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Homework Help: Find pH of 0.1 MNaHCO3

  1. Feb 13, 2009 #1
    1. The problem statement, all variables and given/known data

    Find pH of [tex]0.1M NaHCO_3[/tex]

    [tex] Ka_1 (H_2CO_3) = 4.3 * 10^{ - 7} , Ka_2 (H_2CO_3) = 5.61 * 10^{ - 11}[/tex]

    2. Relevant equations

    [tex]pH = \frac {1}{2}(14 + pKa_1 + logC)[/tex]


    3. The attempt at a solution

    Now [tex]K_h = \frac {Kw}{Ka_1} = 2.3 * 10^{ - 8} > > Ka2[/tex]

    So I neglect dissociation and use salt hydrolysis formula.

    But this gives wrong answer.
     
  2. jcsd
  3. Feb 14, 2009 #2

    Borek

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    Staff: Mentor

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