ and the other with the ability to react with OH-. CH3COOH(aq) + OH-(aq) CH3COO-(aq) + H+(aq) CH3COO-(aq) + H2O(l) CH3COOH(aq) This is how a buffer solution resists changes ion pH Chemistry 21A BUFFER CAPACITY the amount of acid (H+) that can be absorbed by a buffer without causing a significant change in pH. Dr. Dragan Marinkovic ACIDS, BASES and SALTS Buffers Antacids create buffered solutions In blood plasma, the carbonic acid and hydrogen carbonate ion equilibrium buffers the pH. In this buffer, carbonic acid (H2CO3) is the hydrogen-ion donor (acid) and hydrogen carbonate ion (HCO3-) is the hydrogen-ion acceptor (base). H2CO3(aq) Al(OH)3 Wyeth amphojel tablets of aluminum hydroxide CaCO3 Chemistry 21A H+(aq) + HCO3-(aq) Additional H+ is consumed by HCO3and additional OH- is consumed by H2CO3. The value of Ka for this equilibrium is 7.9 × 10-7. The pH of arterial blood plasma is 7.40. If the pH falls below this normal value, a condition called acidosis is produced. If the pH rises above the normal value, he condition is called alkalosis. Dr. Dragan Marinkovic ACIDS, BASES and SALTS Buffers The equilibrium constant will be: In 1 L of solution, there are going to be about 55 moles of water. Kc (a constant) times the concentration of water (another constant) on the left-hand side. The product of those is then given the name Ka. An introduction to pKa pKa bears exactly the same relationship to Ka as pH does to the hydrogen ion concentration: The higher the value for pKa, the weaker the acid. Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Buffers acid Ka (mol dm-3) pKa hydrofluoric acid 5.6 x 10-4 3.3 formic acid 1.6 x 10-4 3.8 acetic acid 1.7 x 10-5 4.8 hydrogen sulfide 8.9 x 10-8 7.1 Phosphoric acid, H3PO4, is an example of a polyprotic acid as it can lose three protons. equilibrium H3PO4 Chemistry 21A H2PO4− + H+ pKa value pKa1 = 2.15 H2PO4− HPO42− + H+ pKa2 = 7.20 HPO42− PO43− + H+ pKa3 = 12.37 Dr. Dragan Marinkovic ACIDS, BASES and SALTS Buffers [base] pKa = pH - log ------------- [acid] pH = -log[H+] or -log[H3O+] same equation Henderson-Hasselbalch equation Relationship between the pH, pKa and the concentrations of acid and base in the buffer. Chemistry 21A [base] pH = pKa + log ------------- [acid] Dr. Dragan Marinkovic ACIDS, BASES and SALTS Buffers If we need to make a buffer solution of a certain pH, we would usually select an acid with the pKa near the desired pH and then adjust the concentration of the acid and the conjugate base (the anion of the acid) to give the desired pH. We can assume that the amount of acid that dissociates is very small and can be neglected. This means that the buffer concentration of the acid and the anion are “equal” to made-up concentrations. Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Buffers Calculate the pH of buffers that contain the acid and conjugate base in following concentrations: a) [HPO42-] = 0.33 M, [PO43-] = 0.52 M pH = pKa + log{[PO43-]/[PO42-]} = 12.66 + log(0.52 M/0.33 M) = 12.66 + 0.20 = 12.86 b) [CH3COOH] = 0.40 M, [CH3COO-] = 0.25 pH = pKa + log[CH3COO-]/[CH3COOH] = 4.74 + log(0.25 M/0.40 M) = 4.74 - 0.2 = 4.54 What ratio of concentrations of NaH2PO4 and Na2HPO4 in solution would give a buffer with pH = 7.65? pH = pKa + log{[PO42-]/[PO4-]} 7.65 = 7.21 + log{[PO42-]/[PO4-]} Chemistry 21A log{[PO42-]/[PO4-]} = 0.44 [PO42-]/[PO4-] = 100.44 = 2.75 Dr. Dragan Marinkovic ACIDS, BASES and SALTS Buffers 1.00 mol of ammonium chloride is added to 2.00 L of 1.00 mol.L-1 NH3(aq) (Kb = 1.8 x 10-5 mol.L-1) a. Calculate the pH of the solution. b. 10.00 mL of the solution is placed into a 1000 mL volumetric flask and made up to the mark with pure water. What is the pH of this solution? [acid] pOH = pKb + log ------------- [base] NH4+(aq) + H2O(l) H3O+(aq) + NH3(aq) [acid] = [NH4+] = 1 mol/2 L = 0.5 mol/L [base] = [NH3] = [NH4OH] = 1 mol/L pOH = -log(1.8x10-5mol/L) + log(0.5/1 molL-1) pOH = 4.74 - 0.30 = 4.44 pH = 14 - pOH = 9.56 Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Buffers 1.00 mol of ammonium chloride is added to 2.00 L of 1.00 mol.L-1 NH3(aq) (Kb = 1.8 x 10-5 mol.L-1) a. Calculate the pH of the solution. b. 10.00 mL of the solution is placed into a 1000 mL volumetric flask and made up to the mark with pure water. What is the pH of this solution? b. Buffer solution diluted 1:100 with pure water, ratio of conjugate acid and base remains unchanged. Therefore pH is unchanged. pH = 9.56 Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Buffers Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Buffers Chemistry 21A Dr. Dragan Marinkovic ACIDS, BASES and SALTS Buffers 1.00 mol of ammonium chloride is added to 2.00 L of 1.00 mol.L-1 NH3(aq) (Kb = 1.8 x 10-5 mol.L-1) a. Calculate the pH of the solution. b. 10.00 mL of the solution is placed into a 1000 mL volumetric flask and made up to the mark with pure water. What is the pH of this solution? c. 10.0 mL of 0.100 mol.L-1 HCl(aq) is added to 100 mL of each of the above solutions. What is the final pH in each case? d. 1.00 mL of 0.100 mol.L-1 NaOH(aq) is added to fresh 100 mL portions of each solution. What is the final pH in each case? [NH4+] = 1.00 mol/2.00 L = 0.500 mol.L-1, [NH3] = 1.00 mol.L-1 (given). neutralization reaction: HCl(aq) + NH3(aq) ==> NH4+(aq) + Cl-(aq) (complete) i.e. 0.0010 mol of NH3 will be converted to [NH4ˆ+] 100 mL of solution contain: amount of NH4+ = 0.500 mol.L-1 x 0.100 L = 0.0500 mol NH4+(aq) + H2O(l) <===> H3O+(aq) + NH3(aq) amount of NH3 = 1.00 mol.L-1 x 0.100 L = 0.100 mol init/mol 0.0500 0.100 ch/mol amount of HCl = 0.100 mol.L-1 x 0.010 L = 0.0010 mol +0.0010 -0.0010 0.0510 final/mol 0.099 pH = pKa - log10(a/b) = -log10{Kw/Kb} - log10{[NH4+]/[NH3]} = 9.26 - log10{0.0051/0.099} = 9.26 + 0.29 => pH = 9.55 Chemistry 21A 14 - 4.74 = 9.26 Dr. Dragan Marinkovic ACIDS, BASES and SALTS Buffers 1.00 mol of ammonium chloride is added to 2.00 L of 1.00 mol.L-1 NH3(aq) (Kb = 1.8 x 10-5 mol.L-1) a. Calculate the pH of the solution. b. 10.00 mL of the solution is placed into a 1000 mL volumetric flask and made up to the mark with pure water. What is the pH of this solution? c. 10.0 mL of 0.100 mol.L-1 HCl(aq) is added to 100 mL of each of the above solutions. What is the final pH in each case? d. 1.00 mL of 0.100 mol.L-1 NaOH(aq) is added to fresh 100 mL portions of each solution. What is the final pH in each case? solution b. We are adding the same amount of HCl to a solution which is 100 times more dilute, i.e. contains 0.0100 times the amounts of NH4+ & NH3: amount of NH4+ = 0.0500 mol x 0.00100 = 0.000500 mol The amounts of HCl and NH3 are equal; we no longer have a buffer, but exact amount of NH3 = 0.100 mol x 0.00100 = 0.00100 mol neutralization, since all the NH3 is converted amount of HCl = 0.100 mol.L-1 x 0.010 L = 0.0010 mol to NH4+ we have 0.00150 mol of NH4+ in 110 mL [NH4+] = 0.00150 mol/0.110 L = 0.0136 mol.L-1 of solution. This is a solution of a weak acid. Ka = Kw/Kb = 1.00 x 10-14/1.8 x 10-5 = 5.6 x 10-10 Chemistry 21A Since [NH4+]>>Ka we can use the approximation [NH4+]>>[H3O+] (sqrt = square root.) [H3O+] = sqrt(C0.Ka) =sqrt(0.00150 x 5.6 x 10-10) mol.L-1 = 9.17 x 10-7 => pH = 6.04 Dr. Dragan Marinkovic ACIDS, BASES and SALTS Buffers 1.00 mol of ammonium chloride is added to 2.00 L of 1.00 mol.L-1 NH3(aq) (Kb = 1.8 x 10-5 mol.L-1) a. Calculate the pH of the solution. b. 10.00 mL of the solution is placed into a 1000 mL volumetric flask and made up to the mark with pure water. What is the pH of this solution? c. 10.0 mL of 0.100 mol.L-1 HCl(aq) is added to 100 mL of each of the above solutions. What is the final pH in each case? d. 1.00 mL of 0.100 mol.L-1 NaOH(aq) is added to fresh 100 mL portions of each solution. What is the final pH in each case? d. (Outline solution only) Similar to c, but a smaller amount of base added. The neutralisation reaction is now: NaOH(aq) + NH4+(aq) ---> Na+(aq) + NH3(aq) + H2O(l) (complete) i. solution a: amount of NH4+ = 0.0500 mol amount of NH3 = 0.100 mol amount of NaOH = 0.00010 mol ii. solution b amount of NH4+ = 0.00050 mol + + NH4 (aq) + H2O(l) <===> H3O (aq) + NH3(aq) amount of NH3 = 0.00100 mol init/mol 0.0500 0.100 ch/mol -0.0001 +0.0001 amount of NaOH = 0.00010 mol still a buffer; final/mol 0.0499 0.100 pH = 9.57 similar to above final [NH +] = 0.00040 mol ; f 4 inal [NH3] =0.00110 mol pH = 9.73 Chemistry 21A Dr. Dragan Marinkovic