Q12 / 160 · AP EAMCET 2022 (07 Jul Shift 2) · Chemistry
At $25^{\circ} \mathrm{C}$, the solubility product of $\mathrm{MCl}$ is $1 \times 10^{-10}$. What is its molar solubility in $0.1 \mathrm{M}$ $\mathrm{NaCl}$ solution at same temperature?
- A$0.1$
- B$0.05$
- Correct answerC$10^{-9}$
- D$10^{-5}$
Solution
$$
K_{\text {sp }} \text { of } M C l=1 \times 10^{-10} \text {. }
$$
Let the molar solubility of $\mathrm{MCl}$ in $0.1 \mathrm{M} \mathrm{NaCl}$ be ' $S$ ' $\mathrm{mol} \mathrm{L}^{-1}$.
$$
\mathrm{MCl} \longrightarrow \mathrm{M}^{+}+\mathrm{Cl}^{-}
$$
The concentration of $\mathrm{Cl}^{-}$will be $(S+0.1) \mathrm{mol} \mathrm{L}^{-1}$, as $0.1 \mathrm{~mol} \mathrm{~L}^{-1}$ are provided by $0.1 \mathrm{M} \mathrm{NaCl}$.
$$
\begin{aligned}
& K_{\text {sp }}=\left[M^{+}\right]\left[\mathrm{Cl}^{-}\right] \\
& K_{\text {sp }}=S \times(S+0.1)=1 \times 10^{-10}
\end{aligned}
$$
As the $K_{\text {sp }}$ is very small, the $S \ll 0.1$ and therefore can be ignored.
$$
S \times 0.1=1 \times 10^{-10} \Rightarrow S=10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}
$$
Hence, the correct answer is option: 'C'.