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Chemistry: Acidic Buffers


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  • Type: Video Tutorial
  • Length: 11:04
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 118 MB
  • Posted: 07/14/2009

This lesson is part of the following series:

Chemistry: Full Course (303 lessons, $198.00)
Chemistry: Final Exam Test Prep and Review (49 lessons, $64.35)
Chemistry: Equilibrium in Aqueous Solution (21 lessons, $31.68)
Chemistry: Buffers (5 lessons, $8.91)

This lesson was selected from a broader, comprehensive course, Chemistry, taught by Professor Harman, Professor Yee, and Professor Sammakia. This course and others are available from Thinkwell, Inc. The full course can be found at The full course covers atoms, molecules and ions, stoichiometry, reactions in aqueous solutions, gases, thermochemistry, Modern Atomic Theory, electron configurations, periodicity, chemical bonding, molecular geometry, bonding theory, oxidation-reduction reactions, condensed phases, solution properties, kinetics, acids and bases, organic reactions, thermodynamics, nuclear chemistry, metals, nonmetals, biochemistry, organic chemistry, and more.

Dean Harman is a professor of chemistry at the University of Virginia, where he has been honored with several teaching awards. He heads Harman Research Group, which specializes in the novel organic transformations made possible by electron-rich metal centers such as Os(II), RE(I), AND W(0). He holds a Ph.D. from Stanford University.

Gordon Yee is an associate professor of chemistry at Virginia Tech in Blacksburg, VA. He received his Ph.D. from Stanford University and completed postdoctoral work at DuPont. A widely published author, Professor Yee studies molecule-based magnetism.

Tarek Sammakia is a Professor of Chemistry at the University of Colorado at Boulder where he teaches organic chemistry to undergraduate and graduate students. He received his Ph.D. from Yale University and carried out postdoctoral research at Harvard University. He has received several national awards for his work in synthetic and mechanistic organic chemistry.

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So can you feel the suspense building? We've just prepared an acidic buffer, and now we're ready to see how this thing actually works. Let's quickly review what we've done. We took 1 mole of formic acid, mole of sodium formate, put them together - and remember that this is going to be an acidic system, because the K[a] of formic acid is much larger than the K[b] of formate. We ended up calculating a pH of 3.5. And again, this is just an arbitrary starting point of 1 mole per liter of formic acid and mole per liter of the formate, but it could have been any other ratio as well, as long as we had significant amounts of both. We'll say more about that in a moment. But right now, let's take our system that we've prepared - and just to remind you again, 1 molar formic acid, molar formate. And now, we're going to treat this thing with acid or base and watch what happens to the pH. So just keep in mind here where we started - pH of 3.5.
Okay, so let's suppose that we add a little bit of strong acid now to it. Let's say it's .1 mole, again per liter of final solution to keep it simple here. We're going to and 1/10 mole per liter of strong acid to this solution. Now, what's going to happen? Remember that a buffer responds to strong acid by the base consuming that acid, the weak base. And so, in this case, formate is going to react with the H plus until effectively all of the H plus is gone. And so, the first thing we need to ask for any buffer problem is, "What are the major species in solution and about what are their concentrations?" Let's think now for a moment about what we've got. We started with 0.5 moles per liter of formate, but we consumed a tenth mole per liter of that formate by using that H plus. So we now have only .4 moles per liter of the formate at equilibrium. On the other hand, by consuming the formate with H plus, we made tenth mole of formic acid. So, whereas we started with 1 molar formic acid, we now have 1.1 moles of formic acid. Okay, before we do anything calculations at all, that is the crucial step to a buffer problem, as far as identifying what happens when you add an acid or base. The first step always is to do the overall crude reaction, if you will. Then we'll go ahead and make our equilibrium calculation. But the main thing going on, we add strong acid, the formate consumes the strong acid. That's the main game that's happening here.
Now, the final question is just what is the pH in this case? All right, well, we now have an initial concentration of .4 moles per liter of the formate. Now again, I'm saying initial, meaning after we added the H plus and it reacted with the formate, but before we let the system come to its new equilibrium. So .4 mole per liter is our concentration of formate now. And again, our concentration of formic acid is 1.1 mole.
Okay, so we set-up our equilibrium expression. Concentration of H plus, once again, is our unknown, x. Our concentration of our formate now is going to be .40 plus x, again, plus whatever little bit of the formic acid dissociates to give us formate. And you'll notice that this is all independent of whether we added strong acid, or strong base or anything like that. The only thing that we needed to know was when we added strong acid, we lost some of the formate, we gained some formic acid. But the way we proceed from here on out is exactly the same that we did before. This is going to be amount of formate plus a little bit more that comes from formic acid. So .40, instead of .5, plus x, just like before, times concentration of H plus divided by concentration of formic acid. 1.1 moles now, instead of 1, minus x. Again, just as we did it before. Once again, the same assumptions are going to apply. We're going to hope that x is very small compared to .4 or 1.1, and we'll check that assumption. We end up then with a calculation of - let's see, that equals to the K[a]. So we're going to assume that that's .4 divided by 1.1 times x. And so, x ends up being 4.9 times 10^-4.
Okay, reality check. That value is very small compared to 1.1. It's very small compared to .4. It's larger than 10^-7. So check, check, check, all of our assumptions work and we end up calculating a pH now of 3.3. So we added a tenth mole of hydrochloric acid to the solution and the pH did change, but only by a tenth of a pH unit. Notice it changed in the right direction. It would make no sense if we added acid and the pH went up. So that's, again, a reality check. Make sure that the pH is going down if you added acid, but notice it doesn't go down very much.
Question: suppose that we added, instead of a tenth mole, 2 moles of HCl. Is there any problem? Well, there's a big problem. If we add 2 moles of HCl, instead of a tenth of a mole, we're going to consume all of the formate. We're going to use up everything that's buffering this system. So initially, the pH is not going to change very much, but, as soon as we start to consume the last bit of formate, the pH is just going to drop like a rock. So this idea is related to the system's buffer capacity. The idea of a buffer capacity is simply you can't just keep adding acid or base forever. You can only buffer the system as much as you've got in reserve here. So the buffer capacity, we'd say, of this system is .5 moles per liter, with respect to addition of acid. But it's a little higher than that - in fact, it's 1 mole per liter with respect to its ability to buffer, with respect to base. So again, there's another idea. The buffer capacity is not necessarily the same for addition of acid as it is for addition of base.
Finally, let's ask, "What happens if we didn't have the buffer?" Let's compare this to something. The pH did drop, after all. What happened if we didn't have the buffer there? If we added a tenth mole of HCl with no buffer, concentration of H plus at equilibrium would be 10^th molar. That means the pH would be 1. So look at that huge difference. Adding that HCl without the buffer there would take our pH from 7 to 1. That would be enough to really mess us up if it was happening in our bodies. But again, that's a dramatic pH change. And certainly in comparison to what happens when we have the buffer system, you can see how little of a pH change this really is.
Now, our next step is to look at what happens when add a base, instead of an acid. So let's return back to our initial concentration of buffer, our initial 1 molar of formic acid and .5 molar of formate. But this time, instead of adding strong acid, we're going to add strong base. Let's suppose that we added .2 moles per liter of sodium hydroxide this time. Once again, our strategy is exactly the same. We ask ourselves, before we do any calculations having to do with equilibria, "What's happening?" Well, what's happening is that the hydroxide is reacting with a formic acid. When that happens, what do we get? We get formate. So how much formate do we have at the end of the day? What we started with, .5, plus .2 moles per liter that come from the hydroxide reacting with formic acid. So that goes up now to .7. What happens to the formic acid? Well, we've lost .2 moles per liter, because of the hydroxide that we added, so it comes down from 1.0 to .8 moles per liter. But all the rest is exactly like we've done before. We just have to use these numbers, instead of the other initial concentrations, and we set-up the problem exactly the same way. So our concentration of H plus will be x, like it was before. Concentration of formic acid now, as our initial concentration, is .8, rather than 1, minus x. And our initial concentration of formate is .7. And notice that we're going to add a little bit more to the formate, because, once again, formic acid is a stronger acid than formate is a base. So it'll increase a little bit and it'll decrease a little bit of the formic acid here. And then all of this, hopefully, we won't care about, because, with any luck, x is going to be small compared to those big numbers. But we'll get to that assumption in a moment.
So we set-up our equilibrium expression. K[a] is equal to H plus concentration times concentration of formate, .7, plus x divided by concentration of formic acid. And then, the crucial assumption - we're going to hope and pray that x is small compared to .7 and .8, but we'll check ourselves. That ends up with x times .7 divided by .8. That turns out to give us a nice simple calculation. X is 2 times 10^-4 molar. Once again, reality check - is that number small compared to .7 or .8? Yes. Is that number big compared to 10^-7, the acid association coming from water? Yes. So that's okay. All of our assumptions worked. We're feeling lucky. Our pH, if we convert this then, is going to be 3.69. Once again, does that make sense? You added base, the pH should go up from its starting value. Remember, before we added any acid or base, the pH was 3.45. So we have increased the pH, meaning we're going a little bit more basic. It's still an acidic solution, but we're less acidic, I should say. But it doesn't change very much. Again, we only move up a couple tenths of a pH unit. Once again, if you do the calculation, without the buffer in there the pH is going to go through a radical change. So you can see, again, how this system responds to either acid or base. As long as we don't exceed the buffer capacity - remember, that's how much the system is able to absorb of acid or base - then our pH is going to go through relatively small changes. And the whole idea of a buffer works.
Once again, a buffer works, because, if you add a small amount of acid, it gets consumed by the weak base. If you add a small amount of base, it gets consumed by the weak acid. And so, the system is protected from dramatic changes in pH.
Equilibrium in Aqueous Solution
Acidic Buffers Page [1 of 2]

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