What are the concentrations of HOAc and OAc- in a 0.2 M “acetate” buffer, pH 5.0? The Ka for acetic acid 1.70x10-5 (pKa 4.77).

Answer 1

I've answered this before. You need two equations. eqn 1 is the HH equation
pH = pKa + log (base)/(acid)
Substitute the number and solve for the ratio of base/acid.
The second equation you need is
(acid) + (base) = 0.2 M
where (acid ) = (HOAc)
(base) = (OAc^-)
Post your work if you get stuck.

Answer 2

Is it correct like that?
Thanks in advance
5=4.77+log(b/a)
b/a=1.698

b+a=0.2
(1.698*a)+a=0.2
A=0.074M
B=0.1259M

Answer 3

That looks OK to me. I obtained b/a = 1.7 but that doesn't change anything with the remainder of the problem. Thanks for showing your work. I suspect I have my calculator set to round to 2 places and your calculator is different. Your work is good.