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March 29, 2017

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a) Calculate the percent ionization of 0.125M lactic acid (Ka=1.4x10^-4) b) Calculate the percent ionization of 0.125M lactic acid in a solution containing 0.00075 M sodium lactate.

  • chemistry - ,

    Lactic acid = HL, sodium lactate = NaL.

    ..........HL ==> H^+ + L^-
    I...... 0.125....0.....0
    C.........-x.....x.....x
    E......0.125-x...x.....x

    Substitute the E line into Ka expression and solve for x = (H^+), the
    % ion = [((H^+)/(HL)]*100 = ?

    for b part, plug into Ka expression the following concns:
    Remember NaL is 100% ionized in solution.
    .........NaL ==> Na^+ + L^-
    I......7.5E-4.....0......0
    C.....-7.5E-4....7.5E-4..7.5E-4
    so final concns are
    (H^+) = x
    (L^-) = x from HL + 7.5E-4 from NaL to make x + 7.5E-4
    (HL) = 0.125-x

    These are plugged into the Ka expression and solve for x = (H^+), then
    %ion = [(H^+)/(HL)]*100 = ?

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