# What is the molar solubility of CaF2 in 0.0022 M Ca(NO3)2? (Ksp = 3.9 x 10-11) CaF2 (s) D Ca2+ (aq) + 2 F- (aq) ?

(Hint: contain common ion Ca2+, concentration of Ca2+ will not be zero at the initial step)

(The answer should be in 3 significant numbers such as 1.50E-03 or 0.00150 for 1.50 x 10-3)

### 1 Answer

Relevance

- BobbyLv 72 months ago
CaF2 = Ca2+ + 2F-

Ksp = [Ca2+] *[F-] ^2 = 3.9 x 10-11

if x mole of CaF2 dissolves

ICE table .......[Ca2+] [F-]

I .......................0.0022 ...0

C.....................+x.........+2x

E.................0.0022 +x ...2x

substituting in Ksp we have

3.9 x 10-11= (0.0022+x) * 4x^2

We assume that x << 0.0022, so that 0.0022 + x = 0.0022

0.0088x^2 = 3.9 x 10-11

x^2 = 3.9 x 10-11 / 0.0088 = 4.43182E-09

x= 6.66 * 10 ^-5

molar solubility of CaF2 in 0.0022 M Ca2+ = 6.66 * 10 ^-5 moles/ liter

there is your answer

Still have questions? Get your answers by asking now.