What is the solubility (in g/L)of barium chromate at 25°C? The solubility product constant for barium chromate is 1.2 × 10^-10 at 25°C. A)0.41 g/L B)3.0 × 10^-8 g/L C)1.5 × 10^-8 g/L D)0.078 g/L E)0.0027 g/L

Question

Quizplus · Accepted Answer

The solubility product expression for barium chromate (BaCrO₄) is Ksp = [Ba²⁺][CrO₄²⁻]. Let s be the solubility in mol/L, then Ksp = s². Solving for s gives s = √(1.2 × 10⁻¹⁰) = 1.1 × 10⁻⁵ mol/L. Converting to g/L using the molar mass of BaCrO₄ (253.3 g/mol) gives solubility = 1.1 × 10⁻⁵ mol/L × 253.3 g/mol = 0.0027 g/L.

What Is the Solubility (In G/L)of Barium Chromate at 25°C