Chemistry Question

Question

lculate the minimum number of grams of AgCl that must be added to 250ml of water in order to form a precipitate. Ksp= 1.8 x 10^(-10) for AgCl

Answer 1

AgCl(s) <==> Ag⁺ + Cl^-

Ksp = [Ag⁺][Cl^-] = 1.8x10^-10  = (x)(x) = x²

x² = 1.8x10^-10

x = 1.34x10^-5 M = [Ag⁺] = [C^l-]

1.34x10^-5 mol/L x 0.250 L = 3.35x10^-6 moles AgCl needed

3.35x10^-6 moles AgCl x 143.3g/mol = 4.8x10^-4 g AgCl