Question

The dissociation constant of $0.05MC{H}_{3}COOH$ is given as $1.8\times {10}^{-5}$. Calculate the percentage dissociation of $C{H}_{3}COOH$ in $0.2MHCl$ solution.

• A. $1.5\times {10}^{-6}M$
• B. $9\times {10}^{-3}M$
• C. $1.8\times {10}^{-4}M$
• D. $6.8\times {10}^{-5}M$

Answer 1

The correct option is B $9\times {10}^{-3}M$

$C{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{-}+H^{+}HCl\to C{l}^{-}+H^{+}CommonionC{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{-}+H^{+}C00C-C\alpha C\alpha 0.2+C\alpha$
Since, $C{H}_{3}COOH$ is a weak acid as compared to $HCl$
$C\alpha$ is very small value as compared to $[H^{+}{]}_{HCl}=0.2M$

$\therefore 0.2+C\alpha \approx 0.2$
and $\therefore C-C\alpha \approx C$

$K_a=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}\left[C{H}_{3}CO{O}^{-}\right]=\frac{1.8\times {10}^{-5}\times 0.05}{0.2}\left[C{H}_{3}CO{O}^{-}\right]=4.5\times {10}^{-6}mol{L}^{-1}C\alpha =4.5\times {10}^{-6}mol{L}^{-1}\alpha =\frac{4.5\times {10}^{-6}mol{L}^{-1}}{0.05mol{L}^{-1}}\alpha =9\times {10}^{-5}$
%$\alpha =(9\times {10}^{-5}\times 100)=9\times {10}^{-3}$