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what is the Ka for a 0.1994M propionic acid solution if pH is 2.795?

CH3CH2CO2H(aq)+H2O(l)<->CH3CH2CO2-(aq)+H3O(aq)

someone please help.. I have a final in two days.

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manda

  1. Call propionic acid HPr.

    HPr ==> H^+ + Pr^-

    Ka = (H^+)(Pr^-)/(HPr)

    initial:
    HPr = 0.1994 M
    H^+ 0
    Pr^- = 0

    final:
    pH = 2.795. Convert to (H^+) by
    pH = -log(H^+) = ??
    Pr^- = same as H^+.
    HPr = 0.1994-(H^+)

    Substitute into Ka expression and solve for Ka.

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    DrBob222

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