Applications of Aqueous Equilibrium (Show work and Answer)

Question

The dissolution process for SrF₂ is:

SrF_2(s) <--> Sr²⁺_(aq) + 2F^-_(aq)

 

The K_sp for SrF₂ = 7.9 E^-10. Determine the molar solubility [SrF₂]:

a. in pure water

b. in 0.20 M NaF solution

c. in 0.20 M Sr(NO₃)₂

Answer 1

SrF₂(s) <--> Sr²⁺(aq) + 2F^-(aq)

Ksp = [Sr²⁺][F^-]²

If we let [Sr²⁺] = x, the [F^-] = 2x...

Ksp = 7.9x10^-10 = (x)(2x)²

4x3 = 7.9x10^-10

x = 5.82x10^-4 M = [Sr²⁺] = molar solubility of SrF₂