question_answer 1)

Calculate the degree of ionization of 0.05 M acetic acid if its \[p{{K}_{a}}\] value is 4.74. How is the degree of dissociation affected when its solution also contains (a) 0.01 M HCl (b) 0.1 M HCl?

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: \[HCl\to {{H}^{+}}+C{{l}^{-}}\] \[[{{H}^{+}}]=0.1M\] \[\underset{\begin{smallmatrix} t\,=\,0 \\ {{t}_{eq}} \end{smallmatrix}}{\mathop{{}}}\,\,\,\,\,\,\underset{\begin{smallmatrix} 0.05 \\ 0.05(1-\alpha ) \\ \approx 0.05 \end{smallmatrix}}{\mathop{C{{H}_{3}}COOH}}\,\rightleftharpoons \underset{\begin{smallmatrix} 0 \\ 0.05\alpha \end{smallmatrix}}{\mathop{C{{H}_{3}}CO{{O}^{-}}}}\,+\underset{\begin{smallmatrix} 0.1 \\ (0.1+0.05\alpha ) \\ =\,0.1 \end{smallmatrix}}{\mathop{{{H}^{+}}}}\,\] \[{{K}_{a}}=\frac{[C{{H}_{3}}CO{{O}^{-}}][{{H}^{+}}]}{[C{{H}_{3}}COOH]}\] \[1.82\times {{10}^{-5}}=\frac{0.05\alpha \times 0.1}{0.05}\] \[\alpha =1.82\times {{10}^{-4}}\]