Question

The solubilities of $AgBr$ in water, in $0.01MCaB{r}_2,0.01MKBr$ and in $0.05MAgN{O}_3$ are $S_1,S_2,S_3$ and $S_4$ respectively. Then the order of solubilities is

• A. $S_1>S_2>S_3>S_4$
• B. $S_1>S_3>S_2>S_4$
• C. $S_2>S_1>S_3>S_4$
• D. $S_4>S_3>S_1>S_2$

Answer 1

Answer: (B)

Solubility of $AgBr$ in $0.01M(CaBr{)}_2\left((S{)}_2\right)\dots \dots$ (i)
$CaB{r}_2$ is a strong electrolyte and dissociates completely to give common ion $B{r}^{-}$
$\underset{0.01M}{CaB{r}_2}\to \underset{0.01M}{C{a}^{2+}}+\underset{0.02Mr}{2B{r}^{-}}$
$AgBr(s)\rightleftharpoons (Ag{)}^{+}((aq)+Br{)}^{-}(aq);(K{)}_{sp}$
$(S{)}_2\left(0(.02+S{)}_2\right)\Rightarrow (S{)}_2\times 0.02=(K{)}_{sp}$
$S_2=50K_{sp}$
Similarly, $KBr$ gives common ion $B{r}^{-}$
And $AgN{O}_3$ gives common ion $A{g}^{+}$and we can calculate $S_3=100K_{sp}\dots \dots$...(iii)
And $S_4=20K_{sp}\dots \dots$ (iv) respectively
From (i), (ii), (iii) and (iv)
$S_1>S_3>S_2>S_4$
(Note: $\sqrt{K_{sp}}>100K_{sp}\because K_{sp}$ order $=10^{-12}$ )