Question

The solubility of magnesium fluoride, , in water is g/L. What is the solubility (in grams per liter) of magnesium fluoride in 0.29 M of sodium fluoride,

Answer 1

Answer:

1.457*10^-8 grams

Explanation:

First we want to find the molar concentration of MgF2. We can do this by dividing 0.016 (the solubility in grams of MgCl2 in a litre of water) by its molar mass (approx. 62.3 grams). Thus, the molar solubility of MgF2 is 2.57*10^-4 M.

Next, we must calculate the Ksp of MgF2. The equilibrium expression is:

MgF2⇄Mg+2F

Thus   $Ksp=[Mg^+][F^-]^2$

This means that, in equilibrium, there are 2.57*10^-4 M of $M^+$ and 5.136*10^-4 M of $F^-$

Plugging in the above information, our Ksp for MgF2 is approximately 6.78*10^-11

Next we will need to use the RICE table. Since there is already 0.29M of NaF dissolved, there is initially 0.29M of $F^-$.

R: MgF2 ⇄ $Mg^{+}$+2$F^-$

I:    N/A        0     0.29M

C:   N/A         +x      +x

--------------------------------------------

E:   N/A         x     0.29+x  

To make calculations easier, we will assume that 0.29+x≈0.29

This means that Ksp=0.29x=6.78*10^-11

Therefore, x≈2.338*10^-10M

Multiply that by 62.3 and we get around 1.457*10^-8 grams.