An aqueous solution is prepared with 0.10text M H_ 2 S and 0.20text M HCl. The equilibrium constants for H_ 2 S rightleftharpoons H^ + + HS^ - is K_1 = 1.0 times 10^ -7 and for HS^ - rightleftharpoons H^ + + S^ 2- is K_2 = 1.2 times 10^ -13 . Calculate the concentration of S^ 2- ions.

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EEJ MAIN Chemistry QUESTION #7115

Question 1

An aqueous solution is prepared with $0.10\text{ M } H_{2}S$ and $0.20\text{ M } HCl$. The equilibrium constants for $H_{2}S \rightleftharpoons H^{+} + HS^{-}$ is $K_1 = 1.0 \times 10^{-7}$ and for $HS^{-} \rightleftharpoons H^{+} + S^{2-}$ is $K_2 = 1.2 \times 10^{-13}$. Calculate the concentration of $S^{2-}$ ions.

$6 \times 10^{-21}$
$5 \times 10^{-19}$
$5 \times 10^{-8}$
$3 \times 10^{-20}$✔️

Correct Answer Explanation

The overall dissociation is $H_2S \rightleftharpoons 2H^{+} + S^{2-}$ with $K_{net} = K_1 \times K_2 = 1.2 \times 10^{-20}$. In the presence of $0.20\text{ M } HCl$ (a strong acid), $[H^{+}]$ is dominated by $HCl$, so $[H^{+}] \approx 0.20\text{ M}$. Using the expression $K_{net} = \frac{[H^{+}]^2 [S^{2-}]}{[H_2S]}$, we get $1.2 \times 10^{-20} = \frac{(0.2)^2 [S^{2-}]}{0.1}$. Solving for $[S^{2-}]$ gives $3 \times 10^{-20}\text{ M}$.

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