Question

A solution contain 0.10 M $H_{2}S$ and 0.3 M HCl.Calculate the concentration of $S^{2-}$ and HS ions in the solution, For $H_{2}S$, $K_{a_1}$ $=$ 1.3$\times {10}^{-7}$ and $K_{a_2}=1.3\times {10}^{-13}$

• A. $1.88\times {10}^{-20}M,4.333\times {10}^{-8}M$
• B. $3.333\times {10}^{-8}M,1.44\times {10}^{-20}M$
• C. $2.44\times {10}^{-20}M,5.33\times {10}^{-13}M$
• D. $3.44\times {10}^{-20}M,1.33\times {10}^{-14}M$

Answer 1

Answer: (A)

$1.88\times {10}^{-20}M,4.333\times {10}^{-8}M$

$H_{2}S\rightleftharpoons \left[H^{+}\right][HS-]$ --------------- 1

$H{S}^{-}\rightleftharpoons \left[H^{+}\right]\left[S^{2-}\right]$ --------------- 2
$K_{a}1=\frac{\left[H^{+}\right]\left[H{S}^{-}\right]}{\left[H_{2}S\right]}$ (For reaction number 1)
Substitute values in the above expression.
$1.3\times {10}^{-7}=\frac{0.3\times \left[H{S}^{-}\right]}{0.1}$
$\left[H{S}^{-}\right]=4.333\times {10}^{-8}$ M
$K_{a}2=\frac{\left[H^{+}\right]\left[S^{2-}\right]}{\left[H{S}^{-}\right]}$ (For reaction number 2)
Substitute values in the above expression.
$1.3\times {10}^{-13}=\frac{0.3\times \left[S^{2-}\right]}{4.333\times {10}^{-8}}$
$\left[H{S}^{-}\right]=1.88\times {10}^{-20}$ M.