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A weak acid, HA, has an ionization constant of 4.34 x 10-8. If we prepare a solution that is 0.273 M in HA what percent of the acid will be ionized

  • chem -

    ...........HA --> H^+ + A^-
    I.......0.273.....0......0
    C.........-x......x......x
    E.......0.273-x...x......x

    Ka = (H^+)(A^-)/(HA)
    Substitute the E line and solve for x = (H^+). Then %ion = [(H^+)/0.273]*100 = ?

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