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# Need help understanding this Kc question? watch

1. A student does a titration to make an ester out of ethanol acid and ethanol. The student adds 1.0 moldm-3 of NaOH to the reaction mixture in a flask.But the NaOH actually used was more concentrated than 1.0 moldm-3. What would be his effect on the kc value?

Here's the equation--> CH3COOH+CH3CH2OH--> CH3COOCH3 +H20

My teacher said that he/she adds less NaOH and thinks there is less acid in the mixture therefore less reactants more products so Kc increases.

I don't really understand this. Shouldn't the equilibrium shift left in the more acidic part? What would be the effect if the NaOH was actually less concentrated?

Please could you explain this concept to me. I have a graded assement on this tomorrow and really need to grasp this. Thank you!
2. What's the NaOH for?

What I learnt was to produce an ester, you need an acid chloride/carboxylic acid + alcohol (warmed under reflux with few drops conc sulphuric acid). If you add NaOH to this, the ester is hydrolysed to alcohol + acid
So the equilibrium would shift right to produce more ester?

So for Kc, it's PRODUCTS/REACTANTS
You now have more reactants, so Kc would decrease?

The Q sounds like the student only found it was more conc after they did the experiment
3. (Original post by Phalange)
What's the NaOH for?

What I learnt was to produce an ester, you need an acid chloride/carboxylic acid + alcohol (warmed under reflux with few drops conc sulphuric acid). If you add NaOH to this, the ester is hydrolysed to alcohol + acid
So the equilibrium would shift right to produce more ester?

So for Kc, it's PRODUCTS/REACTANTS
You now have more reactants, so Kc would decrease?

The Q sounds like the student only found it was more conc after they did the experiment
I forgot to mention that Hcl is included in the mixture. It jsut says on the sheet that it was used to find the Kc.
4. (Original post by Medifield)
I forgot to mention that Hcl is included in the mixture. It jsut says on the sheet that it was used to find the Kc.
So umm what exactly did they titrate? When did they add NaOH and HCl?
5. (Original post by Medifield)
A student does a titration to make an ester out of ethanol acid and ethanol. The student adds 1.0 moldm-3 of NaOH to the reaction mixture in a flask.But the NaOH actually used was more concentrated than 1.0 moldm-3. What would be his effect on the kc value?

Here's the equation--> CH3COOH+CH3CH2OH--> CH3COOCH3 +H20

My teacher said that he/she adds less NaOH and thinks there is less acid in the mixture therefore less reactants more products so Kc increases.

I don't really understand this. Shouldn't the equilibrium shift left in the more acidic part? What would be the effect if the NaOH was actually less concentrated?

Please could you explain this concept to me. I have a graded assement on this tomorrow and really need to grasp this. Thank you!
You are all over the place. It's reallly hard for people to help you if you are not clear.

You carry out an esterification.

Then you analyse the amount of acid in the mixture by titration with NaOH

As both the acid catalyst and the ethanoic acid react with NaOH, the moles of NaOH gve you the total moles of acid (not forgetting that sulphuric acid provides 2 hydrogen ions per molecule)

You have to subtract this from the total initial moles of both acids to get the moles of ethanoic acid used up - this equals the moles of ester formed.

The 'extra' concentration of NaOH cannot affect the actual value of kc as nothing can, BUT it can give you a value for your calculation of kc which is incorrect.

THAT is what the question is trying to get you to show - the effect on the calculated value of ks if the NaOH is actually more concentrated than stated.

Look at it this way:

Less NaOH is needed to titrate the total acid - so you find less moles of acid remaining.

Hence more moles of acid have been used up in the esterification.

Hence more ester seems to be formed.

Hence your calculated value for kc is bigger than the real value.

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Updated: March 29, 2011
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