What percentage of the first ion will have precipitated?

Question

An aqueous solution at room temperature is 0.25 M in both Ba²⁺ and Ca²⁺ ions. Now try to separate these two ions by selective precipitation using SULPHATE. What percentage of the first ion will have precipitated when the second ion begins precipitating? (BaSO₄ Ksp = 1.10 × 10⁻¹⁰ M, CaSO₄ Ksp = 9.1 × 10⁻⁶ M).

Answer 1

An aqueous solution at room temperature is 0.25 M in both Ba²⁺ and Ca²⁺ ions. Now try to separate these two ions by selective precipitation using SULPHATE. What percentage of the first ion will have precipitated when the second ion begins precipitating? (BaSO₄ K_sp = 1.10 × 10⁻¹⁰ M, CaSO₄ K_sp = 9.1 × 10⁻⁶ M).

Step 1: Write the Reactions and K_sp Expressions for each.

BaSO₄(s) ↔ Ba²⁺(aq) + SO₄^2-(aq) K_sp = [Ba²⁺][SO₄^2-] = 1.10 x 10^-10 M

CaSO₄(s) ↔ Ca²⁺(aq) + SO₄^2-(aq) K_sp = [Ca²⁺][SO₄^2-] = 9.1 x 10^-6 M

Step 2: ICE Charts

BaSO₄(s) ↔ Ba²⁺(aq) + SO₄^2-(aq) K_sp = [Ba²⁺][SO₄^2-] = 1.10 x 10^-10 M

0.25 x

(0.25)(x) = 1.10 x 10^-10

x = [SO₄^2-] = 4.4 x 10^-10 M

CaSO₄(s) ↔ Ca²⁺(aq) + SO₄^2-(aq) K_sp = [Ca²⁺][SO₄^2-] = 9.1 x 10^-6 M

0.25 x

(0.25)(x) = 9.1 x 10^-6

x = [SO₄^2-] = 3.64 x 10^-5 M

Since Ba²⁺ requires a much lower concentration of sulfate to precipitate (4.4 x 10^-10 < 3.64 x 10^-5 ), Ba²⁺ is the first ion to precipitate.

Step 3: [Ba²⁺] remaining after [Ca²⁺] begins to precipitate

Ca²⁺ begins to precipitate when SO₄^2- = 3.64 x 10^-5 M.

Use K_sp for BaSO₄ to determine how much Ba²⁺ is remaining at this point.

BaSO₄(s) ↔ Ba²⁺(aq) + SO₄^2-(aq)

K_sp = [Ba²⁺][SO₄^2-] = 1.10 x 10^-10 M

[Ba²⁺](3.64 x 10^-5) = 1.10 x 10^-10

[Ba²⁺]_remaining = 3.02 x 10^-6 M

Step 4: Find the percent

%Ba = amount of [Ba²⁺] precipitate / initial concentration x 100%

[Ba²⁺]_precipitate = [Ba²⁺]_initial - [Ba²⁺]_remaining

[Ba²⁺]_precipitate = 0.25 M - 3.02 x 10^-6 M = 0.249997 M

%Ba = 0.249997 / 0.25 x 100 = 99.999%