#### Transcript Chemistry 1011

Chapter 16. Acid –Base Equilibria . . 1 Equilibria in Solutions of Weak Acids The dissociation of a weak acid is an equilibrium situation with an equilibrium constant, called the acid dissociation constant, Ka based on the equation HA (aq) + H2O (l) H3O+ (aq) + A- (aq) H O A Ka 3 HA 2 Equilibria in Solutions of Weak Acids The acid dissociation constant, Ka is always based on the reaction of one mole of the weak acid with water. If you see the symbol Ka, it always refers to a balanced equation of the form HA (aq) + H2O (l) H3O+ (aq) + A- (aq) 3 Problem The pH of 0.10 mol/L HOCl is 4.23. Calculate Ka for hypochlorous acid. HOCl (aq) + H2O (l) H3O+ (aq) + ClO- (aq) H O ClO Ka 3 HOCl 4 Calculating Equilibrium Concentrations in Solutions of Weak Acids We can calculate equilibrium concentrations of reactants and products in weak acid dissociation reactions with known values for Ka. To do this, we will often use the ICE table technique we saw in the last chapter on equilibrium. 5 Calculating Equilibrium Concentrations in Solutions of Weak Acids We need to figure out what is an acid and what is a base in our system. For example, if we start with 0.10 mol/L HCN, then HCN is an acid, and water is a base. HCN (aq) + H2O (l) H3O+ (aq) + CN- (aq) Ka = 4.9 x 10-10 6 Like our previous equilibrium problems, we then create a table of the initial concentrations of all chemicals, the change in their concentration, and their equilibrium concentrations in terms of known and unknown values. (all in mol/L) Initial conc. Conc. change Equil. conc. HCN (aq) 0.10 -x 0.10 – x + H2O (l) N/A N/A N/A [H O ][CN ] 10 3 K a 4.9x10 [HCN] H3O+ (aq) + 0.0 +x +x CN- (aq) 0.0 +x +x (x)(x) so 4.9x10 (0.10 x) 10 7 (all in mol/L) Initial conc. Conc. change Equil. conc. HCN (aq) 0.10 -x 0.10 – x K a 4.9 x 1010 + H2O (l) N/A N/A N/A H3O+ (aq) + 0.0 +x +x CN- (aq) 0.0 +x +x [H3O ][CN ] (x)(x) 10 so 4.9 x 10 [HCN] (0.10 x) We can ALWAYS solve this equation using the quadratic formula and get the right answer, but it might be possible to do it more simply. 8 [H O ][CN ] (x)(x) 10 3 K a 4.9 x 10 so 4.9 x 1010 [HCN] (0.10 x) Every time we do an weak acid equilibrium problem, divide the initial concentration of the acid by Ka. For this example 0.10 / 4.9 x 10-10 = 2 x 108 9 [H O ][CN ] (x)(x) 10 3 K a 4.9 x 10 so 4.9 x 1010 [HCN] (0.10 x) 0.10 / 4.9 x -10 10 =2x 8 10 Since this value is greater than 100, we can assume that the initial concentration of the acid and the equilibrium concentration of the acid are the same. This assumption will lead to answers with less than 5% error since this pre-check is greater than 100. 10 [H O ][CN ] (x)(x) K a 4.9 x 1010 3 so 4.9 x 1010 [HCN] (0.10 x) The assumption we will make is that x << [HCN]i so [HCN]eqm [HCN]I 4.9 x 10-10 = x2 / 0.10 x2 = (4.9 x 10-10)(0.10) x = 4.9 x 10-11 x = 7.0 x 10-6 mol/L 11 [H O ][CN ] (x)(x) K a 4.9 x 1010 3 so 4.9 x 1010 [HCN] (0.10 x) Based on the assumption we’ve made, at equilibrium x = [H3O+]eqm = [CN-]eqm = 7.0 x 10-6 mol/L (-ve value isn’t physically possible) [HCN]eqm = 0.10 mol/L. 12 . Any time we make an assumption, we MUST check it. We assumed x << [HCN]i To check the assumption, we divide x by [HCN]i and express it as a percentage x 7.0 x 10-6 M -5 7.0 x 10 0.007% HCNi 0.10M 13 As long as the assumption check is less than 5%, then the assumption is valid! If the assumption was not valid, we would have to go back and use the quadratic formula! 14 Remember! H2O (l) + H2O (l) H3O+ (aq) + OH- (aq) is always taking place in water whether or not we have added an acid or base. This reaction also contributes H3O+ (aq) and OH- (aq) to our system at equilibrium 15 Remember! Since at 25 C Kw = [H3O+] [OH-] = 1.0 x 10-14 it turns out that if our acid-base equilibrium we’re interested in gives a pH value between about 6.8 and 7.2 then the auto-dissociation of water contributes a significant amount of [H3O+] and [OH-] to our system and the real pH would not be what we calculated in the problem. 16 Problem Acetic acid CH3COOH (or HAc) is the solute that gives vinegar its characteristic odour and sour taste. Calculate the pH and the concentration of all species present in: a) 1.00 mol/L CH3COOH b) 0.00100 mol/L CH3COOH 17 Problem a) (all in mol/L) Initial conc. Conc. change Equil. conc. CH3COOH (aq) + 1.00 -x 1.00 – x H2O (l) N/A N/A N/A H3O+ (aq) + CH3COO- (aq) 0.0 0.0 +x +x +x +x [H3O ][CH3COO ] (x)(x) 5 K a 1.8 x 10 so 1.8 x 10 [CH3COOH] (1.00 x) 5 Let’s check the initial acid concentration / Ka ratio. 1.00 / 1.8 x 10-5 55000 is larger than 100. 18 Problem a) [H O ][CH COO ] (x)(x) 5 5 3 3 K a 1.8 x 10 so 1.8 x 10 [CH3COOH] (1.00 x) We can probably assume that x << [HAc]i so [HAc]eqm [HAc]i 1.8 x 10-5 = x2 / 1.00 x2 = (1.8 x 10-5)(1.00) x = 1.8 x 10-5 x = 4.2 x 10-3 mol/L (but must be + value since x = [H3O+]) 19 Problem a) So at equilibrium, [H3O+] = [CH3COO-] = 4.2 x 10-3 mol/L [CH3COOH] = 1.00 mol/L. x 4.2 x 10-3 M 4.2 x 10-3 0.42% CH3COOHi 1.00M The assumption was valid and so pH = - log [H3O+] pH = - log 4.2 x 10-3 pH = 2.38 20 Problem b) (all in mol/L) Initial conc. Conc. change Equil. conc. CH3COOH (aq) + 0.0100 0.00100 -x 0.0100 – x 0.00100 H2O (l) N/A N/A N/A H3O+ (aq) + CH3COO- (aq) 0.0 0.0 +x +x +x +x [H O ][CH COO ] (x)(x) 5 5 3 3 K a 1.8 x 10 so 1.8 x 10 [CH3COOH] (0.00100 x) Let’s check the initial acid concentration / Ka ratio. 0.00100 / 1.8 x 10-5 56 is smaller than 100. 21 Problem b) (all in mol/L) Initial conc. Conc. change Equil. conc. CH3COOH (aq) + 0.0100 0.00100 -x 0.0100 – x 0.00100 H2O (l) N/A N/A N/A H3O+ (aq) + CH3COO- (aq) 0.0 0.0 +x +x +x +x [H O ][CH COO ] (x)(x) 5 5 3 3 K a 1.8 x 10 so 1.8 x 10 [CH3COOH] (0.00100 x) We can probably CAN NOT assume that x << [HAc]i so [HAc]eqm [HAc]i 5 [1.8 x 10 ](0.00100 x) x 0 2 5 8 0 x 1.8 x 10 x - 1.8 x 10 2 22 Problem b) (1.8 x 105 ) (1.8 x 105 ) 2 4(1)(-1.8x 108 ) b b 2 4ac x so x 2a 2(1) - 1.8 x 105 3.24 x 1010 7.2 x 108 - 1.8 x 105 3.24 x 1010 7.2 x 108 x or x 2 2 4 - 1.8 x 105 2.69 x 10 - 1.8 x 105 2.69 x 10 4 x or x 2 2 4 4 2.87 x 10 2.51 x 10 x or x 2 2 so x 1.25 x 104 mol/L or x -1.44 x 104 mol/L 23 Problem b) Since [H3O+] = x we must use the positive value, so [H3O+] = [CH3COO-] = 1.3 x 10-4 mol/L [CH3COOH] = 0.00100 mol/L – 1.3 x 10-4 mol/L = 0.00087 mol/L. 24 Problem b) Let’s confirm that x << [HAc]i IS NOT TRUE -4 x 1.3 x 10 M 0.13 13 % CH3COOHi 0.00100M pH = - log [H3O+] pH = - log 4.2 x 10-4 pH = 3.38 25 Problem A vitamin C tablet containing 250 mg of ascorbic acid (C6H8O6; Ka = 8.0 x 10-5 is dissolved in a 250 mL glass of water to give a solution where [C6H8O6] = 5.68 x 10-3 mol/L. What is the pH of the solution? 26 Problem (all in mol/L) Initial conc. Conc. change Equil. conc. C6H8O6 (aq) + 5.68 x 10-3 -x 5.68 x 10-3 – x H2O (l) N/A N/A N/A H3O+ (aq) + C6H7O6- (aq) 0.0 0.0 +x +x +x +x [H O ][C6 H7 O6 ] (x)(x) K a 8.0 x 105 3 so 8.0 x 105 [C6 H8O6 ] (5.68x 10-3 x) Check the initial acid concentration / Ka ratio. 5.68 x 10-3/ 8.0 x 10-5 71 which is not larger than 100 so 5 3 8.0 x 10 (5.68x 10 x) x 0 2 5 7 0 x 8.0 x 10 x - 4.54 x10 2 27 Problem (8.0 x 105 ) (8.0 x 105 ) 2 4(1)(-4.54 x 107 ) b b 2 4ac x so x 2a 2(1) - 8.0 x 105 6.4 x 109 (1.82 x 106 ) - 8.0 x 105 6.4x109 (1.82 x106 ) x or x 2 2 5 3 5 3 - 8.0 x 10 1.35 x 10 - 8.0 x 10 1.35 x 10 x or x 2 2 3 3 1.4 x 10 1.27 x 10 x 3 or x 2 2 4 4 so x 7.1x10 mol/L or x 6.3x10 mol/L 28 Problem Since [H3O+] = x the answer must be the positive value [H3O+] = [C6H7O6-] = 6.3 x 10-4 mol/L [C6H8O6] = (5.68 x 10-3 - 6.3 x 10-4) mol/L = 5.05 x 10-3 mol/L pH = - log [H3O+] pH = - log 6.3 x 10-4 pH = 3.20 29 Degree of ionization The pH of a solution of a weak acid like acetic acid will depend on the initial concentration of the weak acid and Ka. Therefore, we can define a second measure of the strength of a weak acid by looking of the degree (or percent) ionization of the acid. %ionization = [HA]ionized / [HA]initial x 100% 30 Percent ionization In part a) of an earlier problem an acetic acid solution with initial concentration of 1.00 mol/L at equilibrium had [H3O+]eqm = [HA]ionized = 4.2 x 10-3 mol/L %ionization = [HA]ionized / [HA]initial x 100% %ionization = 4.2 x 10-3 mol/L / 1.00 mol/L x 100% %ionized = 0.42% 31 Percent ionization In part b) of an earlier problem an acetic acid solution with initial concentration of 0.00100 mol/L at equilibrium had [H3O+] = [HA]ionized = 1.3 x 10-4 mol/L %ionization = [HA]ionized / [HA]initial x 100% %ionization = 1.3 x 10-4 mol/L / 0.00100 mol/L x 100% %ionization = 13% 32 Figure 33 Equilibria in Solutions of Weak Bases The dissociation of a weak base is an equilibrium situation with an equilibrium constant, called the base dissociation constant, Kb based on the equation B (aq) + H2O (l) BH+ (aq) + OH- (aq) [BH ][OH ] Kb [B] 34 Equilibria in Solutions of Weak Bases The base dissociation constant, Kb is always based on the reaction of one mole of the weak base with water. If you see the symbol Kb, it always refers to a balanced equation of the form B (aq) + H2O (l) BH+ (aq) + OH- (aq) 35 Equilibria in Solutions of Weak Bases Our approach to solving equilibria problems involving bases is exactly the same as for acids. 1. Set up the ICE table 2. Establish the equilibrium constant expression 3. Make a simplifying assumption when possible 4. Solve for x, and then for eq’m amounts 36 Problem Strychnine (C21H22N2O2), a deadly poison used for killing rodents, is a weak base having Kb = 1.8 x 10-6. Calculate the pH if [C21H22N2O2]initial = 4.8 x 10-4 mol/L 37 Problem (all in mol/L) Initial conc. Conc. change Equil. conc. C21H22N2O2 (aq) 4.8 x 10-4 -x 4.8 x 10-4 – x + H2O (l) N/A N/A N/A C21H23N2O2+ (aq) + OH- (aq) 0.0 0.0 +x +x +x +x [C21H 23 N 2O2 ][OH ] (x)(x) 6 K b 1.8 x 10 so 1.8 x 106 [C21H 22 N 2O2 ] (4.8 x 10-4 x) Check the initial base concentration / Kb ratio 4.8 x 10-4 / 1.8 x 10-6 267 which is greater than 100 We are probably good to make a simplifying assumption that x << [C21H22N2O2]i 38 [C 21H 23 N 2 O 2 ][OH ] (x)(x) 6 6 K b 1.8 x 10 so 1.8 x 10 [C 21H 22 N 2O 2 ] (4.8 x 10-4 x) The assumption we will make is that x << [C21H22N2O2]i so [C21H22N2O2]eqm [C21H22N2O2]I 1.8 x 10-6 = x2 / 4.8 x 10-4 x2 = (1.8 x 10-6)(4.8 x 10-4) x = 8.64 x 10-10 x = 2.94 x 10-5 mol/L 39 Problem Since x = [OH-], the answer must be the positive value, x = [C21H23N2O2+] = [OH-] = 2.9 x 10-5 mol/L [C21H22N2O2] = 4.8 x 10-4 mol/L – 2.9 x 10-5 mol/L = 4.5 x 10-4 mol/L. We should check the assumption! x 2.9 x 10-5 M 0.060 6.0% -4 C21H22 N2O2 i 4.8 x 10 M 40 Problem x 2.9 x 10-5 M 0.060 6.0% -4 C21H22 N2O2 i 4.8 x 10 M In this case, the error is more than 5%. I will leave it to you to go back and use the quadratic formula. Compare the two answers 41 Problem To continue towards the answer of the problem AS IF the assumption WERE VALID pOH = - log [OH-] pOH = - log 2.9 x 10-5 pOH = 4.54 pH + pOH = 14.00 pH = 14.00 - pOH pH = 14.00 - (4.54) pH = 9.46 42 Relation Between Ka and Kb The strength of an acid in water is expressed through Ka, while the strength of a base can be expressed through Kb Since Brønsted-Lowry acid-base reactions involve conjugate acid-base pairs there should be a connection between the Ka value and the Kb value of a conjugate acid-base pair. 43 Relation Between Ka and Kb HA (aq) + H2O (l) H3O+ (aq) + A- (aq) H O A Ka 3 HA A- (aq) + H2O (l) OH- (aq) + HA (aq) OH HA A Kb 44 Since these reactions take place in the same beaker at the same time let’s add them together HA (aq) + H2O (l) + A- (aq) + H2O (l) H3O+ (aq) + A- (aq) + OH- (aq) + HA (aq) 2 H2O (l) H3O+ (aq) + OH- (aq) 45 HA (aq) + H2O (l) + A- (aq) + H2O (l) H3O+ (aq) + A- (aq) + OH- (aq) + HA (aq) 2 H2O (l) H3O+ (aq) + OH- (aq) The sum of the reactions is the dissociation of water reaction, which has the ion-product constant for water Kw = [H3O+] [OH-] = 1.0 x 10-14 at 25 °C Closer inspection shows us that H O A OH HA H O OH K HA A Ka x K b 3 3 w 1.0x1014 at 25 C 46 As the strength of an acid increases (larger Ka) the strength of the conjugate base must decrease (smaller Kb) because their product must always be the dissociation constant for water Kw. 47 Strong acids always have very weak conjugate bases. Strong bases always have very weak conjugate acids. Since Ka x Kb = Kw then Ka = Kw / Kb and Kb = Kw / Ka 48 Problem a) – Piperidine (C5H11N) is an amine found in black pepper. Find Kb for piperidine in Appendix C, and then calculate Ka for the C5H11NH+ cation. Kb = 1.3 x 10-3 b) Find Ka for HOCl in Appendix C, and then calculate Kb for OCl-. Ka = 3.5 x 10-8 49 Acid-Base Properties of Salts When acids and bases react with each other, they form ionic compounds called salts. Salts, when dissolved in water, can lead to acidic, basic, or neutral solutions, depending on the relative strengths of the acid and base we derive them from. Strong acid + Strong base Neutral salt solution Strong acid + Weak base Acidic salt solution Weak acid + Strong base Basic salt solution 50 Salts that Yield Neutral Solutions Strong acids and strong bases react to form neutral salt solutions. When the salt dissociates in water, the cation and anion do not appreciably react with water to form H3O+ or OH-. 51 Salts that Yield Neutral Solutions Strong base cations like the alkali metal cations (Li+, Na+, K+) or alkaline earth cations (Ca2+, Sr2+, Ba2+, but NOT Be2+) and strong acid anions such as Cl-, Br-, I-, NO3-, and ClO4- will combine together to give neutral salt solutions with pH = 7. 52 Salts that Yield Neutral Solutions Sodium chloride (NaCl) will dissociate into Na+ and Cl- in water. Cl- has no acidic or basic tendencies. Cl- (aq) + H2O (l) ⇌ no reaction Chloride ions DO NOT HAVE hydrolysis reactions with water since it is the “conjugate” of a strong acid, which makes it very, very weak. 53 Salts that Yield Neutral Solutions Na+ has no acidic or basic tendencies. Na+ (aq) + H2O (l) ⇌ no reaction Sodium ions DO NOT HAVE hydrolysis reactions with water since it is the “conjugate” of a strong base, which makes it very, very weak. 54 Salts that Yield Acidic Solutions The reaction of a strong acid with anions like Cl-, Br-, I-, NO3-, and ClO4with a weak base will lead to an acidic salt solution. The solution is acidic because the anion shows no acidic or basic tendencies, but the cation does, as it is the conjugate acid of a weak base. 55 Salts that Yield Acidic Solutions Ammonium chloride (NH4Cl) will dissociate into NH4+ and Cl- in water. Cl- has no acidic or basic tendencies. Cl- (aq) + H2O (l) ⇌ no reaction Chloride ions DO NOT HAVE hydrolysis reactions with water since it is the “conjugate” of a strong acid, which makes it very, very weak. 56 Salts that Yield Acidic Solutions NH4+ has acidic tendencies. That is: NH4+ (aq) + H2O (l)⇌ NH3 (aq) + H3O+ (aq) Ammonium ions hydrolyze in water because it is the conjugate acid of the weak base NH3, which means ammonium is a weak acid. 57 Salts that Yield Basic Solutions The reaction of a strong base with cations like Li+, Na+, K+, Ca2+, Sr2+, and Ba2+ with a weak acid will lead to an basic salt solution. The solution is acidic because the cation shows no acidic or basic tendencies, but the anion does, as it is the conjugate base of a weak acid. 58 Salts that Yield Basic Solutions Sodium fluoride (NaF) will dissociate into Na+ and F- in water. Na+ (aq) + H2O (l) ⇌ no reaction Sodium ions DO NOT HAVE hydrolysis reactions with water since it is the “conjugate” of a strong base, which makes it very, very weak. 59 Salts that Yield Basic Solutions F- has basic tendencies. That is: F- (aq) + H2O (l)⇌ HF (aq) + OH- (aq) Fluoride ions hydrolyze in water because it is the conjugate base of the weak acid HF, which means fluoride is a weak base. 60 Problem Predict whether the following salt solution is neutral, acidic, or basic and calculate the pH. 0.25 mol/L NH4Br – NH3 has a Kb value of 1.8 x 10-5 NH4+ (aq) 0.25 -x 0.25 – x (all in mol/L) Initial conc. Conc. change Equil. conc. 10 K a 5.56 x10 + H2O (l) N/A N/A N/A [H3O ][NH3 ] [NH4 ] H3O+ (aq) + 0.0 +x +x 10 so 5.56 x10 NH3 (aq) 0.0 +x +x (x)(x) (0.25 x) 61 Problem Initial acid [HA] / Ka ratio is 0.25 / 5.56 x 10-10 4.5 x 108 we can probably assume 0.25 >> x 5.56 x 10-10 = x2 / 0.25 x2 = (5.56 x 10-10)(0.25) x2 = 1.39 x 10-10 x = 1.39 x 10-10 x = 1.18 x 10-5 mol/L 62 Problem Negative answer not physically possible so therefore, [H3O+] = 1.18 x 10-5 mol/L x NH 4 i 1.2 x 10-5 M 4.8 x 10-5 4.8 x 10-3 % 0.25M Since we’ve shown the assumption is valid pH = -log [H3O+] = - log 1.18 x 10-5 = 4.93. 63 Salts that Contain Acidic Cations and Basic Anions If a salt is composed of an acidic cation and a basic anion, the acidity or basicity of the salt solution depends on the relative strengths of the acid and base. 64 Salts that Contain Acidic Cations and Basic Anions If the acid cation is “stronger” than the base anion, it “wins” and the salt solution is acidic. If the base anion is “stronger” than the acid cation, it “wins” and the salt solution is basic. 65 Salts that Contain Acidic Cations and Basic Anions Ka > Kb the acid cation is “stronger” and the salt solution is acidic. Ka < Kb the base anion is “stronger” and the salt solution is basic. Ka Kb the salt solution is close to neutral. 66 Problem Classify each of the following salts as acidic, basic, or neutral: a) KBr b) NaNO2 c) NH4Br d) NH4F Ka for HF = 6.6 x 10-4 Kb for NH3 = 1.8 x 10-5 67 The Common-Ion Effect Solutions consisting of both an acid and its conjugate base are very important because they are very resistant to changes in pH. Such buffer solutions regulate pH in a variety of biological systems. 68 The Common-Ion Effect Let’s consider a solution made of 0.10 moles of acetic acid and 0.10 moles of sodium acetate with a total volume of 1.00 L, making the initial [CH3COOH] = [CH3COO-] = 0.10 mol/L. First we must identify all potential acids and bases in the system. CH3COOH acid CH3COObase Na+ neutral H2O acid or base 69 Point of view of the acid Our reaction will be CH3COOH (aq) + H2O (l) H3O+ (aq) + CH3COO- (aq) Ka = 1.8 x 10-5 (all in mol/L) Initial conc. Conc. change Equil. conc. CH3COOH (aq) + 0.10 -x 0.10 – x H2O (l) N/A N/A N/A H3O+ (aq) + CH3COO- (aq) 0.0 0.10 +x +x +x 0.10 + x Note that the initial concentration of our product CH3COO- is NOT ZERO! 70 [H 3O ][CH 3COO ] (x)(0.10 x) 5 Ka 1.8 x 10 [CH 3COOH] (0.10 x) Let’s check the initial acid concentration / Ka ratio. 0.10 / 1.8 x 10-5 5500 It’s probably safe to assume that x << [HAc]i so [HAc]eqm [HAc]I and x << [Ac-]i so [Ac-]eqm [Ac-]i 1.8 x 10-5 = x (0.10 + x) / (0.10 –x) 1.8 x 10-5 = x (0.10) / (0.10) x = 1.8 x 10-5 mol/L 71 [H 3O ][CH 3COO ] (x)(0.10 x) 5 Ka 1.8 x 10 [CH 3COOH] (0.10 x) At equilibrium, [H3O+] = 1.8 x 10-5 mol/L [CH3COO-] = 0.10 + 1.8 x 10-5 = 0.10 mol/L [CH3COOH] = 0.10 - 1.8 x 10-5 = 0.10 mol/L Assumption was valid! Check for yourself! pH = - log [H3O+] pH = - log 1.8 x 10-5 pH = 4.74 72 If we had started out with only 0.10 mol/L acetic acid, the pH would be found from (all in mol/L) Initial conc. Conc. change Equil. conc. CH3COOH (aq) + 0.10 -x 0.10 – x H2O (l) N/A N/A N/A H3O+ (aq) + CH3COO- (aq) 0.0 0.00 +x +x +x +x [H3O ][CH3COO ] (x)(x) 5 Ka 1.8 x 10 [CH3COOH] (0.10 x) 73 The initial acid concentration / Ka will still be the same, so we can assume x << [HAc]i so [HAc]eqm [HAc]i 1.8 x 10-5 = x2 / (0.10 –x) 1.8 x 10-5 = x2/ (0.10) x = 1.8 x 10-6 mol/L x = 1.3 x 10-3 mol/L (can’t be –ve) pH = - log [H3O+] pH = - log 1.3 x 10-3 pH = 2.89 74 Without the acetate ion the pH of 0.10 M acetic acid is 2.89. With an equal concentration of acetate ion present, the pH of 0.10 M acetic acid – 0.10 M acetate is 4.74 The acetate ion makes a large difference on the equilibrium pH! 75 CH3COOH (aq) + H2O (l) H3O+ (aq) + CH3COO- (aq) Adding the conjugate base (a stress!) to the equilibrium system of an acid dissociation shows the commonion effect, where the addition of a common ion causes the equilibrium to shift. This is an example of Le Chatalier’s Principle. Addition of the weak base to the acid dissociation 76 Problem Calculate the concentrations of all species present, and the pH in a solution that is 0.025 mol/L HCN and 0.010 mol/L NaCN. (Ka of HCN = 4.9 x 10-10) (all in mol/L) Initial conc. Conc. change Equil. conc. HCN (aq) + 0.025 -x 0.025 – x H2O (l) N/A N/A N/A H3O+ (aq) 0.0 +x +x + CN- (aq) 0.010 +x 0.010 + x 77 Problem [H3O ][CN ] (x)(0.010 x) 10 Ka 4.9 x 10 [HCN] (0.025 x) The initial base concentration / Ka ratio is 0.010 / 4.9 x 10-10 2 x 107 It’s probably safe to assume that x << [HCN]i so [HCN]eqm [HCN]I and x << [CN-]i so [CN-]eqm [CN-]i 78 Problem 4.9 x 10-10 = x (0.010 + x) / (0.025 –x) 4.9 x 10-10 = x (0.010) / (0.025) x = 1.2 x 10-9 mol/L So at equilibrium, [H3O+] = 1.2 x 10-9 mol/L [CN-] = 0.010 + 1.2 x 10-9 = 0.010 mol/L [HCN] = 0.025 - 1.2 x 10-9 = 0.025 mol/L. Assumption was valid! Check this for yourself! 79 Problem pH = - log [H3O+] pH = - log 1.2 x 10-9 pH = 8.91 80 Buffer Solutions Solutions that contain both a weak acid and its conjugate base are buffer solutions. These solutions are resistant to changes in pH. 81 Buffer Solutions If more acid (H3O+) or base (OH-) is added to the system, the system has enough of the original acid and conjugate base molecules in the solution to react with the added acid or base, and so the new equilibrium mixture will be very close in composition to the original equilibrium mixture. 82 Buffer solutions A 0.10 molL-1 acetic acid – 0.10 molL-1 acetate mixture has a pH of 4.74 and is a buffer solution! CH3COOH (aq) + H2O (l) H3O+ (aq) + CH3COO- (aq) (all in mol/L) Initial Conc. changes Equil. conc. CH3COOH (aq) + H2O (l) H3O+ (aq) + CH3COO- (aq) 0.10 N/A 0.0 0.10 -x N/A +x +x 0.10 - x N/A +x 0.10 + x [H3O ][CH3COO ] K a 1.8 x 10 [CH3COOH] 5 83 Buffer solutions If we rearrange the Ka expression to solve for + [H3O ] [CH3COOH] 5 [CH3COOH] [H3O ] K a 1.8 x 10 [CH3COO ] [CH3COO ] 84 Buffer solutions [CH3COOH] 5 [CH3COOH] [H3O ] K a 1.8 x 10 [CH3COO ] [CH3COO ] Assume x << [HAc]i so [HAc]eqm [HAc]I and x << [Ac-]i so [Ac-]eqm [Ac-]i, and we should see If [CH3COOH]i = [CH3COO-]i, then [H3O+] = 1.8 x 10-5 M = Ka and pH = pKa = 4.74 85 Buffer solutions What happens if we add 0.01 mol of NaOH (strong base) to 1.00 L of the acetic acid – acetate buffer solution? CH3COOH (aq) + OH- (aq) → H2O (l) + CH3COO- (aq) This reaction goes to completion and keeps occurring until we run out of the limiting reagent OH(all in moles) Initial Change Final (where x = 0.01 due to limiting OH-) CH3COOH (aq) + OH- (aq) → H2O (l) + CH3COO- (aq) 0.10 0.01 N/A 0.10 -x -x N/A +x 0.10 – x 0.01 – x = N/A 0.10 + x = 0.09 0.00 = 0.11 New [CH3COOH] = 0.09 M and new [CH3COO-] = 0.11 M 86 Buffer solutions (all in mol/L) Initial conc. Conc. change Equil. conc. CH3COOH (aq) + 0.09 -x 0.09 – x H2O (l) N/A N/A N/A H3O+ (aq) + CH3COO- (aq) 0.0 0.11 +x +x +x 0.11 + x [H3O ][CH3COO ] (x)(0.11 x) 5 Ka 1.8 x 10 With the assumption smaller [CH3COOH] that x is much (0.09 x) than 0.09 mol[CH (anCOOH] assumption we always need to5 0.09 5 [H3Ocheck ] K aafter 3calculations 1.8 xare 10 done!), we 1.5 xfind 10 M [CH COO ] 0.11 3 Note we’ve made the assumption that x << 0.09 M! pH = - log [H3O+] pH = - log 1.5 x 10-5 pH = 4.82 87 Buffer solutions Adding 0.01 mol of OH- to 1.00 L of water would have given us a pH of 12.0 because there is no significant amount of acid in water for the base to react with. Our buffer solution resisted this change in pH because there is a significant amount of acid (acetic acid) for the added base to react with. 88 Buffer solutions What happens if we add 0.01 mol of HCl (strong acid) to 1.00 L of the acetic acid – acetate buffer solution? CH3COO- (aq) + H3O+ (aq) → H2O (l) + CH3COOH (aq) This reaction goes to completion and keeps occurring until we run out of the limiting reagent H3O+ (all in moles) Initial Change Final (where x = 0.01 due limiting H3O+) CH3COO- (aq) + H3O+ (aq) → H2O (l) + CH3COOH (aq) 0.10 0.01 N/A 0.10 -x -x N/A +x 0.10 – x 0.01 – x = N/A 0.10 + x = 0.09 0.00 = 0.11 New [CH3COOH] = 0.11 M and new [CH3COO-] = 0.09 M 89 Buffer solutions (all in mol/L) Initial conc. Conc. change Equil. conc. CH3COOH (aq) + 0.11 -x 0.11 – x H2O (l) N/A N/A N/A H3O+ (aq) + CH3COO- (aq) 0.0 0.09 +x +x +x 0.09 + x [H3O ][CH3COO ] (x)(0.09 x) 5 Ka 1.8 x 10 With the assumption that x is much smaller than [CH3COOH] (0.110.09 x)mol (an assumption we always need to check after [CH3COOH] 5 0.11 5 are done!), we find [H3O ] K a calculations 1.8 x 10 2.2 x 10 M [CH3COO ] 0.09 Note we’ve made the assumption that x << 0.09 M! pH = - log [H3O+] pH = - log 2.2 x 10-5 pH = 4.66 90 Buffer solutions Adding 0.01 mol of H3O+ to 1.00 L of water would have given us a pH of 2.0 because there is no significant amount of acid in water for the base to react with. Our buffer solution resisted this change in pH because there is a significant amount of base (acetate) for the added acid to react with. 91 . 92 Buffer capacity Buffer capacity is the measure of the ability of a buffer to absorb acid or base without significant change in pH. Larger volumes of buffer solutions have a larger buffer capacity than smaller volumes with the same concentration. Buffer solutions of higher concentrations have a larger buffer capacity than a buffer solution of the same volume with smaller concentrations. 93 Problem Calculate the pH of a 0.100 L buffer solution that is 0.25 mol/L in HF and 0.50 mol/L in NaF. (all in mol/L) Initial conc. Conc. change Equil. conc. HF (aq) + 0.25 -x 0.25 – x H2O (l) N/A N/A N/A H3O+ (aq) 0.0 +x +x + F- (aq) 0.50 +x 0.50 + x [H3O ][F ] (x)(0.50 x) 4 Ka 3.5 x 10 [HF] (0.25 x) With thexassumption x is much smaller Assume << [HCN]i that so [HCN] [HCN]i eqm than 0.25 mol (an assumption we always need -] so [CN-] -] and x << [CN [CN to check after calculations areeqm done!), wei find i 94 Problem [HF] 4 [HF] [H3O ] K a 3.5 x 10 [F ] [F ] 4 0.25 4 3.5 x 10 1.75 x 10 M 0.50 pH = - log [H3O+] pH = - log 1.75 x 10-4 pH = 3.76 95 Problem a) What is the change in pH on addition of 0.002 mol of HNO3? (all in moles) Initial Change Final (where x = 0.002 due to limiting H3O+) F- (aq) 0.050 -x 0.050 – x = 0.048 + H3O+ (aq) → H2O (l) + 0.002 N/A -x N/A 0.002 – x N/A = 0.00 HF (aq) 0.025 +x 0.025 + x = 0.027 New [HF] = 0.27 M and new [F-] = 0.48 M (all in mol/L) Initial conc. Conc. change Equil. conc. HF (aq) 0.27 -x 0.27 – x + H2O (l) N/A N/A N/A H3O+ (aq) + 0.00 +x +x F- (aq) 0.48 +x 0.48 + x [H3O ][F ] (x)(0.48 x) 4 Ka 3.5 x 10 [HF] (0.27 x) 96 Problem [HF] 4 0.27 [H3O ] K a 3.5 x 10 1.97 x 104 M [F ] 0.48 Notice we’ve made the assumption that x << 0.27 M. We should check this! pH = - log [H3O+] pH = - log 1.97 x 10-4 pH = 3.71 97 Problem b) What is the change in pH on addition of 0.004 mol of KOH? (all in moles) Initial Change Final (where x = 0.004 due to limiting OH-) HF (aq) 0.025 -x 0.025 – x = 0.021 + OH- (aq) → H2O (l) + 0.004 N/A -x N/A 0.004 – x N/A = 0.00 F- (aq) 0.050 +x 0.050 + x = 0.054 New [HF] = 0.21 M and new [F-] = 0.54 M (all in mol/L) Initial conc. Conc. change Equil. conc. HF (aq) 0.21 -x 0.21 – x + H2O (l) N/A N/A N/A H3O+ (aq) + 0.00 +x +x F- (aq) 0.54 +x 0.54 + x [H O ][F ] (x)(0.54 x) 4 4 3 K a 3.5x10 so 3.5x10 [HF] (0.21 x) 98 Problem [HF] 4 0.21 [H3O ] K a 3.5 x 10 1.36 x 104 M [F ] 0.54 Notice we’ve made the assumption that x << 0.21 M. We should check this! pH = - log [H3O+] pH = - log 1.36 x 10-4 pH = 3.87 99 The Henderson-Hasselbalch Equation We’ve seen that, for buffer solutions containing members of a conjugate acidbase pair, that [acid] [H3O ] K a [base] [acid] [acid] log K a log log [H3O ] log K a [base] [base] pH = pKa + log [base] / [acid] This is called the HendersonHasselbalch Equation. 100 The Henderson-Hasselbalch Equation If we have a buffer solution of a conjugate acid-base pair, then the pH of the solution will be close to the pKa of the acid. This pKa value is modified by the logarithm of ratio of the concentrations of the base and acid in the solution to give the actual pH. 101 Problem Use the Henderson-Hasselbalch Equation to calculate the pH of a buffer solution prepared by mixing equal volumes of 0.20 mol/L NaHCO3 and 0.10 mol/L Na2CO3. We need the Ka and the concentrations of the acid (HCO3-) and the base (CO32-). Ka = 5.6 x 10-11 (we use the Ka for the second proton of H2CO3!). 102 Problem NOTE: The concentrations we are given for the acid and the base are the concentrations before the mixing of equal volumes! 103 Problem If we mix equal volumes, the total volume is TWICE the volume for the original acid or base solutions. Since the number of moles of acid or base DON’T CHANGE on mixing, the initial concentrations we use will be half the given values. pH = pKa + log [base] / [acid] pH = (-log 5.6 x 10-11) + log (0.05) / (0.10) pH = 10.25 – 0.30 pH = 9.95 104