The conductivity of saturated solution of Ba3(PO4)2 is 1.2 × 10^–5 Ω^–1 cm^–1. The limiting equivale — Electrochemistry Chemistry Question

💡 Solution & Explanation

Equivalent conductivity of Ba3(PO4)2 = 160 + 140 - 100 = 200 Ω^–1 cm^2 eq^–1. The total charge of the cation in Ba3(PO4)2 is n = 6, so molar conductivity Λ_m = 200 × 6 = 1200 Ω^–1 cm^2 mol^–1. Solubility S = (κ × 1000) / Λ_m = (1.2 × 10^–5 × 1000) / 1200 = 10^–5 M. The salt dissociates into 3 Ba^2+ and 2 PO4^3–, so Ksp = (3S)^3 × (2S)^2 = 108 × S^5 = 108 × (10^–5)^5 = 108 × 10^–25 = 1.08 × 10^–23. Therefore, correct answer is B.