Question

$0.02M$ solution of $H_{2}A$ has $pH=4$. If $K_{a_1}$ for the acid is $5\times {10}^{-7}M$, the concentration of $H{A}^{-}$ ion in the solution would be :

• A. $2\times {10}^{-7}M$
• B. $1\times {10}^{-6}M$
• C. $3\times {10}^{-5}M$
• D. $1\times {10}^{-4}M$

Answer 1

The correct option is D $1\times {10}^{-4}M$

Given, $\left[H_{2}A\right]=0.02,pH=4,K_{a1}=5\times {10}^{-7}M$
$pH=4-log\left[H^{+}\right]=4\left[H^{+}\right]={10}^{-4}M$

The dissociation reaction for finding $K_{a1}$ of $H_{2}A$ is :

$H_{2}A\left(aq\right)\rightleftharpoons H^{+}\left(aq\right)+H{A}^{-}\left(aq\right)K_{a1}=\frac{\left[H{A}^{-}\right]\left[H^{+}\right]}{\left[H_{2}A\right]}$
Now by rearranging, the value of $\left[H{A}^{-}\right]$ can be obtained.

$\left[H{A}^{-}\right]=\frac{K_{a1}\times \left[H_{2}A\right]}{\left[H^{+}\right]}\left[H{A}^{-}\right]=\frac{5\times {10}^{-7}\times 0.02}{{10}^{-4}}\left[H{A}^{-}\right]={10}^{-4}M$