Magnesium fluoride dissolves in water to the extent of 8.0 10-2 g/L at 25°C. Calculate the solubility of MgF2(s) in moles per liter, and calculate Ksp for MgF2 at 25°C.

I've been doing this and getting .086. Please help!

nevermind..i got it

wait what is the answer?

To calculate the solubility of MgF2(s) in moles per liter, you can use the given information that it dissolves to the extent of 8.0 * 10^-2 g/L at 25°C.

First, you need to determine the molar mass of MgF2. Magnesium (Mg) has an atomic mass of 24.31 g/mol, and fluorine (F) has an atomic mass of 18.99 g/mol. Since there are two fluorine atoms in MgF2, you can calculate the molar mass as follows:

Molar mass of MgF2 = (Mg atomic mass) + 2 * (F atomic mass)
= 24.31 + 2 * 18.99
= 62.29 g/mol

To convert the solubility from grams per liter to moles per liter, divide the given solubility by the molar mass:

Solubility in moles per liter = (Solubility in grams per liter) / (Molar mass in grams per mole)
= 8.0 * 10^-2 g/L / 62.29 g/mol
= 1.284 * 10^-3 mol/L

So, the solubility of MgF2(s) in moles per liter is approximately 1.284 * 10^-3 mol/L.

To calculate the Ksp (solubility product constant) for MgF2 at 25°C, you need to use the solubility value you obtained and the balanced chemical equation for the dissolution of MgF2:

MgF2(s) ⇌ Mg^2+(aq) + 2F^-(aq)

In this equation, the stoichiometric coefficient for MgF2 is 1, while for Mg^2+ and F^- ions, they are also 1 and 2, respectively.

Considering the equation, the equilibrium expression for the dissolution reaction is as follows:

Ksp = [Mg^2+][F^-]^2

Since MgF2 dissociates into one Mg^2+ ion and two F^- ions, the concentrations of Mg^2+ and F^- ions are equal to the solubility of MgF2(s) calculated previously.

Therefore, Ksp = (Solubility of MgF2)^1 * (Solubility of F^-)^2.
= (1.284 * 10^-3 mol/L)^1 * (1.284 * 10^-3 mol/L)^2
= 1.754 * 10^-9

Hence, the Ksp for MgF2 at 25°C is approximately 1.754 * 10^-9.