Question

At 310 K, the solubility of CaF2 in water is 2.34 ×10-3 g/100 mL. The solubility product of CaF2 is _____ × 10-8 (mol/L)3.
(Given molar mass : CaF2 = 78 g mol-1).

Answer 1

$CaF2 ⇋ Ca^2++2F_{2s}^-$
Ksp = s(2s)2
= 4s3
Solubility(s) = 2⋅34 × 10–3 g/100 mL
= $\frac{2.34×10-3×10}{78}$ mole/lit
= 3×10-4 mole/lit
∴Ksp = 4×(3×10-4)3
= 108×10-12
= 0.0108×10-8(mole/lit)3
∴ x ≈ 0