# What is the molar solubility of magnesium fluoride in a solution that is "0.1 M" sodium fluoride?

Apr 13, 2015

In your case, the molar solubility of magnesium fluoride will be $6.4 \cdot {10}^{- 7} \text{mol/L}$.

You need the value of the solubility product constant, ${K}_{s p}$, for magnesium fluoride; now, there are several values listed for ${K}_{s p}$, so I'll choose one $\to$ $6.4 \cdot {10}^{- 9}$.

If this is not the value given to you, just replace it in the calculations with whatever value you have.

So, the equilibrium equation for the dissociation of magnesium fluoride is

$M g {F}_{2 \left(s\right)} r i g h t \le f t h a r p \infty n s M {g}_{\left(a q\right)}^{2 +} + \textcolor{red}{2} {F}_{\left(a q\right)}^{-}$ $\textcolor{b l u e}{\left(1\right)}$

The key to this problem is the fact that, even before adding the magnesium fluoride, your solution contains fluoride anions, ${F}^{-}$, from the dissociation of the sodium fluoride.

$N a {F}_{\left(s\right)} \to N {a}_{\left(a q\right)}^{+} + {F}_{\left(a q\right)}^{-}$

Notice that 1 mole of sodium fluoride produces 1 mole of fluoride ions in solution; this means that the initial concentration of fluoride ions will be

$\left[{F}^{-}\right] = \left[N a F\right] = \text{0.1 M}$

Construct an ICE table for equation $\textcolor{b l u e}{\left(1\right)}$

$\text{ } M g {F}_{2 \left(s\right)} r i g h t \le f t h a r p \infty n s M {g}_{\left(a q\right)}^{2 +} + \textcolor{red}{2} {F}_{\left(a q\right)}^{-}$
I.......$-$....................0.............0.1
C.....$-$...................(+x)..........(+$\textcolor{red}{2}$x)
E......$-$.....................x...........0.1 + 2x

According to the definition of the solubility product constant, you'll get

${K}_{s p} = \left[M {g}^{2 +}\right] \cdot {\left[{F}^{-}\right]}^{\textcolor{red}{2}}$

$6.4 \cdot {10}^{- 9} = x \cdot {\left(0.1 + 2 x\right)}^{2}$

Since ${K}_{s p}$ is so small, you can approximate $\left(0.1 + 2 x\right)$ with $0.1$ to get

$6.4 \cdot {10}^{- 9} = x \cdot {0.1}^{2} \implies x = \frac{6.4 \cdot {10}^{- 9}}{10} ^ \left(- 2\right) = \textcolor{g r e e n}{6.4 \cdot {10}^{- 7} \text{mol/L}}$

Apr 13, 2015

The molar solubility = $7.4 \times {10}^{- 8} \text{mol/l}$

$M g {F}_{2 \left(s\right)} r i g h t \le f t h a r p \infty n s M {g}_{\left(a q\right)}^{2 +} + 2 {F}_{\left(a q\right)}^{-}$ $\textcolor{red}{\left(1\right)}$

You should be given a value for the solubility product ${K}_{s p}$ which is given by :

${K}_{s p} = \left[M {g}_{\left(a q\right)}^{2 +}\right] {\left[{F}_{\left(a q\right)}^{-}\right]}^{2} = 7.4 \times {10}^{- 10} m o {l}^{3.} {l}^{- 6}$$\textcolor{red}{\left(2\right)}$

From $\textcolor{red}{\left(1\right)}$ you can see from LeChatelier's Principle that increasing $\left[{F}_{\left(a q\right)}^{-}\right]$ will cause the equilibrium to shift to the left thus decreasing the solubility of the $M g {F}_{2}$.

This will also happen if we try to dissolve the salt in a solution which contains an ion which is common, in the case ${F}_{\left(a q\right)}^{-}$.

This is known as "The Common Ion Effect".

The molar solubility of the salt is also equal to $\left[M {g}_{\left(a q\right)}^{2 +}\right]$ as $M g {F}_{2}$ is 1 molar with respect to $M {g}^{2 +}$.

To make things simple we can assume that any ${F}_{\left(a q\right)}^{-}$ from the $M g {F}_{2}$ is tiny in comparison to the ${F}_{\left(a q\right)}^{-}$ from the $N a {F}_{\left(a q\right)}$ so we ignore it.

We'll give$\left[M {g}_{\left(a q\right)}^{2 +}\right]$the symbol $\text{s}$.

Now put the values into $\textcolor{red}{\left(2\right)} \Rightarrow$

$7.4 \times {10}^{- 10} = s \times {\left(0.1\right)}^{2}$

From which :

$s = 7.4 \times {10}^{- 8} \text{mol/l}$