Chemical And Ionic Equilibrium

Chemical and Ionic Equilibrium - Result Question 11

11. 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) $5 \times 10^{-8}$

(b) $3 \times 10^{-20}$

(d) $5 \times 10^{-19}$

(2018 Main)

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Answer:

Correct Answer: 11. (b)

Solution:

Given $\left[H _2 S\right]=0.10 M$

$ [HCl]=0.20$ $ M$ $ So,\left[H^{+}\right]=0.20 M $

$H _2 S \rightleftharpoons H^{+}+HS, K _1=1.0 \times 10^{-7}$

$HS \rightleftharpoons H^{+}+S^{2-}, K _2=1.2 \times 10^{-13}$

It means for,

$ \begin{aligned} & H _2 S \rightleftharpoons 2 H^{+}+S^{2-} \\ & K=K _1 \times K _2=1.0 \times 10^{-7} \times 1.2 \times 10^{-13} \\ &=1.2 \times 10^{-20} \\ &\text{Now }\left[S^{2-}\right]=\frac{K \times\left[H _2 S\right]}{\left[H^{+}\right]^{2}} \quad \text { [according to the final equation]} \\ &=\frac{1.2 \times 10^{-20} \times 0.1 M}{(0.2 M)^{2}} \\ &=\frac{1.2 \times 10^{-20} \times 1 \times 10^{-1} M}{4 \times 10^{-2} M} \\ &=3 \times 10^{-20} M \end{aligned} $