A chemist prepares a solution of magnesium fluoride by measuring out 0.69 mg of magnesium fluoride into a 100.mL volumetric flask.

Question

Calculate the concentration in mol/L of the chemist's magnesium fluoride solution.

 

Answer 1

Magnesium fluoride is NOT very soluble in water, and has a K_sp of approximately 5.2x10^-11.  

MgF₂(s) <==> Mg²⁺ + 2F^-

K_sp = 5.2x10^-11 = [Mg²⁺][2F^-]²

5.2x10^-11 = [x][2x]² = 4x³ 

x = 2.3x10^-4 M

Compare this value to the concentration calculated by dissolving 0.69 mg MgF2 in 100 ml (~1.1x10^-4 M).  This tells you that the solution is not saturated and no precipitate will form.

0.69 mg x 1 mmol/62.3 mg/0.1 L = 0.11 mM = 1.1x10^-4 M.  This is below the Ksp.

Answer 2

I assume you mean that the chemist then filled up the 100mL vol flask?

 

.69mg mg x (1g/1000mg) x (1 mole/62.30 g) x (1/100mL) x (1000mL/L) = ?

 

You can take it from here.