The conductivity of a saturated solution containing AgA (Ksp = 3 × 10^–14) and AgB (Ksp = 1 × 10^–14) is 3.75 × 10^–8 Ω^–1 cm^–1. If the limiting molar conductivity of Ag^+ and A^– ion is 60 and 80 Ω^

Question

The conductivity of a saturated solution containing AgA (Ksp = 3 × 10^–14) and AgB (Ksp = 1 × 10^–14) is 3.75 × 10^–8 Ω^–1 cm^–1. If the limiting molar conductivity of Ag^+ and A^– ion is 60 and 80 Ω^–1 cm^2 mol^–1, respectively, the limiting molar conductivity of B^– (in Ω^–1 cm^2 mol^–1) is

Prashant Kotian · Accepted Answer

Correct answer: C. Let [Ag^+] = x, [A^–] = y, [B^–] = z. By charge balance, x = y + z. Ksp(AgA) = x*y = 3×10^–14 and Ksp(AgB) = x*z = 1×10^–14. Adding gives x(y+z) = x^2 = 4×10^–14, so x = 2×10^–7 M. This implies y = 1.5×10^–7 M and z = 0.5×10^–7 M. Total conductivity κ = (λ_Ag[Ag^+] + λ_A[A^–] + λ_B[B^–]) / 1000. 3.75×10^–8 × 1000 = 60(2×10^–7) + 80(1.5×10^–7) + λ_B(0.5×10^–7). 3.75×10^–5 = 1.2×10^–5 + 1.2×10^–5 + λ_B(0.5×10^–7). 1.35×10^–5 = 0.5×10^–7 * λ_B. λ_B = 13500 / 50 = 270 Ω^–1 cm^2 mol^–1. Therefore, correct answer is C.

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