Asked by Anonymous
A buffer is prepared by dissolving 0.0250 mol sodium nitrite in 250 mL of 0.0410 M
nitrous acid. Assuming no volume change from dissolving, what is the pH of this buffer?
nitrous acid. Assuming no volume change from dissolving, what is the pH of this buffer?
Answers
Answered by
DrBob222
Use the Henderson-Hasselbalch equation.
pH = pKa + log [(base)/(acid)
(acid) = 0.410 M nitrous acid
(base) = 0.0250 mols NaNO3/0.250 L = ?
pKa HNO2 = -log Ka HNO2
Post your work if you get stuck.
pH = pKa + log [(base)/(acid)
(acid) = 0.410 M nitrous acid
(base) = 0.0250 mols NaNO3/0.250 L = ?
pKa HNO2 = -log Ka HNO2
Post your work if you get stuck.
Answered by
DrBob222
00ps. typo. That should be NaNO2 instead of NaNO3.
(acid) = 0.0410 M HNO2 and not 0.410.
(acid) = 0.0410 M HNO2 and not 0.410.
Answered by
connexus helper
oop-
Answered by
DrBob222
Jeez. I said it was a typo. What more do you want? And for your Jeez that's spelled cheese. Do your friends know about this. We could start a club.
Answered by
Anonymous
Your weird be quiet. that was a few days ago. you just don't know how to keep your mouth shut huh?
Answered by
DrBob#1fan
*you're
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.