Question

not sure where I'm going wrong on this buffers question

Answer 1

You need to prepare an acetate buffer of pH 6.21 from a 0.895 M acetic acid solution and a 2.96 M KOH solution. If you have 575 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 6.21? The pKa of acetic acid is 4.76.

Answer 2

my attempt:

using Henderson-Hasselbach

6.21 = 4.76 + log(acetate/acetic acid)

(0.895)(.575) moles of acetic acid --> 0.514625 moles acetic acid

making an ice table gives us 0.514625 - x moles acetic acid remaining after x mol of KOH has been added, and x moles of acetate produced assuming 1:1 reaction

so 6.21 = 4.76 + log(x / (0.514625 - x))

giving x = 0.41685 moles of KOH needed, or 141 mL when taking the 2.96 M concentration into account, but this has been marked wrong >>

Answer 3

also I can't believe I have to keep explaining this but please don't respond unless you are making an attempt to answer the question. Any irrelevant comments are considered spam and will be promptly removed.

Answer 4

ok, nvm, for some reason the software I was using was doing a natural log rather than a base 10 log >_>