Question

Calculate the amount (in mL) of 1.520M NaOH that is required to add the following acetic acid solution to prepare a buffer with the corresponding pH:

30.00mL of a 5.00% (w/v%) acetic acid; the resulting acetate buffer has a pH of 5.75.

pKa of acetic acid = 4.74

So I got this: 5% w/v means 5g HAc/100 mL solution which is (5/60) mols/100 and that is 5/60/0.1L or 0.0833 M.
Then millimols HAc is 30 x 0.0833 = 25.
So acid + base = 25
pH = pKa + log(base)/(acid)
5.75 = 4.74 + log b/a

How do I determine b/a?

equation 2 is a + b = 25

How can I solve for a and b?

Answer 1

You had this in math. It isn't chemistry.
5.75 = 4.74 + log b/a
1.01 = log b/a
10^(b/a) = 1.01
Plug into the calculator; a/b = 10.23

Answer 2

If a/b = 10.23, what is a and what is b?