If pH of a saturated solution of Ba( OH)_2 is 12 , the value of its K_s p is

Given, $\mathrm{pH}$pH of $\mathrm{Ba}(\mathrm{OH})_2=12$Ba(OH)2​=12 $$\begin{aligned} \therefore \quad\left[\mathrm{H}^{+}\right] & =\left[1 \times 10^{-12}\right] \\ \text { and }\left[\mathrm{OH}^{-}\right] & =\frac{1 \times 10^{-14}}{1 \times 10^{-12}}\left[\because\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=1 \times 10^{-14}\right] \end{aligned}$$∴[H+] and [OH−]​=[1×10−12]=1×10−121×10−14​[∵[H+][OH−]=1×10−14]​ $$\begin{aligned} &=1 \times 10^{-2} \mathrm{~mol} / \mathrm{L} \\ & \mathrm{Ba}(\mathrm{OH})_2 \longrightarrow \mathrm{Ba}^{2+}+2 \mathrm{OH}^{-} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2 \\ &=[\mathrm{s}][2 \mathrm{~s}]^2 \\ &=\left[\frac{1 \times 10^{-2}}{2}\right]\left(1 \times 10^{-2}\right)^2 \\ &=0.5 \times 10^{-6}=5.0 \times 10^{-7} \mathrm{M}^3 \end{aligned}$$​=1×10−2 mol/LBa(OH)2​⟶Ba2++2OH−Ksp​=[Ba2+][OH−]2=[s][2 s]2=[21×10−2​](1×10−2)2=0.5×10−6=5.0×10−7M3​