Question

Silver chloride has Ksp = 1.6 x 10-10 and silver chromate has Ksp = 9.0 x 10-12. We have 1.00 liter of a

solution which contains both NaCl (0.10 M) and Na2CrO4 (0.10 M). Solid AgNO3 is added slowly and the solution is stirred well.

a) Which salt precipitates first, AgCl or Ag2CrO4? (Show your work!)

b) What are the concentrations of [Ag+], [Cl-], and [CrO42-] at the point when the second salt just begins to

precipitate?

c) Is this method of selective precipitation a good one for separating the anions in the solution? [Hint: Calculate the percent of the first ion (to precipitate) remaining in solution at the point the

second anion begins to precipitate, in part (b). 99.5+% removed indicates good separation!]

Answer 1

We need to calculate [Ag+] at which each compound will precipitate:
For AgCl:

[Ag+] = Ksp/[Cl-] = 1.6 x 10^-10 / 0.1 M
= 1.6 x 10^-9  M

For Ag2CrO4:

[Ag+]^2 = Ksp/[CrO4 2-] = 9 x 10^-12/ 0.100 M
[Ag+] = 9.54 x 10^-6 M

Since [Ag+]AgCl < [Ag+]Ag2CrO4,

AgCl will precipitate first

b.
From part a, Ag2CrO4 begins to precipitate

when [Ag+] = 9.54 x 10-6 M
So, [Cl-] when [Ag+] = 9.54 x 10-6 M:
[Cl-] = Ksp/[Ag+] = 1.6 x 10^-10 / 9.54 * 10^-6 = 1.67 x 10-5 M