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SO2 in NaOH

  1. Nov 15, 2006 #1

    I'm designing a scrubber and need some help.

    I know there are two reactions involved:

    SO2 + 2 NaOH => Na2SO3 + H2O

    SO2 + NaOH => NaHSO3

    There is a distribution between the two reactions depending on pH.

    How can I be absolutely sure that one reaction will irreversibly occur and the other one doesn't? At which pH should I look to find that all gas has reacted into NaOH?

    The problem is: the first reaction makes the pH descend very quick and the second reaction less quick. So I can't calculate the pH after 100 % absorption of a certain amount of SO2.

    Basically I just want to know how the pH changes after absorbing all the SO2.

  2. jcsd
  3. Nov 15, 2006 #2
    pH will relate to the concentration of H+ ions in solution... you'll have to then probably use things like the concentration of the reactants, solubility constants of the reactants and products, etc. Can you find the pH for just the first step?
  4. Nov 15, 2006 #3


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    I wouldn't write my fundamental reactions that way. Your first reaction is a trimolecular reaction, and as a result, has a very low probability of occurance. The more likely reaction, is bisulfite reaction with sodium hydroxide, making sodium sulfite (bimolecular kinetics).

    So then...

    SO2 + NaOH <=> NaHSO3 --(1)
    NaHSO3 + NaOH <=> Na2SO3 + H2O --(2)

    By inspection, you notice that you are going to have a buffer at each of two pH values (one where [NaHSO3] = [SO2](aq) and the second where [Na2SO3] = [NaHSO3])

    Not sure what exactly that means. But yes, the two reactions have different rate constants and equilibrium constants.

  5. Nov 16, 2006 #4
    Hmm, so this means one molecule of SO2 willl react with 3 molecules of NaOH if the pH is high enough?

    So: if I want to absorb 0,2 mol/s SO2 from the gas (we have a 47 kg/h gas stream at our plant), I need 3 times as much OH- = 0,6 mol/s OH- ions. With an inlet NaOH-liquid pH of 10 = 0,0001 mol/liter OH- ions.

    Then we need to have at least:

    0,6 / 0,0001 = 6000 liter/s of liquid.

    This is 6 m^3/s liquid. When I think about that amount of liquid, it's impossible, so there is something wrong with my calculation.

    Otherwise when you have a pH of 11 you only need 0,6 m^3/s of liquid. This is more acceptable.

    So let's see what a 20 % NaOH has as pH. That is 20 g in 100 g of water.
    Or 0,5 mol NaOH in 0,1 liter. 5 mol/liter OH- ions!

    This means a pH of 14,6? I didn't know that it would be that concentrated?

    So technically I could maintain a pH of 11 by recycling the bottom liquid stream to the feed of pH 14,6.

    What do you think, is this method right?

    PS: there is also a HCl stream in our plant that is mixed into the SO2 stream:

    54 kg/h HCl or 0,4 mol/s HCl (and 47 kg/h SO2).
    To remove all the HCl I need 0,4 mol/s OH- ions. At pH 11 this means an additional 400 l/s liquid which is acceptable. Maybe we can operate at pH 12 with less liquid flow.
    Last edited: Nov 16, 2006
  6. Nov 16, 2006 #5


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    No. Your overall reaction equation (SO2 + 2NaOH -> Na2SO3 + H2O) was stoichiometrically correct, it was just wrong in terms of mechanism involved. Whole process is probably much better approximated with two stepwise reactions, as Gokul wrote.

    Last edited by a moderator: Aug 13, 2013
  7. Nov 17, 2006 #6
    Oh yeah, I read his post wrong... but the calculation method is acceptable right?
  8. Nov 19, 2006 #7
    Does anyone know the pH after which the reaction of SO2 in NaOH is almost 100%.
  9. Nov 21, 2006 #8


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    pH of Na2SO3 is that of a weak base conjugated with HSO3-

    pKa2 = 6.91
  10. Nov 21, 2006 #9

    What is pKa1? I want to see if the first reaction occurs.
    Last edited: Nov 22, 2006
  11. Nov 22, 2006 #10


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    Check my pH calculator. Sulfurous acid is present only in the registered version - but in the free trial you may enter both pKa values to check pH of solutions in question.

    Last edited by a moderator: Aug 13, 2013
  12. Nov 23, 2006 #11
    How do I use that program when I react gas with a liquid solution?

    Like 50-50 HCl/SO2 and NaOH-solution?

    Do I just use the mol/l of gas and the mol/l of NaOH-solution?

    Does this program work when there are multiple components?
  13. Nov 23, 2006 #12


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    mol/L of gas is in this case the same as mol/L of acid. One acid & one base only in the mixture.

    Last edited by a moderator: Aug 13, 2013
  14. Nov 23, 2006 #13
    => So I can't use the pH calculator for a HCl, SO2, NaOH scrubber! So the only way to make a good calculation of pH is to make sure the reactions are irreversibel, like at high pH. That way I can "count" the molecules of OH- without using Ka values.

    Another important question:

    At pH 12, from the graph you can see everything is SO3--. Does this mean that the reaction:

    NaOH + SO2 = NaHSO3 occurs


    NaHSO3 + NaOH = H2O + Na2SO3 occurs?

    Because there isn't any of HSO3- left...

    So if I design my NaOH-scrubber to work above pH 12 I can simplify that 2 OH- molecules react with 1 SO2 molecule.

    Attached Files:

    • so2.pdf
      File size:
      61.1 KB
    Last edited: Nov 23, 2006
  15. Nov 26, 2006 #14


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    You have not mentioned HCl earlier.

    pKa2 for H2SO3 is close to 7 - at pH 7 ratio SO3-2/HSO3- is about 1:1, at pH 8 it is about 10:1, at pH 9 it is about 100:1 (check Henderson-Hasselbalch equation) - so you may assume that at pH 9 there is almost no HSO3-.

    Last edited by a moderator: Aug 13, 2013
  16. Nov 27, 2006 #15
    Okay, so would you think making a scrubber at pH 10 would be equivalent in efficiency as pH 12? For HCl and SO2?

    I mean, reactions will be faster at pH 12 but maybe at pH 10 they are equally fast. How can I estimate that?

    Do you know if there exist equilibrium curves for different pH for a system of SO2-NaOH?
    Last edited: Nov 28, 2006
  17. Nov 27, 2006 #16


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    No idea about speed. Note that starting pH defines scrubber capacity - the higher the pH, the more gases it will be able to absorb.

    Last edited by a moderator: Aug 13, 2013
  18. Dec 10, 2006 #17
  19. Dec 12, 2006 #18


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    http://www.s-a-s.org/journal/96/asv50n1/ASv50n1_sp10.html [Broken]

    "These results indicate that the reversibility of the key reactions become important when the SO2 concentration is low."
    Last edited by a moderator: May 2, 2017
  20. Dec 14, 2006 #19
    Well, that abstract doesn't say at which pH they work so... everything is possible. Of course, low SO2 will give only the first reaction, but when you add more and more OH, le Chatelier says it will make the Na2SO3 reaction more attractive.
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