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Buffers and the Common-Ion Effect (Ch 16) a Buffer Works Through the Common-Ion Effect

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Buffers and the Common-Ion Effect (Ch 16) a Buffer Works Through the Common-Ion Effect Buffers and the Common-ion Effect (Ch 16) A buffer works through the common-ion effect. Acetic acid in water dissociates slightly to produce some acetate ion: - + CH3COOH(aq) + H2O(l) CH3COO (aq) + H3O (aq) acetic acid acetate ion If NaCH3COO is added, it provides a source of - CH3COO ion, and the equilibrium shifts to the left. - CH3COO is common to both solutions. - The addition of CH3COO reduces the % dissociation of the acid. ©2013 McGraw-Hill Ryerson Limited 17-1 Table 17.1 The Effect of Added Acetate Ion on the Dissociation of Acetic Acid - * + [CH3COOH]init [CH3COO ]added % Dissociation [H3O ] pH 0.10 0.00 1.3 1.3x10-3 2.89 0.10 0.050 0.036 3.6x10-5 4.44 0.10 0.10 0.018 1.8x10-5 4.74 0.10 0.15 0.012 1.2x10-5 4.92 [CH COOH] * % Dissociation = 3 dissoc x 100 [CH3COOH]init ©2013 McGraw-Hill Ryerson Limited 17-2 How a Buffer Works The buffer components (HA and A-) are able to consume - + small amounts of added OH or H3O by a shift in equilibrium position. - + CH3COOH(aq) + H2O(l) CH3COO (aq) + H3O (aq) - + Added OH reacts with Added H3O reacts with - CH3COOH, causing a shift to CH3COO , causing a the right. shift to the left. The shift in equilibrium position absorbs the change in + - [H3O ] or [OH ], and the pH changes only slightly. ©2013 McGraw-Hill Ryerson Limited 17-3 Figure 17.3 How a buffer works. Buffer has more HA after Buffer has equal Buffer has more A- after + - - addition of H3O . concentrations of A and HA. addition of OH . + - H3O OH + - - - H2O + CH3COOH ← H3O + CH3COO CH3COOH + OH → CH3COO + H2O 17-4 ©2013 McGraw-Hill Ryerson Limited Figure 17.1 The effect of adding acid or base to an unbuffered solution. A 100-mL sample of The addition of 1 mL of strong acid (left) dilute HCl is adjusted or strong base (right) changes the pH by to pH 5.00. several units. ©2013 McGraw-Hill Ryerson Limited 17-5 Figure 17.2 The effect of adding acid or base to a buffered solution. A 100-mL sample of The addition of 1 mL of strong acid (left) an acetate buffer is or strong base (right) changes the pH very adjusted to pH 5.00. little. The acetate buffer is made by mixing 1 M CH3COOH ( a weak acid) with - 1 M CH3COONa (which provides the conjugate base, CH3COO ). ©2013 McGraw-Hill Ryerson Limited 17-6 Relative Concentrations of Buffer Components - + CH3COOH(aq) + H2O(l) CH3COO (aq) + H3O (aq) [CH COO-][H O+] [CH COOH] K = 3 3 [H O+] = K x 3 a 3 a - [CH3COOH] [CH3COO ] + Since Ka is constant, the [H3O ] of the solution depends on the ratio of buffer component concentrations. [HA] If the ratio increases, [H O+] increases. [A-] 3 [HA] If the ratio decreases, [H O+] decreases. [A-] 3 ©2013 McGraw-Hill Ryerson Limited 17-7 + Sample Problem 17.1 Calculating the Effect of Added H3O or OH- on Buffer pH PROBLEM: Calculate the following pH values(Ka of CH3COOH = 1.8 x 10-5 , assume the additions cause a negligible change in volume.) (a) pH of a buffer solution consisting of 0.50 mol/L CH3COOH and 0.50 mol/L CH3COONa (b) pH after adding 0.020 mol of solid NaOH to 1.0 L of the buffer solution (c) pH after adding 0.020 mol of HCl to 1.0 L of the buffer solution in (a). - PLAN: We can calculate [CH3COOH]init and [CH3COO ]init from the given information. From this we can find the starting pH. For (b) and (c) - + we assume that the added OH or H3O reacts completely with the buffer components. We write a balanced equation in each case, set up + a reaction table, and calculate the new [H3O ]. ©2013 McGraw-Hill Ryerson Limited 17-8 Sample Problem 17.1 SOLUTION: (a) - + Concentration(mol/L) CH3COOH(aq) + H2O(l) CH3COO (aq) + H3O (aq) Initial 0.50 - 0.50 0 Change −x - +x +x Equilibrium 0.50 - x - 0.50 + x x Since Ka is small, x is small, so we assume - [CH3COOH] = 0.50 – x ≈ 0.50 mol/L and [CH3COO ] = 0.50 + x ≈ 0.50 mol/L + [CH3COOH] -5 0.50 -5 x = [H3O ] = Ka x ≈ 1.8x10 x = 1.8x10 mol/L - [CH3COO ] 0.50 -5 Checking the assumption: pH = -log(1.8x10 ) = 4.74 -5 1.8x10 mol/L x 100 = 3.6x10-3% (< 5%; assumption is justified.) 0.50 mol/L ©2013 McGraw-Hill Ryerson Limited 17-9 Sample Problem 17.1 − 0.020 mol − (b) [OH ]added = = 0.020 M OH 1.0 L soln Setting up a reaction table for the stoichiometry: - - Concentration (mol/L) CH3COOH(aq) + OH (aq) → CH3COO (aq) + H2O(l) Initial 0.50 0.020 0.50 - Change -0.020 -0.020 +0.020 - Equilibrium 0.48 0 0.52 - Setting up a reaction table for the acid dissociation, using new initial [ ]: - + Concentration (mol/L) CH3COOH(aq) + H2O(l) CH3COO (aq) + H3O (aq) Initial 0.48 - 0.52 0 Change −x - +x +x Equilibrium 0.48 - x - 0.52 + x x ©2013 McGraw-Hill Ryerson Limited 17-10 Sample Problem 17.1 Since Ka is small, x is small, so we assume - [CH3COOH] = 0.48 – x ≈ 0.48 mol/L and [CH3COO ] = 0.52 + x ≈ 0.52 mol/L + [CH3COOH] -5 0.48 -5 x = [H3O ] = Ka x ≈ 1.8x10 x = 1.7x10 mol/L - [CH3COO ] 0.52 pH = -log(1.7x10-5) = 4.77 ©2013 McGraw-Hill Ryerson Limited 17-11 Sample Problem 17.1 + 0.020 mol + (c) [H3O ]added = = 0.020 M H3O 1.0 L soln Setting up a reaction table for the stoichiometry: - + Concentration (mol/L) CH3COO (aq) + H3O (aq) → CH3COOH(aq) + H2O(l) Initial 0.50 0.020 0.50 - Change -0.020 -0.020 +0.020 - Equilibrium 0.48 0 0.52 - Setting up a reaction table for the acid dissociation, using new initial [ ]: - + Concentration(mol/L) CH3COOH(aq) + H2O(l) CH3COO (aq) + H3O (aq) Initial 0.52 - 0.48 0 Change −x - +x +x Equilibrium 0.52 - x - 0.48 + x x ©2013 McGraw-Hill Ryerson Limited 17-12 Sample Problem 17.1 Since Ka is small, x is small, so we assume - [CH3COOH] = 0.52 – x ≈ 0.52 mol/L and [CH3COO ] = 0.48 + x ≈ 0.48 mol/L + [CH3COOH] -5 0.52 -5 x = [H3O ] = Ka x ≈ 1.8x10 x = 2.0x10 mol/L - [CH3COO ] 0.48 pH = -log(2.0x10-5) = 4.70 ©2013 McGraw-Hill Ryerson Limited 17-13 The Henderson-Hasselbalch Equation - + HA(aq) + H2O(l) A (aq) + H3O (aq) + - [H3O ][A ] + [HA] K = [H3O ] = Ka x a [HA] [A-] [HA] -log[H O+] = -logK – log 3 a [A-] [base] pH = pK + log a [acid] ©2013 McGraw-Hill Ryerson Limited 17-14 Buffer Capacity The buffer capacity is a measure of the “strength” of the buffer, its ability to maintain the pH following addition of strong acid or base. The greater the concentrations of the buffer components, the greater its capacity to resist pH changes. The closer the component concentrations are to each other, the greater the buffer capacity. ©2013 McGraw-Hill Ryerson Limited 17-15 Figure 17.4 The relation between buffer capacity and pH change. When strong base is added, the pH increases least for the most concentrated buffer. This graph shows the final pH values for four different buffer solutions after the addition of strong base. ©2013 McGraw-Hill Ryerson Limited 17-16 Buffer Range The buffer range is the pH range over which the buffer is effective. Buffer range is related to the ratio of buffer component concentrations. [HA] The closer is to 1, the more effective the buffer. [A-] If one component is more than 10 times the other, buffering action is poor. Since log10 = 1, buffers have a usable range within ± 1 pH unit of the pKa of the acid component. ©2013 McGraw-Hill Ryerson Limited 17-17 Sample Problem 17.2 Using Molecular Scenes to Examine Buffers PROBLEM: The molecular scenes below represent samples of four HA/A- buffers. (HA is blue and green, A- is green, and other ions and water are not shown.) (a) Which buffer has the highest pH? (b) Which buffer has the greatest capacity? (c) Should we add a small amount of concentrated strong acid or strong base to convert sample 1 to sample 2 (assuming no volume changes)? ©2013 McGraw-Hill Ryerson Limited 17-18 Sample Problem 17.2 PLAN: Since the volumes of the solutions are equal, the scenes represent molarities as well as numbers. We count the particles of each species present in each scene and calculate the ratio of the buffer components. SOLUTION: [A-]/[HA] ratios: sample 1, 3/3 = 1; sample 2, 2/4 = 0.5; sample 3, 4/4 = 1; and sample 4, 4/2 = 2. ©2013 McGraw-Hill Ryerson Limited 17-19 Sample Problem 17.2 (a) As the pH rises, more HA will be converted to A-. The scene with the highest [A-]/[HA] ratio is at the highest pH. Sample 4 has the highest pH because it has the highest ratio. (b) The buffer with the greatest capacity is the one with the [A-]/[HA] closest to 1. Sample 3 has the greatest buffer capacity. (c) Sample 2 has a lower [A-]/[HA] ratio than sample 1, so we need to increase the [A-] and decrease the [HA]. This is achieved by adding strong acid to sample 1.
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