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3 January, 16:04

For the diprotic weak acid h2a, ka1 = 4.0 * 10-6 and ka2 = 7.2 * 10-9. what is the ph of a 0.0450 m solution of h2a? what are the equilibrium concentrations of h2a and a2 - in this solution?

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  1. 3 January, 16:09
    0
    For the first reaction:

    the ICE table:

    H2A ↔ HA - + H+

    initial 0.045 m 0 0

    change - X + X + X

    Equ (0.045-X) X X

    when Ka1 = [HA-][H+]/[H2A]

    4x10^-6 = X^2 / (0.045-X) by solving for X

    ∴X = 4.2x10^-4

    ∴[ H2a] = 0.045 - (4.2x10^-4) = 0.045 m

    [HA-] = [H+] = 4.2x10^-4

    when PH = - ㏒[H+]

    by substitution:

    PH = - ㏒ (4.2x10^-4)

    = 3.377

    For the second reaction:

    by using ICE table:

    HA - ↔ A^2 - + H+

    when

    Ka2 = [A^2-] [H+] / [HA-]

    by substitution

    7.2 x 10^-9 = [A^2-] * (4.2x10^-4) / (4.2x10^-4)

    ∴[A^2-] = (7.2x10^-9) * (4.2x10^-4) / (4.2x10^-4)

    ∴ [A^2-] = 7.2x10^-9 m
  2. 3 January, 16:26
    0
    That's right how you did it
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