Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Making a buffer solution?
this is a test question i had, for some reason i got really lost on it, and i wanted to know the answer incase it shows up in the future.
use the henderson0 hasselbach equation to perform the following calculation. (The Ka of acetic acid is 1.8* 10^-5).
Calculate the mass of solid sodium acetate required to mix with 100 ml of 0.1 M acetic acid to prepare a pH 4 buffer.
i get a little confused with the equation, because i dont know which conj base and acid concentration to use for th henderson-hasselbach equation.
any help would be appreciated, thank you.
after posting, i tried it again, and i got 4.557 g. now i think i did it right. if anyone could clarify, or show they got the same would be awesome
omg, thank you so much, i had the conj base and acid flipped around, you answered my main question in the first 3 sentances
omg, thank you so much, i had the conj base and acid flipped around, you answered my main question in the first 3 sentances
1 Answer
- Simonizer1218Lv 78 years agoFavorite Answer
The HH equation states pH = pKa + log [salt]/[acid] or + log [conj base]/[acid]
For acetic acid/acetate buffer, obviously the acetic acid is the acid, and the acetate is the conj base
So for pH 4 you have 4 = 4.74 + log [acetate]/[acetic acid]
log [acetate]/[acetic acid] = 4 - 4.74 = -0.74
log acetate - log 0.1 = -0.74
log acetate = -0.74 + log 0.1
log acetate = -1.74
acetate = 0.018 M
molar mass of sodium acetate = 82 g/mole
0.018 mol/L x 0.1 L = 0.0018 moles
0.0018 moles x 82 g/mol = 0.1476 grams
