Question

Strontium sulphate, SrSO4, has a solubility product of 3.2 x 10^-7 Using solution theory, calculate the percentage of the strontium remaining in solution if 100mL of 0.075M strontium chloride solution were to be mixed with 200mL of 0.037M sodium sulphate solution. Explain any assumption used.

Answer 1

Ksp = [Sr+2][SO4-2]

Ksp 3.2*10^-7

SrCl2 + Na2SO4

mmol of Sr+2 = MV = 100*0.075 = 7.5 mmol

mmol of SO4-2 = MV = 200*0.037 = 7.4 mmol

then..

assume Sr+2 is in excess

7.4 mmol of SO4-2 react with 7.4/2 = 3.7 mmol of SR+2

mmol of Sr+2 left = 7.5-3.7 = 3.8 mmol of Sr+2 left

in solution:

Ksp = [Sr+2][SO4-2]

[Sr+2] = mmol/V = (3.8)/(100+200) = 0.012666 M

assume this is in excess

so

3.2*10^-7 = (0.012666) * [SO4-2]

[SO4-2] = (3.2*10^-7) / (0.012666) = 0.000025264 M

the Sr remainig...

assume Sr+2 left is >> than in solution

initial Sr+2 = 7.5

final Sr+2 = 3.8

% Sr+2 in solutoin = 3.8 /7.5*100 = 50.666%