Q63 / 200 · NEET 2012 (Screening) · Chemistry
$\mathrm{pH}$ of a saturated solution of $\mathrm{Ba}(\mathrm{OH})_2$ is 12. The value of solubility product $K_{\text {sp }}$ of $\mathrm{Ba}(\mathrm{OH})_2$ is
- A$3.3 \times 10^{-7}$
- Correct answerB$5.0 \times 10^{-7}$
- C$4.0 \times 10^{-6}$
- D$5.0 \times 10^{-6}$
Solution
Given, $\mathrm{pH}$ of $\mathrm{Ba}(\mathrm{OH})_2=12$
$\begin{aligned}
\therefore \mathrm{pOH} & =14-\mathrm{pH} \\
& =14-12=2
\end{aligned}$
We know that,
$\begin{aligned}
\mathrm{pOH} & =-\log \left[\mathrm{OH}^{-}\right] \\
2 & =-\log \left[\mathrm{OH}^{-}\right] \\
{\left[\mathrm{OH}^{-}\right] } & =\operatorname{antilog}(-2) \\
{\left[\mathrm{OH}^{-}\right] } & =1 \times 10^{-2}
\end{aligned}$
$\mathrm{Ba}(\mathrm{OH})_2$ dissolves in water as
$\begin{aligned}
& \mathrm{Ba}(\mathrm{OH})_2(s) \rightleftharpoons \mathrm{Ba}_s^{2+}+2 \mathrm{OH}_2^{-} \\
& s \mathrm{~mol} \mathrm{~L}^{-1} \\
& \therefore \quad\left[\mathrm{OH}^{-}\right]=2 s=1 \times 10^{-2} \\
& {\left[\mathrm{Ba}^{2+}\right] }=\frac{\left[\mathrm{OH}^{-}\right]}{2}=\frac{1 \times 10^{-2}}{2} \\
& K_{\text {sp }}=\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2
\end{aligned}$
$\begin{aligned}
& =\left(\frac{1 \times 10^{-2}}{2}\right)\left(1 \times 10^{-2}\right)^2 \\
& =0.5 \times 10^{-6}=5 \times 10^{-7}
\end{aligned}$
Hence, the correct answer is option: 'B'.