Looking for classes? Ksquare Career Institute, Bengaluru →
The $K_{sp}$ of $Ag_2CrO_4$, AgCl, AgBr and AgI are, respectively, $1.1\times10^{-12}$, $1.8\times10^{-10}$, $5.0\times10^{-13}$, $8.3\times10^{-17}$. Which one of the following salts will precipitate last if $AgNO_3$ solution is added to the solution containing equal moles of NaCl, NaBr, NaI and $Na_2CrO_4$?
A
AgI
B
AgCl
C
AgBr
D
$Ag_2CrO_4$
Detailed Solution
The salt with the highest solubility precipitates last.
$Ag_2CrO_4$ ($A_2B$ type): $K_{sp} = 4s^3 = 1.1\times10^{-12} \Rightarrow s = (0.275\times10^{-12})^{1/3} \approx 6.5\times10^{-5}$ M
AgCl: $s = \sqrt{1.8\times10^{-10}} \approx 1.3\times10^{-5}$ M
AgBr: $s = \sqrt{5\times10^{-13}} \approx 7\times10^{-7}$ M
AgI: $s = \sqrt{8.3\times10^{-17}} \approx 9\times10^{-9}$ M
$Ag_2CrO_4$ is the most soluble, so it precipitates last.
$Ag_2CrO_4$ ($A_2B$ type): $K_{sp} = 4s^3 = 1.1\times10^{-12} \Rightarrow s = (0.275\times10^{-12})^{1/3} \approx 6.5\times10^{-5}$ M
AgCl: $s = \sqrt{1.8\times10^{-10}} \approx 1.3\times10^{-5}$ M
AgBr: $s = \sqrt{5\times10^{-13}} \approx 7\times10^{-7}$ M
AgI: $s = \sqrt{8.3\times10^{-17}} \approx 9\times10^{-9}$ M
$Ag_2CrO_4$ is the most soluble, so it precipitates last.