Question

The solubility of $Ca{F_2}$ $\left( {{K_{sp}} = 3.4 \times {{10}^{ - 11}}} \right)$ in $0.1\;M$ solution of $NaF$ would be:
A. $3.4 \times {10^{ - 12}}\;{\rm{M}}$
B. $3.4 \times {10^{ - 10}}\;{\rm{M}}$
C. $3.4 \times {10^{ - 9}}\;{\rm{M}}$
D. $3.4 \times {10^{ - 13}}\;{\rm{M}}$
Given:
The concentration of $NaF$ solution is: $\left[ {NaF} \right] = 0.1\;M$
The solubility product of $Ca{F_2}$ is: ${{\rm{K}}_{{\rm{sp}}}} = 3.4 \times {10^{ - 11}}$

Answer 1

We can use the common ion effect to determine the solubility of $Ca{F_2}$ in the given $NaF$ solution.We should also know the way both these compounds dissociate in aqueous solution.

Complete step by step answer
We can write the expression for equilibrium constant for any given chemical equation. When we do so for dissolving a solute in a solvent, the equilibrium constant is known as a solubility product. Let’s assume a general dissolution reaction of a salt ${A_a}{B_b}$ for which the chemical equation can be written as:
${A_a}{B_b}\left( s \right) \to a{A^{m + }}\left( {sol} \right) + b{B^{n - }}\left( {sol} \right)$
We can write the expression for the equilibrium constant as:
$K = \dfrac{{{{\left[ {{A^{m + }}\left( {sol} \right)} \right]}^a}{{\left[ {{B^{n - }}\left( {sol} \right)} \right]}^b}}}{{\left[ {{A_a}{B_b}\left( s \right)} \right]}}$
We do not consider the concentration change for a pure solid. So, we get the expression for solubility product:
${K_{sp}} = {\left[ {{A^{m + }}\left( {sol} \right)} \right]^a}{\left[ {{B^{n - }}\left( {sol} \right)} \right]^b}$
Now, if we consider molar solubility of the salt ${A_a}{B_b}$ to be $X$ we can write:

$${K_{sp}} = {\left[ {aX} \right]^a}{\left[ {bX} \right]^b}\\\ = {a^a}{b^b}{X^{a + b}}$$

So, we can calculate the solubility of the salt $\left( X \right)$ from the value of ${K_{sp}}$ and stoichiometric coefficients but what if one ion is already present in the solution. Then, we will see the common ion effect and solubility would be calculated by taking into account that concentration as well.
Here, initially we have, $0.1\;M$ solution of $NaF$ which means, considering the complete dissolution of salt in the solution, we have:

$NaF$	$\longrightarrow$	$Na^+$	$+$	$F^-$
$0.1\,M$		$0$		$0$
$-0.1\,M$		$+0.1\,M$		$+0.1\,M$
$0$		$0.1\,M$		$0.1\,M$

Now, let’s write the chemical equation for dissolution of $Ca{F_2}$ as follows:
$Ca{F_2} \to C{a^{2 + }} + 2{F^ - }$
Let’s assume that its solubility is $Y$, so we will include the concentration of the common ion, $\left[ {{F^ - }} \right]$ from $NaF$ and write:

$${K_{sp}} = \left[ {C{a^{2 + }}} \right]{\left[ {{F^ - }} \right]^2}\\\ = \left( Y \right){\left( {2Y + 0.1} \right)^2}$$

We are given, ${{\rm{K}}_{{\rm{sp}}}} = 3.4 \times {10^{ - 11}}$, as it is quite smile so we can approximate the second term $\left( {2Y + 0.1 \approx 0.1} \right)$ and write:

$$\left( Y \right){\left( {0.1} \right)^2} = {K_{sp}}\\\ Y = \dfrac{{{K_{sp}}}}{{{\rm{0}}{\rm{.01}}}}$$

We will use the given value ${{\rm{K}}_{{\rm{sp}}}} = 3.4 \times {10^{ - 11}}$ and determine the solubility as follows:

$$\Rightarrow Y = \dfrac{{3.4 \times {{10}^{ - 11}}}}{{{\rm{0}}{\rm{.01}}}}\\\ = 3.4 \times {10^{ - 9}}$$

**Hence, the solubility is determined to be $3.4 \times {10^{ - 9}}$ which makes C to be the correct option.

Note: **
It is very important to write the correct expression for solubility products in terms of solubility by using proper coefficients and powers. Approximations in concentrations are not allowed as the small change on being taken to some power may lead to big changes in the final answer.