Question

If the solubility of \[L{i_3}N{a_3}{(Al{F_6})_2}\] is x mol/L, then its solubility product is equal to:
(A) \[12{x^3}\]
(B) \[18{x^3}\]
(C) \[{x^8}\]
(D) \[2916{x^8}\]

Answer 1

Hint: Solubility is the property of a chemical substance such as solute that can dissolve in a solvent. Both solute and solvent will exist in solid, liquid and gaseous form. The equilibrium constant for solubility can also be referred to as the solubility product.

Complete Solution:
- We can define Solubility as the maximum amount of chemical substance such as solute that can dissolve in a given amount of solvent at a specified temperature.
- The solubility product is kind of an equilibrium constant whose value depends upon the temperature. It is denoted by \[{k_{sp}}\].

Let us now move onto the question.
\[L{i_3}N{a_3}{(Al{F_6})_2}\] dissociates to form 3 moles of \[L{i^ + }\],3 moles of \[N{a^ + }\]and 2 moles of \[AlF_6^{3 - }\].
The solubility product is given as
\[{k_{sp}} = {[aA]^a}{[bB]^b}{[cC]^c}\]
where, A, B and C are the products of the reaction, and a, b and c are the stoichiometric coefficients.
Therefore, the solubility product of \[L{i_3}N{a_3}{(Al{F_6})_2}\] is given as
\[{k_{sp}} = {[3L{i^ + }]^3}{[3N{a^ + }]^3}{[2AlF_6^{3 - }]^2}\]
Given, the solubility is \[x~mol{L^{ - 1}}\].

- Now, substitute the values obtained in equation (1), we get:
\[{k_{sp}} = {[3x]^3}{[3x]^3}{[2x]^2}\]
\[{k_{sp}} = 27{x^3} \times 27{x^3} \times 4{x^2}\]
\[{k_{sp}} = 2916{x^8}\]
Therefore, the solubility product is equal to \[2916{x^8}\].

So, the correct answer is “Option D”.

Note: When calculating the solubility product, we don't consider the concentrations of the solids as their concentrations do not change the expression, any change in their concentration is significant and therefore, is omitted.