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Chemical Equilibrium and Le Chatelier's Principle

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Homework Statement


Without adding any Chromium based compounds, how would you manipulate the above equation to produce more CrO42- ions? Be sure to use chemical equations to show the answer.

Homework Equations



2CrO42- (aq) + 2H3O+ <--> Cr2O72-(aq) + 3H2O (l)

The Attempt at a Solution



So my initial thought here is that we want the reaction to shift left so we must alter the concentration of one of the products on the right. Since adding more Chromium is "not allowed" then it would have to be the water? If this is correct though, I don't know what the equation would look like because if I add more moles of water won't I have to balance all the other species as well to have a balanced chemical equation? Thanks in advance for all the time and help.
 

Answers and Replies

  • #2
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shift left so we must alter the concentration of one of the products on the right.
.... or, on the left. Think about it. You are allowed to add, or remove reactants and products to push an equilibrium around to where you want it.
 
  • #3
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Ah! :) good point! So then, my thoughts becomes; since hydronium is one of the reactants, if we lower the concentration of hydronium by adding sodium hydroxide to neutralize some of the acid then the equilibrium would shift left?
 
  • #4
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Very good.
 
  • #5
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Thank-you for the help. So, would this reaction equation be correct?

2CrO42- + 4H3O + 2NaOH <--> Cr2O7
2- + 7H2O + 2Na
 
  • #6
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Sodium is just a spectator ion, so you don't have to show it formally. All the "action" is in the acid-base-water bookkeeping. You have to add and remove the same amounts to and from both sides of the equation just like algebra.
 
  • #7
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O.k I understand thanks. Does this look better?

2CrO42- + 2H3O+ + NaOH <--> Cr2O72- + 4H2O
 
  • #8
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Try adding OH- to both sides as necessary to "neutralize" one side, then subtract excess water. Forget the sodium ion.
 
  • #9
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Hello,
I'm having problems with the same question...is this right?

2CrO42- + 2H3O + 2OH- <--> Cr2O72- + 3H2O + 2OH-
 
  • #10
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Good first step. 2H3O+ + 2OH- gives you what?
 
  • #11
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4 H2O?
 
  • #12
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Good. Since you've got water on both sides of the equation, you can subtract water from both sides until it disappears from one side or the other.
 
  • #13
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2CrO42- <--> Cr2O72- + 3H2O + 2OH-

thanks, ...so I could just write the equation like this?
 
  • #14
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Wait...but then it wouldn't balance...
 
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  • #16
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Also: if you see the same molecule/ion/element on both sides of the equation, cancel it out. You have

2CrO42- + 2H3O + 2OH- <--> Cr2O72- + 3H2O + 2OH-
You have 2OH- on both sides of the equation, they don't change - so you should remove them. They are just spectators.
 
  • #17
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Hello,

Thanks for all the help. I don't know why Paulham and I are so confused but I don't seem to see the matter clearly either. I understand the theory behind it and how NaOH will remove some H+ from the reactants side shifting the reaction to the left but it's the actual chemical equation and it's accounting that seems to be the problem. I see what Mr. Borek is saying about the OH- being spectator ions but when you remove them you would get

2CrO42- + 2H3O <---> Cr2O72- + 3H2O

which is what the original equation is...so it looks like we did nothing? Thanks again for the time...
 
  • #18
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I'm tired of this: We want to drive it to the left, so add OH-; that gives us water on the left, and water and OH- on the right; subtract three waters from both sides; 2CrO4-2 + H2O ↔ Cr2O7-2 + 2OH-. It's one of life's little headaches when playing with redox reactions deciding whether you're going to balance them in acidic, neutral, or basic media.
 
  • #19
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Oh my goodness, that makes complete sense...to be honest I didn't even clue in that this was a redox reaction. Thank you.
 
  • #20
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The chromate-dichromate equilibrium isn't redox in and of itself, but both species are used as oxidizers in quite a number of other reactions, and the pH question is an endless headache as far as keeping things balanced, as well as dragging in the oxygen hydrogen half-cell potentials for either acid or base conditions.
 
  • #21
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