The molar solubility of MgF2 at 25°C is 1.2 10-3 mol/L. Calculate Ksp.
i got the euqation ksp= [F-]^2[Mg]2+ and plugged in 1.2e-3 and got 1.7e-9 but it is wrong.
also this question
If 0.0968 g of CaF2 dissolves per liter of solution, calculate the solubility-product constant.
how do i do that?
......MgF2 ==> Mg^2+ + 2F^-
I.....solid.....0.......0
C..x dissolves..x.......2x
E......solid....x.......2x
Ksp = (Mg^2+)(F^-)^2
Ksp = (x)(2x)^2
Solve for Ksp. (I'm sure the problem you have is that you didn't double x before you squared.
You can save some time by
Ksp = 4x^3 = 4(1.2E-3)^3 = ?
The CaF2 is the same process.
To calculate the solubility-product constant (Ksp), you need to know the molar solubility of the compound (in moles per liter).
First, let's address the issue with the first question. You correctly identified the equation for the solubility-product constant of MgF2: Ksp = [F-]^2[Mg]2+.
Given that the molar solubility of MgF2 is 1.2 x 10^-3 mol/L, you need to determine the concentration of both F- and Mg2+ ions.
Assuming that MgF2 dissociates completely in water, the concentration of Mg2+ ions is equal to twice the molar solubility because the stoichiometry of the equation is 1:1 (one Mg2+ ion for every one MgF2 unit). So, [Mg2+] = 2 x (1.2 x 10^-3) = 2.4 x 10^-3 mol/L.
Now, we need to find the concentration of F- ions. Since MgF2 dissociates as MgF2 → Mg2+ + 2F-, we know that the concentration of F- ions is twice the concentration of MgF2 because the stoichiometry is 1:2 (2 F- ions for every one MgF2 unit). Therefore, [F-] = 2 x (1.2 x 10^-3) = 2.4 x 10^-3 mol/L.
Plugging these values into the Ksp equation, we get:
Ksp = [F-]^2[Mg2+] = (2.4 x 10^-3)^2 * (2.4 x 10^-3)^2 = 2.0736 x 10^-8
So, the correct value for Ksp is 2.0736 x 10^-8.
Moving on to the second question, you are provided with the mass of CaF2 that dissolves per liter of solution, rather than the molar solubility. To calculate the molar solubility and then the Ksp, you need to convert the mass of CaF2 to moles.
The molar mass of CaF2 is 78.08 g/mol (40.08 g/mol for Ca + 2 x 19.00 g/mol for F).
First, calculate the moles of CaF2:
moles = mass / molar mass = 0.0968 g / 78.08 g/mol = 0.00124 mol
Since the molar solubility is given as per liter of solution, the molar solubility of CaF2 is 0.00124 mol/L.
Now, you can calculate the Ksp using the same equation as before:
Ksp = [F-]^2[Ca2+] = (2[Ca2+]) * [Ca2+] = (2 * 0.00124) * (0.00124) = 6.112 x 10^-6
So, the solubility-product constant for CaF2 is 6.112 x 10^-6.