question_answer 1)

  pH of \[0.08mol\,d{{m}^{-3}}HOCl\] solution is 2.85. Calculate its ionization constant.

Answer:

  \[\underset{{{t}_{\begin{smallmatrix}  0 \\  {{t}_{aq}} \end{smallmatrix}}}}{\overset{{}}{\mathop{{}}}}\,\,\,\underset{\begin{smallmatrix}  C \\  C-C\alpha  \end{smallmatrix}}{\mathop{HOCl}}\,\underset{{}}{\leftrightarrows}\underset{\begin{smallmatrix}  0 \\  C\alpha  \end{smallmatrix}}{\mathop{{{H}^{+}}}}\,+\underset{\begin{smallmatrix}  0 \\  C\alpha  \end{smallmatrix}}{\mathop{OC{{l}^{-}}}}\,\] \[{{K}_{a}}=\frac{[{{H}^{+}}][O{{C}^{-}}]}{[HOCl]}=\frac{C\alpha \times C\alpha }{C-C\alpha }\] \[=\frac{C{{\alpha }^{2}}}{1-\alpha }=C{{\alpha }^{2}}\] \[\alpha =\sqrt{\frac{{{K}_{a}}}{C}}\] \[C\alpha =\sqrt{C{{K}_{a}}}=[{{H}^{+}}]\] \[pH=-\log (0.08\times {{K}_{a}})\] \[2.85=-\frac{1}{2}\log (0.08\times {{K}_{a}})\] \[-5.70=\log (0.08\times {{K}_{a}})\] \[1.99\times {{10}^{-6}}=0.08\times {{K}_{a}}\]                 \[{{K}_{a}}=2.49\times {{10}^{-5}}\]