Question

The solubility product constant for strontium sulfate is 3.44*10^-7. Using the Debye-Hukel limiting law to estimate activity coefficients,

a. Calculate the solubility of strontium sulfate in a solution of 0.100 m KCl

b. Calculate the solubility of strontium sulfate in pure water.

Answer 1

a. Now as SrSO4 is sparingly soulble we can ignore its contribution in the ionic strength of solution

ionic strentgh of the solution of 0.100 m KCl is

i = 1/2( 0.1 1² + 0.11² ) = 0.1 m

activity coefficient of Sr²⁺ and SO₄^2-

^{using debye huckel equation,}

log_A =

for Sr ²⁺   is = 0.5,   for SO₄²⁺ = 0.4

logSr²⁺ = -0.51 (+2)² (0.1)^0.5 / 1+3.30.5(0.1)^0.5 = -(0.51*4*0.31/1.51) = -0.418

or, Sr²⁺ = 0.3819

logSO₄^2- = -0.51 (-2)²(0.1)^0.5/ 1+3.30.4(0.1)^0.5 = -0.51*4*0.31/1.41 = -0.448

SO₄^2- = 0.3564

Ksp = Sr²⁺[Sr²⁺] SO₄^2-[SO₄^2-] = (0.3819S0.3564 S)

S² = Ksp/ 0.38190.3564 =   3.44 10^-7 / 0.136 = 25.29 10^-7

or , S = 15.9010^-4

hence solubility in 0.1 m KCl is 15.9010^-4 mol/L

b. Ionic strength of water is 1.

Ksp of Strontium sulphate = 3.44 10^-7

SrSO₄ Sr²⁺ + SO₄^-2

hence Ksp = [Sr²⁺][SO₄^-2] = s.s = s²

or s= ksp = 5.9010^-4

hence solubility in water is 5.9010^-4 mol/L