Answer:
Phenol
undergoes ionization as:
\[\underset{{{t}_{\begin{smallmatrix}
0 \\
{{t}_{eq}}
\end{smallmatrix}}}}{\overset{{}}{\mathop{{}}}}\,\underset{\begin{smallmatrix}
C \\
C-C\alpha
\end{smallmatrix}}{\mathop{{{C}_{6}}{{H}_{5}}OH}}\,\rightleftharpoons
\underset{\begin{smallmatrix}
Phenolate\,ion \\
0
\\
C\alpha
\end{smallmatrix}}{\mathop{{{C}_{6}}{{H}_{5}}{{O}^{-}}}}\,+\underset{\begin{smallmatrix}
0 \\
C\alpha
\end{smallmatrix}}{\mathop{{{H}^{+}}}}\,\]
\[{{K}_{a}}=\frac{[{{C}_{6}}{{H}_{5}}{{O}^{-}}][{{H}^{+}}]}{[{{C}_{6}}{{H}_{5}}OH]}=\frac{C\alpha
\times C\alpha }{(C-C\alpha )}\]
\[=\frac{C{{\alpha }^{2}}}{1-\alpha }\approx C{{\alpha
}^{2}}\]
\[\alpha
=\sqrt{\frac{{{K}_{a}}}{C}}=\sqrt{\frac{{{10}^{-10}}}{0.05}}=4.47\times
{{10}^{-5}}\]
Concentration of phenolate ion = \[C\alpha \]
\[=0.05\times 4.47\times {{10}^{-5}}\]
\[=2.235\times {{10}^{-8}}M\]
In presence of 0.01 M sodium phenolate, the degree of ionization
may be calculated as:
\[{{C}_{6}}{{H}_{5}}{{O}^{-}}N{{a}^{+}}\to
{{C}_{6}}{{H}_{5}}{{O}^{-}}+N{{a}^{+}}\]
\[[{{C}_{6}}{{H}_{5}}{{O}^{-}}]=0.01=C\alpha \]
Sodium phenolate is strong electrolyte then its phenolate
ion concentration will dominate the concentration of phenolate ion from phenol.
\[{{K}_{a}}=C{{\alpha }^{2}}\]
\[=C\alpha \times \alpha \]
\[{{10}^{-10}}=0.01\times \alpha \]
\[\alpha ={{10}^{-8}}\]
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