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Chemistry

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The Ka of a monoprotic weak acid is 2.04 × 10-3. What is the percent ionization of a 0.181 M solution of this acid?

  • Chemistry - ,

    ............HA ==> H^+ + A^-
    initial..0.181.....0......0
    change......-x.....x......x
    equil.....0.181-x..x.......x

    Ka = (H^+)(A^-)/(HA)
    Substitute ICE chart into Ka expressiion and solve for H^+; then
    %ion = [(H^+)/(0.181)]*100 = ?

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