Chemical And Ionic Equilibrium Result Question 6
6. If $K_{\text {sp }}$ 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, $0.1$ M of $\mathrm{AgNO}_3$ is added to this solution after which let the solubility of $\mathrm{Ag}_2 \mathrm{CO}_3$ becomes $S^{\prime}$.
$\therefore {\left[\mathrm{Ag}^{+}\right]} =S+0.1 \text { and }\left[\mathrm{CO}_3^{2-}\right]=S^{\prime} $
$K_{\mathrm{sp}} =(S+0.1)^2\left(S^{\prime}\right)\hspace {15mm} …(i) $
$\text { Given, } K_{\text {sp }} =8 \times 10^{-12}$
$\because K_{\text {sp }}$ is very small, we neglect $S^{\prime}$ against $S$ in Eq. (i)
$\begin{aligned} \because K_{\text {sp }} & =(0.1)^2 S^{\prime} \\ or\hspace {5mm}8 \times 10^{-12} & =0.01 S^{\prime} \\ or\hspace {5mm}S^{\prime} & =8 \times 10^{-12} \times 10^2=8 \times 10^{-10} \mathrm{M} \end{aligned}$
Thus, molar solubility of $\mathrm{Ag}_2 \mathrm{CO}_3$ in $0.1$ M
$\mathrm{AgNO}_3$ is $8 \times 10^{-10} \mathrm{M}$
BETA
