Question

Sodium sulfate is slowly added to a solution containing 0.0500 M Ca2 (aq) and 0.0350 M Ag (aq). What will be the concentration of Ca2 (aq) when Ag2SO4(s) begins to precipitate? Solubility-product constants, Ksp, can be found here.

1. Calculate [Ca^2+] = _____M

2. What percentage of the Ca2+ (aq) can be precipitated from the Ag+ (aq) by selective precipitation? ______%

Answer 1

Sodium sulfate is slowly added to a solution containing 0.0500 M Ca2 (aq) and 0.0350 M Ag (aq). What will be the concentration of Ca2 (aq) when Ag2SO4(s) begins to precipitate? Solubility-product constants, Ksp, can be found here.

1. Calculate [Ca^2+] = _____M

2. What percentage of the Ca2+ (aq) can be precipitated from the Ag+ (aq) by selective precipitation? ______%

The reaction equations and their respective solubility equilibrium equations are

Ag₂SO₄(s) ⇄ 2 Ag⁺(aq) + SO₄²⁻(aq)

CaSO₄(s) ⇄ Ca²⁺(aq) + SO₄²⁻(aq)
=>

Ksp₁ = [Ag⁺]²∙[SO₄²⁻]
Ksp₂ = [Ca²⁺]∙[SO₄²⁻]
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Ksp Ag2SO4 = 1.20x10^-5
Ksp CaSO4 = 4.93x10^-5

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From solubility equilibrium for silver sulfate follows the sulfate concentration at which Ag₂SO₄ starts to precipitate:
[SO₄²⁻] = Ksp₁ / [Ag⁺]²
= 1.20×10⁻⁵ / (3.50×10⁻² )²
= 0.979 ×10⁻² M
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The calcium ion concentration at this sulfate concentration is:
[Ca²⁺] = Ksp₂ / [SO₄²⁻]
= 4.93×10⁻⁵ / 0.979 ×10⁻²
= 5.035×10⁻³ M
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That means due to precipitation of calcium sulfate the concentration of calcium ions in solution has decreased by:
∆[Ca²⁺] = 5.00×10⁻² M - 5.035×10⁻³ M =0.044965 = 4.4965×10⁻² M

So the fraction of initial amount of Ca²⁺ which has been separated as precipitate is:
α = 4.4965×10⁻² / 5.00×10⁻² M = 0.8993 = 89.93%