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Homework Help: Please help! Calculate the % hydrolysis for 1 M Na2CO3 if [OH-] = 5.4x10^-(4).

  1. Mar 18, 2010 #1
    Calculate the % hydrolysis for 1 M Na2CO3 if [OH-] = 5.4x10^-(4)
     
  2. jcsd
  3. Mar 18, 2010 #2

    Borek

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    Done.

    Ignoring template and not showing any effort will lead you nowhere.
     
  4. Mar 18, 2010 #3
    Huh? But I asked my question. This is the only information I was given.

    No effort? But I'm stuck! I'm showing some effort by asking for some help here on ONE problem!
     
  5. Mar 18, 2010 #4
    1. The problem statement, all variables and given/known data

    Calculate the % hydrolysis for 1 M Na2CO3 if [OH-] = 5.4x10^-(4).

    Not sure what to put for the other fields.
     
  6. Mar 18, 2010 #5

    Borek

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    Do you know any reaction equations? What is hydrolysis?
     
    Last edited by a moderator: Aug 13, 2013
  7. Mar 18, 2010 #6
    Calculate the theoretical % hydrolysis for 1 M solution of Na2CO3

    1. The problem statement, all variables and given/known data
    Calculate the theoretical % hydrolysis for 1 M solution of Na2CO3


    2. Relevant equations
    CO3^(2-) + 2H2O -> H2CO3 + 2OH^(-1)

    or

    CO3^(2-) + H2O -> HCO3^(-1) + OH^(-1)
    HCO3^(-1) + H2O -> H2CO3 + OH^(-1)

    Chart values for the Ka1 and Ka2 of H2CO3:
    Ka1 = 4.3x10^(-7)
    Ka2 = 5.6x10^(-11)

    3. The attempt at a solution

    Kb1 = {[HCO3^(-1)][OH^(-1)]}/[CO3^(-2)] = Kw/Ka2 = [10^(-14)]/[5.6x10^(-11) = 1.78x10^(-4)

    Kb2 = {[H2CO3][OH^(-1)]}/[HCO3^(-1)] = Kw/Ka1 = [10^(-14)]/[4.3x10^(-7)] = 2.32x10^(-8)
     
  8. Mar 18, 2010 #7

    Redbelly98

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    Moderator's note: merged two threads.
     
  9. Mar 19, 2010 #8

    Borek

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    Re: Calculate the theoretical % hydrolysis for 1 M solution of Na2CO3

    You know equilibrium concentration of OH- - can you use it to calculate concentration of HCO3-?

    How is hydrolysis percentage defined?
     
    Last edited by a moderator: Aug 13, 2013
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