Can't understand equilibrium contstant example

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pondering-soul
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I understand that ethanol and ethanoic acid mixed make an equilibrium mixture and that an acid catalyst can be used but I get lost on the 3rd paragraph.

"The mixture may be analysed by titrating the ethanoic acid with standard alkali (allowing for the amount of acid catalyst added)."

when titrating, are we just putting the whole equilibrium mixture with the standard alkali or are we seperating just the ethanoic acid somehow and then titrating?

how does titrating the mixture against standard alkali solution allow us to work out the amount of acid catalyst added?

and how does it not significantly disturb the equilbrium mixture? i get that the reversible reaction is much slower than a titration reaction but how does that relate to not disturbing the mixture?

thanks for anyone in advance for trying to help, I know its a long post :/
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Trumbles
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"how does titrating the mixture against standard alkali solution allow us to work out the amount of acid catalyst added?"

You're misreading this. In the example, you made the mixture so you know how much strong acid was put in. The point is that you have to work out how much alkali is needed to neutralise the, let's say, 10ml of 1M sulphuric acid catalyst before you can work out how much alkali was used up in neutralising the ethanoic acid. It's the amount of ethanoic acid at equilibrium that you're trying to calculate.

"and how does it not significantly disturb the equilbrium mixture? i get that the reversible reaction is much slower than a titration reaction but how does that relate to not disturbing the mixture?"

You're trying to calculate the number of moles of ethanoic acid at equilibrium. A reasonable concern is that as soon as you start reducing the amount of ethanoic acid present by neutralising it, you're no longer at equilibrium: the ester will be breaking down to release acid faster than it's being replaced (ester formation removes acid).
So you might worry that the amount of ethanoic acid you've neutralised would equal the amount of ethanoic acid at equilibrium PLUS the amount of ethanoic acid formed after you'd shifted equilibrium by adding alkali. This would throw your calculation off: it would look like there was more ethanoic acid present at equilibrium than there actually is.
The text reassures you that so long as you do the titration over minutes and not days, the ester doesn't have time to react much with water and produce much more acid.

Does that help?
Last edited by Trumbles; 9 months ago
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username5195672
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nvm, i thought this was abt buffers
Last edited by username5195672; 9 months ago
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pondering-soul
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(Original post by pondering-soul)
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I understand that ethanol and ethanoic acid mixed make an equilibrium mixture and that an acid catalyst can be used but I get lost on the 3rd paragraph.

"The mixture may be analysed by titrating the ethanoic acid with standard alkali (allowing for the amount of acid catalyst added)."

when titrating, are we just putting the whole equilibrium mixture with the standard alkali or are we seperating just the ethanoic acid somehow and then titrating?

how does titrating the mixture against standard alkali solution allow us to work out the amount of acid catalyst added?

and how does it not significantly disturb the equilbrium mixture? i get that the reversible reaction is much slower than a titration reaction but how does that relate to not disturbing the mixture?

thanks for anyone in advance for trying to help, I know its a long post
(Original post by Trumbles)
"how does titrating the mixture against standard alkali solution allow us to work out the amount of acid catalyst added?"

You're misreading this. In the example, you made the mixture so you know how much strong acid was put in. The point is that you have to work out how much alkali is needed to neutralise the, let's say, 10ml of 1M sulphuric acid catalyst before you can work out how much alkali was used up in neutralising the ethanoic acid. It's the amount of ethanoic acid at equilibrium that you're trying to calculate.

"and how does it not significantly disturb the equilbrium mixture? i get that the reversible reaction is much slower than a titration reaction but how does that relate to not disturbing the mixture?"

You're trying to calculate the number of moles of ethanoic acid at equilibrium. A reasonable concern is that as soon as you start reducing the amount of ethanoic acid present by neutralising it, you're no longer at equilibrium: the ester will be breaking down to release acid faster than it's being replaced (ester formation removes acid).
So you might worry that the amount of ethanoic acid you've neutralised would equal the amount of ethanoic acid at equilibrium PLUS the amount of ethanoic acid formed after you'd shifted equilibrium by adding alkali. This would throw your calculation off: it would look like there was more ethanoic acid present at equilibrium than there actually is.
The text reassures you that so long as you do the titration over minutes and not days, the ester doesn't have time to react much with water and produce much more acid.

Does that help?
Wow, I get it now. Yeah that does help thanks a lot bro!
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