The solubility product of BaSO₄ is 1 × 10⁻¹⁰ at 298 K. The solubility of BaSO₄ in 0.1 M K₂ SO₄(aq) solution is ___________ × 10⁻⁹ g L⁻¹ (nearest integer).
Given: Molar mass of BaSO₄ is 233 g mol⁻¹

Solution

Chemistry ma’am · 4 steps

The 0.1 M sulphate is already there, so put that in Ksp and only the barium is left to find.

1. Common ion. K₂SO₄ dissolves completely and gives [SO₄²⁻] = 0.1 M. If s mol of BaSO₄ dissolves per litre, it adds s of Ba²⁺ and s more of SO₄²⁻.

   Kₛₚ = [Ba²⁺][SO₄²⁻] = s (0.1 + s) ≈ 0.1 s

2. Solve for s. s is tiny next to 0.1, so 0.1 + s ≈ 0.1. Then s = (1 × 10⁻¹⁰)/0.1 = 1 × 10⁻⁹ mol L⁻¹.
3. Convert to grams. Multiply by the molar mass: 1 × 10⁻⁹ × 233 = 233 × 10⁻⁹ g L⁻¹.
4. So. The solubility is 233 × 10⁻⁹ g L⁻¹, so the answer is 233.

The common slip: 1. Stopping at the molar solubility gives 1, because 1 × 10⁻⁹ is in mol L⁻¹. Multiply by 233 to get grams per litre.

The idea to keep. With a common ion, put its concentration straight into Kₛₚ and drop the small s beside it. Then check the units the question asks for: mol per litre or grams per litre.

Answer from the NTA final answer key. Solution by PrepWiser.