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Homework Help: Finding minimum concentrations in a buffer solution

  1. Mar 28, 2010 #1
    1. The problem statement, all variables and given/known data

    "Calculate the minimum concentrations of formic acid and sodium formate that are needed to prepare 500.0 mL of a pH 3.80 buffer whose pH will not change by more than 0.10 unit if 1.00 mL of 0.100 M strong acid or strong base is added."

    Ka of formic acid = 1.8E-4
    pKa of formic acid = 3.74

    2. Relevant equations

    Henderson-Hasselbalch equation: pH = pKa + log([base]/[acid])

    3. The attempt at a solution

    So...the initial concentration of H3O+ in the pH 3.80 solution would be 10^-3.80 = 1.58E-4 M. In a 500 mL solution, it is: 500 mL x 1.58E-4 mol / mL = 0.079 mmol H3O+. The thing is, I'm not sure how to apply this.

    Alternatively, knowing the pH and pKa of the acid, I can use the Henderson-Hasselbalch equation: 3.80 = 3.74 + log([base]/[acid]). The resulting ratio is 0.871. I am unsure how I would be able to use this to calculate the concentrations of acid and salt (base). If we are adding .100 M strong acid (or base), the resulting pH would be 3.70 (or 3.90), correct? Using 3.70 as the pH in the H-H equation, the resulting ratio is 0.912.

    I'm just really lost. Thanks for any help!
  2. jcsd
  3. Mar 30, 2010 #2


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    Be a bit precise to not lose marks and risk also confusing yourself. You mean to say here that the 500 ml solution contains 0.079 mmoles of H3O+ (which I will call H+). It is still 1.58E-4 M (that is its H+ concentration) or rather molarity. I personally would rather think in molarities, so think of adding 2ml to 1l and divide by 2 at the end. Also to avoid any thing else superfluous work out how to get to exactly pH 3.7.

    To be asking you a slightly tricky, though perfectly and typically practical question like this, they are testing for a bit of insight instead of blind application of a formula, to see if you can see you do have in the problem all the information needed to solve the question (and without making it more complicated than necessary).

    You have worked out the ratio [base]/[acid]. Which is equal to what?

    To [Na+]/[HCOO-] in your original pH3.8 solution.

    You know the ratio of these two things, but you do not know their absolute amounts because the latter can be anything (as long as they in that ratio and also are well above your proton concentration of the order of 10-4M).

    The only thing you know the absolute amounts of is the 2ml of strong acid or base to be added.

    Maybe you can start from your acid base ratio at pH 3.7 as you did, but I see clearer to just ask again what is [H+] , this time at pH 3.7?

    Those extra protons compared to those at pH 3.8 have to be balanced by something extra of opposite charge, which is actually the Cl- (or other anion) of the strong acid you add. And that number are in your added 2 ml. Likewise when you go in the opposite direction and increase the pH - well you work it out.

    Hopefully this gets you moving, not time to finish this off nicely at this instant.
  4. Mar 31, 2010 #3


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    I have to thank you for the fact that it is after some decades of believing that that thing was the Henderson Hasselbach equation I whilst doing your problem looked it up and found the name was Hasselbalch as you have written, and that I am far from the only one who thought that. I think it is OK to outline how I got result.

    First I work out [H+] at pH 3.8 as you have and at 3.7.

    Then from the equilibrium equations I work out [HCOO-]/[HCOOH] as you have, but for both pH's (I get slightly different from you, please check). Except I wouldn't bother with Henderson-Hasselbalch, you have got the equilibrium constant.

    I then express in both cases [HCOO-] as a fraction of total of both forms of formate, i.e.

    [HCOO-]/([HCOO-] + [HCOOH]).

    (The denominator is constant, being the total amount of formate plus formic acid, and this stays constant as we change pH by adding acid or alkali, though we don't know what it is yet.)

    The difference between the two is the fraction of all formate that changed from HCOO- to HCOOH when we added our 0.2 mmoles of strong acid. I get that to be 0.057 of the formate. But a litre of 0.057 of the formate has to equal the 0.2 mmoles of acid, we added. Actually to be accurate we have also to add in the H+ increase in going from pH3.8 to 3.7 which is not negligible, this is what is mentioned in the first line of calculation above and I calculate 4.1X10-5M or 0.041 mM. In short, electroneutrality of change demands the Cl- added equal the HCOO- that disappears + the H+ that appears. (The only other substantial presence here, the Na+ doesn't change.) So the total formate has to be (0.2+0.041)/0.057 = about 4.2 mM. I am quite error-prone so see if you get similar. You know the ratio base/acid so you can calculate how much of each you need to make a 500ml buffer solution.

    I hope the idea makes sense, as natural way to make this calc. The concentration for a change 0.1 pH unit the other way when adding strong base should be practically the same. If you do an exact calculation I think you will find them very slightly different, perhaps you can explain that. You may find in your books something about 'buffering capacity' with formulae, which puts the present kind of calc. into a formula, but you need to be able to do this without that.

    Notice you have added 0.2 mmole/l of strong acid but the change in [H+] is only about 4X10-5M. That is what a buffer does, mop up most of the added H+.
  5. Mar 31, 2010 #4


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    SM posted this question both here and on chemicalforums. It happened I have answered it there first with hints, but SM have not visited the site since.

    Not that s/he visted PF to check if anyone answered...

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