Answer:
To calculate the solubility product constant (Ksp) for magnesium fluoride (MgF2) at 25 degrees C, you can use the given information:
Given:
Mass of solute (MgF2) = 8.0 x 10^-2 g
Volume of solvent (water) = 1 liter
First, we need to calculate the molar concentration (molarity) of MgF2 in the solution:
Molar mass of MgF2 = 24.31 g/mol (for Mg) + 2 * 18.99 g/mol (for F) = 62.29 g/mol
Molarity (M) = Mass of solute (in grams) / Molar mass of MgF2
M = 8.0 x 10^-2 g / 62.29 g/mol
Next, convert molarity to moles per liter:
Moles per liter (mol/L) = Molarity (M)
Now, we can write the dissolution equation for MgF2:
MgF2 (s) ⇌ Mg²⁺ (aq) + 2F⁻ (aq)
Since 1 mole of MgF2 produces 1 mole of Mg²⁺ ions and 2 moles of F⁻ ions, the equilibrium concentration of Mg²⁺ will be equal to the molarity (M) of MgF2, and the equilibrium concentration of F⁻ will be twice the molarity (2M) of MgF2.
Now, we can write the Ksp expression for MgF2:
Ksp = [Mg²⁺] * [F⁻]^2
Ksp = M * (2M)^2
Ksp = 4M^3
Finally, substitute the value of M:
Ksp = 4 * (8.0 x 10^-2 g / 62.29 g/mol)^3
Calculate the Ksp to find the answer