Question

The solubilities of $AgBr$ in water, in $0.01 \,M \,CaBr _{2}, 0.01 \,M \, KBr$ and in $0.05 \, M \, AgNO _{3}$ 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 $0.01 M ( CaBr )_{2}\left(( S )_{2}\right) \ldots \ldots$ (i)
$CaBr _{2}$ is a strong electrolyte and dissociates completely to give common ion $Br ^{-}$
$\underset{0.01 M }{ CaBr _{2}} \rightarrow \underset{0.01 M }{ Ca ^{2+}}+\underset{0.02 Mr }{2 Br ^{-}}$
$AgBr ( s ) \rightleftharpoons( Ag )^{+}(( aq )+ Br )^{-}( aq ) ;( K )_{ sp }$
$( S )_{2}\left(0(.02+ S )_{2}\right) \Rightarrow( S )_{2} \times 0.02=( K )_{ sp }$
$S _{2}=50 \,K _{ sp }$
Similarly, $KBr$ gives common ion $Br ^{-}$
And $AgNO _{3}$ gives common ion $Ag ^{+}$and we can calculate $S _{3}=100 K _{ sp } \ldots \ldots$...(iii)
And $S _{4}=20 \, K _{ sp } \ldots \ldots$ (iv) respectively
From (i), (ii), (iii) and (iv)
$S _{1}> S _{3}> S _{2}> S _{4}$
(Note: $\sqrt{ K _{ sp }}>100 \,K _{ sp } \because K _{ sp }$ order $=10^{-12}$ )