Question

Arrange in increasing order of solubility of $AgBr$ in the given solutions.
(i) $0.1MN{H}_3$
(ii) $0.1MAgN{O}_3$
(iii) $0.2MNaBr$
(iv) Pure water

• A. $\left(iii\right)<\left(ii\right)<\left(iv\right)<\left(i\right)$
• B. $\left(iii\right)<\left(ii\right)<\left(i\right)<\left(iv\right)$
• C. $\left(iii\right)<\left(ii\right)=\left(i\right)<\left(iv\right)$
• D. $\left(i\right)<\left(iii\right)<\left(iv\right)<\left(ii\right)$

Answer 1

The correct option is A $\left(iii\right)<\left(ii\right)<\left(iv\right)<\left(i\right)$

Order of solubility : Complex formation > Pure water > Common ion effect.
$AgBr$ forms complex $\left[Ag\right(N{H}_3{)}_2{]}^{+}$ in $N{H}_3$ so solubility is maximum in $N{H}_3\left(aq\right)$.
In $AgN{O}_{3}andNaBr$, there are common ions with AgBr. So their solubility will be less.