Question

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

Answer 1

Hint: 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 $S{{r}^{2+}}$ = $\left[ S{{r}^{2+}} \right]$ = $16\times {{10}^{-3}}M$
Reduced to concentration of $S{{r}^{2+}}$= $\left[ S{{r}^{2+}} \right]$ = $2.5\times {{10}^{-3}}M$
Thus, $\left[ S{{r}^{2+}} \right]$ precipitated = $\left( 16-2.5 \right)\times {{10}^{-3}} = 13.5\times {{10}^{-3}}M$
Now, as the reaction takes place as:
$S{{r}^{2+}}+2{{F}^{-}}\to Sr{{F}_{2}}$
So, concentration of ${{F}^{-}}$ needed for the precipitation = $\left[ {{F}^{-}} \right]$ = $2\times 13.5\times {{10}^{-3}}=27\times {{10}^{-3}}M$
Also, we have:
${{K}_{s{{p}_{Sr{{F}_{2}}}}}}=8\times {{10}^{-10}}=\left[ S{{r}^{2+}} \right]{{\left[ {{F}^{-}} \right]}^{2}}$
Thus, ${{\left[ {{F}^{-}} \right]}^{2}} = \dfrac{8\times {{10}^{-10}}}{2.5\times {{10}^{-3}}} = 3.2\times {{10}^{-7}}M$
So, $\left[ {{F}^{-}} \right]=5.65\times {{10}^{-4}}M$ i.e. the concentration of ${{F}^{-}}$ which will also appear in the solution state.
Thus, concentration of ${{F}^{-}}$ needed = $\left[ {{F}^{-}} \right]=\left( 27\times {{10}^{-3}} \right)+\left( 5.65\times {{10}^{-4}} \right) = 0.0275M$
Now, we know that molar mass of NaF = 42 g/mol
Thus, NaF required for 1 L of solution = $0.0275M\times 42g=1.157g$
As stated, NaF required for 100 ml of solution = $\dfrac{1.157}{10} = 0.1157g$
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, ${{K}_{sp}}$ 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.