pH of a saturated solution of Ba( OH)_2 is 12. The value of solubility product K_ sp of Ba( OH)_2 is

Given, $\mathrm{pH}$pH of $\mathrm{Ba}(\mathrm{OH})_2=12$Ba(OH)2​=12 $$\begin{aligned} \therefore \mathrm{pOH} & =14-\mathrm{pH} \\ & =14-12=2 \end{aligned}$$∴pOH​=14−pH=14−12=2​ We know that, $$\begin{aligned} \mathrm{pOH} & =-\log \left[\mathrm{OH}^{-}\right] \\ 2 & =-\log \left[\mathrm{OH}^{-}\right] \\ {\left[\mathrm{OH}^{-}\right] } & =\operatorname{antilog}(-2) \\ {\left[\mathrm{OH}^{-}\right] } & =1 \times 10^{-2} \end{aligned}$$pOH2[OH−][OH−]​=−log[OH−]=−log[OH−]=antilog(−2)=1×10−2​ $\mathrm{Ba}(\mathrm{OH})_2$Ba(OH)2​ dissolves in water as $$\begin{aligned} & \mathrm{Ba}(\mathrm{OH})_2(s) \rightleftharpoons \mathrm{Ba}_s^{2+}+2 \mathrm{OH}_2^{-} \\ & s \mathrm{~mol} \mathrm{~L}^{-1} \\ & \therefore \quad\left[\mathrm{OH}^{-}\right]=2 s=1 \times 10^{-2} \\ & {\left[\mathrm{Ba}^{2+}\right] }=\frac{\left[\mathrm{OH}^{-}\right]}{2}=\frac{1 \times 10^{-2}}{2} \\ & K_{\text {sp }}=\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2 \end{aligned}$$​Ba(OH)2​(s)⇌Bas2+​+2OH2−​s mol L−1∴[OH−]=2s=1×10−2[Ba2+]=2[OH−]​=21×10−2​Ksp ​=[Ba2+][OH−]2​ $$\begin{aligned} & =\left(\frac{1 \times 10^{-2}}{2}\right)\left(1 \times 10^{-2}\right)^2 \\ & =0.5 \times 10^{-6}=5 \times 10^{-7} \end{aligned}$$​=(21×10−2​)(1×10−2)2=0.5×10−6=5×10−7​