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