Question

The molar solubility, ${\text{s}}$ of ${\text{B}}{{\text{a}}_3}{\left( {{\text{P}}{{\text{O}}_4}} \right)_2}$ in terms of ${{\text{K}}_{{\text{sp}}}}$ is:
A. ${\text{s = }}{\left( {{{\text{K}}_{{\text{sp}}}}} \right)^{\dfrac{1}{2}}}$
B. ${\text{s = }}{\left( {{{\text{K}}_{{\text{sp}}}}} \right)^{\dfrac{1}{5}}}$
C. ${\text{s = }}{\left( {\dfrac{{{{\text{K}}_{{\text{sp}}}}}}{{27}}} \right)^{\dfrac{1}{5}}}$
D. ${\text{s = }}{\left( {\dfrac{{{{\text{K}}_{{\text{sp}}}}}}{{108}}} \right)^{\dfrac{1}{5}}}$

Answer 1

Hint: Solubility may be considered to be an equilibrium between solid and ion solution. Any ionic solid is completely ionized in aqueous solution, once it actually dissolves. Solubility product is the mathematical product of its dissolved ion concentrations raised to the power of their stoichiometric coefficients.

Complete step by step solution:
Solubility of a substance is the amount of that substance that will dissolve in a given amount of solvent. The equation to find the solubility product is called solubility product principle. Most salts dissociate into ions when they dissolve. Although all compounds dissolve in water to some extent, some compounds are classified as insoluble. The solubility of these insoluble compounds can be described in terms of their solubility product constant, ${{\text{K}}_{{\text{sp}}}}$.
The reaction depicting the dissociation equilibrium for ${\text{B}}{{\text{a}}_3}{\left( {{\text{P}}{{\text{O}}_4}} \right)_2}$ is:
${\text{B}}{{\text{a}}_3}{\left( {{\text{P}}{{\text{O}}_4}} \right)_2} \rightleftharpoons 3{\text{B}}{{\text{a}}^{2 + }} + 2{\text{P}}{{\text{O}}_4}^{3 - }$
$\,\,\, \,\,\,\,$${\text{s}}$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$3{\text{s}}$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$2{\text{s}}$
${\text{s,3s,2s}}$ are the concentrations of each ion at equilibrium.
This equilibrium system may be described by the mass-action expression. Solubility product can be calculated by the following expression:
Solubility product, ${{\text{K}}_{{\text{sp}}}} = {\left[ {{\text{B}}{{\text{a}}^{2 + }}} \right]^3}{\left[ {{\text{P}}{{\text{O}}_4}^3} \right]^2}$
We know that the molar solubility is the number of moles of solute which are dissolved in one litre of solution. Each mole of salt produces three moles of ${\text{B}}{{\text{a}}^{2 + }}$ ions, thus $3{\text{s}}$, and two moles of ${\text{P}}{{\text{O}}_4}^{3 - }$, thus $2{\text{s}}$.
Substituting the values in the above equation, we get
${{\text{K}}_{{\text{sp}}}} = {\left[ {{\text{B}}{{\text{a}}^{2 + }}} \right]^3}{\left[ {{\text{P}}{{\text{O}}_4}^3} \right]^2} = {\left( {3{\text{s}}} \right)^3} \times {\left( {2{\text{s}}} \right)^2} = 27{{\text{s}}^3} \times 4{{\text{s}}^2} = 108{{\text{s}}^5}$
i.e. ${{\text{K}}_{{\text{sp}}}} = 108{{\text{s}}^5}$
Simplifying, we get
${{\text{s}}^5} = \dfrac{{{{\text{K}}_{{\text{sp}}}}}}{{108}}$
Taking root on both sides, we get
${\text{s = }}{\left( {\dfrac{{{{\text{K}}_{{\text{sp}}}}}}{{108}}} \right)^{\dfrac{1}{5}}}$
Now the molar solubility of ${\text{B}}{{\text{a}}_3}{\left( {{\text{P}}{{\text{O}}_4}} \right)_2}$, ${\text{s = }}{\left( {\dfrac{{{{\text{K}}_{{\text{sp}}}}}}{{108}}} \right)^{\dfrac{1}{5}}}$.

Thus the correct option is D.

Note: ${{\text{Q}}_{{\text{sp}}}}$ is the ion-product expression for a slightly soluble ionic compound. When the solution is saturated, the system is at equilibrium and the ionic product will be equal to the solubility product. Solubility product value of a salt indicates how far the dissolution proceeds at equilibrium.