Question

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

• A. $0.01N$
• B. $4.45\times 10^{-5}$
• C. $8.0\times 10^{-5}$
• D. unpredictable

Answer 1

Answer: (B)

$H_{2}A{H}^{+}+H{A}^{-}$as
$pH=4\left[H^{+}\right]=10^{-4}M$
$K_{a_1}=\frac{\left[H^{+}\right]\left[H{A}^{-}\right]}{\left[H_{2}A\right]}$
Or $\frac{10^{-4}\times \left[H{A}^{-}\right]}{10^{-2}}=4.45\times 10^{-7}$
Or $\left[H{A}^{-}\right]=4.45\times 10^{-5}$