Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: briteyellowness on September 07, 2005, 11:27:05 AM
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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!
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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.
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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