Let the solubility of CaF2 be s mol dm-³

CaF2 → Ca²+ + 2F-
□s□□□□ s □□ 2s

Ksp = [Ca²+][F-]²
Ksp = (s)(2s)² = 4s³


So 4.0 x 10^(-11) = 4s³

s³ = 10^(-11)
s = ³√(10^(-11))

s ≈ 2.1544 × 10-⁴
= 2.15 × 10-⁴ (3s.f)


Solubility of CaF2 = 2.15 ×10-⁴ mol dm-³

Let the solubility of CaF2 in 0.025 mol dm-³ NaF solution be x mol dm-³

NaF → Na+ + F-
Since NaF ≡ F- , [F-] = 0.025 mol dm-³


Ksp = [Ca²+][F-]²
Ksp = (x)(2x + 0.025)²

So 4.0 x 10^(-11) = (x)(2x + 0.025)²

Since [F-] from NaF ≫ [F-] fromCaF2, by the approximation method,


4.0 x 10^(-11) ≈ (x)(0.025)²

x ≈ 4.0 x 10^(-11) ÷ (0.025)²

x ≈ 6.40 x 10^-8

Solubility of CaF2 in the NaF
= 6.40 ×10^(-8) mol dm-³

We can see that the solubility in (a) is much lower than in (b)

(a) is about 3360 times as large as (b)

This is due to the common ion effect (you can add in explanation using Le Chatelier's Principle)

I don't quite understand the approximation part for (b)

NaF is fully soluble in water (all sodium salts are) , but CaF2 is only sparingly soluble.

So the solubility of F- from the CaF2 is very much smaller than the solubility of F- from NaF. (0.025≫x)

The total solubility of F- in the mixture is the sum of the two. (solubility basically means the maximum concentration possible)

We are trying to dissolve CaF2 in the 0.025M NaF solution and solving for solubility of CaF2.

Since it is much smaller, the solubility of F from CaF2 can be taken as negligible so 2x + 0.025 ≈ 0.025

This is called the approximation method. Basically we avoid having to solve for quadratic equations when we do so.


The solubilities calculated with the approximation will give very close results to calculating without approximation.