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## 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- ?

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