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

The equilibrium expression for the acidity constant, K_a, for the reaction is:

$K_a = \frac{[\text{H}_3O^+][\text{HCOO}^-]}{[\text{HCOOH}]}$

To determine the concentration of H_3O^+ ions, we first establish the initial concentrations:

• Initial concentration of HCOOH = ( \frac{0.500 \text{ mol}}{2.00 \text{ L}} = 0.250 \text{ M} )
• Initial concentration of HCOO^- = ( \frac{0.100 \text{ mol}}{2.00 \text{ L}} = 0.050 \text{ M} )

Assuming that HCOOH only partially dissociates, we let 'x' represent the change in concentration at equilibrium. Thus, at equilibrium:

• [HCOOH] = 0.250 - x
• [HCOO^-] = 0.050 + x
• [H_3O^+] = x

Substituting into the K_a expression:

$K_a = \frac{x(0.050+x)}{(0.250-x)}$

Given that HCOOH is a weak acid, we assume that x is much smaller than 0.250. We can simplify:

$K_a = 1.8 \times 10^{-4} = \frac{x(0.050)}{0.250}$

Solving for 'x':
$x = \frac{(1.8 \times 10^{-4})(0.250)}{0.050} = 9.0 \times 10^{-4} \text{ M}$

Thus, the concentration of H_3O^+ ions is approximately 9.0 \times 10^{-4} \text{ M}.

The pH can be calculated using the concentration of H_3O^+ ions:

$\text{pH} = -\log[\text{H}_3O^+] = -\log(9.0 \times 10^{-4}) \approx 3.05$

Solution B will have a higher pH than Solution A.

This is because Solution A contains only methanoic acid, while Solution B contains both methanoic acid and the salt sodium methanoate. The presence of sodium methanoate increases the concentration of HCOO^- ions, which shifts the equilibrium:

$\text{HCOOH}(aq) \rightleftharpoons \text{HCOO}^-(aq) + \text{H}_3O^+(aq)$

towards the left, reducing the concentration of H_3O^+ ions, and thus raising the pH of the solution.

Therefore, due to the buffer effect, Solution B has the higher pH compared to Solution A.