# Determination of the Kc for the ethanoic acid acid/ethyl ethanoate equilibrium help?

If I've started with the following volumes:
Ethanoic Acid - 30cm3
Ethanol - 30cm3
Ethyl Ethanoate - 0cm3
HCl (concentration: 1.0 mol dm3) - 10cm3
Water - 30cm3

We were asked to work out the moles of each substance before the equilibrium was achieved; the values achieved were out of 100cm3 (total volume of reaction mixture) so they had to be by 10 to have them out of 1dm3.

Then they were reacted together to reach equilibrium. Afterwards, 1cm3 of the solution was taken out and put into a conical flask with 25cm3 of deionized/distilled water.

It took 7cm3 of Sodium Hydroxide to turn the solution alkaline. Does that mean there is 7x10-4mol within that 1cm3 and would that have to be multiplied by a thousand to get it in terms of dm3 like before with the 100cm3? So, from this, how would I calculate the concentrations of each substance at equilibrium? And hence calculate the value of Kc? (I'm basically clueless with this since we were just given this to work out without any explanation, so any form of help would be greatly appreciated!!)

*another thing, originally we had to work out the moles using the volumes and densities:
Ethanol: 0.79gcm3
Ethanoic Acid: 1.05gcm3
Ethyl Ethanoate: 0.92gcm3
Water: 1gcm3
And then had to work out concentrations (don't know why that was needed) but with the HCl, even though it states the concentration being 1moldm3, we somehow ended up with 0.1moldm3, it's just one giant mess*
Original post by TheDiracEquation
If I've started with the following volumes:
Ethanoic Acid - 30cm3
Ethanol - 30cm3
Ethyl Ethanoate - 0cm3
HCl (concentration: 1.0 mol dm3) - 10cm3
Water - 30cm3

We were asked to work out the moles of each substance before the equilibrium was achieved; the values achieved were out of 100cm3 (total volume of reaction mixture) so they had to be by 10 to have them out of 1dm3.

Then they were reacted together to reach equilibrium. Afterwards, 1cm3 of the solution was taken out and put into a conical flask with 25cm3 of deionized/distilled water.

It took 7cm3 of Sodium Hydroxide to turn the solution alkaline. Does that mean there is 7x10-4mol within that 1cm3 and would that have to be multiplied by a thousand to get it in terms of dm3 like before with the 100cm3? So, from this, how would I calculate the concentrations of each substance at equilibrium? And hence calculate the value of Kc? (I'm basically clueless with this since we were just given this to work out without any explanation, so any form of help would be greatly appreciated!!)

*another thing, originally we had to work out the moles using the volumes and densities:
Ethanol: 0.79gcm3
Ethanoic Acid: 1.05gcm3
Ethyl Ethanoate: 0.92gcm3
Water: 1gcm3
And then had to work out concentrations (don't know why that was needed) but with the HCl, even though it states the concentration being 1moldm3, we somehow ended up with 0.1moldm3, it's just one giant mess*

You have to work out the moles of each component BEFORE so that you calculate the moles of each component AFTER by finding out how much acid (moles) has been used up.

The equation is:

acid + alcohol --> ester + water

The coefficients are all 1, so the number of moles of acid used up equals
the moles of ester and water formed.

The only complication is the use of an acid catalyst, which must be factored into the total acid titration (equals moles of HCl + moles of ethanoic acid)

This is simply done by subtracting the moles of HCl from the total moles acid (obtained from the NaOH titration results)
may i ask why wouldn't the reactant side of the equation be CH3COOH + C2H5OH + HCl + H2O while the product side be water and a salt (I don't know how to write the product side) (I am doing the same question so it would be great if someone could help) thanks
(edited 1 year ago)