JEE PYQ: Ionic Equilibrium Question 29
Question 29 - 2019 (12 Jan Shift 1)
Type: MCQ
If $K_{sp}$ of $\text{Ag}_2\text{CO}_3$ is $8 \times 10^{-12}$, the molar solubility of $\text{Ag}_2\text{CO}_3$ in 0.1 M $\text{AgNO}_3$ is:
(a) $8 \times 10^{-12}$ M
(b) $8 \times 10^{-11}$ M
(c) $8 \times 10^{-10}$ M
(d) $8 \times 10^{-13}$ M
Show Answer
Answer: (b)
Solution
As $\text{AgNO}_3$ dissociates completely, in 0.1 M $\text{AgNO}_3$ solution, $[\text{Ag}^+] = 0.1$ M.
$\text{Ag}_2\text{CO}_3 \rightleftharpoons 2\text{Ag}^+ + \text{CO}_3^{2-}$
$K_{sp} = [\text{Ag}^+]^2[\text{CO}_3^{2-}]$
$8 \times 10^{-12} = (0.1 + 2s)^2 \times s$
Since $2s \ll 0.1$: $(0.1)^2 \times s = 8 \times 10^{-12}$
$0.01 \times s = 8 \times 10^{-12}$
$s = 8 \times 10^{-10}$ M
BETA
