Chemistry 104

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What is the percent ionization of a solution prepared by dissolving 0.0286 mol of chloroacetic acid in 1.60 L of water? For chloroacetic acid ,Ka=1.4x10^-3

  • Chemistry 104 -

    Chloroacetic acid's formula is too long to write. Let's just call it HAcl.
    (HAcl) = 0.0286mol/1.60L = about 0.018 but you do it more accurately.
    ........HAcl ==> H^+ + Acl^-
    I.......0.018.....0......0
    C........-x.......x.......x
    E......0.018-x....x......x

    Ka = (H^+)(Acl^-)/(HAcl)
    Substitute Ka and the E line into Ka expression and solve for x = (H^+). Then
    %ionization = (H^+)/(0.018)
    Don't forget to redo the 0.018 number.

  • Chemistry 104 -

    What you mean redo 0.018?

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