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 $0.1MHCl$ solution.

• A. $1.8\times {10}^{-6}mol{L}^{-1}$
• B. $1.8\times {10}^{-4}mol{L}^{-1}$
• C. $2.8\times {10}^{-6}mol{L}^{-1}$
• D. $2.4\times {10}^{-4}mol{L}^{-1}$

Answer 1

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

$C{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{-}+H^{+}HCl\to C{l}^{-}+H^{+}CommonionC{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{-}+H^{+}0.01000.01-xx0.1+x$
Here x is very small value so x can be neglect.
$\therefore 1+x=0.1$
$K_a=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}\left[C{H}_{3}CO{O}^{-}\right]=\frac{1.8\times {10}^{-5}\times 0.01}{0.1}\left[C{H}_{3}CO{O}^{-}\right]=1.8\times {10}^{-6}mol{L}^{-1}$