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A diprotic acid solution H2A has a molarity of .95 M. the concentrations of the species present at equilibrium are as follows: [H+] = .25 M, [HA-]=.25M, [A^2-] =4.6x10^-4 M. what are the first and second ionization constants for this acid?

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    H2A ==> H^+ + HA^-
    HA^- ==> H^+ + A^-2

    k1 = (H^+)(HA^-)/(H2A)
    (H^+) = 0.25 from the problem.
    (HA^-) = 0.25 from the problem.
    (H2A) = 0.95-.25 = ??
    Substitute and solve for k1.

    k2 = (H^+)(A^=)/(HA^-).
    (H^+) = 0.25 from the problem.
    (HA^-) = 0.25 from the problem.
    (A^=) = 4.6E-4. Note (H^+) and (HA^-) cancel (although you CAN put those numbers in and crunch them on the calculator) but the bottom line is k2 = (A^=) = 4.6E-4.

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