Practice question

Given: The K_{sp of PbF₂ is 3.3 × 10⁻⁸. What is its solubility in 0.01 M NaF ? These values define the system as per NCERT data. Formula: K_{sp = [Pb²+][F-]², [F-] = 0.01, 3.3 × 10⁻⁸= S · (0.01)². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: S = 3.3 × 10⁻⁸/ 0.0001 = 3.3 × 10⁻⁴ M. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.