- #1
frozonecom
- 63
- 0
Homework Statement
A solution of [itex]AgNO_{3}[/itex] is added to a solution containing 0.100 M [itex]Cl^{-}[/itex] and 0.100 [itex]CrO_{4}^{2-}[/itex].
What will be the concentration of the less soluble compund when the more soluble one begins to precipitate?
Homework Equations
Ksp AgCl = [itex]1.82 x 10^{-10}[/itex]
Ksp Ag2CrO4 = [itex]1.2 x 10^{-12}[/itex]
The Attempt at a Solution
So, by calculating for their molar solubilities, I would know which one would precipitate first (which one is more soluble or less soluble)(2x)^2 (x) = Ksp Ag2CrO4
x=Molar solubility of Ag2CrO4 = 6.69 x 10^-5 M
(x)(x) = Ksp AgCl
x= Molar solubility of AgCl = 1.35 x 10^-5 M
Thus, Ag2CrO4 is more soluble and AgCl is the less soluble compound.
Now, how will I find the concentration of the less soluble compound when the more soluble one begins to precipitate?
Please guide me. Here's my attempt for a solution.
The more soluble compound, Ag2CrO4 will begin to precipitate at this Ag+ concentration
[Ag+]^2 [CrO42-] = Ksp Ag2CrO4
[Ag+] = sqrt( Ksp Ag2CrO4 / [CrO42-] ) = 3.46 x 10^-6 M
Now, how would I find the concentration of AgCl in the solution? Again, here is my attempt:
I think, I should substitute the Ag+ concentration at the formula
[Ag+][Cl-] = Ksp AgCl
, But, is the Cl- concentration that I will get equal to the concentration of AgCl in the solution? I'm very confused. :(. Anyway, here's an attempt:
[Cl-] = Ksp AgCl / [Ag+] = 5.26 x 10^-5 M
I mean, the answer CAN be plausible since the concentration seemingly decreased. Is it correct guys?
I think what I am having problem with is that why would the concentration of AgCl be equal to the equilibrium conc of Cl- ?
Last edited: