a. i - VCE - SSCE Chemistry - Question 6 - 2011 - Paper 1

The equilibrium expression for the acidity constant, $K_a$, for the dissociation of methanoic acid is given by:

$K_a = \frac{[H_3O^+][HCOO^-]}{[HCOOH]}$

To determine the concentration of $H_3O^+$ ions in the solution:

1. Initial Concentrations:

   • Initial concentration of methanoic acid (HCOOH): $C_{HCOOH} = \frac{0.500 \text{ mol}}{2.00 \text{ L}} = 0.250 \, M$
   • Initial concentration of sodium methanoate (HCOONa): $C_{HCOO^-} = \frac{0.100 \text{ mol}}{2.00 \text{ L}} = 0.050 \, M$

2. Assumptions:

   • At equilibrium, the concentration of $HCOOH$ remains approximately 0.250 M (given, as it is a weak acid).
   • Because both HCOOH and HCOO^- are in equilibrium and $HCOO^-$ acts as a base, we can assume: $HCOO^- + H_3O^+ \rightleftharpoons HCOOH + H_2O$

Using the equilibrium concentrations:

• Let $x$ be the change in concentration since $HCOO^-$ will react:
• $[HCOO^-] = 0.050 - x$
• Setting $[H_3O^+] = x$ at equilibrium gives: $K_a = \frac{x(0.050 - x)}{0.250}$ For weak acid approximations and small values of $x$, this simplifies to: $K_a = \frac{x \cdot 0.050}{0.250}$
• For methanoic acid, $K_a \approx 1.77 \times 10^{-4}$ (from reference data): $1.77 \times 10^{-4} = \frac{x \cdot 0.050}{0.250}$
• Solving for x: $x = \frac{(1.77 \times 10^{-4}) \cdot 0.250}{0.050} = 8.85 \times 10^{-4}$

Therefore, the concentration of $H_3O^+$ is: $[H_3O^+] \approx 8.85 \times 10^{-4} \, M$

To calculate the pH of the solution, we use the formula:

$pH = -\log[H_3O^+]$

Substituting the value we calculated:

$pH = -\log(8.85 \times 10^{-4}) \approx 3.05$

Solution B has a higher pH compared to Solution A. This can be justified by considering the equilibria involved in both solutions:

• Solution A contains only methanoic acid, which partially dissociates in solution:

$HCOOH \rightleftharpoons H_3O^+ + HCOO^-$

• Solution B contains both methanoic acid and its salt (sodium methanoate), which provides more $HCOO^-$ ions:

$HCOO^- + H_3O^+ \rightleftharpoons HCOOH + H_2O$

This reaction shifts the equilibrium to favor the formation of $HCOOH$, resulting in lower concentration of $H_3O^+$ ions in Solution B. Thus, the pH of Solution B is higher than that of Solution A.