NEET2018Chemistry-Ionic Equilibrium
NEET 2018 Chemistry Solubility Product MCQ Question
Type: MCQ-numerical-Medium-Class 11
The solubility of BaSO₄ in water is 2.42 × 10⁻³ g L⁻¹ at 298 K. The value of its solubility product (Ksp) will be (Given : molar mass of BaSO₄ = 233 g mol⁻¹)
A
1.08 × 10⁻¹⁰ mol² L⁻²
B
1.08 × 10⁻¹² mol² L⁻²
C
1.08 × 10⁻¹⁴ mol² L⁻²
D
1.08 × 10⁻⁸ mol² L⁻²
Correct Answer
Option A
Detailed Explanation
The solubility in mol/L is calculated as 2.42 × 10⁻³ g/L ÷ 233 g/mol = 1.04 × 10⁻⁵ mol/L. Ksp = (1.04 × 10⁻⁵)² = 1.08 × 10⁻¹⁰ mol² L⁻².
Found an issue with this question?