Q. $AgNO_3$ solution is added to the solution containing equal moles of $NaCl, NaBr, NaI$ and $Na_2 CrO_4 $ the $Ksp $ of $Ag_2CrO_4 , AgCl, AgBr$ and $AgI$ are respectively, $1.1 \times 10^{-12} , 1.8 \times 10^{-10} , 5.0 \times 10^{-13}, 8.3 \times 10^{-17}. $ Which one of the following salts will precipitate at the last?
Equilibrium
Solution:
Salt
$K_{sp}$
Solublity
$Ag_2CrO_4$
$1.1 \times 10^{-12} = 4s^3$
$ s = \sqrt[3]{\frac{K_{sp}}{4}} = 0.65 \times 10^{-4}$
$AgCl$
$1.8 \times 10^{-10} = s^2$
$s = \sqrt{K_{sp}} = 1.34 \times 10^{-5}$
$AgBr$
$5 \times 10^{-13} = s^2$
$s = \sqrt{K_{sp}} = 0.71 \times 10^{-6}$
$AgI$
$8.3 \times 10^{-17} = s^2$
$s = \sqrt{K_{sp}} = 0.9 \times 10^{-8}$
Solubility of $Ag_2 CrO_4 $ is highest thus, it will be precipitated at last.
| Salt | $K_{sp}$ | Solublity |
|---|---|---|
| $Ag_2CrO_4$ | $1.1 \times 10^{-12} = 4s^3$ | $ s = \sqrt[3]{\frac{K_{sp}}{4}} = 0.65 \times 10^{-4}$ |
| $AgCl$ | $1.8 \times 10^{-10} = s^2$ | $s = \sqrt{K_{sp}} = 1.34 \times 10^{-5}$ |
| $AgBr$ | $5 \times 10^{-13} = s^2$ | $s = \sqrt{K_{sp}} = 0.71 \times 10^{-6}$ |
| $AgI$ | $8.3 \times 10^{-17} = s^2$ | $s = \sqrt{K_{sp}} = 0.9 \times 10^{-8}$ |