Question

The dissociation constant of $0.01MC{H}_{3}COOH$ is $1.8\times {10}^{-5}$ then calculate $C{H}_{3}CO{O}^{-}$ concentration in 0.1 M $HCl$ solution.

• A. $1.8\times {10}^{-6}mol{L}^{-1}$
• B. $1.8\times {10}^{-8}mol{L}^{-1}$
• C. $3.8\times {10}^{-6}mol{L}^{-1}$
• D. $3.8\times {10}^{-8}mol{L}^{-1}$

Answer 1

The correct option is A $1.8\times {10}^{-6}mol{L}^{-1}$

Given, $K_a=1.8\times {10}^{-5},\left[C{H}_{3}COOH\right]=0.01M,\left[HCl\right]=0.1$
The dissociation of acid is given by
$C{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{-}+H^{+}HCl\rightleftharpoons C{l}^{-}+H^{+}$
$commonion$
$C{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{-}+H^{+}0.01000.01-xxx$
$H_{overall}^{+}=[H^{+}{]}_{C{H}_{3}COOH}+[H^{+}{]}_{HCl}=x+0.1$
Since $C{H}_{3}COOHisaweakacid$ So, x i.e. dissociation is a very small value as compared to 0.1
So, $0.1+x\approx 0.1$
$k_a=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}\left[C{H}_{3}CO{O}^{-}\right]=\frac{1.8\times {10}^{-5}\times 0.01}{0.1}\left[C{H}_{3}CO{O}^{-}\right]=1.8\times {10}^{-6}mol{L}^{-1}$
Hence, the concentration of $C{H}_{3}CO{O}^{-}$ in 0.1 M $HCl$ solution is $1.8\times {10}^{-6}mol{L}^{-1}$