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Question: The solubility of Lead Fluoride \((Pb{{F}_{2}})\) is \(2.5\times {{10}^{-3}}\)mol/L. What is the sol...

The solubility of Lead Fluoride (PbF2)(Pb{{F}_{2}}) is 2.5×10−32.5\times {{10}^{-3}}mol/L. What is the solubility in 0.10M NaF solution?
A. 6.3×10−6M6.3\times {{10}^{-6}}M
B. 1.6×10−6M1.6\times {{10}^{-6}}M
C. 5.0×10−2M5.0\times {{10}^{-2}}M
D. 1.6×10−5M1.6\times {{10}^{-5}}M

Explanation

Solution

To solve this question firstly we have to write the balanced equilibrium equation for the dissolution reaction that is given in the question and then solubility product expression for that reaction. To find the solubility product we have a formula that is,Ksp=[A+]a[B−]b{{K}_{sp}}={{[{{A}^{+}}]}^{a}}{{[{{B}^{-}}]}^{b}} where Ksp{{K}_{sp}} is the solubility product constant.

Complete answer:
From your chemistry lessons you have learned about the solubility of a substance and about the solubility product constant.
Solubility denoted by s is defined as the tendency of a solute to get dissolved in the solvent so they can form the solution. Solubility of an ionic compound in the water varies greatly. Ionic compounds dissociate in water to form cation and anions.
Solubility product constant is defined as the equilibrium constant in which the dissolution of solid substance takes place in aqueous solution. The symbol of solubility product constant is Ksp{{K}_{sp}}. The expression of solubility product constant for a balanced equilibrium equation for the dissolution reaction is:
AaBb⇌aA+bB{{A}_{a}}{{B}_{b}}\rightleftharpoons aA+bB
Ksp=[A+]a[B−]b{{K}_{sp}}={{[{{A}^{+}}]}^{a}}{{[{{B}^{-}}]}^{b}}
Where,Ksp{{K}_{sp}} = solubility product constant
[A+][{{A}^{+}}] = cation in the aqueous solution
[B−][{{B}^{-}}] = anion in the aqueous solution
And a, b = relative concentration of A and B.
In the question the solubility (s) for PbF2Pb{{F}_{2}} is given, So to find the solubility product constant we have write the equilibrium dissolution equation for PbF2Pb{{F}_{2}},
PbF2⇌Pb2++2F−Pb{{F}_{2}}\rightleftharpoons P{{b}^{2+}}+2{{F}^{-}}
Now, the solubility constant for this reaction is given as,
Ksp=[Pb2+][F−]2{{K}_{sp}}=[P{{b}^{2+}}]{{[{{F}^{-}}]}^{2}}
Ksp=(s)×(2s)2{{K}_{sp}}=(s)\times {{(2s)}^{2}}…….. (1)
Here s is the solubility and the value of solubility is given as 2.5×10−32.5\times {{10}^{-3}}mol/L. So, after putting the value of s in the equation (1) we get:
Ksp=(2.5×10−3)×(2×2.5×10−3)2{{K}_{sp}}=(2.5\times {{10}^{-3}})\times {{(2\times 2.5\times {{10}^{-3}})}^{2}}
∴Ksp=6.4×10−8\therefore {{K}_{sp}}=6.4\times {{10}^{-8}}
Now we have to find the value of solubility for NaF. So, the dissociation reaction of NaF will be written as :
NaF⇌Na++F−NaF\rightleftharpoons N{{a}^{+}}+{{F}^{-}}
0.10M 0.10M 0.10M (Concentration is given as 0.10 M)
So, the solubility constant for NaF will be:
Ksp=(s)×(2s+0.10)2{{K}_{sp}}=(s)\times {{(2s+0.10)}^{2}}
Here, the value of Ksp{{K}_{sp}} will be 6.4×10−86.4\times{{10}^{-8}}.So, after putting the value of Ksp{{K}_{sp}}in equation (2) we will get:
s=6.3×10−6Ms=6.3\times {{10}^{-6}}M

Thus the correct option will be (A).

Note:
Solubility product constant depends upon temperature as the value of temperature increases value increases because the solubility increases. The value of Solubility is different for each salt. The value of solubility depends mainly on salvation enthalpy of the ion and lattice enthalpy of salt. And the salvation enthalpy of the ions are always negative which refers to the release of energy during the reaction.
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