Question

A solution contains $0.10xM$ and $0.2xM$ $HCl$. Calculate the concentration of $\left[S^{2-}\right]$ ions in the solution.

Given: For $H_{2}S,K_{a1}=1.0\times {10}^{-7},K_{a2}=1.3\times {10}^{-13}$

• A. $\left[S^{2-}\right]=1.44\times {10}^{-20}M$
• B. $\left[S^{2-}\right]=3.32\times {10}^{-5}M$
• C. $\left[S^{2-}\right]={6.5\times 10}^{-15}M$
• D. $\left[S^{2-}\right]={3.25\times 10}^{-20}M$

Answer 1

The correct option is D $\left[S^{2-}\right]={3.25\times 10}^{-20}M$

Since $HCl$ is a strong acid, we will consider the $H^{+}$ is given by $HCl$ such that $\left[H^{+}\right]=0.2$ $M$

$H_{2}S⟶H{S}^{-}+H^{+}$

$K_1=\frac{\left[H{S}^{-}\right]\left[H^{+}\right]}{\left[H_{2}S\right]}={10}^{-7}$

$\therefore \frac{\left[H{S}^{-}\right]\times 0.2}{0.1}={10}^{-7}$

$\therefore \left[H{S}^{-}\right]=5\times {10}^{-8}$

$H{S}^{-}⟶H^{+}+S^{2-}$

$K_2=\frac{\left[S^{2-}\right]\left[H^{+}\right]}{\left[H{S}^{-}\right]}=1.3\times {10}^{-13}$

$\left[S^{2-}\right]=\frac{\left[H{S}^{-}\right]}{\left[H^{+}\right]}\times 1.3\times {10}^{-13}=\frac{5\times {10}^{-8}\times 1.3\times {10}^{-13}}{2\times {10}^{-1}}$

$\left[S^{2-}\right]=3.25\times {10}^{-20}$ $M$