The problems are on the buffer solutions worksheet. Try them on your own first, then compare with the answers below.
Quick answer: A buffer is a solution of a weak acid and its conjugate base that resists pH change. These worked answers correct the 20 problems of the AP Chemistry worksheet on buffer solutions, from equal amounts of acid and base to capacity limit: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the reaction first, then the equilibrium step, which is the order that earns AP credit. Final values are in bold. Check your significant figures: pH values are given to two decimal places.
Level: AP Chemistry · Curriculum: AP Chem Unit 8
Problem 1 Equal amounts of acid and base
[A⁻]/[HA] = 1 and log₁₀(1) = 0, so pH = pKa = 4.74.Problem 2 Unequal buffer ratio
pH = 4.74 + log₁₀(0.40/0.20) = 4.74 + 0.30 = 5.04.Problem 3 Spot the buffers
A buffer needs a weak acid with its conjugate base, or a weak base with its conjugate acid. (a) No: HCl is strong. (b) Yes. (c) Yes. (d) No: NaOH is strong. Buffers: (b) and (c).Problem 4 Choose a buffer pair
The best buffer has a pKa within 1 unit of the target pH. 9.25 is only 0.05 away from 9.2, while 4.74 is 4.5 away. Use NH₄⁺/NH₃.Problem 5 Equivalence volume
n(HCl) = 0.0250 × 0.100 = 2.50 × 10⁻³ mol. V(NaOH) = 2.50 × 10⁻³ / 0.0500 = 0.0500 dm³ = 50.0 cm³.Problem 6 pKa from half-equivalence
1. At half-equivalence [HA] = [A⁻], so pKa = 4.76.2. Ka = 10⁻⁴·⁷⁶ = 1.7 × 10⁻⁵.
Problem 7 Ammonia buffer
pH = pKa + log₁₀([NH₃]/[NH₄⁺]) = 9.25 + log₁₀(0.30/0.20) = 9.43.Problem 8 Adding base to a buffer
1. n(OH⁻) = 0.0050 × 1.0 = 5.0 × 10⁻³ mol.2. HA = 0.0200 − 0.0050 = 0.0150 mol; A⁻ = 0.0200 + 0.0050 = 0.0250 mol.
3. pH = 4.74 + log₁₀(0.0250/0.0150) = 4.96.
Problem 9 Adding acid to a buffer
1. NH₃ + H⁺ → NH₄⁺.2. NH₃ = 0.025 − 0.0050 = 0.020 mol; NH₄⁺ = 0.025 + 0.0050 = 0.030 mol.
3. pH = 9.25 + log₁₀(0.020/0.030) = 9.07.
Problem 10 Making a buffer of given pH
1. log₁₀ratio = 4.90 − 4.74 = 0.16, so ratio = 1.45.2. n = 1.445 × 0.200 = 0.289 mol.
3. m = 0.289 × 82.0 = 23.7 g.
Problem 11 Partial neutralisation
1. n(HA) = 0.0500 × 0.200 = 0.0100 mol; n(OH⁻) = 0.0200 × 0.100 = 2.00 × 10⁻³ mol.2. A⁻ = 2.00 × 10⁻³ mol; HA = 0.0100 − 0.0020 = 8.0 × 10⁻³ mol.
3. pH = 4.74 + log₁₀(0.0020/0.0080) = 4.14.
Problem 12 Before equivalence, strong acid
n(H⁺) left = 2.50 × 10⁻³ − 2.00 × 10⁻³ = 5.0 × 10⁻⁴ mol in 45.0 cm³. [H⁺] = 0.0111 mol/dm³, so pH = 1.95.Problem 13 Equivalence of ethanoic acid
1. 2.50 × 10⁻³ mol in 50.0 cm³ gives 0.0500 mol/dm³.2. Kb = 1.0 × 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.56 × 10⁻¹⁰.
3. [OH⁻] = √(5.56 × 10⁻¹⁰ × 0.0500) = 5.27 × 10⁻⁶ mol/dm³; pOH = 5.28; pH = 8.72.
Problem 14 Choose the indicator
(a) The steep section runs from about pH 3 to 11, so bromothymol blue (or any of the three) works. (b) The equivalence point is near pH 8.7, which lies in phenolphthalein’s range: phenolphthalein. Methyl orange would change colour too early.Problem 15 Comparing capacity
1. A: A⁻ = 0.48 mol, HA = 0.52 mol; pH = 4.74 + log₁₀(0.48/0.52) = 4.71.2. B: A⁻ = 0.030 mol, HA = 0.070 mol; pH = 4.74 + log₁₀(0.030/0.070) = 4.37.
3. Buffer A changed by 0.03 and B by 0.37. The more concentrated buffer A has the greater capacity.
Problem 16 Complete titration of methanoic acid
(a) x = √(1.8 × 10⁻⁴ × 0.100) = 4.243 × 10⁻³ mol/dm³ (4.2 % of 0.100, valid); pH = 2.37.(b) Half-equivalence: pH = pKa = 3.74.
(c) [HCOO⁻] = 0.0500 mol/dm³; Kb = 5.56 × 10⁻¹¹; [OH⁻] = 1.667 × 10⁻⁶ mol/dm³; pH = 8.22.
(d) Excess OH⁻ = 5.0 × 10⁻⁴ mol in 55.0 cm³ = 9.091 × 10⁻³ mol/dm³; pH = 11.96.
Problem 17 Derive the buffer equation
Rearrange: [H₃O⁺] = Ka × [HA]/[A⁻]. Take −log₁₀ of both sides: pH = −log₁₀Ka − log₁₀([HA]/[A⁻]) = pKa + log₁₀([A⁻]/[HA]). With a ratio of 0.10: pH = 4.74 + log₁₀(0.10) = 4.74 − 1 = 3.74, one unit below pKa.Problem 18 Buffer from ammonia and HCl
1. 9.00 = 9.25 + log₁₀ratio, so ratio = 10⁻⁰·²⁵ = 0.562.2. Let n mol HCl convert NH₃ to NH₄⁺: (0.0500 − n)/n = 0.562, so n = 0.0500/1.562 = 0.0320 mol.
3. V = 0.0320/0.50 = 64.0 cm³.
Problem 19 Unknown weak acid
1. n(NaOH) = 0.0315 × 0.100 = 3.15 × 10⁻³ mol = n(HA); c = 3.15 × 10⁻³ / 0.0250 = 0.126 mol/dm³.2. 15.75 cm³ is half of 31.5 cm³, so pKa = 4.20 and Ka = 10⁻⁴·²⁰ = 6.3 × 10⁻⁵.
3. Half the acid has been converted, so [HA] = [A⁻] and log₁₀(1) = 0.
Problem 20 Capacity limit
1. The new pH is 4.64, so log₁₀ratio = −0.10 and ratio = 10⁻⁰·¹⁰ = 0.7943.2. (0.30 − n)/(0.30 + n) = 0.7943, so n = 0.30 × (1 − 0.7943)/(1 + 0.7943) = 0.0344 mol.
3. V = 0.0344 / 1.0 = 34 cm³.
