1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dicarboxylic acid titration

  1. Aug 23, 2014 #1

    nmr

    User Avatar

    Malonic acid can ionise in two stages as it is a dibasic acid.
    The values of pKa1 and pKa2 of malonic acid are 2.85 and 5.70 respectively. Calculate the pH of the first and second equivalence points of the titration of 10cm^3 of malonic acid of concentration 0.1 M with sodium hydroxide of concentration 0.1 M.

    my attempt at a solution:
    maximum buffering capacity before first endpoint: pH=pKa1 = 2.85

    [OH-] = [(Kw/Ka1)*(0.05)]^(0.5)
    therefore pOH = 6.23
    pH = 14-6.23 = 7.77

    im stuck here as the answer given for the pH of the first endpoint is 3.50 and I have no idea how to work it out
     
  2. jcsd
  3. Aug 23, 2014 #2

    Borek

    User Avatar

    Staff: Mentor

    First endpoint pH given as an answer is wrong, so don't worry if you can't reproduce it.

    At first endpoint you have a solution of an amphiprotic salt.
     
  4. Aug 23, 2014 #3

    nmr

    User Avatar

    so is my pH value correct ?
     
  5. Aug 23, 2014 #4

    Borek

    User Avatar

    Staff: Mentor

    Nope.
     
  6. Aug 23, 2014 #5

    nmr

    User Avatar

    how would i go about solving for the pH then?
     
  7. Aug 23, 2014 #6

    Borek

    User Avatar

    Staff: Mentor

    I have pointed you in the right direction in my first answer.

    If you plan to ignore hints please tell, so that I don't waste my time answering.
     
  8. Aug 23, 2014 #7

    nmr

    User Avatar

    ah ok sorry
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Dicarboxylic acid titration
  1. Amino Acid Titration (Replies: 3)

  2. Amino Acid Titration (Replies: 7)

Loading...