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HELP - equilibrium constant calculations for esterification

Just come back to my chemistry coursework after revision and holidays. Does anyone know if you have to take into consideration the water used to dilute the reactants in an equilibrium reaction when you are calculating the equilibrium constant? If so, how?!
I'm reacting short chained alcohols and carboxylic acids to produce esters and water.

Please, someone, help??:frown:
Thanks
Reply 1
Avatar for Eau
Eau
15
You probably don't need to take into account of water because its dissociation constant is very very small.
no, for esterification you DO need to consider the concentration of the water in the mixture.
It does take part in the equilibrium.
Reply 3
Avatar for Dashbrook
Dashbrook
OP
i thought as much, any help on how to do it though?

Thanks
how to do what? the whole experiment?
Reply 5
Avatar for Dashbrook
Dashbrook
OP
nope, done the experiment and calculated how much acid was left at the end of the experiment and therefore how much ester was producted. However I need to calculate Kc and i dont know how to take the extra water into account when i'm doing that.
you already have the mass of the water you started with from the mass of the HCl solution (or whatever other acid) you added. (subtract the mass of HCl contained)
The extra moles of water will be equal to the moles of acid used up in the esterification.

I am assuming that you titrated the mixture both before and after the equilibrium was established to find the total moles of acid present (= moles of acid catalyst + moles of ethanoic acid)
Reply 7
Avatar for Dashbrook
Dashbrook
OP
Thanks :biggrin: . Have attempted to sort it out but going to my chem teacher tomorrow and getting it checked.

Cheers,

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