Question

$0.01M$ Solution of $H_{2}A$ has $pH=4$ . If $K_{a_1}$ for the acid is $4.45\times {10}^{-7}$ , the concentration of $H{A}^{-}$ ion in the solution would be :

• A. $1.0\times {10}^{-2}M$
• B. $4.45\times {10}^{-5}M$
• C. $8.0\times {10}^{-5}M$
• D. Unpredictable

Answer 1

The correct option is B $4.45\times {10}^{-5}M$

Given, $\left[H_{2}A\right]=0.01,pH=4,K_{a1}=4.45\times {10}^{-7}$
The dissociation reaction for the weak dibasic acid is :
$H_{2}A\rightleftharpoons H^{+}+H{A}^{-}{K}_a=\frac{\left[H{A}^{-}\right]\left[H^{+}\right]}{\left[H_{2}A\right]}$
Now by rearranging, the value of $\left[H{A}^{-}\right]$ can be obtained.

$\left[H{A}^{-}\right]=\frac{K_a\times \left[H_{2}A\right]}{\left[H^{+}\right]}\left[H{A}^{-}\right]=\frac{4.45\times {10}^{-7}\times 1\times {10}^{-2}}{1\times {10}^{-4}}\left[H{A}^{-}\right]=4.45\times {10}^{-5}M$