Question

Silver chromate is sparingly soluble in aqueous solutions. The Ksp of Ag2CrO4 is 1.12× 10^–12. What is the solubility (in mol/L) of silver chromate in 1.50 M potassium chromate aq solution? In 1.50 M silver nitrate aq solution? In pure water?

I'm assuming I need to write the solubility expression. If I have this right, Ksp=[2Ag]^2[CrO4] which becomes 1.12x10^-12=4s^3. But how to I then incorporate the questions it's asking for? Where would I plug in potassium chromate if there is no potassium in the original expression?

Answer 1

in potassium chromate `s=4,3 xx 10^-7` mol/L
in silver nitrate `s=5 xx 10^-13` mol/L
in distilled water`s=6.5 xx 10^-5` mol/L

Explanation:

`Ag_2CrO_4 = 2 Ag^+ + CrO_4^2-`
if s are the mol of salt that solubilize you have solubilized 2s mol of `Ag^+` and s mol of chromate
`Kps =[Ag^+]^2 xx [CrO_4^2-]= (2s)^2 xx s = 4 s^3= 1,12 xx 10^-12`
hence `s = root(3)((Kps)/4) = 0.65 xx 10^-4= 6.5 xx 10^(-5)` that is the solubility in mol/L in pure water

in a solution of one same ion you have also the ions of the new salt so you have for a solution of potassium chromate 1,50 M
`Kps =[Ag^+]^2 xx [CrO_4^2-]= (2s)^2 xx (s+1,5) = 4 s^3= 1,12 xx 10^-12`
since s is smallcompared with 1,5 you can write:
`Kps =[Ag^+]^2 xx [CrO_4^2-]= (2s)^2 xx 1,5) = 6 s^2= 1,12 xx 10^-12`
hence `s = root(2)((Kps)/6)=0.43 xx 10^-6 = 4.3 xx 10^-7` mol/L that is about 100 times less soluble of the solution in pure water.

in a solution of silver nitrate 1,5 M you have
`Kps =[Ag^+]^2 xx [CrO_4^2-]= (2s+ 1.5)^2 xx s = 4 s^3= 1,12 xx 10^-12`
but s << 1.5, so you can write:
`Kps =[Ag^+]^2 xx [CrO_4^2-]= (1.5)^2 xx s = 2.25 s= 1,12 xx 10^-12`
`s= 0.5 xx 10^-12= 5 xx 10^-13` mol/L that is about 1 milion times less soluble than in pure water