Question

The solubility product of a salt having formula ${M_2}{X_3}$ is $2.2 \times {10^{ - 20}}$. If the solubility of an another salt having formula ${M_2}X$ is twice the molar enthalpy of ${M_2}{X_3}$, the solubility product of ${M_2}X$ is:
A.$3 \times {10^{ - 12}}$
B.$9.16 \times {10^{ - 5}}$
C.$4.58 \times {10^{ - 5}}$
D.$2.76 \times {10^{ - 18}}$

Answer 1

Hint:When we have to calculate the molarity of an ion and its solubility product is known, we find the solubility of the ion in the solution as it is equal to the concentration of the ion in the given solution. The solubility product or ${{\text{K}}_{{\text{sp}}}}$ represents the equilibrium between a salt and its dissociated ions in a solution.

Complete step by step answer:
The solubility product constant of a salt is the equilibrium constant for the dissolving of a solid salt into water making an aqueous solution. When the solid substance is in equilibrium with its saturated solution, the product of the concentrations of the constituent ions present in the solution gives the solubility product. The solubility product can also be expressed in terms of the solubility of ions formed by dissociation in the solution.
The dissolution of ${M_2}{X_3}$ is given as,
${M_2}{X_3}(s) \rightleftharpoons 2M\left( {aq} \right) + 3X\left( {aq} \right)$
Considering the concentration of M as x, concentration of X is $\dfrac{3}{2}x$.
We know that solubility products are given as the product of the concentrations or solubilities of the constituent ions. Hence, ${K_{sp}} = {\left( x \right)^2}{\left( {3x/2} \right)^3} = 2.2 \times {10^{ - 10}}$
$ \Rightarrow {x^3} = 6.5 \times {10^{ - 11}}$
$\therefore x = 4 \times {10^{ - 4}}$
So, the molar solubility of M is $\dfrac{1}{2}\left( {4 \times {{10}^{ - 4}}} \right) = 2 \times {10^{ - 4}}$
The solubility of ${M_2}X$ is given as twice the molar enthalpy of ${M_2}{X_3}$
${M_2}X \rightleftharpoons 2M(aq) + X(aq)$
Let the solubility of X as y.
${K_{sp}} = {\left( {2y} \right)^2}\left( y \right) = 4{y^3}$
${K_{sp}} = 4{\left( {4 \times {{10}^{ - 4}}} \right)^3} = 3 \times {10^{ - 12}}$

Hence, the correct answer is A.

Note:
The equilibrium constant is in general numerically calculated by allowing the reaction under consideration to proceed to equilibrium and then measuring the concentrations of each constituent present in the reaction mixture. Since the concentrations of the constituents of the reaction mixture are measured at equilibrium, the equilibrium constant always has the same value for a given reaction irrespective of the initial amount taken of the reactants.