Selective Precipitation Chemistry Problem

AI Thread Summary
The discussion revolves around a chemistry problem involving the selective precipitation of silver compounds, specifically AgCl and Ag2CrO4. It is established that AgCl is less soluble than Ag2CrO4, with respective molar solubilities calculated as 1.35 x 10^-5 M and 6.69 x 10^-5 M. The confusion arises regarding the concentration of Cl- when Ag2CrO4 begins to precipitate, with participants clarifying that the problem's wording may have contributed to misunderstandings about the equilibrium concentrations of ions versus compounds. The correct approach involves calculating Cl- concentration using the Ksp of AgCl at the point of Ag2CrO4 precipitation. The discussion concludes with a suggestion to revise educational materials to prevent similar confusion in the future.
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Homework Statement


A solution of AgNO_{3} is added to a solution containing 0.100 M Cl^{-} and 0.100 CrO_{4}^{2-}.

What will be the concentration of the less soluble compund when the more soluble one begins to precipitate?

Homework Equations


Ksp AgCl = 1.82 x 10^{-10}
Ksp Ag2CrO4 = 1.2 x 10^{-12}

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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Anyone here who can help me? Our test didn't include anything like this (thank god), but anyway, I still feel the need to learn this kind of stuff. It looks simple but I still can't be certain about this.

Please help.
 
Sorry, somehow I missed your post earlier.

What you did is correct. I guess the wording of the problem is wrong - it is a classic question, asked every year everywhere, and it always asks about concentration of the ion (Cl- here) and not "compound". Compound is ambiguous in this context which is why you are confused.
 
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Borek said:
Sorry, somehow I missed your post earlier.

What you did is correct. I guess the wording of the problem is wrong - it is a classic question, asked every year everywhere, and it always asks about concentration of the ion (Cl- here) and not "compound". Compound is ambiguous in this context which is why you are confused.

Oh. Truth be told, I was wacking my head finding a similar problem in textbooks & internet and like you said, it's concerned only on the final equilibrium concentration of an ion. Thanks.

I guess it's time to suggest to our institute to revise their laboratory manual, or at least their self-assesment questions. It has become more of a confusion rather than a help.

Again, thank you very much. :)
 
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