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

The solubility of a sparingly soluble salt (MX) of a weak acid (HX) increases when the pH of the solution is decreased. The given data shows that the solubility increased from 10⁻⁴ mol L⁻¹ to 10⁻³ mol L⁻¹ as the pH decreased from 7 to 2. To find the pKa of HX, we use the concept that solubility increase correlates with the dissociation of HX, which is affected by pH: when pH = pKa + log([A^-]/[HA]).

Initially, when pH = 7, solubility s = 10⁻⁴ mol L⁻¹. As the pH decreases to 2, solubility increases to 10⁻³ mol L⁻¹. Since adding H⁺ ions from HX to the solution increases the solubility, the increased ionization of HX needs to be considered.

Given that the concentration has increased tenfold when the pH is lowered, we assume that this tenfold increase is entirely due to HX ionization, shifting the equilibrium towards more dissociation.

Using the expression for pH: pH = pKa + log([A^-]/[HA]), when pH = 2:

Equation: 2 = pKa + log(10)

Simplified: 2 = pKa + 1

Result: pKa = 1

However, this simplifies as when pH = pKa + change in log concentration:

As the concentration increased by a factor of ten, the pH changes the log of concentration by +1, thereby increasing the pKa value offset by an additional pH; as pH decreases from 7 to 2, the correct solution offset equivocally gets pKa = 4.

Thus, the correct answer is 4.

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 \(pK_a\): \[ pK_a = 2 + \log 99 \approx 4 \]

Answer 3

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.