Exams › JEE Advanced › Chemistry
The specific conductance of a saturated AgCl solution at 25 deg C is 1.26 * 10⁻⁶ ohm⁻¹ cm⁻¹. The contribution of water to conductance is negligible. The molar conductivities at infinite dilution of Ag+ and Cl- ions are 62 and 76 ohm⁻¹ cm² mol⁻¹ respectively. Calculate the solubility product (Ksp) of AgCl at 25 deg C.
- 8.3 * 10⁻¹¹
- 6.2 * 10⁻¹²
- 4.6 * 10⁻¹⁰
- 2.5 * 10⁻¹⁴
Correct answer: 8.3 * 10⁻¹¹
Solution
Lambdaₘ at infinite dilution for AgCl = 62 + 76 = 138 S cm² mol⁻¹. Molar conductivity of the saturated solution ≈ Lambdaₘ(inf) (since very dilute). Solubility c = kappa / Lambdaₘ = 1.26*10⁻⁶ / 138 mol/cm³. Convert: c = 9.13*10⁻⁹ mol/cm³ = 9.13*10⁻⁶ mol/L. Ksp = c² = (9.13*10⁻⁶)² ≈ 8.3*10⁻¹¹ mol² L⁻².
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