[briteyellowness]

okay, i know that the equation is

pH = pKa + log (A^-/[HA])

so if there was acetic acid, then the conjugate base would be acetate.  Can I still use this equation if something else happened like I added Hydroxide ions?  That isn't the CB right?  so I can't put that in as the Conjugate Base Concentration?

so if 100 mL of .1 M NaOH was added to 150 mL of .2 acetic acid, can i find the pH?  The pKa of acetic acid is 4.76.  

Should i use the equation an ICE equation where i can add the OH that will deplete the H and make more acetate?  I'm not quite sure how I should go about doing that.  Thanks a lot!

[sdekivit]

sure, u can stll use the HH-equation, but you need to know that the OH(-) ions will react with acetic acid according the following reaction:

HAc + OH(-) --> Ac(-) + H2O

So you can calculate how many HAc and Ac(-) there are in the solution after adding hydroxide ions.

[briteyellowness]

wait.. so what would that be?  finding the Kb of acetate ion so then i can calculate how much acetate is formed?  so then i know the concentration of HAc and OH and I shall just find the Concentration of Acetate?
And then with the HH eq, I just plug in that number for Acetate and use the given concentration for HAc?


and/or

since HAc is a weak acid, i can't just say that since we added .01 mol of NaOH, only .01 moles of HAc forms Acetate right?


oooh!

kay, with the equation, that Ka = [H+][Ac-]/[HAc]
and plugging in 10^-4.76 for the Ka
and .2 for HAc

then i figure out that Ac- is .0018 M

so
Hac + OH-  ---> Ac-          + H20
.03      0           2.8*10^-4    0
+.01   +.01       -.01

which gives me Hac = .04 mol
Ac- =.. a negative number i'm doing something wrong.


wait.. so is it
Hac   --->   H+           +      Ac-
.03            2.8*10^-4   2.8*10^-4
-.01           +.01               +.01     (because the OH will keep on combinging with the H+ to form H20, thus doing this?)


Hac = .02
Ac- = .01028

and plugging this into the HH we get pH  = 4.47