For a sparingly soluble salt AB₂, the equilibrium concentrations of A²⁺ ions and B⁻ ions are 1.2 × 10⁻⁴ M and 0.24 × 10⁻³ M, respectively. The solubility product of AB₂ is :

Solution

Chemistry ma’am · 4 steps

The ion concentrations are already given, so plug them into Kₛₚ and square [B⁻] once.

1. Write the equilibrium. AB₂ ⇌ A²⁺ + 2B⁻. The coefficient 2 on B⁻ becomes a power of 2 in the expression.

   Kₛₚ = [A²⁺][B⁻]²

2. Use the given values. These are the actual equilibrium concentrations, so put them in directly. Do not multiply [B⁻] by 2 again. Write 0.24 × 10⁻³ as 2.4 × 10⁻⁴, so [B⁻]² = 5.76 × 10⁻⁸.
3. Multiply. Kₛₚ = (1.2 × 10⁻⁴) × (5.76 × 10⁻⁸) = 6.912 × 10⁻¹².
4. So. The solubility product is about 6.91 × 10⁻¹², which is option (c).

The idea to keep. If the question gives you ion concentrations, use them as they are. The factor of 2 is already inside [B⁻]. Use (2s)² only when you start from the solubility s.

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