A-LevelChemistryEquilibria9701/41 · Oct/Nov 2021 · Q210 Marks

Solution Y is hydrochloric acid, HCl(aq). Solution Z is aqueous 4-chlorobutanoic acid, Cl(CH2)3CO2H(aq). The pKa of Cl(CH2)3CO2H(aq) is 4.52. The pH of both solutions is 4.00.

(a)(i)Write an expression for the Ka of Cl(CH2)3CO2H(aq).[1]
(a)(ii)Write a mathematical expression to describe the relationship between Ka and pKa.[1]
(a)(iii)Calculate [H⁺] in solutions Y and Z.[1]
(a)(iv)Calculate the ratio [HCl] dissolved in solution Y ────────────── [Cl(CH2)3CO2H] dissolved in solution Z[2]
(b)A buffer solution of pH 5.00 is produced by adding sodium propanoate to 5.00 g of propanoic acid in 100 cm³ of distilled water. Calculate the mass of sodium propanoate that must be used to produce this buffer solution. The Ka of propanoic acid is 1.35 × 10⁻⁵ moldm⁻³. [Mr: propanoic acid, 74.0; sodium propanoate, 96.0][3]
(c)Some dilute sulfuric acid is mixed with a small sample of the buffer solution described in (b). The final pH of the mixture is close to 1. Explain this observation.[2]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.

[1] Ka = [H+][Cl(CH2)3 CO2-]/[Cl(CH2)3 CO2H]
[1] pKa = - logKa OR Ka = 10-pKa
[1] [H+] = 10–4.0 = 1 x 10-4
[1] [HCl] = 1 × 10-4 ecf 2(a)(iii)
[1] Ka= 10-4.52 = 3.02 × 10-5 [Cl(CH2)3CO2H] = (1 × 10-4)2 / 3.02 × 10-5 = 3.3 x 10-4 ecf
[1] [HCI] / [Cl(CH2)3 CO2H] = 1×10-4 / 3.3×10-4 = 0.302 min 2sf ecf
[1] [H+] = 10-5 OR 1 × 10-5 Ka = [H+][A-]/ [HA] OR pH = pKa + log [A-] / [HA]
[1] moles of A¯ = 0.0912 ecf
[1] mass of sodium propanoate = 0.0912 × 96 = 8.76 min 2sf ecf
[1] all of the (sodium) propanoate (ion) has been protonated/converted to (propanoic) acid / neutralised
[1] H+ is in excess / H⁺ is 0.1 mol dm-³ (from the H2SO4)

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