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Question: How many times solubility of \[Ca{{F}_{2}}\] is decreased in \(4X{{10}^{-3}}M\) KF (aq) solution as ...

How many times solubility of CaF2Ca{{F}_{2}} is decreased in 4X10−3M4X{{10}^{-3}}M KF (aq) solution as compared to pure water at 25oC{{25}^{o}}C ?
Given: KSP(CaF2)=3.2×10−11{{K}_{SP}}(Ca{{F}_{2}})=3.2\times {{10}^{-11}}
(A) 50
(B) 100
(C) 500
(D) 1000

Explanation

Solution

The solubility product is an equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solution dissolves in solution. The more soluble substance means it has a higher solubility product value. Solubility product is only applicable to the sparingly soluble solution.

Complete step by step solution:
Given the solubility product of CaF2Ca{{F}_{2}}, KSP=3.2×10−11{{K}_{SP}}=3.2\times {{10}^{-11}}
Let us consider, the solubility of calcium difluoride CaF2Ca{{F}_{2}} be ‘a’ moles per liter
CaF2⇌Ca2++2F−Ca{{F}_{2}}\rightleftharpoons C{{a}^{2+}}+2{{F}^{-}}
Solubility product of CaF2Ca{{F}_{2}}, KSP=[Ca+2][F−]2=3.2×10−11{{K}_{SP}}=[C{{a}^{+2}}]{{[{{F}^{-}}]}^{2}}=3.2\times {{10}^{-11}}--- (1)
The concentrations of [Ca2+][C{{a}^{2+}}] and [F−][{{F}^{-}}] are ‘a’ and ‘2a’ respectively, then
a×(2a)2=3.2×10−11a\times {{(2a)}^{2}}=3.2\times {{10}^{-11}}
Therefore, a = 2×10−4molL−12\times {{10}^{-4}}mol{{L}^{-1}}
In presence of 4×10−3M4\times {{10}^{-3}}M KF (aq) solution, consider the solubility is ‘b’,
Then the equation (1) changes with respect to KF aqueous solution,
b×(4×10−3×2b)2=3.2×10−11b\times {{(4\times {{10}^{-3}}\times 2b)}^{2}}=3.2\times {{10}^{-11}}
Therefore, b = 2×10−6molL−12\times {{10}^{-6}}mol{{L}^{-1}}
Solubility decreases in 4×10−3M4\times {{10}^{-3}}M KF (aq) solution as compared to pure water,
= ba\dfrac{b}{a} = 2×10−42×10−6\dfrac{2\times {{10}^{-4}}}{2\times {{10}^{-6}}} =100
Hence, 100 times solubility CaF2Ca{{F}_{2}} is decreased in 4×10−3M4\times {{10}^{-3}}M KF (aq) solution as compared to pure water at25oC{{25}^{o}}C.

So, the correct answer is option B.

Note: Solubility products cannot be used for normally soluble compounds like sodium chloride, silver nitrate, etc. interactions between the ions in the solution interfere with the simple equilibrium. The solubility product is a value that the solution is saturated, if there is any solid present, cannot dissolve any more solid than the solution said to be saturated.