Practice question

Given: The K_{sp of BaCrO₄ is 1.2 × 10⁻¹⁰. What is its solubility in 0.01 M Na₂CrO₄ ? These values define the system as per NCERT data. Formula: K_{sp = [Ba²+][CrO₄²-], [CrO₄²-] = 0.01, 1.2 × 10⁻¹⁰= S · 0.01. This is standard NCERT relation. Substitution & Calculation: S = 1.2 × 10⁻⁸M. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.