Chemical and Ionic Equilibrium 1 Question 10
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11. An aqueous solution contains $0.10 \mathrm{M} \mathrm{H}{2} \mathrm{~S}$ and $0.20 \mathrm{M} \mathrm{HCl}$. If the equilibrium constants for the formation of $\mathrm{HS}^{-}$from $\mathrm{H}{2} \mathrm{~S}$ is $1.0 \times 10^{-7}$ and that of $\mathrm{S}^{2-}$ from $\mathrm{HS}^{-}$ions is $1.2 \times 10^{-13}$ then the concentration of $\mathrm{S}^{2-}$ ions in aqueous solution is :
======= ####11. An aqueous solution contains $0.10 \mathrm{M} \mathrm{H}{2} \mathrm{~S}$ and $0.20 \mathrm{M} \mathrm{HCl}$. If the equilibrium constants for the formation of $\mathrm{HS}^{-}$from $\mathrm{H}{2} \mathrm{~S}$ is $1.0 \times 10^{-7}$ and that of $\mathrm{S}^{2-}$ from $\mathrm{HS}^{-}$ions is $1.2 \times 10^{-13}$ then the concentration of $\mathrm{S}^{2-}$ ions in aqueous solution is :
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed (a) $5 \times 10^{-8}$
(b) $3 \times 10^{-20}$
(c) $6 \times 10^{-21}$
(d) $5 \times 10^{-19}$
(2018 Main)
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Solution:
- Given $\left[\mathrm{H}_{2} \mathrm{~S}\right]=0.10 \mathrm{M}$
$$ [\mathrm{HCl}]=0.20 \mathrm{M} \mathrm{So},\left[\mathrm{H}^{+}\right]=0.20 \mathrm{M} $$
$\mathrm{H}{2} \mathrm{~S} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HS}, K{1}=1.0 \times 10^{-7}$
$\mathrm{HS} \rightleftharpoons \mathrm{H}^{+}+\mathrm{S}^{2-}, K_{2}=1.2 \times 10^{-13}$
It means for,
$$ \begin{aligned} & \mathrm{H}{2} \mathrm{~S} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{S}^{2-} \ & K=K{1} \times K_{2}=1.0 \times 10^{-7} \times 1.2 \times 10^{-13} \ &=1.2 \times 10^{-20} \ &\left.\rho^{2-}\right]=\frac{K \times\left[\mathrm{H}_{2} \mathrm{~S}\right]}{\left[\mathrm{H}^{+}\right]^{2}} \quad \text { [according to the final } \ &=\frac{1.2 \times 10^{-20} \times 0.1 \mathrm{M}}{(0.2 \mathrm{M})^{2}} \ &=\frac{1.2 \times 10^{-20} \times 1 \times 10^{-1} \mathrm{M}}{4 \times 10^{-2} \mathrm{M}} \ &=3 \times 10^{-20} \mathrm{M} \end{aligned} $$
Now $\quad\left[\mathrm{S}^{2-}\right]=\frac{K \times\left[\mathrm{H}_{2} \mathrm{~S}\right]}{\left[\mathrm{H}^{+}\right]^{2}} \quad$ [according to the final equation]
