Question

In which of the given solutions will $AgBr$ have maximum solubility?

• A. Pure water
• B. ${10}^{-3}M$ $NaBr$
• C. ${10}^{-3}M$ $N{H}_{4}OH$
• D. ${10}^{-3}M$ $HBr$

Answer 1

The correct option is C ${10}^{-3}M$ $N{H}_{4}OH$

$AgBr$ will have maximum solubility in ${10}^{-3}MN{H}_{4}OH$.

Adding a common ion decreases solubility, as the reaction shifts towards the left to relieve the stress of the excess product.

In $N{H}_{4}OH$, due to the formation of a coordination complex $AgBr$ is more soluble.
$AgBr\to A{g}^{+}+B{r}^{-}\stackrel{N{H}_3}{\to }\left[Ag\right(N{H}_3{)}_2]Br$