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Quantity of buffer solution needed to resist pH change

  1. Jun 17, 2012 #1
    Hello there,

    I've got a question here which asks me to calculate how much (in mol/L) of buffer solution to add to 1 litre of solution to resist the change away from a pH of 7.0.

    a) if there was an increase of 3.99367 x 10-8 of [H+] ions

    Because the question is talking about pH in a swimming pool, I'm going to assmume that the buffer solution is chlorine.

    I have no idea how to approach this?

    any assitance will be appreciated :smile:
     
  2. jcsd
  3. Jun 17, 2012 #2

    Borek

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    Staff: Mentor

    The way it is worded now it doesn't make sense and can't be solved. First of all, answer depends on the buffer identity (more precisely - on the pKa of the acid used), and I don't like the idea of chlorine being called "a buffer solution". Then, when you add acid to the buffered solution pH goes down - always. Magnitude of the change depends on the amount of acid and on the buffer capacity, so as long as you are not told by how much pH can change, you can't give any answer.
     
  4. Jun 17, 2012 #3
    No :rofl:

    Well, I just assumed that Chlorine could react with the extra hyrogen ions in solution, by forming hydrochloric acid, thus, lowering the concentration of [H+] ions?
     
  5. Jun 17, 2012 #4

    Borek

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    If anything, Cl2 reacts with water undergoing disproportionation and producing two acids, so it lowers the pH.
     
  6. Jun 18, 2012 #5
    Heh... I didn't actually think about that.

    How about something like Sodium Carbonate? Na2CO3

    The sodiums will ionise in solution and whats left should be Carbonate, which will form Carbonic Acid? will that increase the pH?
     
  7. Jun 18, 2012 #6

    Borek

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    Now you are just throwing compounds around. Yes, solution of sodium carbonate has pH above 7. Does it change the fact question as stated doesn't make sense?
     
  8. Jun 18, 2012 #7
    Well I figured that the disociation of carbonic acid was what put in hydrogen ions in solution to begin with (in an earlier part of the same question), I just needed to do the reverse to push the pH back. Plus, Soda Ash is something used in Pools...

    What I'm trying to figure out is: How much Sodium Carbonate do I need to neutralize 3.99367 x 10-8 of H+ ions

    I know that Carbonic acid is di-protic
    and the pka values are
    pKa1 = 6.367
    pKa2 = 10.329

    Although I'm not sure what units they'd be in, and I'm pretty sure this is for the forward reaction
     
  9. Jun 18, 2012 #8

    Borek

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    Technically you can answer that they react 2:1 so you need 2x10-8M of carbonate. However, it doesn't work this way in the real solution, and the problem is still so vague there is no way to make it solvable.
     
  10. Jun 18, 2012 #9
    why is it 2:1 ?
     
  11. Jun 18, 2012 #10

    Borek

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    Staff: Mentor

    Write reaction equation, the simplest and the most obvious one.
     
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