Question

Use the solubility product to calculate the solubility of Ag 2 CrO 4 (2Ag + + CrO 4 2 - ) in water expressed as

(a) moles per liter (b) g/100 mL

Answer 1

(a) Ag2CrO4(s) ---> 2Ag^+(aq) + CrO4^2-(aq)

Ksp = [Ag^+]^2 [CrO4^2-] = 2.6 x 10^-12

Note: Since the concentrations of pure solids and liquids do not change, the constant concentration of Ag2CrO4 is included in Ksp value.

(b) Ag2CrO4(s) ---> 2Ag^+(aq) + CrO4^2-(aq)
If x M dissolves ........ 2x M ............. x M

Ksp = [Ag^+]^2 [CrO4^2-] = 2.6 x 10^-12
Ksp = (2x)^2.(x) = 2.6 x 10^-12
4x^3 = 2.6 x 10^-12
x^3 = 6.5x10^-13
x = 8.66x10^-5 M

Solubility in term of Ag^4 = 2x M = 2x8.66x10^-5
= 1.73x10^-4 M (mol / L)

(c) x M calculated as 8.66x10^-5 M represents the solubility of Ag2CrO4 in moles per liter ( = 1000 mL).
For 100 mL, solubility becomes
8.66x10^-5 x (100 / 1000) = 8.66x10^-6 mol

Molar mass of Ag2CrO4 = 332 g/mol
Mass of Ag2CrO4 dissolved = 8.66x10^-6 mol x 332 g/mol
= 0.02875 g = 2.875 mg = 0.002875 g/100mL