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Kc Titration Question

Hi, please could someone help me with this question?
For some reason, I feel stumped on the first question.

The titration result I got was 25.5cm3

Here are the instructions:
You are going to measure the equilibrium constant Kc for the reaction below.
CH3COOH + CH3CH2CH2OH CH3COOCH2CH2CH3 + H2O
ethanoic acid + propan-1-ol propyl ethanoate + water
This reaction is very slow even in the presence of sulfuric acid which is used as a catalyst.
1) Place 5.0 cm3 of ethanoic acid from a burette into a clean, dry boiling tube.
2) Place 5.0 cm3 of propan-1-ol from a burette into the same boiling tube.
3) Place 2.0 cm3 of 1.00 mol dm-3 sulfuric acid into the same boiling tube.
4) Seal the end of the tube tightly with cling film or parafilm.
5) Keeping your thumb and some fingers on the cling film / parafilm, swirl the tube carefully to ensure thechemicals mix well.6) Leave the tube in a boiling rack for a week.

Part 2
1) Transfer the contents of your boiling tube into a volumetric flask using deionised water. Wash all thecontents of the tube into the volumetric flask and then make it up to 250 cm3 in the normal way. You needto ensure very thorough shaking throughout the process.
2) Set up a burette containing 0.200 mol dm-3 sodium hydroxide solution.
3) Pipette 25.0 cm3 of the solution from the volumetric flask into a conical flask, add 3-4 drops ofphenolphthalein and titrate against the sodium hydroxide solution.
4) Repeat until you have concordant results. Record all your results in a suitable table.

Here are the questions:
1) Use your titration results to find the total moles of H in the 250 cm3 solution.
2) Calculate the moles of H ions in the original 2.0 cm3 of 1.00 mol dm-3 sulfuric acid (remember that thereare two H ions for each H2SO4 unit).
3) Calculate the moles of ethanoic acid at equilibrium (this is the difference between the total moles of H found via titration and the moles of H from the sulfuric acid catalyst).
CH3COOH CH3CH2CH2OH CH3COOCH2CH2CH3 H2O
moles at start
change in moles
moles at equilibrium
4) Calculate A, the moles of ethanoic acid in the original 5.0 cm3. (density of ethanoic acid = 1.05 g cm-3).
5) Calculate B, the moles of propan-2-ol in the original 5.0 cm3. (density of propan-1-ol = 0.803 g cm-3)
6) Calculate x, the moles of ethanoic acid that reacted.7) Calculate the moles of each substance at equilibrium.8) Write an expression for Kc.
9) Calculate Kc and state its units (or that there are no units if there are none)

If anyone could help that would be amazing, thank you xx
(edited 7 months ago)
What have you tried so far?
Original post by TypicalNerd
What have you tried so far?

Hey, so for q1 it took 25.55cm^3 of NaOH to neutralise the 25cm^3 of the solution. Ik the concentration is 0.2 moldm^-3 so moles of NaOH is 0.2 x 0.0255 = 0.02555. This is where I am stuck - usually I would use stoichemtry to find the moles but I think I am being out of by trying to find the H ions which is making me over complicate the question.
Original post by Pineapple27272
Hey, so for q1 it took 25.55cm^3 of NaOH to neutralise the 25cm^3 of the solution. Ik the concentration is 0.2 moldm^-3 so moles of NaOH is 0.2 x 0.0255 = 0.02555. This is where I am stuck - usually I would use stoichemtry to find the moles but I think I am being out of by trying to find the H ions which is making me over complicate the question.

So (1 is asking how many H^+ ions were in the 250 cm^3.

As a sort of ionic equation:

NaOH + H^+ —> Na^+ + H2O

From that, you can deduce that for every one mole of NaOH that reacts, you have 1 mole of H^+ ions, so the moles of sodium hydroxide used equals the moles of H^+ ions in the portion of the solution you titrated.

What fraction of the 250 cm^3 of solution is 25 cm^3 and so can you use that and your answer for the number of moles of NaOH added in the titration to get your final answer?
Ahh I've got it now!! Thank you :wink:

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