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Question: The molarity of \[M{g^{2 + }}\]ions in a saturated solution of \[M{g_3}{\left( {P{O_4}} \right)_2}\ ...

The molarity of Mg2+M{g^{2 + }}ions in a saturated solution of Mg3(PO4)2 ]whosesolubilityproductis\[1.08 ×10−13M5M{g_3}{\left( {P{O_4}} \right)_2}\ ]whose solubility product is \[1.08{\text{ }} \times {10^{ - 13}}{M^5} is,
A .1.0 ×10−3M1.0{\text{ }} \times {10^{ - 3}}M
B. 2.0×10−3M{\text{ }}2.0 \times {10^{ - 3}}M
C. 3.0×10−3M{\text{ }}3.0 \times {10^{ - 3}}M
D.  4.0×10−3M{\text{ }}4.0 \times {10^{ - 3}}M

Explanation

Solution

When a partly soluble salt dissolves into water it forms a dynamic equilibrium between the hydrated ions and its solid salt molecule. Due to this equilibrium the amount of solid salt hydrolyzed remains unchanged after achieving an amount of hydration. Due to this a new term is introduced which is the solubility product (Ksp)\left( {{K_{sp}}} \right).

Complete step by step answer:
According to the principle, Solubility product (Ksp)\left( {{K_{sp}}} \right) is a mathematical term, which can be calculated by multiplication of the molar concentrations of solvated ions( when water is solvent then it is hydrated ion) of the salt with the power of their stoichiometric coefficient. Therefore, with increasing the value of molar concentrations of the ions the solubility product value also increases and vice-versa.
With increasing the temperature solubility of the salt increases as well as the concentrations of solvated ions. Therefore, with increasing temperature the value of solubility product increases and vice-versa.
Now the given salt, let assume the concentration of the hydrated salt Mg3(PO4)2M{g_3}{\left( {P{O_4}} \right)_2} is x moles per liter. Then the expression of solubility product is,

Mg3(PO4)2⇄3Mg+2+2PO4−2  x 3x 2x Ksp=[Mg+2]3[PO4−2]2  =[3x]3[2x]2  M{g_3}{\left( {P{O_4}} \right)_2} \rightleftarrows 3M{g^{ + 2}} + 2P{O_4}^{ - 2} \\\ {\text{ }}x{\text{ }}3x{\text{ }}2x \\\ {K_{sp}} = {\left[ {M{g^{ + 2}}} \right]^3}{\left[ {P{O_4}^{ - 2}} \right]^2} \\\ {\text{ }} = {\left[ {3x} \right]^3}{\left[ {2x} \right]^2} \\\

Now given, Ksp=1.08×10−13M5{K_{sp}} = 1.08 \times {10^{ - 13}}{M^5}
Therefore, calculate the value of x,

 Ksp=[Mg+2]3[PO4−2]2 or,Ksp =[3x]3[2x]2 or,1.08×10−3M5=[3x]3[2x]2 or,1.08×10−3M5=108x5 or,1.08×10−3M1085=x5 or,x=9.88×10−4M  {\text{ }}{K_{sp}} = {\left[ {M{g^{ + 2}}} \right]^3}{\left[ {P{O_4}^{ - 2}} \right]^2} \\\ or,{K_{sp}}{\text{ }} = {\left[ {3x} \right]^3}{\left[ {2x} \right]^2} \\\ or,1.08 \times {10^{ - 3}}{M^5} = {\left[ {3x} \right]^3}{\left[ {2x} \right]^2} \\\ or,1.08 \times {10^{ - 3}}{M^5} = 108{x^5} \\\ or,{\dfrac{{1.08 \times {{10}^{ - 3}}M}}{{108}}^5} = {x^5} \\\ or,x = 9.88 \times {10^{ - 4}}M \\\

Therefore, the concentration of Mg2+M{g^{2 + }}is

3x = {\text{ }}3 \times 9.88 \times {10^{ - 4}}M \\\ = {\text{ }}3.0 \times {10^{ - 3}}M \\\ $$. **The correct option is C.** **Note:** According to Le Chatelier’s principle , if a common ion is added in the equilibrium of solution of an immiscible solute, the solubility of that solute will decrease. Therefore, with increasing temperature the value of solubility products increases and vice-versa.