Question
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
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.
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.
Its oops, not the way you spelled it bob. Jeez do the teachers know about this
oop-
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.
Your weird be quiet. that was a few days ago. you just don't know how to keep your mouth shut huh?
*you're
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