Silver chloride has Ksp=1.6x10^-10 and silver cheomate has Ksp=9.0x10^-12. We have 1.00 liter of a solution contains both nacl(0.10M) and Na2cro4(0.10M) solid AgNo3 is added slowly and the solution is stirred well.

Question

a) which precipitate first Agcl or AgCro4?
B)what are the concentration of [Ag^+],[Cl-] and [CrO4^2-] at the point when the second salt just begins to precipitate?
C) is the method of selective precipitatation a good one for separating the anions in the solution?[hint calculate the percent of the fist ion (to precipitate) remaining in solution at the point the second anion begins to precipitate,in part (b) 99.5+% removed indicates good seperation!]

Answer 1

I divided ksp of each one /M and then found which precipitate first is that right can you pls explain parts b and c thanks!

Answer 2

First you need to be sure part a is right. Did you get AgCl pptng first? Ksp/0.1 is Ag^+ needed for pptn AgCl but sqrt (Ksp/0.1) is Ag^+ needed for pptn of Ag2CrO4.
Adding AgNO3 drop wise to the mixed solution will ppt AgCl first.

B. AgCl will continue to ppt (with no Ag2CrO4) until the Ksp for Ag2CrO4 is reached. What is the (Ag^+) at that point. (Ag^+)(CrO4^2-) = 9E-12
Since CrO4 is 0.1, then Ag^+ (from above) is sqrt(9E-12/0.1) = 9.48E-6
Knowing that is Ag^+ you can calculate Cl^- from Ksp AgCl. Of course the CrO4^2- is 0.1 (ok, it's 1 ion less than that).

C.
You have the initial Ag^+ from part A. You know Ag^+ when Ag2CrO4 just begins to ppt (from part B).
Calculate the percentage Ag still in solution and if it is 0.5% or less then the two can be separated this way.