A saturated solution of MgF2 at 23 degrees C was prepared by dissolving solid MgF2 in 1 L water. The Ksp of MgF2 is 1.5e-5.
a.)Calculate the mass of MgF2 dissolved.
b.) When .1 mols of solid KF was dissolved in the MgF2 solution, precipitation was observed. Find the mass of the additional precipitate.
I think I get a, I just found the molarity of Mg2+ using the Ksp and then converted to grams; but, I have no idea how to do b. Thanks in advance.
The b part is a common ion problem; i.e., F^- is the ion common to both KF and MgF2.
For part b, Ksp = (Mg^2+)(F^-)^2
The common ion shifts the solubility to the left. MgF2 ==> Mg^2+ + 2F^-
because of Le Chatlier's Principle. The new solubility will = new (Mg^2+) and you can convert that to g MgF2 in the 1L. Subtraction from the initial amount will give you the amount pptd.
I'm sorry but I am still so confused on pretty much everything. How do you calculate the new solubility? And how do you convert that to g MgF2 in 1L? And what is the initial amount that you subtract from?
To calculate the mass of MgF2 dissolved in the saturated solution, you need to use the solubility product constant (Ksp) and the molar mass of MgF2. The Ksp expression for MgF2 is:
Ksp = [Mg2+][F-]^2
Since MgF2 dissociates into one Mg2+ ion and two F- ions, we can assume that the concentration of F- ions is twice the concentration of Mg2+ ions. Let's denote the concentration of MgF2 that has dissolved as "x M".
a) Calculation of the mass of MgF2 dissolved:
1. Write the Ksp expression using the concentration "x":
Ksp = (x)(2x)^2 = 4x^3
2. Substitute the value of Ksp (1.5e-5) into the equation:
1.5e-5 = 4x^3
3. Solve for x:
x^3 = (1.5e-5) / 4
x^3 = 3.75e-6
x ≈ 0.015 M
4. Calculate the molar mass of MgF2:
Molar mass of MgF2 = atomic mass of Mg + 2 * atomic mass of F
= 24.31 g/mol + 2 * 19.00 g/mol
= 24.31 g/mol + 38.00 g/mol
= 62.31 g/mol
5. Calculate the mass of MgF2 dissolved:
Mass = moles * molar mass
Mass = 0.015 mol/L * 62.31 g/mol
Mass ≈ 0.94 g
The mass of MgF2 dissolved in the saturated solution is approximately 0.94 grams.
b) To find the mass of the additional precipitate when 0.1 moles of solid KF is dissolved in the MgF2 solution:
1. Determine the stoichiometry of the reaction between MgF2 and KF:
MgF2 (s) + KF (aq) → MgF2 (aq) + KF (s)
From the balanced equation, we can see that for every mole of KF added, one mole of solid MgF2 will precipitate. Therefore, the mass of the additional precipitate will be equal to the molar mass of MgF2 (62.31 g/mol) multiplied by 0.1 moles:
Mass = 0.1 mol * 62.31 g/mol
Mass = 6.23 g
The mass of the additional precipitate is 6.23 grams.