$p O H=p K_{b}+\log \frac{[\text { salt }]}{\text { loase }}$
$p H+p O H=14$
$K_{b}=1 \times 10^{-10},[\text { salt }]=[b a s e]$
$p O H=-\log K_{b}+\log \frac{[\text {salt}]}{[b a s e]}$
$\therefore \quad p O H=-\log \left(1 \times 10^{-10}\right)+\log 1=10$
$p H+p O H=14$$\left[\because \text { concentration of }\left[B^{-}\right]=[H B]\right.$
$p H=14-10=4$
$[$આપેલ : પાણીમાં $Ca ( OH )_{2}$ નો દ્રાવ્યતા ગુણાકાર $=5.5 \times 10^{-6}$ છે.$]$
$\begin{array}{*{20}{c}}
{C{H_3}C{H_2}OH{\text{ }} + {\text{ }}{H_3}{O^ + }\, \to \,C{H_3}C{H_2} - {O^ + } - H\, + {H_2}O} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,H\,\,\,\,\,\,}
\end{array}$