⇒Ionic Equilibrium

For a monoacidic base,

$\begin{array}{rl}& \left[{\mathrm{OH}}^{-}\right]=\mathrm{c}\times \alpha \\ & \left[{\mathrm{OH}}^{-}\right]=0.04\times \frac{3}{100}\\ & \left[{\mathrm{OH}}^{-}\right]=1.2\times {10}^{-3}\mathrm{mol}{\mathrm{L}}^{-1}\end{array}$

$\underset{0.005\mathrm{M}}{\mathrm{NaOH}}⟶\underset{0.005\mathrm{M}}{{\mathrm{Na}}^{+}}+\underset{0.005\mathrm{M}}{{\mathrm{OH}}^{-}}$

$\begin{array}{rl}\mathrm{pOH}& =-{\log }_{10}[\mathrm{OH}]\\ & =-{\log }_{10}[0.005]\\ & =-{\log }_{10}(5\times {10}^{-3})\\ & =-[\log {10}^{-3}+\log 5]\\ & =-[-3+0.699]\\ & =-[-2.301]\\ & =2.301\\ \mathrm{pOH}& +\mathrm{pH}=14\\ \therefore \mathrm{pH}& =14-2.301\\ & =11.699\\ & \approx 11.7\end{array}$

$\begin{array}{ll}& {\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]\\ \therefore & {10}^{-14}=0.05\times \left[{\mathrm{OH}}^{-}\right]\\ & [\mathrm{OH}]=\frac{{10}^{-14}}{0.05}\\ & \left[{\mathrm{OH}}^{-}\right]=2.0\times {10}^{-13}\mathrm{M}\end{array}$

Perchloric acid is a strong monobasic acid.

Hence, $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=1.36\times {10}^{-2}\mathrm{M}$

$\begin{array}{rl}\therefore \mathrm{pH}& =-{\log }_{10}\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\\ & =-{\log }_{10}[1.36\times {10}^{-2}]\\ & =-{\log }_{10}1.36-{\log }_{10}{10}^{-2}\\ & =-{\log }_{10}1.36+2\\ & =2-0.1335\\ \therefore \mathrm{pH}& =1.86\end{array}$

$\begin{array}{ll}& \mathrm{pOH}=11\\ & \mathrm{pH}+\mathrm{pOH}=14\\ \therefore & \mathrm{pH}=14-\mathrm{pOH}=14-11=3\\ & \mathrm{pH}=-{\log }_{10}\left[{\mathrm{H}}^{+}\right]\\ \therefore & \left[{\mathrm{H}}^{+}\right]={10}^{-\mathrm{pH}}={10}^{-3}\mathrm{M}\end{array}$