solubility?

Question

Calculate the solubility in grams per liter (note that Ksp is in terms of moles and liters) of Ag2CrO4
Ksp = 9.00 x 10−12

Answer 1

Ag2CrO4(s) <==> 2 Ag+1(aq) + CrO4-2(aq)

let x be the solubility of the Ag2CrO4
then [Ag+1] = 2x
and [CrO4-2] = x

Ksp = 9.00 x 10^-12 = (2x)^2 (x)
9.00 x 10^-12 = 4 x^3
2.25 x 10^-12 = x^3
1.31 x 10^-4 = x = solubility of Ag2CrO4

the two important things to remember to solve this correctly:
1) the silver concentration is twice that of the chromate because Ag2CrO4 has 2 Ag ions and 1 CrO4 ion
2) you square the concentration of the silver in the equilibrium expression because of the coefficient of 2 in front of the silver in the equation at the top

Answer 2

For this compound,

Ksp = [Ag+]^2 X [CrO42-]

Since each time a formula unit of this compound dissolves it produces 2 Ag+ ions for every Cr042- ion, let [CrO42-]=x. Then, [Ag+] = 2x

So, Ksp = (2x)^2 x= 4x^3 = 9.00 X 10^-12

Solving for x will give you the molar solubility or the solubility in terms of moles/L. Using the molar mass of the compound you can convert mol/L to g/L.

Answer 3

0.0435, this is the answer in grams per liter