1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted