Question

The solubility product (KSP) for strontium fluoride (MW = 125.62 g/mol) at 25C is 4.33x10-9. Strontium fluoride dissociates per SrF2  Sr2+ + 2F- Assuming that the activity coefficients for all of the dissolved ionic species are one, the solubility (mg/L) of this compound in water at 25C is most nearly A. 7.6 B. 9.3 C. 130 D. 240

Answer 1

solution -

SrF2 ----- > Sr^2+ + 2F^-

                  x            2x

ksp = [Sr62+][F^-]^2

4.33*10^-9 = [x][2x]^2

4.33*10^-9 = 4x^3

4.33*10^-9 /4 = x^3

1.0825*10^-9 = x^3

taking cube root of both sides we get

0.001027 mol /l= x

now lets calculate the mass from the moles

0.001027 mol * 125.62 g per mol = 0.130 g/L

now lets convert this g to mg

0.130g/L * 1000 mg/ 1 g = 130 mg/L

so the answer is option C that is 130 mg/L is the solubility of the SrF2