The solubility of PbCrO4 in water at 25°C is measured to be ×1.710−4gL . Use this information to calculate Ksp for PbCrO4 .?

Molar mass of PbCrO₄ = (207.2 + 52.0 + 16.0×4) g/mol = 323.2 g/mol
Solubility of PbCrO₄ = (1.7 × 10⁻⁴ g/L) / (323.2 g/mol) = 5.26 × 10⁻⁷ mol/L

PbCrO₄(s) ⇌ Pb²⁺(aq) + CrO₄²⁻(aq)
1 mole of PbCrO₄ dissociates to give 1 mole of Pb²⁺ ions and 1 mole of CrO₄²⁻ ions.
At equilibrium, [Pb²⁺] = [CrO₄²⁻] = 5.26 × 10⁻⁷ mol/L

Ksp for PbCrO₄ = (5.26 × 10⁻⁷)² = 2.77 × 10⁻¹³ = 2.8 × 10⁻¹³ (to 2 sig. fig.)