What is the Ka for chloroacetic acid (C2H3ClO2) which has a pH of 1.52 in a 0.112 M solution? (4 marks) 1.
The Correct Answer and Explanation is :
To determine the acid dissociation constant (Ka) for chloroacetic acid (C₂H₃ClO₂) in a 0.112 M solution with a pH of 1.52, we can follow these steps:
1. Calculate the concentration of hydrogen ions ([H⁺]):
The pH of a solution is related to the concentration of hydrogen ions by the equation:
pH=−log[H+]\text{pH} = -\log[\text{H}^+]
Rearranging to solve for [H⁺]:
[H+]=10−pH[\text{H}^+] = 10^{-\text{pH}}
Substituting the given pH value:
[H+]=10−1.52≈0.0302 M[\text{H}^+] = 10^{-1.52} \approx 0.0302 \, \text{M}
2. Set up the dissociation equilibrium expression:
Chloroacetic acid dissociates in water as follows:
HA⇌H++A−\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-
The acid dissociation constant (Ka) is given by:
Ka=[H+][A−][HA]K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}
3. Determine the concentrations at equilibrium:
Let the initial concentration of chloroacetic acid be 0.112 M.
At equilibrium, the concentration of H⁺ ions is 0.0302 M.
Since each mole of HA dissociates to produce one mole of H⁺ and one mole of A⁻, the concentration of A⁻ at equilibrium is also 0.0302 M.
The concentration of undissociated HA at equilibrium is:
[HA]=0.112 M−0.0302 M=0.0818 M[\text{HA}] = 0.112 \, \text{M} – 0.0302 \, \text{M} = 0.0818 \, \text{M}
4. Calculate Ka:
Substitute the equilibrium concentrations into the Ka expression:
Ka=(0.0302 M)(0.0302 M)0.0818 MK_a = \frac{(0.0302 \, \text{M})(0.0302 \, \text{M})}{0.0818 \, \text{M}}
Ka≈0.000912040.0818K_a \approx \frac{0.00091204}{0.0818}
Ka≈1.11×10−5K_a \approx 1.11 \times 10^{-5}
Therefore, the acid dissociation constant (Ka) for chloroacetic acid in this solution is approximately 1.11 × 10⁻⁵.
This calculation demonstrates how the pH of a solution can be used to determine the strength of a weak acid by calculating its Ka.