Question

The solubility product, \(K_{sp}\) of a sparingly soluble salt, \(\mathrm{AX_2}\), is \(3.2 \times 10^{-14}\) mol\(^3\) lit\(^{-3}\). Its solubility is

• A. $4 \times 10^{-3}$ mol lit$^{-1}$
• B. $2 \times 10^{-2}$ mol lit$^{-1}$
• C. $2 \times 10^{-5}$ mol lit$^{-1}$
• D. $8 \times 10^{-4}$ mol lit$^{-1}$
• E. $2 \times 10^{-3}$ mol lit$^{-1}$

Hint:

For $\mathrm{AX_2}$: - $K_{sp} = 4s^3$ - Solve cube root carefully

Answer 1

The Correct Option is C

Concept: Dissociation: \[ \mathrm{AX_2 \rightleftharpoons A^{2+} + 2X^-} \] Let solubility = $s$: \[ [A^{2+}] = s,\quad [X^-] = 2s \] \[ K_{sp} = [A^{2+}][X^-]^2 = s(2s)^2 = 4s^3 \]

Step 1: Substitute value.
\[ 3.2 \times 10^{-14} = 4s^3 \]

Step 2: Solve for $s$.
\[ s^3 = \frac{3.2 \times 10^{-14}}{4} = 0.8 \times 10^{-14} = 8 \times 10^{-15} \] \[ s = (8 \times 10^{-15})^{1/3} = 2 \times 10^{-5} \]