question_answer 1)

The ionization constant of propanoic acid is\[1.32\times {{10}^{-5}}\].Calculate the degree of ionization of the acid in its 0.05 M solution and also its\[pH\]. What will be its degree of ionization if the solution is 0.01 M in HCl also?

Answer:

\[\underset{\begin{smallmatrix} t=0 \\ {{t}_{eq}} \end{smallmatrix}}{\overset{{}}{\mathop{{}}}}\,\,\,\,\underset{\begin{smallmatrix} C \\ C-C\alpha \end{smallmatrix}}{\mathop{C{{H}_{3}}C{{H}_{2}}COOH}}\,\rightleftharpoons \underset{\begin{smallmatrix} 0 \\ C\alpha \end{smallmatrix}}{\mathop{C{{H}_{3}}C{{H}_{2}}CO{{O}^{-}}}}\,+\underset{\begin{smallmatrix} 0 \\ C\alpha \end{smallmatrix}}{\mathop{{{H}^{+}}}}\,\] \[{{K}_{a}}=\frac{[C{{H}_{3}}C{{H}_{2}}COOH][{{H}^{+}}]}{[C{{H}_{3}}C{{H}_{2}}COOH]}=\frac{C\alpha \times C\alpha }{C-C\alpha }\] \[=\frac{C{{\alpha }^{2}}}{1-\alpha }\approx C{{\alpha }^{2}}\] \[\alpha =\sqrt{\frac{{{K}_{a}}}{C}}=\sqrt{\frac{1.32\times {{10}^{-5}}}{0.05}}\] \[=0.0163\] \[pH=-\log [C\alpha ]\] \[=-\log [0.05\times 0.0163]=3.088\] If the solution contains 0.01 M HCl also then, \[[{{H}^{+}}]=C\alpha =0.01M\] \[\therefore \] \[{{K}_{a}}=C{{\alpha }^{2}}=C\alpha \times \alpha \] \[1.32\times {{10}^{-5}}=0.01\times \alpha \] \[\alpha =1.32\times {{10}^{-3}}\]