Answer:
2.06 × 10⁻¹⁰
Step-by-step explanation:
Let's consider the solution of a generic compound AB₂.
AB₂(s) ⇄ A²⁺(aq) + 2B⁻(aq)
We can relate the molar solubility (S) with the solubility product constant (Kps) using an ICE chart.
AB₂(s) ⇄ A²⁺(aq) + 2B⁻(aq)
I 0 0
C +S +2S
E S 2S
The solubility product constant is:
Kps = [A²⁺] × [B⁻]² = S × (2S)² = 4 × S³ = 4 × (3.72 × 10⁻⁴)³ = 2.06 × 10⁻¹⁰