Question

At T(K), \(K_{\text{sp}}\) of two ionic salts MX\(_2\) and MX is \(5 \times 10^{-13}\) and \(1.6 \times 10^{-11}\) respectively. The ratio of molar solubility of MX\(_2\) and MX is

• A. 12.5
• B. 1.25
• C. 6.25
• D. 7.50

Hint:

For a salt of type MX\(_2\), use \(K_{\text{sp}} = 4s^3\), and for MX, use \(K_{\text{sp}} = s^2\). Always simplify before calculating ratios.

Answer 1

The Correct Option is A

Step 1: Use relation between \(K_{\text{sp}}\) and solubility Let \(s_1\) be solubility of MX\(_2\) and \(s_2\) be solubility of MX. For MX\(_2\): \[ \text{MX}_2 \rightleftharpoons \text{M}^{2+} + 2\text{X}^- \Rightarrow K_{\text{sp}} = [\text{M}^{2+}][\text{X}^-]^2 = s_1 . (2s_1)^2 = 4s_1^3 \Rightarrow s_1 = \sqrt[3]{\frac{K_{\text{sp1}}}{4}} = \sqrt[3]{\frac{5 \times 10^{-13}}{4}} \] For MX: \[ \text{MX} \rightleftharpoons \text{M}^+ + \text{X}^- \Rightarrow K_{\text{sp}} = s_2^2 \Rightarrow s_2 = \sqrt{1.6 \times 10^{-11}} \] Step 2: Take the ratio of solubilities \[ \frac{s_1}{s_2} = \frac{\sqrt[3]{\frac{5 \times 10^{-13}}{4}}}{\sqrt{1.6 \times 10^{-11}}} \approx \frac{4.957 \times 10^{-5}}{3.99 \times 10^{-6}} \approx 12.5 \] Step 3: Final Answer
\[ \text{Ratio} = 12.5 \]