Chemical and Ionic Equilibrium - Result Question 67
####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) $3 \times 10^{-20}$
(c) $6 \times 10^{-21}$
(d) $5 \times 10^{-19}$
(2018 Main)
Show Answer
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,
$$ \begin{aligned} & H _2 S \rightleftharpoons 2 H^{+}+S^{2-} \ & K=K _1 \times K _2=1.0 \times 10^{-7} \times 1.2 \times 10^{-13} \ &=1.2 \times 10^{-20} \ &\left.\rho^{2-}\right]=\frac{K \times\left[H _2 S\right]}{\left[H^{+}\right]^{2}} \quad \text { [according to the final } \ &=\frac{1.2 \times 10^{-20} \times 0.1 M}{(0.2 M)^{2}} \ &=\frac{1.2 \times 10^{-20} \times 1 \times 10^{-1} M}{4 \times 10^{-2} M} \ &=3 \times 10^{-20} M \end{aligned} $$
Now $\quad\left[S^{2-}\right]=\frac{K \times\left[H _2 S\right]}{\left[H^{+}\right]^{2}} \quad$ [according to the final equation]
