What is the solubility of silver iodide in grams per milliliter?

Question

What is the solubility of silver iodide in grams per milliliter at a temperature at which the Ksp of AgI is 1.45 * 10^-16?

Answer 1

AgI(s) <==> Ag⁺(aq) + I^-(aq)

The Ksp is based on the molar concentration of the substance (moles / liter), so we will first find the solubility in moles per liter, and then convert to grams per milliliter.

Ksp = [Ag⁺][I^-]

1.45x10^-16 = (x)(x)

x2 = 1.45x10^-16

x = 1.20x10^-8 M = [Ag⁺] = [I^-]

Converting this to grams / ml, we have...

molar mass AgI = 235 g / mole

1.20x10^-8 mol / L x 235 g / mol = 2.83x10^-6 g / L x 1 L / 1000 ml = 2.83x10^-9 g / ml