A-LevelChemistryEquilibria9701/42 · Oct/Nov 2023 · Q211 Marks

Benzoic acid, C6H5COOH, is a weak acid. The K₂ of benzoic acid is 6.31 × 10–5 moldm-3 at 298 K. A 1.00dm³ buffer solution is made at 298 K containing 1.00g of C6H5COOH and a slightly greater mass of sodium benzoate, C6H5COO¯Na*. This buffer solution has a pH of 4.15.

(a)Define buffer solution.[1]
(b)(i)Write equations to show how this solution acts as a buffer solution when the named substances are added to it: dilute aqueous sodium hydroxide[1]
(b)(ii)dilute aqueous nitric acid.[1]
(c)Calculate the H+ concentration and the C6H5COOH concentration in the buffer solution described. Use the expression for the Ka of C6H5COOH to calculate the concentration of C6H5COO-Na+ in the buffer solution. Show your working and give each answer to a minimum of three significant figures. [H+] = ............................ moldm-3 [C6H5COOH] = ............................ mol dm-3 [C6H5COO-Na+] = ............................ mol dm-3[3]
(d)A 10.0 cm³ sample of the buffer solution is mixed with 10.0cm³ of 1.00moldm-³ KOH. Both solutions are at 298 K. A reaction is allowed to occur without stirring. Two observations are recorded: • the temperature, after the reaction is complete, is fractionally above 298 K • the pH, after the reaction, is greater than 13. Explain these two observations.[2]
(e)(i)Magnesium benzoate, Mg(C6H5COO)2, has a solubility in water of less than 1.00gdm-3 at 298 K. Ksp = [Mg2+][C6H5COO®]² = 1.76 × 10–7 at 298 K Calculate the solubility of Mg(C6H5COO)2 in water at 298 K. Give your answer in gdm-3. Show your working. [Mr: Mg(C6H5COO)2, 266.3] solubility = ............................ gdm-³[2]
(e)(ii)An excess of Mg(C6H5COO)2 is added to a sample of 0.50 moldm¬³ MgSO4 at 298K. State whether the equilibrium concentration of Mg(C6H5COO)2 is higher than, the same as, or lower than your answer to (i). Explain your answer. The concentration is ............................ the concentration in (i). explanation ..........................................................................................................................[1]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.

[1] (a) resists pH change when small amount of acid or alkali is added
[1] (b)(i) C6H5COOH + NaOH → C6H5COO-Na⁺ + H2O
[1] (b)(ii) C6H5COO-Na⁺ + HNO3 → C6H5COOH + NaNO3
[1] (c) [H+] = 7.08 × 10-5
[1] (c) [C6H5COOH] = 8.20 × 10®³
[1] (c) [C6H5COO-Na⁺] = 7.31 × 10–3
[1] (d) neutralisation is exothermic
[1] (d) All of the C6H5COOH has reacted AND excess KOH
[1] (e)(i) If [X] = [Mg2+] = [Mg(C6H5COO¯)2] 4X3 = 1.76 x 10-7 so X = 3.53 × 10-3
[1] (e)(i) 0.940 g dm-³
[1] (e)(ii) lower than AND common ion effect