I just need help with someone to check my answer/help a little, but no one is answering me!!!

Pipette 20.0cm^3 of 0.0100moldm^-3 CH3COOH(aq) into a conical flask.
Excess 0.0100moldm^-3 KOH(aw) from burette.
Total 50.00cm^3 alkali added.
CH3COOH(aw) + KOH(aq) --> CH3COOK(aw) + H2O(l)

Draw the pH curve.

So here is my working out so far, (finding end pH).

n acid used = 0.0100*20/1000 = 0.0002
n alkali used = 0.0100*50/1000 = 0.0005
Ratio 1:1, n alkali reacted = 0.0002
Equialance point = 0.0002/0.0100 = 0.02

End pH - as ratio 1:1, alkali reacted = 0.0002
So n alkali solution at end, .0.0005-0.0002=0.0003
Concetration alkali at end - 0.0100*(30/75) = 0.004
[H+] = Kw/[OH-]
= 1* 10^-14/0.004
=

Is this right? I have just started this topic so I am a little unsure.
Please help, I still need to find the start pH etc.

This is OK:
n acid used = 0.0100*20/1000 = 0.0002
n alkali used = 0.0100*50/1000 = 0.0005
Ratio 1:1, n alkali reacted = 0.0002

This is not needed:
Equialance point = 0.0002/0.0100 = 0.02

This is OK:
End pH - as ratio 1:1, alkali reacted = 0.0002
So n alkali solution at end, .0.0005-0.0002=0.0003
Concetration alkali at end - 0.0100*(30/75) = 0.004

This is incorrect:
[H+] = Kw/[OH-]
= 1* 10^-14/0.004
=

Is this right?

=================

Since KOH is a strong base, we focus only on it and ignore the presence of the CH3COOK in the solution.

pOH = -log 0.004 = 2.40

pH + pOH = pKw

pH + 2.40 = 14.00

pH = 11.60

===================

"I still need to find the start pH"

0.0100moldm^-3 CH3COOH(aq)

Please go here:

http://www.chemteam.info/AcidBase/Ka-Sol…

for an explanation of what to do

pH = 3.376 for your solution.

The answer is too lengthy to fit into the Yahoo Answers format. If you email me at vailhp@muohio.edu, I will send the solution in its entirety.