If the solubility product of AB₂ is 3.20 × 10 –11 M³, then the solubility of AB₂ in pure water is _____ × 10 –4 mol L –1 .
[Assuming that neither kind of ion reacts with water]

Solution

Chemistry ma’am · 3 steps

Two B⁻ ions per formula unit make Kₛₚ = 4s³, then a cube root gives s.

1. Set up the ions. Let s mol of AB₂ dissolve per litre. The equilibrium AB₂ ⇌ A²⁺ + 2B⁻ gives [A²⁺] = s and [B⁻] = 2s.

   Kₛₚ = [A²⁺][B⁻]² = s(2s)² = 4s³

2. Solve for s. 4s³ = 3.20 × 10⁻¹¹, so s³ = 8.0 × 10⁻¹². The cube root of 8 is 2 and the cube root of 10⁻¹² is 10⁻⁴, so s = 2 × 10⁻⁴ mol L⁻¹.
3. So. The solubility of AB₂ in pure water is 2 × 10⁻⁴ mol L⁻¹, so the answer is 2.

The common slip: 3.17. Writing Kₛₚ = s³ and forgetting that [B⁻] = 2s gives s = 3.17 × 10⁻⁴, so 3.17 instead of 2.

The idea to keep. For a salt AₓB_y, each ion's concentration is its coefficient times s, and that coefficient is raised to the power too. For a 1:2 salt that gives Kₛₚ = 4s³.

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