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chemistry

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A 0.010 M solution of a weak monoprotic acid is 3.0% dissociated. What is the equilibrium constant, Ka, for this acid?

  • chemistry -

    HA = H^+ + A^-

    Ka = (H^+)(A^-)/(HA)

    initial:
    HA = 0.01
    H^+ = 0
    A^- = 0

    change:
    H^+ = x
    A^- = x
    HA = 0.01 - x

    equilibrium:
    x = 0.01 x 0.03 = ??
    Substitute and solve for Ka.

  • chemistry -

    3.0 * 10^-5

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