chemistry

A student prepares a 0.20 mol/L aqueous solution of ascorbic acid and meausres its pH as 2.40. Based on theis evidence, what is the Ka of ascorbic acid?

H+ = 0.00398
C- is the same
(HC) = 0.2 = 0.00398

(0.00398)(0.00398) / 0.0199

=0.000796

Hi DrBob222 you told me you got 8.0 x 10^-5 how did you get this as an answer?

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asked by Dustin
  1. 0.00398*0.00398/(0.2-0.00398) =
    (0.00398)(0.00398)/0.196 = 8.08 x 10^-5

    Note the 0.0199 in the denominator should have been 0.196. I try to proof these things but sometimes I miss them like the denominator above. Check my 8.08 x 10^-5

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    posted by DrBob222
  2. I am still getting the wrong answer. YOu are positive this is the answer? Perhaps my computer system is wrong?

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    posted by Ross
  3. I'm usually never sure of anything. I found information on the internet and in The Merck Index on ascorbic acid. It has two pka values. pK1 is 4.17 (k1=6.7 x 10^-5) and pK2 is 11.6 (k2=2.5 x 10^-12). You aren't titrating this acid, you are making a 0.2 M solution, placing that in water, and measuring the pH. That means ONLY the stronger acid will be measured because the weaker acid is about 30,000,000 times weaker than the stronger one and it will have neglibible effect on the total hydrogen ion. So I forget the weaker acid, and we treat it as a monorotic acid.
    HC ==> H^+ + C^-
    pH = 2.40 makes (H^+) = 0.00398
    If (H^+) = 0.00398, then anion is 0.00398, and the unionized ascrobic acid is 0.2 - 0.00398 = 0.196
    Ka = (H^+)(C^-)/(HC) =
    (0.00398)(0.00398)/(0.196) = 8.08 x 10^-5 for pK1 value of 4.09 (which compares favorably with the 4.17 listed in the books.)I don't see any error I've made which is why I went back over it in detail. Of course you know than some errors just keep popping up when we check these things. I'd be interested in knowing how to solve the problem if this is wrong. One extra note. In The Merck Index, they list 5 mg/mL and 50 mg/mL as being pH = 3 and 2 respectively. That is about 0.28 M which is about what your problem has.

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    posted by DrBob222

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