Question

A sample of AgCl was treated with 5.00 mL of 1.5 M Na₂CO₃ solution to give Ag₂CO₃. The remaining solution contained 0.0026 g of Cl⁻ per litre. Calculate the solubility product of AgCl ($K_{sp}$(Ag₂CO₃) = 8.2 × 10⁻¹²).

Answer 1

0.01 moles of Na2CO3 produces 0.01 moles of Ag2CO3 in 5 ml solutionAg2CO3 ⇔ 2 Ag+ + CO32-                         2S            SKsp = [Ag+]2 [CO32-]       = [2S]2 [S]      = 4 S34S3 = 8.2 × 10-12S3 = 8.2 × 10-124 = 2.05 × 10-12S = 1. 27 × 10-4[Ag+] = 1. 27 × 10-4[Cl-] = 0.00335.5 = 8.45 × 10-5Ksp of AgCl = [Ag+] [Cl-]   = 1. 27 × 10-4 ×  8.45 × 10-5   = 1.073 × 10-8