That is correct! The correct expression for the solubility product constant, Ksp, of magnesium fluoride would be:
Ksp = (Mg^2+)(F^-)^2
Since the concentration of fluoride ions, [F^-], is 2x, you can substitute it into the equation:
Ksp = (Mg^2+)(2x)^2
Simplifying further:
Ksp = 4x^3
And since the given value for Ksp is 6.4E-9, you can set up the equation:
6.4E-9 = 4x^3
Solving for x:
x^3 = (6.4E-9)/(4)
x^3 = 1.6E-9
x = (1.6E-9)^(1/3)
x = 1.2E-3 M
So, the concentration of magnesium ions, [Mg^2+], and therefore the solubility of magnesium fluoride, is 1.2E-3 M.
To calculate the solubility in grams per liter, you need to convert the concentration into moles per liter by multiplying by the molar mass of MgF2:
Mass of MgF2 (g/L) = (1.2E-3 mol/L)(62.3 g/mol)
Mass of MgF2 (g/L) = 7.5E-2 g/L
Therefore, the solubility of magnesium fluoride in water is 7.5E-2 grams per liter.