$25^{\circ}C$ তাপমাত্রায় $Ca(OH)_{2}$ এর $K_{sp} = 3.2 \times 10^{-14}$ হলে দ্রবণের pH মান কত?

$\mathrm{Ca}(\mathrm{OH})_{2} \rightleftharpoons \mathrm{Ca}^{2}+2 \mathrm{OH}$ $S~~~~~~~~~~~~~~~~~~~S~~~~~~~~~~~~~2S$ $\begin{array}{l} K_{S P}=4 s^{3} \\ \Rightarrow 3.2 \times 10^{-14}=4 s^{3} \\ \Rightarrow S=\sqrt[3]{\frac{3.2 \times 10^{-14}}{4}} \\ \Rightarrow S=2.5 \times 10^{-5}\end{array}$

$\begin{aligned} \therefore p O H & =-\log \left(2.5 \times 10^{-5}\right) \\ & =4.7 \\ \therefore P H & =14-4.7 \\ & =9.3\end{aligned}$