Are you using the Henderson-Hasselbalch equation or just Ka for CH3COOH? And show me how you obtained 1.41:1.00M. I can't get that.
(pure CH3COOH) Ka = 1.8 x 10-5
sodium acetate NaCH3COO
CH3COOH --> CH3COO + H
Ka= [H+][CH3COO-]/[CH3COOH] = 1.8 x10^-5
pH = pKa + log [base]/[acid]
5.0 (chosen pH of buffer) = 4.85 + log [base]/[acid]
.15 = log [base]/[acid]
10^0.15 = [1.41 M]/[1 M]
OK. But the acid:base ratio is not 1.41:1. If pH you want from the problem is 5.0, then (H^+)= 1E-5 and
Ka = (H^+)(CH3COO^-)/(CH3COOH) then
(CH3COOH)/(CH3COO^-) = (H^+)/Ka
acid/base = (1E-5/1.8E-5) = 0.555. But let me show you a much easier way to do it since you typed in the HH equation.
pH = pKa + log (base/acid)
5.0 = 4.745 + log B/A
0.255 = Log B/A
B/A = 1.80 or A/B = 0.555 but I never use either of those as a number although the calculator calculates that number during the following.
Now, how many mL of the 17.4 stuff will you use? 17.4 M x # mL used = ?? millimoles. Then plug that into the HH equation and solve for base (which will be in millimoles if you used acid in millimoles).
5.00 = 4.745 + log(mmoles base)/(mmoles acid) and solve for mmoles base. I get something like 3.13 mmoles base. Convert mmoles to moles, then to grams of CH3COONa. To prepare the buffer you measure the acetic acid with a pipet (or graduated cylinder probably is good enough), add in the grams sodium acetate, stir until dissolved and you have it. Your instructor may have put some limitations on the preparation; for example, you may know how much acetic acid is to be used.