question_answer 1)

The pH of 0.1 M solution of HCNO (cyanic acid) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Answer:

HCNO is a weak acid it undergoes ionisation as follows: \[\underset{\begin{smallmatrix} t=0 \\ {{t}_{eq}} \end{smallmatrix}}{\overset{{}}{\mathop{{}}}}\,\,\,\,\,\underset{\begin{smallmatrix} C \\ C-C\alpha \end{smallmatrix}}{\mathop{HCNO}}\,\rightleftharpoons \underset{\begin{smallmatrix} 0 \\ C\alpha \end{smallmatrix}}{\mathop{{{H}^{+}}}}\,+\underset{\begin{smallmatrix} 0 \\ C\alpha \end{smallmatrix}}{\mathop{\overset{-}{\mathop{C}}\,NO}}\,\] \[[{{H}^{+}}]=C\alpha \] \[\therefore \] \[C\alpha =\text{Antilog }\!\![\!\!\text{ -pH }\!\!]\!\!\text{ }\] \[0.1\times \alpha =\text{Antilog }\!\![\!\!\text{ -2}\text{.34 }\!\!]\!\!\text{ }\] Degree of ionisation, \[\alpha \] = 0.0457 \[{{K}_{a}}=\frac{C\alpha \times C\alpha }{1-\alpha }\] \[=\frac{C{{\alpha }^{2}}}{1-\alpha }=\frac{0.1\times {{(0.00457)}^{2}}}{1-0.0457}\] \[=2.188\times {{10}^{-4}}\]