Question

Given ${M_2}S{O_4}$ ( ${M^ + }$ is a monovalent metal ion) has a ${K_{sp}}$ of $1.2 \times {10^{ - 5}}$ at $298K$ . The maximum concentration of ${M^ + }$ ion that could be attained in a saturated solution of this solid at \[298K\] is.
A. $3.46 \times {10^{ - 3}}M$
B. $2.89 \times {10^{ - 3}}M$
C. $2.8 \times {10^{ - 3}}M$
D. $7 \times {10^{ - 3}}M$

Answer 1

Hint: We must remember that the solubility is characterized as a property of a substance called solute to get broken up in a dissolvable to form a solution. The dissolvability of ionic mixes (which disassociate to shape cations and anions) in water shifts to an incredible arrangement. A few mixes are profoundly dissolvable and may even ingest dampness from the climate though others are exceptionally insoluble.

Complete step by step answer:
The given reaction is,
${M_2}S{O_4} \rightleftarrows 2{M^ + } + S{O_4}^{2 - }$
Let us assume the solubility as S.
Now we calculate the solubility product ${K_{sp}}$ of the reaction,
${K_{sp}} = {\left[ {2S} \right]^2}\left[ S \right]$
It is given ${K_{sp}}$ is $1.2 \times {10^{ - 5}}$
Substitute the given value in the above equation,
$1.2 \times {10^{ - 5}} = 4{S^3}$
$12 \times {10^{ - 6}} = 4{S^3}$
$3 \times {10^{ - 6}} = {S^3}$
$S = 1.4 \times {10^{ - 2}}M$
The solubility of ${M^ + }$ is $2S$ so its value is $2.8 \times {10^{ - 2}}M$.

So, the correct answer is Option C.

Note: Now we discuss about the essentialness of Solubility Product:
At the point when a salt is disintegrated in a dissolvable the solid powers of fascination of solute (grid enthalpy of its particles) should be overwhelmed by the associations among particles and the dissolvable.
The solvation enthalpy of particles is consistently negative which implies that energy is delivered during this cycle.
The idea of the dissolvable decides the measure of energy delivered during solvation that is solvation enthalpy.
Non-polar solvents have a little estimation of solvation enthalpy, implying that this energy isn't adequate to beat the grid enthalpy.
So the salts are not broken up in non-polar solvents. Henceforth, for salt to be broken up in a dissolvable, its solvation enthalpy ought to be more noteworthy than its grid enthalpy.
Solvency relies upon temperature and it is distinctive for each salt.