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

The equilibrium expression for the acidity constant, K_a, is given by:

$$K_a = \frac{[\text{H}_3\text{O}^+][\text{HCOO}^-]}{[\text{HCOOH}]}$$

To find the concentration of H3O^+ ions, we first note the initial concentrations:

• Initial concentration of HCOOH is: $\frac{0.500 \text{ mol}}{2.00 \text{ L}} = 0.250 \text{ M}$

• Initial concentration of HCOONa is: $\frac{0.100 \text{ mol}}{2.00 \text{ L}} = 0.050 \text{ M}$

Using the assumptions provided, we set up the equilibrium:

• Let the concentration of H3O^+ at equilibrium be x.
• Thus, the equilibrium concentrations are:
  • [HCOOH] = 0.250 M (approximately equal to initial)
  • [HCOO^-] = 0.050 M + x (approximating HCOO^- contribution from dissociation)

Applying the equilibrium expression:

$$K_a = \frac{[\text{H}_3\text{O}^+][\text{HCOO}^-]}{[\text{HCOOH}]} \implies K_a = \frac{x(0.050 + x)}{0.250}$$

Assuming ( K_a ) for methanoic acid is approximately 1.8 x 10^-4, we can solve:

$$1.8 \times 10^{-4} = \frac{x(0.050)}{0.250}\ x = 3.6 \times 10^{-3} \text{ M}$$

The pH is calculated using the concentration of H3O^+:

$$pH = -\log([\text{H}_3\text{O}^+]) = -\log(3.6 \times 10^{-3}) \approx 2.44$$

Solution A contains only methanoic acid, whereas Solution B contains both methanoic acid and sodium methanoate. The presence of sodium methanoate introduces a common ion effect which shifts the equilibrium:

$$\text{HCOOH}(aq) \rightleftharpoons \text{HCOO}^-(aq) + \text{H}_3\text{O}^+(aq)$$

According to Le Chatelier's Principle, the addition of HCOO^- from sodium methanoate will shift the equilibrium to the left, reducing the concentration of H3O^+ in Solution B relative to Solution A. Thus, Solution B will exhibit a higher pH due to the mitigated acidity from the common ion effect.