Question

Calculate the molar solubility of SrF2 (Ksp = 4.3×10−9) in the following substances.

1.2×10⁻²M  Sr(NO3)2

1.2×10⁻²M  NaF

Answer 1

Answer – Given, Ksp = 4.3×10^-9

molar solubility of SrF₂ in 1.2×10⁻²M   Sr(NO₃)₂

we know, [Sr(NO₃)₂] = [Sr²⁺] = 1.2*10^-2 M

we know, Ksp expression for the SrF₂

Ksp = [Sr²⁺] [F^-]²

4.3×10^-9 = (1.2*10^-2 M)(2x)²

4x² = 4.3×10^-9 / 1.2*10^-2 M

      = 3.58*10^-7

So, x² = 8.95*10^-8

x = 3.0*10^-4 M

molar solubility of SrF₂ in 1.2×10⁻²M   Sr(NO₃)₂ is 3.0*10^-4 M

molar solubility of SrF₂ in 1.2×10⁻²M   NaF

we know, Ksp expression for the SrF₂

Ksp = [Sr²⁺] [F^-]²

4.3×10^-9 = (x) (1.2*10^-2)²

x = 4.3×10^-9 / (1.2*10^-2)²M

      = 3.00*10^-5

So, molar solubility of SrF₂ in 3.00*10^-5 M