The pH of a 0.1 M solution of NH₄OH (having dissociation constant $\left.{\mathrm{K}}_{\mathrm{b}}=1.0\times {10}^{-5}\right)$ is equal to

${\mathrm{K}}_{\mathrm{b}}={10}^{-5},{\mathrm{pK}}_{\mathrm{b}}=5$

NH₄OH is a week base

$\begin{array}{l}\therefore \left[\begin{array}{c}\mathrm{OH}\end{array}\right]=\mathrm{C\alpha }\\ \left[\begin{array}{c}\ominus \\ \mathrm{OH}\end{array}\right]=\sqrt{{\mathrm{K}}_{\mathrm{b}}\times \mathrm{C}}\\ \mathrm{pOH}=\frac{1}{2}\left({\mathrm{pK}}_{\mathrm{b}}-\log \mathrm{C}\right)=\frac{1}{2}(5-\log 0.1)=\frac{1}{2}(5+1)=3\\ \mathrm{pH}=14-\mathrm{pOH}=14-3=11\end{array}$