2nd Year College Chem, Buffers & pH!?

What is the pH at the point in a titration at which 60.00 mL of 0.100 M potassium hydroxide, KOH has been added to 100.00 mL of .100M acetic acid, CH3COOH

I think its 4.52, but it could be 4.92. Any inputs! Please explain if you disagree with me.

How I solved it.
[OH]- = 6/10 * 1.8 x 10^5- = 4.52

Answer:
no, but u were close!

lets go thru it.

first lets write the net ionic reaction.

CH3COOH + OH- --> H2O + CH3COO-

now we have to change the values of OH- (KOH) and acetic acid to moles. (M x L)

you have 0.1L x 0.1M acid = 0.01 moles

you have 0.06L x 0.1M OH- = 0.006 moles

lets set up a reaction chart

CH3COOH + OH- -->CH3COO- + H2O

0.01mol.0.006mol .0
-0.006...-0.006..+0.006
0.004...0.....0...

now we have a buffer. the buffer equation is

[H+] = Ka x moles acid / moles base

[H+] = 1.8 x 10^ -5 x 4/6 (or 0.004 / 0.006)

[H+]= 1.2 x 10^ -5

-log of that = 4.92 is the pH.

if u need further help please contact me and ill be glad to help.

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2nd Year College Chem, Buffers & pH!?

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