Answer:
\[p{{K}_{a}}=-\log
{{K}_{a}}=4.74\]
\[\therefore
\] \[{{K}_{a}}=\text{Antilog}(-4.74)\]
\[=1.82\times
{{10}^{-5}}\]
The degree of ionization of acetic acid may be calculated as:
\[\alpha
=\sqrt{\frac{{{K}_{a}}}{C}}=\sqrt{\frac{1.82\times {{10}^{-5}}}{0.05}}\]
= 0.01908
(a) In presence of 0.01 M HCl:
\[\underset{\begin{smallmatrix}
t=0 \\
{{t}_{eq}}
\end{smallmatrix}}{\overset{{}}{\mathop{{}}}}\,\,\,\,\,\underset{\begin{smallmatrix}
0.05 \\
0.05(1-\alpha )
\\
\approx 0.05
\end{smallmatrix}}{\mathop{C{{H}_{3}}COOH}}\,\rightleftharpoons
\underset{\begin{smallmatrix}
0 \\
0.05\alpha
\end{smallmatrix}}{\mathop{C{{H}_{3}}CO{{O}^{-}}}}\,+\underset{\begin{smallmatrix}
0.01 \\
(0.01+0.05\alpha )
\\
\approx 0.1
\end{smallmatrix}}{\mathop{{{H}^{+}}}}\,\]
\[{{K}_{a}}=\frac{[C{{H}_{3}}CO{{O}^{-}}][{{H}^{+}}]}{[C{{H}_{3}}COOH]}\]
\[1.82\times {{10}^{-5}}=\frac{0.05\alpha
\times 0.01}{[C{{H}_{3}}COOH]}\]
\[\alpha =1.82\times {{10}^{-3}}\]
(b) In presence of 0.1 M HCl:
\[HCl\to {{H}^{+}}+C{{l}^{-}}\]
\[[{{H}^{+}}]=0.1M\]
\[\underset{\begin{smallmatrix}
t\,=\,0 \\
{{t}_{eq}}
\end{smallmatrix}}{\mathop{{}}}\,\,\,\,\,\,\underset{\begin{smallmatrix}
0.05 \\
0.05(1-\alpha )
\\
\approx 0.05
\end{smallmatrix}}{\mathop{C{{H}_{3}}COOH}}\,\rightleftharpoons
\underset{\begin{smallmatrix}
0 \\
0.05\alpha
\end{smallmatrix}}{\mathop{C{{H}_{3}}CO{{O}^{-}}}}\,+\underset{\begin{smallmatrix}
0.1 \\
(0.1+0.05\alpha )
\\
=\,0.1
\end{smallmatrix}}{\mathop{{{H}^{+}}}}\,\] \[{{K}_{a}}=\frac{[C{{H}_{3}}CO{{O}^{-}}][{{H}^{+}}]}{[C{{H}_{3}}COOH]}\]
\[1.82\times {{10}^{-5}}=\frac{0.05\alpha
\times 0.1}{0.05}\]
\[\alpha =1.82\times {{10}^{-4}}\]
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