question_answer 1)

The ionisation constant of HF is\[3.2\times {{10}^{-4}}\]. Calculate the degree of dissociation of HF in its 0.02 M solution. Calculate the concentration of all species present\[{{H}_{3}}{{O}^{+}},{{F}^{-}}\] and HF in the solution and its pH.

Answer:

\[\underset{{{t}_{eq}}}{\overset{{}}{\mathop{t=0}}}\,\underset{\underset{C-C\alpha }{\mathop{C}}\,}{\mathop{HF}}\,\rightleftharpoons \underset{\begin{smallmatrix} 0 \\ C\alpha \end{smallmatrix}}{\mathop{{{H}^{+}}}}\,+\underset{\underset{C\alpha }{\mathop{0}}\,}{\mathop{{{F}^{-}}}}\,\] \[{{K}_{a}}=\frac{[{{H}^{+}}][{{F}^{-}}]}{[HF]}\frac{C\alpha \times C\alpha }{C-C\alpha }\] \[=\frac{C{{\alpha }^{2}}}{1-\alpha }\approx C{{\alpha }^{2}}\] \[\alpha =\sqrt{\frac{{{K}_{a}}}{C}}=\sqrt{\frac{3.2\times {{10}^{-4}}}{0.02}}=0.126\] \[[{{H}^{+}}]=[{{F}^{-}}]=C\alpha \] \[0.02\times 0.126=2.5\times {{10}^{-3}}\] \[[HF]=C-C\alpha =0.02(1-0.126)=0.0175\] \[pH=-\log [{{H}^{+}}]=-\log [2.5\times {{10}^{-3}}]\] \[=2.602\]