April 28, 2024, 09:43:15 PM
Forum Rules: Read This Before Posting


Topic: half-neutralisation and pH  (Read 6093 times)

0 Members and 1 Guest are viewing this topic.

Ice-cream

  • Guest
half-neutralisation and pH
« on: May 28, 2005, 09:00:40 AM »
1. When you add 0.2M NaOH solution to 40ml of 0.2M solution of a weak acid, where on the titration curve would pH=pKa?

(What I know at the moment is that for a weak acid, pH will = pKa when half the volume of base need to reach equivalent point is added - half-neutralisation, so my answer would be at 20ml but does any1 actually agree with me? Do the concentrations affect anything??)

Also can some1 explain to me why pH will = pKa in that situation? I just know for a fact that that's the case but I don't actually know WHY that's so.

2. 40ml of 0.2M nitric acid solution is added to 50ml 0.1M sodium hydroxide solution. What is the pH of the solution?

(What I think is that since both at strong acids and bases, they will dissociate completely so since there are 0.008mols HNO3 and 0.005mols NaOH, the HNO3 would be in excess and converting 0.003 mols of HNO3 to concentration, there would be 0.03333M HNO3. therefore pH is 1.48. Can any1 tell me if it's correct? coz I really don't want to do any other questions similar unless I know I'm doing the correct thing.)

Thanks guys!


Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27665
  • Mole Snacks: +1801/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re:half-neutralisation and pH
« Reply #1 on: May 28, 2005, 10:42:27 AM »
What I know at the moment is that for a weak acid, pH will = pKa when half the volume of base need to reach equivalent point is added - half-neutralisation, so my answer would be at 20ml but does any1 actually agree with me? Do the concentrations affect anything??

You are right. As for concentration - as long as the solution is reasonably concentrated, it holds. Once it is diluted you must take autoionization of water into account.

Quote
Also can some1 explain to me why pH will = pKa in that situation? I just know for a fact that that's the case but I don't actually know WHY that's so.

Search for Henderson-Hasselbalch (buffer) equation.

Quote
What I think is that since both at strong acids and bases, they will dissociate completely so since there are 0.008mols HNO3 and 0.005mols NaOH, the HNO3 would be in excess and converting 0.003 mols of HNO3 to concentration, there would be 0.03333M HNO3. therefore pH is 1.48. Can any1 tell me if it's correct? coz I really don't want to do any other questions similar unless I know I'm doing the correct thing.

Haven't check the numbers, buy in general you are correct. If strong acid and base are mixed identify limiting reagent, what is left defines pH.
« Last Edit: May 28, 2005, 10:44:28 AM by Borek »
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Sponsored Links