Question

Solubility of a salt ${{M}_{2}}{{X}_{3}}$ is $Ymol/lt$. The solubility product of the salt will be
A. $6{{Y}^{4}}$
B. $36{{Y}^{5}}$
C. $64{{Y}^{4}}$
D. $108{{Y}^{5}}$

Answer 1

Hint: Solubility product depends on the number of ions formed in a reaction at equilibrium. A salt of type ${{M}_{2}}{{X}_{3}}$can dissociate into two M cations and three X anions.

Complete Step by Step Answer:
Solubility product of a species is defined as the product of the concentration of the cation and anion formed from a species at equilibrium. At this stage the coefficients before the ions exist as a power to the concentration of the corresponding cation and anion.
${{M}_{2}}{{X}_{3}}$ dissociate at equilibrium as follows-
${{M}_{2}}{{X}_{3}}\rightleftharpoons 2{{M}^{+3}}+3{{X}^{2-}}$

Thus solubility product is given as follows:
$K_{sp}={{(2S)}^{2}}\times {{\left( 3S \right)}^{3}}$
$K_{sp}=4{{S}^{2}}\times 27{{S}^{3}}$
Putting the value of S as Y we get:
$K_{sp}=108{{Y}^{5}}$; where $K_{sp}$ is defined as the solubility product and $S$ is defined as the solubility for the cation and anion. M cation has 2 as coefficient and power and X anion has three as coefficient and power on solubility.

Thus the solubility product of salt of ${{M}_{2}}{{X}_{3}}$is $108{{Y}^{5}}$ .
Thus the correct option is D.

Note: The solubility product defines the soluble ability of a salt in a solution. Presence of common ions can decrease the solubility product of a salt and consequently decrease its solubility and help to form precipitate.