Question

On decreasing the 𝑝H from 7 to 2, the solubility of a sparingly soluble salt (MX) of a weak acid (HX) increased from 10⁻⁴ mol L⁻¹ to 10⁻³ mol L⁻¹. The 𝑝Ka of HX is

• A. 3
• B. 4
• C. 5
• D. 2

Answer 1

The Correct Option is B

Correct option is:(B) 4.

1. Initially, the solubility of the sparingly soluble salt MX is 10−410−4 mol L−1−1 at pH=7.
2. As the pH is decreased to 2, the solubility of MX increases to 10−310−3 mol L−1−1.

The increase in solubility with decreasing pH suggests that HX is undergoing a dissociation process as the pH decreases. This is characteristic of a weak acid (HX) undergoing dissociation:

HX⇌H++X−

The solubility of MX is directly related to the concentration of HX that dissociates into H+ and X−. Lowering the pH shifts the equilibrium to the right, leading to an increase in the concentration of H+, which in turn increases the solubility of MX.

From the given information, we can conclude that the pKa of HX is 44. This is because at pH=pKa, the concentrations of the dissociated +H+ and the undissociated HX are equal, resulting in a solubility of 10−410−4 mol L−1−1. As the pH drops below pKa, the concentration of +H+ exceeds the concentration of undissociated HX, leading to an increase in solubility.

Answer 2

The Correct Answer is  option  \(\mathbf{(B) \, 4}\) \[ \text{MX} \rightleftharpoons \text{M}^{\oplus} + \text{X}^{\ominus} \] \[ \text{X}^{\ominus} + \text{H}^{\oplus} \rightleftharpoons \text{HX} \] \[ S = \sqrt{\frac{K_{sp}}{1 + \frac{H^{\oplus}}{K_a}}} \] \[ 10^{-4} = \sqrt{\frac{K_{sp}}{1 + \frac{10^{-7}}{K_a}}} \quad \text{...(1)} \] \[ 10^{-3} = \sqrt{\frac{K_{sp}}{1 + \frac{10^{-2}}{K_a}}} \quad \text{...(2)} \] Dividing Equation (1) by Equation (2): \[ \frac{10^{-4}}{10^{-3}} = \sqrt{\frac{1 + \frac{10^{-7}}{K_a}}{1 + \frac{10^{-2}}{K_a}}} \] \[ 10^{-2} = \sqrt{\frac{1 + \frac{10^{-7}}{K_a}}{1 + \frac{10^{-2}}{K_a}}} \] Squaring both sides: \[ 10^{-4} + \frac{10^{-7}}{K_a} = 1 + \frac{10^{-7}}{K_a} \] Rearranging terms: \[ 10^{-4} - 10^{-7} = \frac{10^{-2}}{K_a} \] \[ 10^{-4} = 0.99 \frac{10^{-2}}{K_a} \] Solving for \(K_a\): \[ K_a = \frac{10^{-4}}{0.99} \times 10^{-2} \] \[ K_a \approx \frac{1}{99} \times 10^{-2} \] Calculating the logarithm for \(pK_a\): \[ pK_a = 2 + \log 99 \approx 4 \]