[Defiledxhalo]

Below is the question:
Consider the titration of a 20.0 mL sample of 0.105 M HC₂H₃O₂ with 0.125 M NaOH. Determine the initial pH levels.
So I know the pH = -log[acid], so I assumed the way to go about this problem is just pH = -log(0.105), which gives me 0.979, which is wrong. The book says the answer is 2.86.
Could someone help explain how I would set up this problem? Thanks!

[sjb]

Below is the question:
Consider the titration of a 20.0 mL sample of 0.105 M HC₂H₃O₂ with 0.125 M NaOH. Determine the initial pH levels.
So I know the pH = -log[acid], so I assumed the way to go about this problem is just pH = -log(0.105), which gives me 0.979, which is wrong. The book says the answer is 2.86.
Could someone help explain how I would set up this problem? Thanks!

Is C₂H₃O₂H a strong acid?

[mugabo daniel]

pH = -log[acid], this is not correct rather pH = -log[H+]
in this case [acid] = [H+] since bascity of the acid is 1
but 20 ml of 0.105 M contain 0.0021 moles of H+
      pH = - log[0.0021]
           = 2.68
this is what i get, please guys correct me if am wrong.

[grs35]

ok first thing you need to know is that acetic acid is a weak acid and you need to write down the acid dissociation reaction.

acetic acid+h20 --> h30+ +c2h3o2

i dont know how much you know about titration, but you have to set up an ICE table. the initial concentration of acetic acid is 0.105 M and the rest is zero.

so, you write the equilibrium expression as x^2=1.8.10^-5X(0.105-x)

since the reaction goes against the overall reaction, you can use small x approximation. if you solve for X you will find 1.375X10^-3 M this is the concentration of H30+

negative log of it will give you the pH which is 2.86169...

[AWK]

In this approximation showing more than 2 significant digits is a serious error (since used Ka=1.8^.10^-5).
Note - in logaritms significant digits are after decimal point.

[mugabo daniel]

thanks for that approach guys. guess i got confused somewhere.