le chatelier's principle question

[SSerenity]

I scanned the question here (with the alleged solution)

https://www.dropbox.com/s/2mhdk3mzzn5gyr7/New Doc 3.pdf

I don't understand why the answer is B. Since HS- and H2S have small dissociation constants, the addition of protons to a this solution will favor the removal of S2- from solution. This in turn will drive the reaction to the RIGHT, not the left!

Can someone please explain this?

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[monkeyvokes]

I scanned the question here (with the alleged solution)

https://www.dropbox.com/s/2mhdk3mzzn5gyr7/New Doc 3.pdf

I don't understand why the answer is B. Since HS- and H2S have small dissociation constants, the addition of protons to a this solution will favor the removal of S2- from solution. This in turn will drive the reaction to the RIGHT, not the left!

Can someone please explain this?

reaction 1 is h2s --> hs + H, and has ka1
reaction 2 is hs --> s + H and has ka2

adding protons shifts the reaction to the left because you are adding a product.

 

[monkeyvokes]

I scanned the question here (with the alleged solution)

https://www.dropbox.com/s/2mhdk3mzzn5gyr7/New Doc 3.pdf

I don't understand why the answer is B. Since HS- and H2S have small dissociation constants, the addition of protons to a this solution will favor the removal of S2- from solution. This in turn will drive the reaction to the RIGHT, not the left!

Can someone please explain this?

the fact that adding protons favors the removal of S2- is an example of la chatlier's principle in effect. removing S2- due to the addition of protons is a left shift.

 

[SSerenity]

reaction 1 is h2s --> hs + H, and has ka1
reaction 2 is hs --> s + H and has ka2

adding protons shifts the reaction to the left because you are adding a product.

I agree, that adding protons will shift the equilibrium to the left, but only in the above reaction ^ !

But in the PDF, Reaction 1 has S2- on the right side of the arrow (products side). By adding protons, you are essentially removing the product (forming HS-/H2S). Because we are losing product, the equilibrium should shift to the right to replace the lost product correct?

 

[chizledfrmstone]

I agree, that adding protons will shift the equilibrium to the left, but only in the above reaction ^ !

But in the PDF, Reaction 1 has S2- on the right side of the arrow (products side). By adding protons, you are essentially removing the product (forming HS-/H2S). Because we are losing product, the equilibrium should shift to the right to replace the lost product correct?

To me it sounds like you're thinking too far ahead.

If you add protons you have and excess of H+ right?

You want to maintain Ka1 so the reaction will do that by making more reactants. It goes to the left.

Ask yourself why are we losing product...

 

[SSerenity]

This is the reaction I thought Reaction 1 was referring to, (stated in the passage)

So adding protons will convert the S2- to its acid form, effectively lowering [S2-]. I'm sorry, I still don't understand why this would drive it to the left.

 

[monkeyvokes]

This is the reaction I thought Reaction 1 was referring to, (stated in the passage)

So adding protons will convert the S2- to its acid form, effectively lowering [S2-]. I'm sorry, I still don't understand why this would drive it to the left.

oohh yeah I didn't see those two reactions. You're right, it seems like you'd form H2S and shift the reaction to the right. Where did you find this question?

 

[SSerenity]

I found this in one of the Gold Standard Exams. I've been finding a lot of mistakes in their exams lately

 

[monkeyvokes]

I found this in one of the Gold Standard Exams. I've been finding a lot of mistakes in their exams lately

Well, the price of gold hasn't been very stable lately.