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What is the percent ionization of a 1.38 mol/L weak acid if its Ka = 2.7 x 10^-6?

Thanks again!

  • chemistry -

    Write the ionization equation. I would call the weak acid HA.
    Write the Ka expression.
    Set up an ICE chart if that will help.
    Let y = (H^+), then (A^-) will be y and the unionized HA will be 1.38 - y.
    Substitute these variables into the Ka expression and solve for y = (H^+).
    Then percent ionization =
    [(H^+)/(HA)]*100 = ??
    Post your work if you need further help.

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