What is the pH when acetic acid is titrated with NaOH?

Question

A 100 ml sample of .200 M acetic acid (Ka=1.8x10^-5) with 100 ml of .200 M NaOH. What is the pH?

The answer is 8.87 but im unsure of the process. I though that you set up an equation like conc. of products over the conc. of the reactants and set it equal to Ka but it didn't work out.

Its finals week and i appreciate everyone that helps out here, you're all great. Thank You!

Answer 1

Bless you, bless you.

The reaction the occurs is:
OH- + HOAc = OAc- + H2O
Since we have equal moles of NaOH and HOAc, they would react completely with each other to form OAc-. In other words, this would be the same problem as if OAc- had been the only chemical added to water.

The concentration of this would be 0.04 moles (the moles formed by the reaction) / 0.2 L, or 0.2 M. We can set up the Ka eqtn. which is:
[OAc-][H+]/[HOAc] = 1.8x10-5
To get the equilibruim eqtn corresponding to the above reaction equation, multiply num and den by [OH-] and note [H+][OH-] = Kw. Then
[OAc-][Kw]/[HOAc][OH-] = 1.8x10-5, and finally
[OAc-]/[HOAc][OH-]= 1.8 x 10^9
As mentioned, the math proceeds is AS IF ONLY OAc- was initially added. For each x moles of this reacted, x moles of HOAc and OH- are formed. Moreover, x< 0.2 / x^2 = 1.8x10^9 and
x^2 = 1.11x10-10
x= 1.05x10-5= [OH-]
We can find pOH = 5- log 0.05=4.94
and pH= 9.06
While it is nice to have product conc. over reactant concs. , it is not necessary as long as you have set the equation up correctly and done the math right. If it bothers you, just "flip" the equation over.