Question

The Ksp for silver chromate (Ag2CrO4) is 1.1x10-12. Using the same initial set up of 10mL of 1M silver nitrate is combined with 25mL of 0.1M sodium chromate, what is the mass of the silver chromate precipitate in grams?

Answer 1

Let solubility of silver chromate = s

      Ag₂CrO4 ==> 2Ag⁺ + CrO4^-2

                               2s      s

Ksp = [Ag+]²[CrO4^-2] = [2s]²[s]= 4s³

Ksp = 1.1x10^-12 = 4s³

s³ = 0.275 x10^-12

s =6.503 X 10-5
The maximum amount of Ag+ that can be present in the solution = solubility = 2s = 1.301 X 10^-4

The maximum amount of chromate that can be present in the solution = solubility = s =6.503 X 10-5

[Ag+] after addition of 1M silver nitrate = Molarity of AgNO3 X volume of AgNO3 / total volume = 1 X 10 / 10+25 mL = 0.2857 M

[CrO4^-2] = Molarity of sodium chromate X volume / total volume = 0.1 X 25 / 35 = 0.0714 M

The Ag+ precipitated from the dissolution of 1M silver nitrate = (0.2857 - 1.301 X 10^-4) = 0.2856 M

The [CrO4^-2] precipitated = 0.0714 - 6.503 X 10-5 = 0.0713

[CrO4^-2] is limiting reagent, so [Ag2CrO4] precipitated = 0.0713 M

Mass of [Ag2CrO4] precipitated = Molarity X molar mass X volume = 0.0713 X 331.73 X 0.035L = 0.828 grams