\(\therefore \mathrm{K}_{\mathrm{a}}=(x)(x) / 0.05-x=1.0 \times 10^{-10}\) (Given)
or \(x^2 / 0.05=1.0 \times 10^{-10}\)
or \(x^2=5 \times 10^{-12}\)
or \(x=2.23 \times 10^{-6} \mathrm{M}\)
In presence of \(0.01 \mathrm{M} \mathrm{C}_6 \mathrm{H}_5 \mathrm{ONa}\), suppose \(y\) is the amount
of phenol dissociated, then at equilibrium
\({\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}\right]=0.05-y \approx 0.05 }\)
\({\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{O}^{-}\right]=0.01+y \approx 0.01 \mathrm{M},\left[\mathrm{H}^{+}\right]=y \mathrm{M} }\)
\(\mathrm{K}_{\mathrm{a}}=(0.01)(y) / 0.05=1.0 \times 10^{-10}\) (Given)
or \(y=5 \times 10^{-10}\)
\(\alpha=\frac{y}{\mathrm{c}}=\frac{5 \times 10^{-10}}{5 \times 10^{-2}}=10^{-8}\)