Titration of a weak acid and strong base...

[Aerlinn]

You can hide this advertisement by registering.

For instance if we had ethanoic acid (in vinegar) titrated against a NaOH standard in the burette, and the vinegar is diluted to, eg. 250ml, the volume of the volumetric flask before taking out aliquots... why is the
vinegar be diluted?

Using the example above, if I were to draw one of those titration curves, with the 'volume of acid' on the horizontal axis and the increasing pH on the vertical axis, because it is a strong base added to a weak acid, would the curve begin at the bottom, go up, go steeper, and then go up very slowly for the broad end point? Or would it start off going up steeply for the broad end point? (using phenolphthalein indicator)

I get this reaction:
CH3COOH(aq) + NaOH(aq) <---> CH3COONa(aq) + H2O(l)
Because the ethanoic acid doesn't ionise completely... The reaction doesn't run to completion, does it? Then, how would we be expected to do the calculations?

Some quick help= before tomorrow!! (URGENT =S)

[Ea22.007]

i think it actually does go to completion, the titration coz
when u add the base it will react with the dissociated H+ ions to form water.
then u are actually removing H+ from the solution so therefore u would be affecting another equilibrium
: CH3COOH <---> CH3COO- + H+

as u see if u remove the H+ more ethanoic acid will lose their H+ which would then also be neutralized by the NaOH u are titrating with so eventually all of the ethanoic acid would be reacted with.
So the calculations should be the same as usual i think.

[Aerlinn]

Isn't it bases that dissociate?
Acids like ethanoic acid would go:
CH3COOH(aq) + H20(l) <----> CH3COO-(aq) + H30+(aq)
I think.
You'd get some CH3COO- and H30+ ions in solution, and some CH3COOH...
Then when you add the NaOH what happens... ? *puzzled*
Help!! =S
So mainly, questions are:
1. The reactions for the titration. I know I've written out the overall reaction in my post above, but I don't know the stuff in between...
ie. First reaction: CH3COOH(aq) + H20(l) <----> CH3COO-(aq) + H30+(aq)
Second?

2. Why the vinegar is diluted

3. What the titration curve looks like...

[xiao1985]

Ea is right...

first you have the ionisation of CH3COOH equation:

CH3COOH + H2O <-----> CH3COO - + H3O+ (1)
note the equilibrium... very important

Now, the tiration really goes like H3O+ + OH- <----> 2H2O (2)
(well the equillibrium can be considered as a single arrow, since it almost goes to completion any way)

Now, the more NaOH you add, the more H+ will be reacted... and more that happens, the more equilibrium will shift to the right for eqn (1)... which means it is true that ethanoic acid doesn 't ionise completely, but since H+ are constantly removed through neutralisation... which means when you added enough OH- (or NaOH) you effectively push the equilbrium to the completion...

now i remember mention this else where... dilution makes diminishes non-systematic errors...

say for eg, you incurr an error of 0.05 mL upon reading the burette... (i mean, honestly, you eye sight is only as good to a certain extend...)

if you used non-diluted vinegar, you will need say 2.5mL of vinegar to reach end point. and the 0.05 error will be an error of 2%... whereas, after dilution, you will now need 25 mL to achieve end point... but because your ability to read burette doesn't change, the error will still be 0.05mL.. .but now, you incurred a 0.2% error... a 10 fold decrease in inaccuracy, isn't that nice?!

titration curve... start at very high pH (NaOH in conical flask) then slowly drops until the end point, which there will be a sharp drop of pH, and plataeus at low (but not very low) pH... (buffering of CH3COOH) then you should have like a "step" in the plateau.... (since CH3COOH is not a strong acid nor is it concentrated)

[Aerlinn]

I see It's all good, except 'dilution makes diminishes non-systematic errors...' In the burette... or... everywhere?
I think you oddly swapped it around in the second bit, and thought the vinegar went in the burette, and the NaOH went in the flasks, but it was really the other way around ('cause I guess you'd have to add an awful large amount of vinegar to neutralise the NaOH if we had vinegar in da burette)... So my question was really, why dilute the vinegar that goes in the conical flasks?
And the titration curve would go the opposite way me thinks, going up...?