AS & A Level Chemistry 32.1 Alcohols: Exam Style Questions Paper 4

Question

(a) Aqueous solutions of methanoic acid, HCOOH, and propanoic acid, CH₃CH₂COOH, are mixed together. An equilibrium is set up between two conjugate acid–base pairs.

(i) Define conjugate acid–base pair.

(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.

(b) (i) Write an expression for the acid dissociation constant, \(K_a\), for butanoic acid, CH₃CH₂CH₂COOH.

(ii) The \(pK_a\) of CH₃CH₂CH₂COOH is 4.82. A solution of CH₃CH₂CH₂COOH(aq) has a pH of 3.25. Calculate the concentration, in mol/dm³, of CH₃CH₂CH₂COOH in this solution.

(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 mol/dm³ CH₃COONa. The buffer solution has pH 5.70. The Ka of CH₃COOH is 1.78 × 10⁻⁵ moldm⁻³. Calculate the initial concentration, in mol/dm³, of CH₃COOH used.

(d) 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: \(O_2 + 4H^+ + 4e^– \to 2H_2O\).
In this fuel cell, the overall cell reaction is the same as that for the complete combustion of HCOOH.
(i) Deduce the half‑equation for the reaction at the anode.

(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.

▶️ Answer/Explanation

Solution

(a)(i) A conjugate acid–base pair consists of two species that differ by the presence or absence of a proton (\(H^+\)). For example, \(HCOOH\) (acid) and \(HCOO^-\) (base) form a conjugate pair.

(a)(ii) The stronger acid is HCOOH (lower pKa = 3.75). The equilibrium equation is: \[ HCOOH + H_2O \rightleftharpoons H_3O^+ + HCOO^- \] Explanation: The stronger acid donates a proton to water, forming hydronium ions (\(H_3O^+\)) and the conjugate base (\(HCOO^-\)).

(b)(i) The \(K_a\) expression for butanoic acid is: \[ K_a = \frac{[H^+][CH_3CH_2CH_2COO^-]}{[CH_3CH_2CH_2COOH]} \]

(b)(ii) Given pH = 3.25, \([H^+] = 10^{-3.25} = 5.62 \times 10^{-4} \ \text{mol/dm}^3\). Using \(K_a = 10^{-4.82} = 1.51 \times 10^{-5}\): \[ 1.51 \times 10^{-5} = \frac{(5.62 \times 10^{-4})^2}{[CH_3CH_2CH_2COOH]} \] Solving gives \([CH_3CH_2CH_2COOH] = 0.0209 \ \text{mol/dm}^3\). Explanation: The concentration is derived from the \(K_a\) expression, assuming \([H^+] \approx [CH_3CH_2CH_2COO^-]\).

(c)(i) A buffer solution resists changes in pH when small amounts of acid or base are added. It contains a weak acid and its conjugate base (or vice versa).

(c)(ii) Using the Henderson-Hasselbalch equation: \[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \right) \] Substituting pH = 5.70, \(K_a = 1.78 \times 10^{-5}\) (pKa = 4.75), and \([CH_3COO^-] = 0.125 \times \frac{400}{1000} = 0.05 \ \text{mol}\): \[ 5.70 = 4.75 + \log \left( \frac{0.05}{[\text{CH}_3\text{COOH}]} \right) \] Solving gives \([CH_3COOH] = 0.0562 \ \text{mol/dm}^3\). Explanation: The initial concentration is calculated using the buffer equation and mole ratios.

(d)(i) The anode half-reaction is: \[ HCOOH \to CO_2 + 2H^+ + 2e^- \] Explanation: Methanoic acid is oxidised to CO₂, releasing protons and electrons.

(d)(ii) Total charge \(Q = 3.75 \ \text{A} \times 2400 \ \text{s} = 9000 \ \text{C}\). Moles of electrons = \(9000 / 96485 = 0.0933 \ \text{mol}\). Since \(O_2 + 4e^- \to 2H_2O\), moles of \(O_2 = 0.0933 / 4 = 0.0233 \ \text{mol}\). Volume at RTP = \(0.0233 \times 24000 = 560 \ \text{cm}^3\). Explanation: The volume of oxygen is calculated using Faraday’s laws and stoichiometry.

AS & A Level Chemistry 32.1 Alcohols: Exam Style Questions Paper 4

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