If $\mathrm{pH}$ of a saturated solution of $\mathrm{Ba}(\mathrm{OH})_2$ is 12 , the value of its $K_{s p}$ is

If $\mathrm{pH}$ of a saturated solution of $\mathrm{Ba}(\mathrm{OH})_2$ is 12 , the value of its $K_{s p}$ is
  1. $4.00 \times 10^{-6} \mathrm{M}^3$
  2. $4.00 \times 10^{-7} \mathrm{M}^3$
  3. $5.00 \times 10^{-6} \mathrm{M}^3$
  4. $5.00 \times 10^{-7} \mathrm{M}^3$

Solution

Given, $\mathrm{pH}$ of $\mathrm{Ba}(\mathrm{OH})_2=12$ $\begin{aligned} \therefore \quad\left[\mathrm{H}^{+}\right] & =\left[1 \times 10^{-12}\right] \\ \text { and }\left[\mathrm{OH}^{-}\right] & =\frac{1 \times 10^{-14}}{1 \times 10^{-12}}\left[\because\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=1 \times 10^{-14}\right] \end{aligned}$ $\begin{aligned} &=1 \times 10^{-2} \mathrm{~mol} / \mathrm{L} \\ & \mathrm{Ba}(\mathrm{OH})_2 \longrightarrow \mathrm{Ba}^{2+}+2 \mathrm{OH}^{-} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2 \\ &=[\mathrm{s}][2 \mathrm{~s}]^2 \\ &=\left[\frac{1 \times 10^{-2}}{2}\right]\left(1 \times 10^{-2}\right)^2 \\ &=0.5 \times 10^{-6}=5.0 \times 10^{-7} \mathrm{M}^3 \end{aligned}$

Asked in: NEET 2010 (Screening)

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