\(p O H=p K_{b}+\log \frac{[\text { salt }]}{\text { loase }}\)
\(p H+p O H=14\)
\(K_{b}=1 \times 10^{-10},[\text { salt }]=[b a s e]\)
\(p O H=-\log K_{b}+\log \frac{[\text {salt}]}{[b a s e]}\)
\(\therefore \quad p O H=-\log \left(1 \times 10^{-10}\right)+\log 1=10\)
\(p H+p O H=14\)\(\left[\because \text { concentration of }\left[B^{-}\right]=[H B]\right.\)
\(p H=14-10=4\)