Question

An aqueous solution contains 0.10 M $H_{2}S$ and 0.20 M HCl. If the equilibrium constants for the formation of $HS^{-}$ from $H_{2}S$ is $1.0\times 10^{- 7}$ and that of $S^{2 -}$ from $HS^{-}$ ions is $1.2\times 10^{- 13}$ then the concentration of $S^{2 -}$ ions in aqueous solution is

• A. $5\times 10^{- 19}$
• B. $5\times 10^{- 8}$
• C. $3\times 10^{- 20}$
• D. $6\times 10^{- 21}$

Answer 1

Answer: (C)

$\underset{\underset{= 0.1}{\left(\right. 0.1 - x \left.\right)}}{H_{2} S \left(\right. a q \left.\right)}\rightleftharpoons\underset{\underset{\sim eq 0.2}{\left(\right. 2 x + 0.20 \left.\right)}}{2 H^{+}}+\underset{x}{S^{2 -}},$

$K_{a}=K_{a_{1}}=1.2\times 10^{- 20}$

$1.2 \times 10^{-20}=\frac{(0.2)^{2} \times\left[ S ^{2-}\right]}{0.1}$

$\left[\right.S^{2 -}\left]\right.=3\times 10^{- 20}$ '