Question

$0.1$ mole arsenic acid $\left(H_{3}As{O}_4\right)$ is dissolved in a 1L buffer solution of $pH=8$. Which of the following options is/are correct? For arsenic acid: $K{a}_1=2.5\times {10}^{-4},K{a}_2=5\times {10}^{-8},K{a}_3=2\times {10}^{-13}$.
(‘<<’ sign denotes that the higher concentration is 100 times greater than the lower one or more)

• A. $\left[H_{3}As{O}_4\right]<<\left[H_{2}As{O}_4^{-}\right]$
• B. $\left[H_{2}As{O}_4^{-}\right]<<\left[HAs{O}_4^{2-}\right]$
• C. $\left[HAs{O}_4^{2-}\right]<<\left[H_{2}As{O}_4^{-}\right]$
• D. $\left[As{O}_4^{3-}\right]<<\left[HAs{O}_4^{2-}\right]$

Answer 1

Answer: (A), (D)

The correct options are
A $\left[H_{3}As{O}_4\right]<<\left[H_{2}As{O}_4^{-}\right]$
D $\left[As{O}_4^{3-}\right]<<\left[HAs{O}_4^{2-}\right]$

Since $H_{3}As{O}_4$ is a polybasic acid and given that$\left[H^{+}\right]={10}^{-8}$
it will ionise like
$H_{3}As{O}_4\leftrightharpoons H_{2}As{O}_4^{-}+H^{+}$
$K{a}_1=\frac{\left[H_{2}As{O}_4^{-}\right]\left[H^{+}\right]}{\left[H_{3}As{O}_4\right]}$
$\frac{\left[H_{3}As{O}_4\right]}{\left[H_{2}As{O}_4^{-}\right]}=\frac{\left[H^{+}\right]}{K{a}_1}=\frac{{10}^{-8}}{2.5\times {10}^{-4}}$
$\left[H_{3}As{O}_4\right]<<\left[H_{2}As{O}_4^{-}\right]$
similarly for second ionisation
$\frac{\left[HAs{O}_4^{2-}\right]}{\left[H_{2}As{O}_4^{-}\right]}=\frac{\left[H^{+}\right]}{K{a}_2}=\frac{{10}^{-8}}{5\times {10}^{-8}}=\frac{1}{5}$
for third ionisation
$\frac{\left[HAs{O}_4^{2-}\right]}{\left[As{O}_4^{3-}\right]}=\frac{\left[H^{+}\right]}{K{a}_3}=\frac{{10}^{-8}}{2\times {10}^{-13}}=5\times {10}^4$
Hence
$\left[As{O}_4^{3-}\right]<<\left[HAs{O}_4^{2-}\right]$