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Question: How much NaF should be added to 100 ml of solution having 0.016 M in \(S{{r}^{2+}}\) ions to reduce ...

How much NaF should be added to 100 ml of solution having 0.016 M in Sr2+S{{r}^{2+}} ions to reduce its concentration to 2.5×10−3M2.5\times {{10}^{-3}}M ? (KspSrF2=8×10−10)\left( {{K}_{s{{p}_{Sr{{F}_{2}}}}}}=8\times {{10}^{-10}} \right)
(A) 0.098 g
(B) 0.168 g
(C) 0.177 g
(D) 0.115 g

Explanation

Solution

These are based on the concepts of equilibrium and solubility equilibria. As, the name suggests equilibrium refers to the balance within the system of reactions or a reaction.

Complete Solution :
Let us solve the illustration likewise;
Given data-
Concentration of Sr2+S{{r}^{2+}} = [Sr2+]\left[ S{{r}^{2+}} \right] = 16×10−3M16\times {{10}^{-3}}M
Reduced to concentration of Sr2+S{{r}^{2+}}= [Sr2+]\left[ S{{r}^{2+}} \right] = 2.5×10−3M2.5\times {{10}^{-3}}M
Thus, [Sr2+]\left[ S{{r}^{2+}} \right] precipitated = (16−2.5)×10−3=13.5×10−3M\left( 16-2.5 \right)\times {{10}^{-3}} = 13.5\times {{10}^{-3}}M
Now, as the reaction takes place as:
Sr2++2F−→SrF2S{{r}^{2+}}+2{{F}^{-}}\to Sr{{F}_{2}}
So, concentration of F−{{F}^{-}} needed for the precipitation = [F−]\left[ {{F}^{-}} \right] = 2×13.5×10−3=27×10−3M2\times 13.5\times {{10}^{-3}}=27\times {{10}^{-3}}M
Also, we have:
KspSrF2=8×10−10=[Sr2+][F−]2{{K}_{s{{p}_{Sr{{F}_{2}}}}}}=8\times {{10}^{-10}}=\left[ S{{r}^{2+}} \right]{{\left[ {{F}^{-}} \right]}^{2}}
Thus, [F−]2=8×10−102.5×10−3=3.2×10−7M{{\left[ {{F}^{-}} \right]}^{2}} = \dfrac{8\times {{10}^{-10}}}{2.5\times {{10}^{-3}}} = 3.2\times {{10}^{-7}}M
So, [F−]=5.65×10−4M\left[ {{F}^{-}} \right]=5.65\times {{10}^{-4}}M i.e. the concentration of F−{{F}^{-}} which will also appear in the solution state.
Thus, concentration of F−{{F}^{-}} needed = [F−]=(27×10−3)+(5.65×10−4)=0.0275M\left[ {{F}^{-}} \right]=\left( 27\times {{10}^{-3}} \right)+\left( 5.65\times {{10}^{-4}} \right) = 0.0275M
Now, we know that molar mass of NaF = 42 g/mol
Thus, NaF required for 1 L of solution = 0.0275M×42g=1.157g0.0275M\times 42g=1.157g
As stated, NaF required for 100 ml of solution = 1.15710=0.1157g\dfrac{1.157}{10} = 0.1157g
Therefore, 0.1157 g of NaF is required to be added in 100 ml of solution.
So, the correct answer is “Option D”.

Note: Here, Ksp{{K}_{sp}} is the solubility product constant which is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level of dissolution of solute in the solution.