Question

Strontium,Sr2+, is a heavy metal tht frequently causes concern when it is present in water. Some preliminary calculations are necessary to determine whether it is feasible to precipitate Sr2+ as the sulfate. Given that 10-4 mole/ liter of Sr2+ and 10^-4 mole/liter of SO42- are originally present and that pK = 7.9 FOR SrS04

1) How much Sr2+ will precipitate if 1x10^-3 mole Na2SO4 is added perliter of solution?

2) How much SO42- and Sr2+ remian after precipitation?

3) Assuming that the residulas calculated in part b) are satisfactory, what factors should be studied experimentally to determine the feasibility of the process?

1. Strontium, Sr is a heavy metal that frequently causes concern when it is present in water. Some preliminary calculations are necessary to determine whether it is feasible to precipitate Srt as the sulfate. Given that 10 4 mole/ liter of Sr and 10 molelhiter of so are originally present and that pKso 7.8 for SrSO (a) How much Sr will precipitate (moles/liter if 1 x 10-3 mole NaaSO is added per liter of solution? (b) How much SO and Sr remain after precipitation? (c) Assuming that the residuals calculated in part (b) are satisfactory, what factors should be studied experimentally to determine the feasibility of the process?

Answer 1

SrSO₄ (s) < -------------> Sr²⁺ (aq.) + SO₄^2- (aq.)

K = [Sr²⁺][SO₄^2-]

10^-7.9 = (S) (0.0001+0.001)

S = 1.14 * 10^-5 M

Initial moles of Sr²⁺ = 0.0001 mol

Final moles of Sr²⁺ = 0.0000114 mol

moles of Sr²⁺ that is precipitated = 0.0001 - 0.0000114 = 0.0000886 mol

(b) Remaining Sr²⁺ = 0.0000114 mol and SO₄^2- = 0.0011 mol

(c) Because the K of SrSO4 is small it is feasible to precipitate Sr²⁺ as its sulphate.