Answer:
4.0 × 10⁻⁹
Explanation:
Let's consider the solution of copper(I) chloride.
CuCl₂(s) ⇄ Cu²⁺(aq) + 2 Cl⁻(aq)
We can relate the solubility (S) with the solubility product constant (Ksp) using an ICE chart.
CuCl₂(s) ⇄ Cu²⁺(aq) + 2 Cl⁻(aq)
I 0 0
C +S +2S
E S 2S
The Ksp is:
Ksp = [Cu²⁺] × [Cl⁻]² = S × (2S)² = 4 S³
Ksp = 4 × (1.00 × 10⁻³)³ ≈ 4.0 × 10⁻⁹