A-LevelChemistryEquilibria9701/42 · May/June 2024 · Q613 Marks

Aqueous solutions of methanoic acid, HCOOH, and propanoic acid, CH₃CH₂COOH, are mixed together. An equilibrium is set up between two conjugate acid-base pairs. Define conjugate acid-base pair.

(a)(i)Aqueous solutions of methanoic acid, HCOOH, and propanoic acid, CH₃CH₂COOH, are mixed together. An equilibrium is set up between two conjugate acid-base pairs. Define conjugate acid-base pair.[1]
(a)(ii)The pKₐ of HCOOH is 3.75 and of CH₃CH₂COOH is 4.87. Complete the equation for the Brønsted-Lowry equilibrium between the stronger of these two acids and water. ............................ + H₂O ⇌ ............................ + ............................ [1][1]
(b)(i)Write an expression for the acid dissociation constant, Kₐ, for butanoic acid, CH₃CH₂CH₂COOH. Kₐ = ................................... [1][1]
(b)(ii)The pKₐ of CH₃CH₂CH₂COOH is 4.82. A solution of CH₃CH₂CH₂COOH(aq) has a pH of 3.25. Calculate the concentration, in moldm⁻³, of CH₃CH₂CH₂COOH in this solution. concentration of CH₃CH₂CH₂COOH = ................................... moldm⁻³ [2][2]
(c)(i)Define buffer solution.[2]
(c)(ii)A buffer solution containing a mixture of CH₃COOH and CH₃COONa is prepared as follows. A solution of 600 cm³ of CH₃COOH is mixed with 400 cm³ of 0.125 moldm⁻³ CH₃COONa. The buffer solution has pH 5.70. The Kₐ of CH₃COOH is 1.78 × 10⁻⁵moldm⁻³. Calculate the initial concentration, in moldm⁻³, of CH₃COOH used. concentration of CH₃COOH = ................................... mol dm⁻³ [3][3]
(d)(i)A fuel cell is an electrochemical cell that can be used to generate electrical energy by using oxygen to oxidise a fuel. Methanoic acid, HCOOH, is being investigated as a fuel in fuel cells. When the cell operates, HCOOH is oxidised to carbon dioxide. The half-equation for the reaction at the cathode is: O₂ + 4H⁺ + 4e⁻ → 2H₂O. In this fuel cell, the overall cell reaction is the same as that for the complete combustion of HCOOH. Deduce the half-equation for the reaction at the anode.[1]
(d)(ii)Calculate the volume, in cm³, of oxygen used when a current of 3.75A is delivered by the cell for 40.0 minutes. Assume the cell operates at room conditions. volume of oxygen = ................................... cm³ [2][2]

Mark scheme

Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.

[1] species / molecules / pair that differ (by the presence or absence) of a H+ ion / proton
[1] HCOOH + H2O ⇌ H3O+ + HCOO-
[1] Ka = [H+] [CH3CH2CH2COO¯] / [CH3CH2CH2COOH]
[2] M1 Ka = 10–4.82 = 1.51 × 10–5 OR [H+] = 10–3.25 = 5.62 × 10–4
M2 [HA] = (10-3.25)2 ÷ 1.51 × 10–5 = 0.021 / 0.0209 / 0.02089 (mol dm³) ecf min 2sf
[2] M1 a solution that resists / opposes / minimises changes in pH
M2 when small amounts of acid / H+ and base / alkali/ OH- are added to it
[3] Method 1
M1 [H+] = 10-5.70 OR 2.00 × 10–6 OR 1.99526 × 10–6
M2 moles CH3COONa = 0.400 × 0.125 = 5.00 × 10-2
moles CH3COOHeqm = (5.00 × 10–2 × 2.00 × 10–6) ÷ 1.78 × 10–5 OR 5.60 × 10–3
M3 [CH3COOH]initial = 5.60 × 10-3 × 1000/600
[CH3COOH]initial = 0.00933–0.00936 (mol dm³) ecf min 2sf
Method 2
M1 pKa = -log(1.78 × 10–5) OR 4.75
M2 pH = pKa + log[B] / [HA]
5.7 = 4.75 + log (0.05/(0.6x))
0.95 = log (0.05 / (0.6x))
M3 100.95 = (0.05 / (0.6x))
5.348x = 0.05
x = 0.00935-0.00937 (mol dm-³) ecf min 2sf
[1] HCOOH → CO2 + 2H+ + 2e-
[2] M1 Q = 3.75 × 40 × 60 OR 9000 C AND use of 96500 OR 1.6 × 10-19 × 6.02 × 1023
M2 moles of oxygen = 9000 ÷ 386000 = 0.0233
volume of O2 = 0.0233 × 24000 = 559.6 / 559.8/560 cm³ ecf min 2sf
ALLOW [use of Q ÷ (1.6 × 10–19 × 6.02 × 1023)] = 560.6 cm³