Question

The solubility product of NiS is $4.9 \times 10^{-5}$ at 298 K . Calculate its solubility in moldm$^{-3}$ at the same temperature?

• A. $1.69 \times 10^{-3}$
• B. $7.0 \times 10^{-3}$
• C. $2.45 \times 10^{-3}$
• D. $6.18 \times 10^{-3}$

Hint:

For salts of type $AB$, Solubility $S = \sqrt{K_{sp}}$.

Answer 1

The Correct Option is B

Step 1: Dissociation Equillibrium
$\text{NiS}_{(s)} \rightleftharpoons \text{Ni}^{2+}_{(aq)} + \text{S}^{2-}_{(aq)}$
If solubility is $S$, then $[\text{Ni}^{2+}] = S$ and $[\text{S}^{2-}] = S$.
Step 2: Formula
$K_{sp} = [\text{Ni}^{2+}][\text{S}^{2-}] = S^2$
Step 3: Calculation
$S = \sqrt{K_{sp}} = \sqrt{4.9 \times 10^{-5}}$
$S = \sqrt{49 \times 10^{-6}}$
$S = 7.0 \times 10^{-3} \text{ mol dm}^{-3}$.
Final Answer: (B)