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Finding acid dissociation rate constant from titration

  1. Jul 28, 2016 #1
    1. The problem statement, all variables and given/known data
    When a weak acid HA is titrated 75% of its endpoint, it has a pH of 5.25. Calculate its Ka.

    2. Relevant equations
    Ka = [H+ ][A- ]/[HA]
    [H+] = 10^-pH
    3. The attempt at a solution
    I found [H+ ] from the pH to be 10^-5.25 M. From the dissociation equation of HA, we know there is a one-to-one molar ratio between [H+ ] and [A- ], so [A- ] = 10^-5.25. We know that 75% of the original HA has been titrated with the reaction: HA(aq) + OH-(aq) <------> A-(aq) + H2O (l), so there is 3 times the amount of A- than HA (75%:25%).
    Ka = [10^-5.25][10^-5.25] / [10-^5.25 / 3] Ka = 1.69x10^-5

    Im just not sure if I'm doing this right
     
  2. jcsd
  3. Jul 29, 2016 #2

    epenguin

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    Homework Helper
    Gold Member

    It looks right to me You could have simply said that the ratio[A- ]/[HA] is 3, so Ka = 3×10-5.25.

    It would be better to put in more arithmetical steps – then if you make a mistake you still get nearly full credit.

    The phrase I have bolded is definitely not true, and contradicts the rest of what you say.. Also this is nothing to do with rate constants.
     
    Last edited: Jul 30, 2016
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