Calculate the molar solubility of AgCl in a 4.00−L solution containing 10.0 g of dissolved CaCl2.

Answer 1

You'll need the Ksp for AgCl to solve this problem. I find a value of 1.8x10^-10.

Using this value and a molar mass of CaCl₂ = 111 g/mol, we can solve the problem (Common Ion).

AgCl(s) <==> Ag⁺(aq) + Cl^-(aq)

Ksp = [Ag⁺][Cl^-]

From 10.0 g CaCl₂ / 4.00 L we find the [Cl^-], the common ion.

10.0 g CaCl₂ x 1 mol/111 g x 2 mol Cl-/mol CaCl₂ / 4.00 L = 0.045 M Cl^- (the contribution of Cl^- from AgCl is considered negligible)

1.8x10^-10 = [Ag⁺][0.045]

[Ag⁺] = 4x10^-9 M = molar solubility of AgCl under these conditions