Question

If the ionic product of $M(OH)_2$ is $5 \times 10^{-10}$, then the molar solubility of $M(OH)_2$ in $0.1$ M NaOH is}

• A. $5 \times 10^{-12}$ M
• B. $5 \times 10^{-8}$ M
• C. $5 \times 10^{-10}$ M
• D. $5 \times 10^{-9}$ M
• E. $5 \times 10^{-16}$ M

Hint:

In common ion problems, always check whether added ion dominates equilibrium — it drastically reduces solubility.

Answer 1

The Correct Option is B

Concept: For sparingly soluble salts: \[ M(OH)_2 \rightleftharpoons M^{2+} + 2OH^- \] Solubility product: \[ K_{sp} = [M^{2+}][OH^-]^2 \] In presence of common ion (NaOH), $[OH^-]$ is already high, reducing solubility.

Step 1: Let solubility = $s$. \[ [M^{2+}] = s \]

Step 2: Hydroxide concentration. From NaOH: \[ [OH^-] = 0.1 \text{ M} \] Contribution from salt (2s) is negligible compared to 0.1.

Step 3: Apply $K_{sp$.} \[ K_{sp} = s (0.1)^2 \] \[ 5 \times 10^{-10} = s \times 10^{-2} \]

Step 4: Solve for $s$. \[ s = \frac{5 \times 10^{-10}}{10^{-2}} = 5 \times 10^{-8} \] But careful: since $K_{sp} = 5 \times 10^{-10}$ already includes squared OH, correct substitution: \[ s = \frac{5 \times 10^{-10}}{(0.1)^2} = \frac{5 \times 10^{-10}}{10^{-2}} = 5 \times 10^{-8} \]