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About this Lesson
 Type: Video Tutorial
 Length: 10:17
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 111 MB
 Posted: 07/15/2009
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: Reactions of Acids and Bases (3 lessons, $4.95)
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 http://www.thinkwell.com/student/product/chemistry. 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, oxidationreduction 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 electronrich 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 moleculebased 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.
About this Author
 Thinkwell
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11/14/2008
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We're now pretty comfortable with the idea that if we take a weak acid and put it in water that an equilibrium is established between that weak acid and its conjugate base. We know how to calculate the Ph for that solution. We know how to calculate the concentration of the weak base equilibrium as well as the weak acid. We know how to do the same exact type of problem if we start with a weak base and put that in aqueous solution.
Now, we're going to continue with the theme of looking at equilibria in aqueous solution, in general, but now look at situations where we involve a reaction of an acid and a base. In particular, we're going to look at what happens when we put a weak acid which a strong base or with a weak base. In fact, we'll look at all of these combinations of acids and bases.
We'll start with the simplest case, which is actually a strong acid and a strong base. We'll make sure that we're feeling with comfortable with that neutralization reaction, except we'll quantify the final equilibrium that we get and then we'll go onto weak acids and weak bases. So once again, our theme now will be the reaction of an acid with a base in aqueous solution.
So Case 1 would be then a strong acid and a strong base. In this case, we'll look at hydrochloric acid which we know, when it dissolves in water, fully dissociates and gives us H plus and CL minus. And sodium hydroxide, which again, when it dissolves in water, fully dissociates to give us sodium plus and hydroxide. So overall, the reaction we're expecting between a strong acid and a strong base is a neutralization reaction giving us water and also giving us the spectator ions that were not involved in the reactions  sodium and the chloride, in this particular case.
Our first step then in describing this is to write out the individual ions and solution and identify the reaction. So here we go. H[3]O plus and CL minus, that would be when we dissolve the HCL in the water, and then sodium plus and hydroxide. And those combined together again to give us water and sodium plus and chloride minus.
Okay, now what we're going to do is remove the spectator ions from this and, again, write down just the actual reaction that we're interested in. And it simplifies just to H[3]O+ plus hydroxide goes to two water. Now, let me just remind you that when we write down H[3]O+ that this is a salvated proton. You may see in other textbooks or your instructor may write H+ instead of H[3]O+ or H+ aqueous. It all means exactly the same thing. But if we're going to write down formally H[3]O+, we have to account for that extra water in our product. So, again, you may see this final reaction as H+ plus OH goes to one H[2]O. And it will mean exactly the same thing as this, just to clarify that point.
So once again, what we're interested in is this reaction. And we haven't seen that equilibrium but we've seen something very similar to it. We've seen the opposite equilibrium. We've seen that H[2]O goes to H[3]O+ plus OH, or rather 2H[2]O goes to H[3]O+ plus hydroxide. Now, that's just the autoionization of water or the dissociation of water. We know that that's K[W]. It has a special name. The equilibrium constant at room temperature is 1.0 times 10^14. So we have a value for the reverse of this reaction and we remember from chemical equilibria that if we take that reaction and reverse it, what we must do to the K is take the reciprocal of the K. So in this case, the equilibrium describing this reaction as written is one over the equilibrium constant for the reverse reaction, and that's again K[w]. And so, in this case, we end up with 1 x 10^14. Well, what does this tell us? 1 x 10^14 is a huge number. It says that this equilibrium must lie very far to the right and therefore, again, essentially we have complete neutralization as long as we started with equal amounts of acid and base of the H[3]O+ and the hydroxide.
We could ask, "All right, what's the concentration of H[3]O+ or of hydroxide at equilibrium?" We know how to calculate that because that's the exactly the same thing as if we'd started just with water and let this system run to equilibrium. And we know then that governed by this equilibrium  or if we wanted to write it in the reverse and flip the K to give us K[w]  we know that the concentration of these must be 10^7 each if our final equilibrium constant is 1 x 10^14. And just to clarify that for you, with this written as is, remember that the equilibrium constant for this thing is to find as concentration of products, but we just have a liquid on the product side over concentration of reactants, and if each of these is 10^7 molar, then that turns out to be one times 10^14.
So again, this may look different, but you've seen this before. This is just the exact reverse of water dissociating to give hydroxide and H[3]O plus. Now the more interesting problem comes when we look at the reaction of a weak acid and a strong base. In this case, let's consider formic acid, this is a model for formic acid. Formic acid, in particular again, has got this OH bond that's very easily cleaved and the resulting negative charge here is resonance stabilized. So we have an acidic molecule, but it's not a strong acid. We know that just when it's dissolved in water it's primarily in the form of formic acid, not of formate and H+.
Nonetheless, when we combine that with a strong base, the question is will that, in fact, go to formate plus hydroxide? This would be the reaction we'd be worried about. And where, in fact, does this equilibrium lie? Well, the answer lies entirely with how weak is this acid and how strong is this base? There are some acids that are going to be so weak, for instance, methane. It's going to be so weak that whether you throw a strong base at it or not, you're not going to get any significant reaction. So we have to now turn to the numbers, to K[a], that gives us a quantitative explanation of just how strong this acid and just how strong this base is to see if this reaction actually will occur.
Now K[a], again, is what describes how strong this acid is. 1.8 x 10^4 tells us it's a weak acid. And we're going to, now, want to write out the equation that we do know, that this number describes, and see if we can get from that equation to the overall chemical reaction that we're interested in. Okay, once again, we're trying to get an equilibrium constant for this reaction.
Okay, well what we do know is formic acid reacts, to some degree, to give us H[3]O+ and formate and that is described by the equilibrium constant K[a]. We have a value for that. How do we go from here down to the equation in green, which is the one we just looked at? Well, we know that H[3]O can combine with hydroxide, and remember, that's what we're adding to give us water. And we know that equilibrium. That's just . Once again, we just saw this equilibrium, if you remember. So if we combine those two steps together, we end up with the reaction we're interested in. So, remember now, when we combine reactions to get a new reaction, we must take the product of those equilibrium constants. K[a] divided by K[w] is going to be our answer. In other words, K[a] x gives us will describe that equilibrium then. Plugging in numbers, K[a] is 1.8 x 10^4; K[w] is 1 x 10^14. This gives us a K of 1.8 x 10^10.
So in this particular case, this equilibrium lies far to the right. Remember that large K tells us so. The equilibrium lies far to the right. So this is a strong enough weak acid. Let me say that again, this is a strong enough weak acid to react with this strong base for the reaction to go far to this side. But again, that is not a general result. Sometimes this will work. Sometimes it won't. It depends just how weak this acid actually is. In this case, again we calculated that this is the dominant species. So if you were presented with a problem like you mix these guys together in equal amounts. You establish this equilibrium. What is the Ph? You can do this? Because you know that the dominant species in solution now will be sodium formate. This will all be consumed and sodium formate in water is a problem you've dealt with before. That's a K[b] problem. You know the K[b] for sodium formate. You know how much sodium formate you have. And we already know how to calculate what the Ph would be. If you want a little bit of a review on that, then go back to the earlier tutorials on how to do K[b] problems.
So, again, just summarizing then, when we're dealing with a reaction of an acid and a base, what we need to do first is assess where the equilibrium lies. Is it on the left or the right side? And in some cases, it's not easy to decide that by inspection. We need to actually go to the numbers and look at the K[a] for just how strong that acid is and combine that with our knowledge of how good the base is that we're interested in and that's going to tell us where the equilibrium lies.
Equilibrium in Aqeous Solution
Reactions of Acids and Bases
Strong AcidStrong Base and Weak AcidStrong Base Reactions Page [2 of 2]
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