posted by Lily

Propanoic acid has a Ka of 1.3 x 10^-5. What is the % ionization in a 3.0M solution?

  1. DrBob222

    Let's call propanoic acid HPr. Then
    ...........HPr ==> H^+ + Pr^-

    Ka = (H^+)(Pr^-)/(HPr)
    Substitute from the ICE chart into the Ka expression and solve for x = (H^+).
    Then %ionization = [(H^+)/3.0)]*100 = ?

  2. Lily

    thank you so much!

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