June 26, 2024, 03:24:49 AM
Forum Rules: Read This Before Posting


Topic: pH of a solution of an acid  (Read 2319 times)

0 Members and 1 Guest are viewing this topic.

Offline Cassowary

  • New Member
  • **
  • Posts: 5
  • Mole Snacks: +0/-0
pH of a solution of an acid
« on: October 16, 2011, 04:31:40 PM »
I've been trying to answer a problem:


The pH of an aqueous solution obtained by dissolving 0.49 g of o-nitrobenzoic acid in 707 mL of pure H2O is ____."

The answer is supposed to be 2.54, but I can't seem to get that answer. The values I have for the molecular weight of o-nitrobenzoic is 167.12 g/mol, the pKa as 2.19. I've been using the Henderson–Hasselbalch equation.

Can anyone tell me what I'm doing wrong? Or is the supposed answer wrong?

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27712
  • Mole Snacks: +1804/-411
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: pH of a solution of an acid
« Reply #1 on: October 16, 2011, 05:49:38 PM »
HH equation doesn't make much sense for this question. There is no buffer here, just a solution of the weak acid.

2.54 doesn't look correct to me. I got about 0.3 less.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline Cassowary

  • New Member
  • **
  • Posts: 5
  • Mole Snacks: +0/-0
Re: pH of a solution of an acid
« Reply #2 on: October 16, 2011, 06:29:43 PM »
Blah, didn't mean HH, sorry, I meant whatever this process is: http://en.wikipedia.org/wiki/Weak_acid

I end up getting 1.21...

pKa = 2.19 -> Ka = 0.006456

x is [H+]

Ka = (x^2)/(0.004147-x) ->

x^2 + (Ka)(x) - (Ka)(0.004147) -> quadratic

And I end up with 1.21.

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27712
  • Mole Snacks: +1804/-411
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: pH of a solution of an acid
« Reply #3 on: October 17, 2011, 02:47:50 AM »
Ka = (x^2)/(0.004147-x) ->

x^2 + (Ka)(x) - (Ka)(0.004147) -> quadratic

And I end up with 1.21.

Check your math.

Sorry, I was wrong yesterday, 2.54 seems to be the correct answer.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Sponsored Links