Question

A solution was prepped by adding 0.7949 grams of silver (1) nitrate and 1.4721 grams of strontium (2) nitrate to a 250 ml flask and diluting it to the mark. Sodium sulfate (s) was added to the solution (no change in volume b/c negligible)

a. What percipitate will form first?

b. What is the conc. of the cation from the least soluble species when the second percipitate begins to fall out?

c. What is the max amount of pure compound (in grams) that can be recovered during this selective percipitation?

Answer 1

0.7949 grams of silver (1) nitrate is 170 g/mol = 0.0047 mole in 0.25L is 0.019 M

1.4721 grams of strontium (2) nitrate is 211.6 g/mol = 0.0069 mole in 0.25 L is 0.028 M

Ksp for strontium (2) sulfate is 3.44 x 10⁻⁷

Ksp for Ag(1) sulfate is 1.4×10^–5

SrSO_{4 $\rightleftharpoons$} Sr²⁺(aq) + SO₄^2-(aq)

Ksp = [Sr²⁺][SO₄^2-]

3.44 x 10⁻⁷ = 0.028 x [SO₄^2-]

[SO₄^2-] = 3.44 x 10⁻⁷/ 0.028

[SO₄^2-] = 1.23 x 10^-5 M at this concentration of sulfate Sr²⁺ will pecipitate

Ag₂SO_{4 $\rightleftharpoons$} 2Ag⁺(aq) + SO₄^2-(aq)

Ksp = [Ag⁺]²[SO₄^2-]

1.4×10^–5 =0.019² x [SO₄^2-]

[SO₄^2-] = 0.0388 M at this concentration of sulfate Ag⁺ will pecipitate

a) Sr²⁺ will precipitate first

b) When Ag+ begins to precipitate Sr²⁺ will be

Ksp = [Sr²⁺][SO₄^2-]

3.44 x 10⁻⁷ = [Sr²⁺] x 0.0388

[Sr²⁺] = 3.44 x 10⁻⁷ /0.0388

[Sr²⁺] = 8.87 x 10^-6 M when silver begins to precipitate

So the amount of Sr that presipitates is 0.028 x 0.25 - 8.87 x 10^-6 x 0.25 = 0.00689 moles of SrSO4 precipitates which is 0.00689 x 183.68 = 1.267 g