Answer:
The ionization
bromoacetic acid will take place as,
\[\underset{{{t}_{\begin{smallmatrix}
0 \\
{{t}_{eq}}
\end{smallmatrix}}}}{\overset{{}}{\mathop{{}}}}\,\underset{\begin{smallmatrix}
C \\
C-C\alpha
\end{smallmatrix}}{\mathop{BrC{{H}_{2}}COOH}}\,\rightleftharpoons
\underset{\begin{smallmatrix}
0 \\
C\alpha
\end{smallmatrix}}{\mathop{BrC{{H}_{2}}CO{{O}^{-}}}}\,+\underset{\begin{smallmatrix}
0 \\
C\alpha
\end{smallmatrix}}{\mathop{{{H}^{+}}}}\,\]
\[[{{H}^{+}}]=C\alpha =0.1\times 0.132\]
\[=0.0132M\]
\[pH=-{{\log }_{10}}[{{H}^{+}}]\]
\[pH=-\log [0.0132]\]
\[=1.879\]
Dissociation constant may be calculated as:
\[{{K}_{a}}=\frac{C\alpha \times C\alpha
}{C-C\alpha }=\frac{C{{\alpha }^{2}}}{1-\alpha }\]
\[\approx C{{\alpha }^{2}}\]
\[{{K}_{a}}=0.1\times {{(0.132)}^{2}}\]
\[=1.74\times {{10}^{-3}}\]
\[p{{K}_{a}}=-{{\log }_{10}}{{K}_{a}}\]
\[=-\log 1.74\times {{10}^{-3}}\]
= 2.758
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