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PH=pKa+log([A-]/[HA]) add lots of water and the pH still remains constant?

  1. Aug 28, 2011 #1
    Hey.. so we have this formula pH=pKa+log([A-]/[HA]).. If we add lots of water, the concentration of both A- and HA decrease proportionally so the pH doesn't change..

    But this is weird. If the volume increases by large amounts in an acidic solution, shouldn't at least HA decrease in concentration and A- increase to keep up the pH?
     
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  3. Aug 28, 2011 #2

    wukunlin

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    actually water contains both proton donors and receivers (I don't remember their actual names, far too long since I did my last chem paper)

    the more water you add the closer the concentration of proton donors and receivers will be closer to that of water's so adding more water will bring the pH closer to water's (ie, 7)
     
  4. Aug 28, 2011 #3
    yeah, it's called OH and H3O.

    This is what I'd expect but the ratio between [A-][HA] remains constant??
     
  5. Aug 28, 2011 #4

    Borek

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    No, the more diluted the solution, the closer the pH to 7, the closer the ratio to 10pH-pKa. See acid base titration indicators for explanation from where the equation comes from.

    Compare also Henderson-Hasselbalch equation page - it shows results of pH calculation for diluted buffer solutions.
     
  6. Aug 28, 2011 #5
    well this is weird, because in my textbook they simplify it and claim the pH doesn't change.. which is pretty annoying

    hmm how so? The ratio should remain constant since both the concentrations decrease at a steady rate. can you pls explain closer concerning this?




    While I realize that the [H3O+] concentration would decrease if the volume gets bigger while the amount of [HA] moles remains constant in the solution, I can't seem to find a math answer to this.

    As for your 1st link, it's stuff about titration-explanation about the indicator-acids reacting with water.. your second link merely shows that the pH will start closing to 7 when we add volume.. no mathematical explanation..
     
  7. Aug 28, 2011 #6

    Borek

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    So they should not write textbooks, as they don't know enough, or they overgeneralize in an unacceptable way. When you dilute long enough you end with almost pure water, so pH is close to 7.

    In reality more like slightly below 6, as in typical lab there is no such thing as pure water, but that's another problem.

    You are forgetting that both HA and A- can react with water.

    No, it will tend to get close to 7. If your buffer has pH 10, that means increasing [H3O+]

    Why do you think number of moles of HA is constant? It is not. Solution is in equilibrium, which doesn't mean constant [HA]/[A-] ratio, it means constant Ka and Kw. Not to mention fact hat solution obeys mass and charge balances all the time.

    Yes, and - as I wrote - it explains the problem. Indicator is a weak acid, equation derived there (actually just a rearranged Ka definition) shows how the ratio of acid (HInd is equivalent to HA) to conjugate base depends on the pH and pKa. It can be applied to any weak acid, not just indicators.

    For math you will have to go through 3rd or 4th degree polynomials. In the end you will find out that when you dilute the solution, pH tends to [itex]\sqrt {K_w}[/itex].
     
  8. Aug 28, 2011 #7
    allright, thank you 4 the help ;)

    I was talking about an acidic solution btw, as I wrote in one of the previous posts. But yeh, [HA] concentration should decrease the more the solution gets delluted (not remain constant, sorry my mistake), no? this is for an acidic solution ofc
     
  9. Aug 28, 2011 #8

    Borek

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    Well, the more you dilute, the lower the concentration, that's obvious.

    What is interesting here is the ratio of HA/A - and that will reach the value described by the equation mentioned earlier, no matter how much the solution is diluted. This is not necessarily an obvious result.
     
  10. Aug 29, 2011 #9
    ah yes I meant the ratio [A-]/[H3O+] will become bigger and bigger until a certain point the more the solution is dilliuted right?

    sorry it was really late yesterday
     
  11. Aug 29, 2011 #10

    Borek

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    No, this one will tend to zero, as [H3O+] in diluted solution is constant, while [A-] gets lower and lower.
     
  12. Aug 29, 2011 #11
    doesn't the acid release more H3O to compensate the lowering H3O concentration?

    Why'd [A-] get lower faster than H3O aka [HA]? HA has a bigger Ka than A- has Kb, since the solution is acidic?
     
  13. Aug 29, 2011 #12

    Borek

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    It can't release more H+ than there are HA present. Imagine starting with 1M HA solution and diluting it 1010 times. What is the final pH? Does it matter what was the initial HA concentration?

    H3O+ AKA HA? These are two separate things, see above. And you were already told how to calculate what the final [HA]/[A-] ratio will be, no matter what the initial concentration was.
     
  14. Aug 29, 2011 #13
    no, I didn't mean it would get deluded to 10^10

    Well I said *until a certain point* [A-] will decrease in concentration slower than [HA], because [HA] will constantly release H3O at a faster rate than A- will capture H+. HA will release H+ and A- will capture H+ due to the le chatelier principle

    btw I don't mean to sound annoying, so sorry if I do..
     
  15. Aug 29, 2011 #14

    Borek

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    I have a feeling it depends on pKa. For relatively strong acids it can be true, but the weaker the acid, the less likely it becomes. But it is just intuition, besides, we are so far from the original problem I have no idea what you are really thinking about.

    I am attaching image with concentrations of HA and A- for acetic acid, see if that fits your thinking.
     

    Attached Files:

  16. Aug 30, 2011 #15
    ok thx 4 the help
     
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