Question

Silver chromate is sparingly soluble in aqueous solutions. The ${K_{sp}}$ of $A{g_2}Cr{O_4}$ is $1.12 \times {10^{ - 12}}$ . What is the solubility (in mol/L) of silver chromate in 1.50 M potassium chromate aqueous solution? In 1.50 M silver nitrate aqueous solution and in pure water?

Answer 1

In this problem we are given the solubility constant ${K_{sp}}$ . The value of ${K_{sp}}$ is given as the product of the solubilities of the respective ions present in the salt, raised to their respective stoichiometries. The solutions given to us are silver chromate, potassium chromate, silver nitrate in water.

Complete answer:
We are given silver chromate having the chemical formula $A{g_2}Cr{O_4}$ . The ${K_{sp}}$ is given as $1.12 \times {10^{ - 12}}$ .
1. The first solvent given to us is water, only silver chromate and water is present in the solution. Therefore, consider the solubilities of ions of silver chromate in water to be ‘s’. The dissociation can be given as:
$A{g_2}{\text{Cr}}{{\text{O}}_4}{\text{ }} \rightleftharpoons {\text{ 2A}}{{\text{g}}^ + }{\text{ }} + {\text{ Cr}}O_4^{2 - }$

T=0	a	-	-
T=equilibrium	$a - s$	$2s$	$s$

The ${K_{sp}}$ can be given as the product of the solubilities of products. Mathematically it can be shown as: ${K_{sp}} = {(2s)^2}(s) = 1.12 \times {10^{ - 12}}$
${K_{sp}} = 4{s^3} = 1.12 \times {10^{ - 12}}$
Therefore, $s = \sqrt[3]{{\dfrac{{1.12 \times {{10}^{ - 12}}}}{4}}}$
$s = 0.65 \times {10^{ - 4}} = 6.5 \times {10^{ - 5}}mol/L$
This is the solubility of silver chromate in water.
2. The next solvent we are given is 1.50M potassium chromate. We can see that in both salts Chromate ion is common. In this question we are given that $A{g_2}Cr{O_4}$ and Potassium chromate, in here the common ion effect will come into play. The common ion in both the solutions is $CrO_4^{2 - }$ . Hence the concentration of $CrO_4^{2 - }$ will increase in the resultant solution.
The dissociation of 1.50 M potassium chromate can be given as: ${K_2}Cr{O_4} \rightleftharpoons 2{K^ + } + CrO_4^{2 - }$ . Therefore, the concentration of $CrO_4^{2 - }$ will also be 1.50M. The dissociation of silver chromate can hence be given as
$A{g_2}{\text{Cr}}{{\text{O}}_4}{\text{ }} \rightleftharpoons {\text{ 2A}}{{\text{g}}^ + }{\text{ }} + {\text{ Cr}}O_4^{2 - }$

T=0	a	-	-
T=equilibrium	$a - s$	$2s$	$s + 1.50$

The ${K_{sp}}$ is given as $1.12 \times {10^{ - 12}}$ . Mathematically can be given as: : ${K_{sp}} = {(2s)^2}(s + 1.50) = 1.12 \times {10^{ - 12}}$
The value of s is very small compared to 1.50M
${K_{sp}} = {(2s)^2}(1.50) = 1.12 \times {10^{ - 12}}$
${K_{sp}} = 4 \times {s^2} \times 1.50 = 1.12 \times {10^{ - 12}}$
${K_{sp}} = 6{s^2} = 1.12 \times {10^{ - 12}}$
Therefore, $s = \sqrt[2]{{\dfrac{{1.12 \times {{10}^{ - 12}}}}{6}}} = 0.43 \times {10^{ - 6}}mol/L$
This is the solubility in potassium chromate.
3. The next solvent we are given is 1.50M silver nitrate. We can see that in both the salts silver ions are common. In this question we are given that $A{g_2}Cr{O_4}$ and silver nitrate, in here the common ion effect will come into play. The common ion in both the solutions is $A{g^ + }$ . Hence the concentration of $A{g^ + }$ will increase in the resultant solution.
The dissociation of 1.50 M potassium chromate can be given as: $AgN{O_3} \rightleftharpoons A{g^ + } + NO_3^ -$. Therefore, the concentration of $A{g^ + }$ will also be 1.50M. The dissociation of silver chromate can hence be given as:
$A{g_2}{\text{Cr}}{{\text{O}}_4}{\text{ }} \rightleftharpoons {\text{ 2A}}{{\text{g}}^ + }{\text{ }} + {\text{ Cr}}O_4^{2 - }$

T=0	a	-	-
T=equilibrium	$a - s$	$2s + 1.50$	$s$

The ${K_{sp}}$ is given as $1.12 \times {10^{ - 12}}$ . Mathematically can be given as: : ${K_{sp}} = {(2s + 1.50)^2}(s) = 1.12 \times {10^{ - 12}}$
The value of s is very small compared to 1.50M
${K_{sp}} = {(1.50)^2} \times s = 1.12 \times {10^{ - 12}}$
${K_{sp}} = 2.25s = 1.12 \times {10^{ - 12}}$
Therefore, $s = \dfrac{{1.12 \times {{10}^{ - 12}}}}{{2.25}} = 0.5 \times {10^{ - 12}} = 5 \times {10^{ - 13}}mol/L$
This is the solubility in silver nitrate.

Note:
The order of solubilities in different solvents can be given as: $water > potassium{\text{ }}chromate > silver{\text{ }}nitrate$. More the value of ${K_{sp}}$ more will be the solubility in that respective solvent. If we are given two or more solutions having at least one common ion, always the common ion effect will come into play, and an increase in the concentration of that common ion is observed. If concentration of one ion is increased, the system will try to decrease the concentration and the equilibrium will move forward.