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Homework Help: Buffer Question - With addition of OH-

  1. Feb 4, 2010 #1
    1. The problem statement, all variables and given/known data
    You add 100 mL of 1 M H3A
    and 4 mL of 5 M NaOH all in 896mL water
    Ka=6.309x10^-10

    What is pH of solution made?
    H3A+H2O---->H2A-+H3O+


    2. Relevant equations

    ka=[h2a][H+]/[H3A]

    3. The attempt at a solution
    Concentrations initially:
    [H3A]=100mL * 1 M / (896+4+100) mL = 0.1 M H3A
    [NaOH]=4mL *5 M / (896+4+100) mL = .02 M NaOH

    Then, using ice table:
    H3A + H2O -----> H2A + H3O+
    I .1 M 0 0
    C -x +x +x
    E .1-x x x

    ka=[h2a][H+]/[H3A]
    6.3x10^-10=[x^2]/[.1-x] assume -x is negligable.
    so solving for x i get 7.94x10-6 M H30+

    however,
    then looking at my initial amount of NaOH:
    OH- + H+ --> H2O
    init .02 7.94e-6
    chang -7.94e-6 -7.94e-6
    fin .02 0
    OH = is around .02 M still?
    So basically ive gone no where and I do not understand where im going wrong in this problem...
     
  2. jcsd
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