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Topic: Solubility with increased pressure.  (Read 5174 times)

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Offline Fmeub

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Solubility with increased pressure.
« on: January 16, 2008, 12:11:38 PM »
This is a question from an exam paper whose memo I don't have.

Solubility product of Ag2CrO4 is 1.1 x 10-11 at 25 degrees Celsius and pressure 1 bar. Calculate the silver ion equilibrium concentration at 25 degrees Celsius and pressure of 3 bars after placing 20g of siver chromate in a beaker with:
a) 100 mL pure water
b) 100 mL of 0.005 M potassium chromate.

The answers I got is:
a) 1.77 x 10-5 M
b) 4.70 x 10-6 M

In b) the assumption was made that [CrO42-] = 0.005 M.

Are my answer incorrect and should I have taken the pressure into account? It doesn't seem to make sense to incorporate the pressure when one is not working with a gas.               

Offline Arkcon

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Re: Solubility with increased pressure.
« Reply #1 on: January 16, 2008, 12:34:42 PM »
For solids and liquids, pressure has a minimal effect and is usually ignored.  You can find a formula for pressure dependence at wikipedia {here}.  Seems like a pretty obscure point to put on an exam, if the prof hasn't made a point of teaching it.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline Fmeub

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Re: Solubility with increased pressure.
« Reply #2 on: January 16, 2008, 01:14:21 PM »
 Thank you!  :D
Does your calculations agree with my answers?

Offline Borek

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Re: Solubility with increased pressure.
« Reply #3 on: January 16, 2008, 02:14:25 PM »
Show your work, either I have a dumb moment or you are twice wrong.
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Offline Fmeub

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Re: Solubility with increased pressure.
« Reply #4 on: January 16, 2008, 03:37:33 PM »
I have reconcidered it. Does a concentration of 2.80 x 10-5 M seem a better answer for a)  ?

For b) I'm not sure. This is what I've done:

From the stoichiometric ratio, [CrO42-] = 0.5[Ag+].

In the solution then
[CrO42-] = 0.5 [Ag+] + 0.005 M (from potassium dichromate)

Assume the dissociation of potassium dichromate surpresses the solubility of silver dichromate, so that [CrO4] = 0.005.
In other words 0.5[Ag+] << 0.005.

 [Ag+]2 (0.005) = 1.1 x 10-11
Therefore  [Ag+] = 4.69 x 10-5 M
i.e. 4.69 x 10-6 M in 100 mL.

OR should the final answer be 4.69 x 10-4 M in 100 mL. Should I multiply or divide by ten? I'm confused. ???

Another thing that I'm not sure about is whether I should use the the "20g" given or not. With the exam paper there wasn't any periodic table or molar mass indicated, and it's not as if I memorize elements' molar mass to four decimal places.





Offline Borek

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Re: Solubility with increased pressure.
« Reply #5 on: January 16, 2008, 04:47:12 PM »
2.8x10-4M and 4.69x10-5M. Equilbrium concentration is intensive property and doesn't depend on the sample volume!
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Offline Fmeub

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Re: Solubility with increased pressure.
« Reply #6 on: January 17, 2008, 05:02:55 AM »
My heartfelt gratitude for reminding me... ;)

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