Question

Consider a solution that is 1.3×10−2 M in Ba2+ and 2.0×10−2 M in Ca2+. Ksp(BaSO4)=1.07×10−10 Ksp(CaSO4)=7.10×10−5 If sodium sulfate is used to selectively precipitate one of the cations while leaving the other cation in solution, which cation will precipitate first?

Answer 1

Answer:

Explanation:

Ksp(BaSO4)=1.07×10−10

BaSO₄ → Ba²⁺ + SO₄²⁻

1.07×10⁻¹⁰ = ( Ba²⁺) × ( SO₄²⁻)

but Ba²⁺ = 1.3×10⁻² M

1.07×10⁻¹⁰  = 1.3×10⁻² M × ( SO₄²⁻)

( SO₄²⁻)  = 1.07×10⁻¹⁰  / 1.3×10⁻² = 0.823 × 10⁻⁸ M

while Ksp(CaSO4)=7.10×10−5

CaSO₄ → Ca²⁺ + SO₄²⁻

7.10×10⁻⁵ = 2.0×10⁻² × ( SO₄²⁻)

( SO₄²⁻)  = 7.10×10⁻⁵  /  2.0×10⁻² = 3.55 × 10⁻³ M

comparing the concentration of sulfate ions, Ba²⁺ cation will precipitate first because the Ba²⁺ requires 0.823 × 10⁻⁸ M sodium sulfate which less compared the about needed by CaSO₄