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Buffer solutions

  1. Nov 3, 2009 #1
    1. The problem statement, all variables and given/known data

    If you began with 8.5 mL of 0.35 M NaHCO3, how many mL of 0.10 M HCl would you need to add to get a pH 7 buffer?

    2. Relevant equations

    pH = pKa + log([base]/[acid]) or pH = pKa + log(mmol base/mmol acid)

    3. The attempt at a solution

    I have tried making a reaction table, but I get stuck at the beginning since I don't know how much HCl ( or H+ since the Cl^- doesn't react with water) I am starting with. By the way, the equation I have written is: HCO3^- + H^+ ---> H2CO3 (don't know if this is correct). I have also tried using the Henderson-Hasselbalch equation for buffers but that didn't give me the correct answer. Are both these methods the incorrect approach or did I maybe make a calculation error. I know I haven't shown any work, so I am just looking for some direction on how to start. Any help is much appreciated.
     
  2. jcsd
  3. Nov 3, 2009 #2

    symbolipoint

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    Interesting problem, and somewhat typical:
    Can you construct a reasonably analytical equation for the Kb quantity? Sodium bicarbonate is alkaline and will therefore give an alkaline solution if alone in water.
     
  4. Nov 3, 2009 #3

    Borek

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    At the beginning there is no H+ and you can assume concentration of HCO3- to be just 0.35M. Then assume protonation goes to completion and amount of H2CO3 equals amount of added HCl.

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    methods
     
  5. Nov 3, 2009 #4
    So is that saying that if I set up a Kb expression, I should get:

    Kb= [H2CO3][OH^-]/[HCO3^-]= 2.33e-8
     
    Last edited: Nov 3, 2009
  6. Nov 3, 2009 #5

    symbolipoint

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    Yes, that was what I suggested. Now, could you find some refinements to that expression? The bracketted quantities are MOLAR concentrations. You may want to view the situation using hydroxide concentration as being important and hydronium less important.
     
  7. Nov 4, 2009 #6

    Borek

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    Why are you approaching it from Kb, instead of using just a Henderson-Hasselbalch equation with pH/pKa pair? That'll be much easier and much faster. Find [HCO3-]/[H2CO3] ratio and calculate amount of HCl assuming - as I told you earlier - that it equals amount of [H2CO3].

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  8. Nov 4, 2009 #7

    symbolipoint

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    A couple of my posts here are based on confusion with polyfunctional acids and bases. I spent a few minutes restudying some of this today from an old analytical textbook, and I'm still somewhat confused. I'm clear enough with monoprotic ones, but still not well clear with polyprotic/functional ones.

    Main reason I chose Kb as an approach is that bicarbonate alone in water is alkaline.
     
  9. Nov 4, 2009 #8

    Borek

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    If the difference between pKas is around three or more, you can safely ignore the other dissociation steps in calculations. Why? As you know Henderson-Hasselbalch equation is just a modified version of the acid dissociation constant definition. The same definition can be rearranged to this form:

    [tex]\frac {[HA]} {[A^-]} = 10^{pK_a-pH}[/tex]

    pretty useful when you want to check what is ratio of forms. Obviously, if pH is 3 units from pKa, one of the forms has concentration 1000 times larger, which means the other one can be ignored. Buffer pH is usually close to pKa, hence 3 units idea. Even if you are 'only' 2 units from pKa, that still means 100 excess of one form.

    Kb can be used:

    [tex]pOH = pK_b + \log \frac {[B^+]} {[BOH]}[/tex]

    (that's again just version of Henderson-Hasselbalch equation), but it is not necessary.

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  10. Nov 4, 2009 #9

    symbolipoint

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    Wow! One way to handle this is to calculate theoretical points for a titration curve; lengthy, but seems easier than trying to develop an algebraic equation in order to find how much volume of the acid to add to give the desired pH.
     
  11. Nov 5, 2009 #10

    Borek

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    This is somewhat unexpected, but it is possible to derive the equation that will give amount of titrant that has to be added if you want solution of a given pH - this is plug 'n chug equation, complicated, but relatively simple in use. At the same time it is not possible to derive similar equation that will allow calculation other way around - find pH for a given volume. That is, equation will take form of some relatively high degree polynomial, and value we are looking for will be a root - not an easy thing to find if we have 4th or higher degree polynomial.

    See titration curve calculation at my site.

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