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Common ion effect - calculations.

  • Thread starter j3llzang
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  • #1
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Homework Statement


When 50.0 mL of a 2.0 × 10-4 M Ag2CrO4 solution was added to 50.0 mL of a Na2CrO4 solution a ppt formed (Ag2Cro4). What was the initial concentration of this Na2CrO4 solution?
Ksp Ag2CrO4 = 1.1 × 10-12


Homework Equations


Given A2 (aq) + B(aq) -> AB(s)
Ksp = [A]n

The Attempt at a Solution


I tried using the "ICE" Table, but it didn't help:
Ag2CrO -> 2Ag + CrO4
I 2.0x10-4 2.0x10-4 ?
C +x(?) -2x(?) -x(?)
E 2.0x10-4(unchanged) 2.75x10-5

I found [CrO4] at eqb using Ksp equation. (2.75x10-5M)
The initial [CrO4] was the [CrO4] from Ag2CrO4 plus Na2CrO4, right?
but why is it that [CrO4] at equilibrium is smaller than that of the initial [CrO4] in the initial Ag2CrO4 concentration?
Can anyone solve this problem???
Explanation would be nice :)

Thanks!!~
 

Answers and Replies

  • #2
Borek
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What was the final concentration of Ag+ after solutions has been mixed? (Don't think about Ksp, it was just diluted).

Can you calculate concentration of chromate now?
 
  • #3
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What was the final concentration of Ag+ after solutions has been mixed? (Don't think about Ksp, it was just diluted).

Can you calculate concentration of chromate now?
I like how everyone answers the questions with another question. lol

As I stated above, I did find the final concentration of Ag+, which came out to be 2.0x10-4M
 
  • #4
Borek
Mentor
28,412
2,814
I like how everyone concentrates on the first part of the answer but ignores the other part, one that leads to the solution.

So you have a 2.0x10-4M solution of Ag[sup+[/sup] and you know concentration of CrO42- is high enough to start precipitation. What is concentration of CrO42-? Where did it came from?
 
  • #5
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Hmm.. concentration of CrO4 comes from Na2CrO4...
Well, I used algebra to find the initial concentration of CrO4, and ended up with a negative value.
Here's what I did:
I declared x as [CrO4]init,
so [CrO4] @ eqb = 1.0x10-4 + x

Since Ksp = [Ag]^2[CrO4],
[CrO4] (at eqb?) = 1.1x10-12 / (2.0x10-4)^2
x = -7.25x10^5 M = [CrO4] initial.

I have no clue as to what this value actually means....
Any idea?
Thanks for your help :D
 
  • #6
Borek
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28,412
2,814
LOL, unless I am missing something that's not your fault. Question is wrong. There is no such thing as 2x10-4M solution of Ag2CrO4, no wonder you are getting negative concentrations.

Check what is silver chromate solubility.

On the second thought, there is a slight chance that question is OK and you were expected to take into account chromate protonation. But I strongly doubt it.
 
  • #7
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oh my gosh, that's so funny! I thought my chemistry teacher was evil enough to give us a question that had a negative answer :)
Well, solubility of Silver Chromate is 1.1x10-12, so I guess the question had it right.
But how do you know that 2.0x10-4 M solution doesn't exist?
Thanks!
(ps: I don't think we know what protonation is yet...)
 
  • #8
Borek
Mentor
28,412
2,814
Well, solubility of Silver Chromate is 1.1x10-12, so I guess the question had it right.
This is not solubility, this is Ksp. Solubility is a concentration of a saturated solution.
 

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