Question

Calculate the molar solubility and solubility in grams/liter for Copper(I) iodide

Answer 1

K_sp CuI = 1.27 x 10^-12

CuI (s) Cu⁺ (aq) + I^- (aq)

Let molar solubility of CuI be 's'

[Cu⁺] = s

[I^-] = s

K_sp CuI = [Cu⁺][I^-]

1.27 x 10^-12 = (s) * (s)

s² = 1.27 x 10^-12

s = (1.27 x 10^-12)^1/2

s = 1.13 x 10^-6 M

Molar solubility of CuI = s = 1.13 x 10^-6 M

solubility = (molar solubility) * (molar mass CuI)

solubility = (1.13 x 10^-6 M) * (190.45 g/mol)

solubility = 2.15 x 10^-4 g/L