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Double Dissaciation Problem

  1. Mar 12, 2005 #1
    Hello again,

    Yes, it's me with another disassociation problem.

    If you have .0213M of ascorbic acid, what is the pH of a solution containing the second disassociated form?


    HC6H7O6 <--> H + C6H7O6

    8.0e-5 = x^2 / (.0213 - x)
    x = 1.265e-3M

    C6H7O6 <---> H + C6H6O6
    .02M - x (1.265-3 + x) (1.265-3 + x)


    1.6e-12 = (1.265e-3+x)^2/(.02-x)

    Now I apparently set up the second part wrong as I am getting imaginary molarities and pHs. My chemistry teacher said I was not supose to square the rpoducts like that.

    I do not understand why
  2. jcsd
  3. Mar 12, 2005 #2


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    Science Advisor
    Homework Helper

    The initial concentration of the conjugate anion is 0. Also the initial concentration of the acid for the second Ka should be the same as the hydronium concentration.
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