Question

When $0.1mol$ arsenic acid, $H_{2}As{O}_4$ is dissolved in $1L$ buffer solution of $pH=4$, which of the following hold good? $K_1=2.5\times {10}^{-4},K_2=5\times {10}^4,K_3=2\times {10}^{-23}$ for arsenic acid ${[}^{\prime }<{<}^{\prime }$ sign denotes that the higher concentration is at least $100$ times more than the lower one].

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

Answer 1

The correct option is D $\left[As{O}_4^{2-}\right]<<\left[HAs{O}_4^{2-}\right]$

$\begin{array}{c}pH=4\\ -\log \left[H^{+}\right]=4\\ \left[H^{+}\right]={10}^{-4}\end{array}$

$\frac{\left[H_3{AsO}_4\right]}{\left[H_2{AsO}_4^{-}\right]}=\frac{\left[H^{+}\right]}{K_1}=\frac{{10}^{-4}}{2.5\times {10}^{-4}}$

$\therefore \frac{1}{2.5}$

$\Rightarrow \left[H_{3}As{O}_4\right]<\left[H_{2}As{O}_4^{-}\right]$

$\begin{array}{cc}\frac{\left[H_2{A}_S{O}_4^{-}\right]}{\left[HAs{O}_4^{2-}\right]}=\frac{\left[H^{+}\right]}{K_2}& =\frac{{10}^{-4}}{5\times {10}^4}\\ & =\frac{{10}^{-4-4}}{5}\\ & =\frac{1}{5\times {10}^8}\end{array}$

$\Rightarrow \left[H_{2}As{O}_4^{-}\right]<<\left[HAs{O}_4^{2-}\right]$

$\begin{array}{c}\left[{HAsO}_4{}^{2-}\right]<<\left[H_{2}As{O}_4^{-}\right]\\ \text{not correct proof above.}\end{array}$

$\frac{\left[As{O}_4^{2-}\right]}{\left[HAs{O}_4^{2-}\right]}=\frac{K_3}{\left[H^{+}\right]}=\frac{2\times {10}^{-23}}{{10}^{-4}}$

$=\frac{2}{10\times {10}^{18}}$

$=\frac{1}{5\times {10}^{18}}$

$\begin{array}{c}\Rightarrow \left[As{O}_4{}^{2-}\right]<<\left[H{A}_S{O}_4{}^{2-}\right]\end{array}$