question_answer 78)

The degree of ionization of 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the \[p{{K}_{a}}\] of bromoacetic acid.

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