Question

Solubility products of the salts of types $MX,M{X_2},{M_3}X$ at temperature $T$ are $4 \times {10^{ - 8}},3.2 \times {10^{ - 14}},27 \times {10^{ - 16}}$ . Solubility of the salts at $T$are in the order.
A. $MX > M{X_2} > {M_3}X$
B. ${M_3}X > M{X_2} > MX$
C. $M{X_2} > {M_3}X > MX$
D. $MX > {M_3}X > M{X_2}$

Answer 1

Hint: Solubility product (${K_{sp}}$) of a substance is defined as the mathematical product of the dissolved ion concentrations raised to the power of their corresponding stoichiometric coefficients. The ${K_{sp}}$of a substance is different for different substances (or compounds).

Complete answer:
As per the question, we have been provided with three different types of salts which produce different concentrations of ions when dissolved in an aqueous solution or any kind of solvent. The solubility product of a salt/compound is dependent on the temperature. But, here, as all the dissociations take place at the same temperature, thus, this factor will be neglected. The temperature will not influence any of the solubility products of the various salts provided. Let us discuss each of the salts one by one:
(i) $MX$ type of salt on dissociation produces two ions. The reaction can be written as:
$MX \to {M^ + } + {X^ - }$
The value of solubility product for $MX$ type of salt is equal to:
${K_{sp}} = {S^2}$
Substituting the values, we have:
$4 \times {10^{ - 8}} = {S^2}$
Thus, the value of solubility is equal to:
$S = \sqrt {4 \times {{10}^{ - 8}}} = 2 \times {10^{ - 4}}$
(ii) $M{X_2}$ type of salt on dissociation produces three ions. The reaction can be written as:
$M{X_2} \to {M^ + } + 2{X^ - }$
The value of solubility product for $M{X_2}$ type of salt is equal to:
${K_{sp}} = 4{S^3}$
Substituting the values, we have:
$3.2 \times {10^{ - 14}} = 4{S^3}$
Thus, the value of solubility is equal to:
$S = \sqrt[3]{{\dfrac{{3.2 \times {{10}^{ - 14}}}}{4}}} = 2 \times {10^{ - 5}}$
(iii) ${M_3}X$ type of salt on dissociation produces four ions. The reaction can be written as:
${M_3}X \to 3{M^ + } + {X^ - }$
The value of solubility product for ${M_3}X$ type of salt is equal to:
${K_{sp}} = 27{S^4}$
Substituting the values, we have:
$27 \times {10^{ - 16}} = 27{S^4}$
Thus, the value of solubility is equal to:
$S = \sqrt[4]{{\dfrac{{27 \times {{10}^{ - 16}}}}{{27}}}} = 1 \times {10^{ - 4}}$

Thus, the correct option is D. $MX > {M_3}X > M{X_2}$ .

Note:
The solubility product usually increases with an increase in temperature due to the increased solubility. Solubility can be defined as a property of the solute to get dissolved in a solvent in order to form a solution. The greater the value of solubility product of a substance, the greater will be the tendency of the salt to form free dissolved ions in the solution.