Chemical and Ionic Equilibrium 2 Question 6
6. If $K_{\mathrm{sp}}$ of $\mathrm{Ag}{2} \mathrm{CO}{3}$ is $8 \times 10^{-12}$, the molar solubility of $\mathrm{Ag}{2} \mathrm{CO}{3}$ in $0.1 \mathrm{M} \mathrm{AgNO}_{3}$ is
(2019 Main, 12 Jan II)
(a) $8 \times 10^{-12} \mathrm{M}$
(b) $8 \times 10^{-13} \mathrm{M}$
(c) $8 \times 10^{-10} \mathrm{M}$
(d) $8 \times 10^{-11} \mathrm{M}$
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Answer:
Correct Answer: 6. (c)
Solution:
- Let the solubility of $\mathrm{Ag}{2} \mathrm{CO}{3}$ is $S$. Now, $0.1 \mathrm{M}$ of $\mathrm{AgNO}{3}$ is added to this solution after which let the solubility of $\mathrm{Ag}{2} \mathrm{CO}_{3}$ becomes $S^{\prime}$.
$$ \begin{array}{lll} \therefore & {\left[\mathrm{Ag}^{+}\right]} & =S+0.1 \text { and }\left[\mathrm{CO}{3}^{2-}\right]=S^{\prime} \ & K{\text {sp }} & =(S+0.1)^{2}\left(S^{\prime}\right) \ \text { Given, } & K_{\text {sp }} & =8 \times 10^{-12} \end{array} $$
$\because K_{\text {sp }}$ is very small, we neglect $S^{\prime}$ against $S$ in Eq. (i)
$$ \begin{aligned} & \therefore \quad K_{\mathrm{sp}}=(0.1)^{2} S^{\prime} \ & \text { or } \quad 8 \times 10^{-12}=0.01 S^{\prime} \ & \text { or } \quad S^{\prime}=8 \times 10^{-12} \times 10^{2}=8 \times 10^{-10} \mathrm{M} \end{aligned} $$
Thus, molar solubility of $\mathrm{Ag}{2} \mathrm{CO}{3}$ in $0.1 \mathrm{M}$
$\mathrm{AgNO}_{3}$ is $8 \times 10^{-10} \mathrm{M}$.
