Selective Precipitation Chemistry Problem

[frozonecom]

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- ?

 

[frozonecom]

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.

 

[Borek]

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.

 

[frozonecom]

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. :)

 

[DeeNos]

Dear student,

Thank you for your detailed response and attempt at solving the problem. I would like to provide some guidance and clarification on your solution.

Firstly, your approach of calculating the molar solubility of both compounds and comparing them to determine which one is more soluble is correct. However, the units for molar solubility are in mol/L, not M. So the molar solubility of Ag2CrO4 would be 6.69 x 10^-5 mol/L and the molar solubility of AgCl would be 1.35 x 10^-5 mol/L.

Now, to find the concentration of Ag+ when Ag2CrO4 begins to precipitate, you have correctly used the Ksp equation for Ag2CrO4. However, your calculation for the concentration of Ag+ is incorrect. It should be:

[Ag+] = sqrt( Ksp Ag2CrO4 / [CrO42-] ) = sqrt( 1.2 x 10^-12 / 0.100) = 3.46 x 10^-6 mol/L

Note that the concentration of CrO42- is given in M, so it needs to be converted to mol/L by multiplying by the volume of the solution.

Next, to find the concentration of AgCl when Ag2CrO4 begins to precipitate, you can use the same approach of solving for [Ag+] and then substituting it into the Ksp equation for AgCl. This will give you the concentration of AgCl in mol/L. To convert it to M, you can divide by the volume of the solution.

So, the final concentration of AgCl would be:

[AgCl] = Ksp AgCl / [Ag+] = (1.82 x 10^-10) / (3.46 x 10^-6) = 5.26 x 10^-5 mol/L

To convert it to M, you can divide by the volume of the solution. This is the concentration of AgCl when Ag2CrO4 begins to precipitate.

I hope this helps to clarify your solution. Keep up the good work as a scientist and keep practicing and asking questions. Best of luck with your studies!