Chemistry - Ionic Equilibrium Question with Solution | TestHub

Given at $330\mathrm{K}$ $$\begin{array}{rl}{\mathrm{K}}_w& =10^{-13.6}\\ \text{ i.e. }{\mathrm{pK}}_w& =\mathrm{pH}+\mathrm{pOH}\\ \because {\mathrm{pOH}}^2& =-\log \left[{\mathrm{OH}}^{-}\right]\\ 13.6& =\mathrm{pH}+\mathrm{pOH}\\ \mathrm{pOH}& =-\log 10^{-4}\\ 13.6& =\mathrm{pH}+4\\ \therefore \mathrm{pH}& =13.6-4\\ & =9.6\end{array}$$