Chemical And Ionic Equilibrium Result Question 36
36. The $K_{\mathrm{sp}}$ of $\mathrm{Ag}_2 \mathrm{CrO}_4$ is $1.1 \times 10^{-12}$ at $298$ K . The solubility (in $\mathrm{mol} / \mathrm{L}$ ) of $\mathrm{Ag}_2 \mathrm{CrO}_4$ in a $0.1 $ $\mathrm{M} $ $\mathrm{AgNO}_3$ solution is
(a) $1.1 \times 10^{-11}$
(b) $1.1 \times 10^{-10}$
(c) $1.1 \times 10^{-12}$
(d) $1.1 \times 10^{-9}$
(2013 Adv.)
Show Answer
Answer:
Correct Answer: 36. ( b )
Solution:
- PLAN: In presence of common ion (in this case $\mathrm{Ag}^{+}$ion) solubility of sparingly soluble salt is decreased.
Let solubility of $\mathrm{Ag}_2 \mathrm{CrO}_4$ in presence of $0.1$ M
$ \begin{aligned} & \mathrm{AgNO}_3=x \\ & \mathrm{Ag}_2 \mathrm{CrO}_4 \rightleftharpoons 2 \underset{2x} {\mathrm{Ag}^{+}}+ \underset{x}{\mathrm{CrO}_4^{2-}} \\ & \mathrm{AgNO}_3 \rightleftharpoons \underset{0.1}{\mathrm{Ag}^{+}}+\underset{0.1}{\mathrm{NO}_3^{-}} \\ \end{aligned} $
Total $\left[\mathrm{Ag}^{+}\right]=(2 x+0.1) \mathrm{M} \approx 0.1 \mathrm{M}$
$ \begin{aligned} \text { as } x & < < < 0.1 \mathrm{M} \\ {\left[\mathrm{CrO}_4^{2-}\right] } & =x \mathrm{M} \end{aligned} $
Thus, $\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CrO}4^{2-}\right]=K{\text {sp }}$
$ \begin{aligned} (0.1)^2(x) & =1.1 \times 10^{-12} \\ x & =1.1 \times 10^{-10} \mathrm{M} \end{aligned} $
BETA
