Question

The solubility product of $A{g_2}Cr{O_4}$ is $32 \times {10^{ - 12}}$. What is the concentration of $CrO_4^ -$ ions in that solution?
A.$2 \times {10^{ - 4}}m/s$
B.$16 \times {10^{ - 4}}m/s$
C.$8 \times {10^{ - 4}}m/s$
D.$8 \times {10^{ - 8}}m/s$

Answer 1

To calculate the concentration of an ion when the solubility product of its compound is given, we can determine the solubility of the respective ion, as the solubility is equal to the concentration of the ion.

Complete step by step answer:
The solubility product constant (${K_{sp}}$) is the equilibrium constant for the dissolution of a solid substance into an aqueous solution of it. This further tells us that the solid compound when in equilibrium with its saturated solution, the product of concentrations of ions of both the products are equal to the solubility product constant.
Solubility products then can be expressed in terms of the solubility of the products formed. It is defined as a property of a substance in which solute is dissolved in a solvent in order to form a solution. The chemical equation involved in the reaction is-
$A{g_2}Cr{O_4} \rightleftharpoons 2A{g^ + } + CrO_4^{2 - }$
Let us consider the solubility of $A{g_2}Cr{O_4}$ and $CrO_4^{2 - }$is S for both and for $A{g^ + }$is 2S after equilibrium.
${K_{sp}} = {[A{g^ + }]^2}[CrO_4^{2 - }]$
${K_{sp}} = {(2S)^2}S = 4{S^3}$
$S = {(\dfrac{{{K_{sp}}}}{4})^{\dfrac{1}{3}}}$
Putting the value of ${K_{sp}}$ in the above formula, we get
$= {(\dfrac{{32 \times {{10}^{ - 12}}}}{4})^{\dfrac{1}{3}}}$
= $2 \times {10^{ - 4}}m/s$ or $2 \times {10^{ - 4}}M$

Hence, the correct option is (A).

Note:
One must associate the stoichiometric coefficient of every ion or compound with the solubility product while solving for concentration and solubility. The higher the value of ${K_{sp}}$, higher will be the solubility.