posted by Alexis .
Can a buffer solution with pH = 4.70 be prepared using water, 6.0 M HC2H3O2 and 6.0 M NaOH? Justify your answer. Ka = 1.8 x 10-5 for HC2H3O2 , K = 1.0 x 10-14 .
The Henderson-Hasselbalch equation is
pH = pKa + log [(base)/(acid)]
pKa = 4.74 from Ka.
Generally it is easy to prepare a buffer within 1 pH of the pKa. You can work through the calculation if you wish; for example, if equal amounts of acid and base are chosen, then (B) = (A), the ratio is 1 and log of that term is 0 and pH = 4.74 which is very close to the desired pH.
After I went to bed I realized that part of my answer might be confusing. By (B)= (A), I am referring to base = acetate ion an acid = acetic acid. To GET equal amount of acetate ion and acetic acid, one reacts X amount of NaOH and twice that amount of acetic acid. That produces X amount of acetate ion, leaves an equal amount of acetic acid, and none of the NaOH you started with. I hope this helps clear things up.