Question

3. What is the molar Solubility of CaF2 in 0.250 M Ca(NO3)2? (Ksp of CaF2 = 3.2 x 10-8)

4. How many grams of Ag2SO4 will be soluble in 100 ml of water?

(Ksp of Ag2SO4 = 4.1 x 10-7)

5. How many grams of Ag2SO4 will be soluble in 250 ml of 0.100 M Na2SO4?

(Ksp of Ag2SO4 = 4.1 x 10-7)

Answer 1

3) Solution-

Calcium fluoride dissolves and dissociates as given below
CaF₂(s) ⇄ Ca²⁺(aq) + 2 F⁻(aq)

Let us assume x be the molar solubility of calcium fluoride.

When we add x moles to one Liter water all salt molecules dissociate to one calcium ion and two fluoride ions.

The solution already has [Ca2=] = 0.250

now;
Ksp = [Ca2+][F-]^2 = 3.2 x 10^-8

at equilibrium :
[Ca2+] = x + 0.250
[F-]= 2x

3.2 x 10^-11 = (x + 0.25)(2x)^2

x is very small compared to 0.25 so the x inside the brackets can be eliminated and the equation becomes

.25 * 4x^2 = 3.2 x 10^-8
x^2 = 3.2 * 10^-8
x^2 = 2.15 * 10 ^-10
x = 1.78 * 10 ^-4

Hence the molar solubility is 1.78 * 10 ^-4 moles / liter