Question

Three sparingly soluble salts that have the same solubility products are given below.
I) ${{\text{A}}_{\text{2}}}{\text{X}}$
II) ${\text{AX}}$
III) ${\text{A}}{{\text{X}}_{\text{3}}}$
Their solubilities in a saturated solution will be such that:
A) III > II > I
B) III > I > II
C) II > III > I
D) II > I > III

Answer 1

Hint: We know that the solubility product of any salt at any temperature is the product of the molar concentration of its constituent ions. The concentration of ions is raised to the number of ions produced on dissociation of one molecule of the salt.


Complete solution:
We know that the solubility product of any salt at any temperature is the product of the molar concentration of its constituent ions. The concentration of ions is raised to the number of ions produced on dissociation of one molecule of the salt.
We are given that the three sparingly soluble salts ${{\text{A}}_{\text{2}}}{\text{X}}$, ${\text{AX}}$ and ${\text{A}}{{\text{X}}_{\text{3}}}$ have same solubility products.
We know that the solubility of a salt at any temperature is calculated from its solubility product.
We are given a salt ${{\text{A}}_{\text{2}}}{\text{X}}$. The salt dissociates as follows:
${{\text{A}}_{\text{2}}}{\text{X}} \rightleftharpoons 2{{\text{A}}^ + } + {{\text{X}}^{2 - }}$
The solubility product of the salt ${{\text{A}}_{\text{2}}}{\text{X}}$ is given as follows:
${{\text{K}}_{{\text{SP}}}} = {[{{\text{A}}^ + }]^2}[{{\text{X}}^{2 - }}]$
Where ${{\text{K}}_{{\text{SP}}}}$ is the solubility product.
For the salt ${{\text{A}}_{\text{2}}}{\text{X}}$,
${{\text{K}}_{{\text{SP}}}} = {[{{\text{A}}^ + }]^2}[{{\text{X}}^{2 - }}]$
${{\text{K}}_{{\text{SP}}}} = {\left( {2s} \right)^2} \times s$
${{\text{K}}_{{\text{SP}}}} = 4{s^3}$
$s = {\left( {\dfrac{{{{\text{K}}_{{\text{SP}}}}}}{4}} \right)^{{\text{1/3}}}}$
Where $s$ is the solubility of the ions.
We are given a salt ${{\text{A}}_{\text{2}}}{\text{X}}$. The salt dissociates as follows:
${\text{AX}} \rightleftharpoons {{\text{A}}^ + } + {{\text{X}}^ - }$
The solubility product of the salt ${\text{AX}}$ is given as follows:
${{\text{K}}_{{\text{SP}}}} = [{{\text{A}}^ + }][{{\text{X}}^ - }]$
Where ${{\text{K}}_{{\text{SP}}}}$ is the solubility product.
For the salt ${\text{AX}}$,
${{\text{K}}_{{\text{SP}}}} = [{{\text{A}}^ + }][{{\text{X}}^ - }] = s \times s = {s^2}$
$s = \sqrt {{{\text{K}}_{{\text{SP}}}}} $
Where $s$ is the solubility of the ions.
We are given a salt ${\text{A}}{{\text{X}}_{\text{3}}}$. The salt dissociates as follows:
${\text{A}}{{\text{X}}_{\text{3}}} \rightleftharpoons {{\text{A}}^{3 + }} + 3{{\text{X}}^ - }$
The solubility product of the salt ${\text{A}}{{\text{X}}_{\text{3}}}$ is given as follows:
${{\text{K}}_{{\text{SP}}}} = [{{\text{A}}^{3 + }}]{[{{\text{X}}^ - }]^3}$
Where ${{\text{K}}_{{\text{SP}}}}$ is the solubility product.
For the salt ${\text{A}}{{\text{X}}_{\text{3}}}$,
${{\text{K}}_{{\text{SP}}}} = [{{\text{A}}^{3 + }}]{[{{\text{X}}^ - }]^3}$
${{\text{K}}_{{\text{SP}}}} = s \times {\left( {3s} \right)^3}$
${{\text{K}}_{{\text{SP}}}} = 27{s^4}$
$s = {\left( {\dfrac{{{{\text{K}}_{{\text{SP}}}}}}{{27}}} \right)^{{\text{1/4}}}}$
Where $s$ is the solubility of the ions.
Thus, $\sqrt {{{\text{K}}_{{\text{SP}}}}} $ > ${\left( {\dfrac{{{{\text{K}}_{{\text{SP}}}}}}{4}} \right)^{{\text{1/3}}}}$ > ${\left( {\dfrac{{{{\text{K}}_{{\text{SP}}}}}}{{27}}} \right)^{{\text{1/4}}}}$.
Thus, solubilities in a saturated solution are ${\text{AX}}$ > ${{\text{A}}_{\text{2}}}{\text{X}}$ > ${\text{A}}{{\text{X}}_{\text{3}}}$. Thus, solubilities in a saturated solution are II > I > III.

Thus, the correct option is (D)


Note: The solubility product is calculated using the concentration of the ions in which the salt has dissociated. Solubility factor depends on various factors such as temperature, pressure and nature of the electrolyte. The concentration of ions is affected by these factors and thus, the solubility product gets affected.