Question

Solubility product $\left(K_{sp}\right)$ values of the following sparingly soluble salts at ${25}^{\circ }C$ is :

Salts $K_{sp}$
$AgCl1.8\times {10}^{-10}AgI1.1\times {10}^{-16}PbCr{O}_{4}1.8\times {10}^{-14}AgBr3.3\times {10}^{-13}$
The least soluble and the most soluble salts respectively are:

• A. $PbCr{O}_4$ and $AgCl$
• B. $AgI$ and $AgCl$
• C. $AgI$ and $PbCr{O}_4$
• D. $AgI$ and $AgBr$

Answer 1

The correct option is B $AgI$ and $AgCl$

Here, all the salts are of $AB$ type:
The equilibrium relation of all the undissolved salts with their respective ions is of the type:
$AB\left(s\right)\rightleftharpoons A^{+}\left(aq\right)+B^{-}\left(aq\right)$

When the stoichiometry of the precipitated salts is the same, then the salts are precipitated in the increasing order of their $K_{sp}$ value.
So, the least soluble salt is $AgI$ and the most soluble is $AgCl$