Chemistry - Ionic Equilibrium Question with Solution | TestHub

Key Idea (i) For basic buffer $$\mathrm{pOH}={\mathrm{pK}}_{\mathrm{b}}+\log \frac{\text{ [salt }]}{[\text{ base }]}$$ (ii) $\mathrm{pH}+\mathrm{pOH}=14$ Given, ${\mathrm{K}}_{\mathrm{b}}=1\times 10^{-10}$, [salt] = [base] $$\begin{array}{rl}\mathrm{pOH}& =-\log {\mathrm{K}}_{\mathrm{b}}+\log \frac{[\text{ salt }]}{[\text{ base }]}\\ \therefore \mathrm{pOH}& =-\log \left(1\times 10^{-10}\right)+\log 1\\ & =10\\ \mathrm{pH}+\mathrm{pOH}& =14\\ \mathrm{pH}& =14-10=4\end{array}$$