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Acids, Bases & Equilibrium · Acedra
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Acids, Bases & Equilibrium
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0:00
Hello everyone, Victor is here and in this video I wanna talk about the acids and bases and acid-base equilibrium. So before we jump into the acid-base equilibrium, let's talk about what acids and bases are to begin with. There are 3 main definitions. The first one is the Arrhenius definition. And according to the Arrhenius definition, acid is something that donates H plus into the solution while base is something that donates an OH minus into the solution.
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This definition is very limited and it's only applicable for Arrhenius solutions which is obviously not the case in organic chemistry. In organic we rarely deal with Arrhenius solutions which means that we need to look at something else. The next definition which is significantly more useful is a Bronsted-Lowry definition. And according to the Bronsted-Lowry definition, the acid is what donates H plus into the solution so acid is the H plus donor while the base is the H plus hydronium acceptor.
1:01
And that definition is significantly wider and way more applicable because now things that have nothing to do with the OH groups can all of a sudden be bases. So for instance if in Arrhenius definition the acid would be something like HCl, the base has to be something like sodium hydroxide. But in the Bronsted-Lowry definition acid can still be HCl but the base can be an OH minus not sodium hydroxide just the OH minus because OH minus will be accepting the proton.
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Or we can have let's say NH3 as a base. Or we can have let's say CH3O minus as a base etc. We have a whole world of species that we are going to see in organic chemistry that simply don't even exist in aqueous solutions. Something like let's say H minus the hydride anion which is a perfectly fine Bronsted-Lowry base doesn't even exist in aqueous solution it's a little bit too strong for that. Finally we have one more definition and that is a Lewis definition. And according to the Lewis acid base
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theory acid is the electron acceptor while base is going to be electron donor. So in the Lewis theory let's say the H plus by itself is going to be an acid because H plus can accept electrons while H minus for instance is going to be the donor of electron density so H minus can be a Lewis base or NH3 can be a Lewis base or OH minus can be a base whatever has the electron pair can be a base according to the Lewis theory. Now if we are not specifying what type of acid base chemistry we
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are dealing with typically we are always going to be talking about the Bronsted-Lowry definition so for the rest of this video whenever I'm talking about acid base equilibrium and acid base reactions I am always going to be talking about the Bronsted-Lowry definition. I am going to make another video about the Lewis acids and bases so here we'll omit that completely. So I've mentioned that we are going to be talking about the acid base equilibrium here. So what exactly is acid base equilibrium?
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Let's look at a classic example that you probably remember from general chemistry where I have H2O reacting with something like HCl giving me H3O plus and Cl minus. All of things that I wanna point out here from the nomenclature perspective.
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Water in this case is accepting the proton from my HCl. I can show it with curved arrows like this, the electrons from the oxygen going to go to the hydrogen and then hydrogen is going to break off from the chlorine. Since water is the acceptor of the proton, water is going to be our base.
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Likewise, since HCl is going to be the donor of said proton, HCl here is going to be an acid. There is also a very important relationship between the starting materials or reagents in our reaction and the products in the reaction. Whatever has accepted the proton, which in our case is H2O plus, now becomes the conjugate acid and whatever has lost the proton, in my case that's going to be Cl minus, is going to be a conjugate base. So remember, the term conjugate, either base or acid, always
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refers to the product of the Brusted Lowry acid base reaction and you cannot use the term conjugate acid or conjugate base outside of this equilibrium with no context. If I just say that, you know, H2O plus is a conjugate acid, well, I have to mean that that is a conjugate acid in an equilibrium where it grabbed the proton from something else. So the words conjugate don't really have much meaning outside of the equilibrium when it comes to the acid base reactions.
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One other thing that I want to point out here before we go further is the use of curved arrows. I am going to be using curved arrows throughout this entire video and when I am using the curved arrows, we are always going to start our curved arrow from the source of the electron density, in this case that was an electron pair on the oxygen, then I am going to point those electrons towards where they are going and this way I have made my oxygen-hydrogen bond. When I have the curved arrow starting
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from the bond between hydrogen and chlorine like I have in my HCl and then those electrons are pointing at the chlorine atom, well, now that means that I am breaking that bond and I am going to end up with an electron pair on chlorine in addition to everything that we had there to begin with. Let me know in the comments below if curved arrows is something that you are struggling with so I will make another tutorial specifically how we deal with curved arrows then.
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But for now, let's look at a couple more examples of Bronsted-Lowry bases and acids and their reactions. So here I have an example where I have the acetic acid, this piece is on the left, this one reacting with sulfuric acid, this piece is on the right. And what I am seeing is that this oxygen has accepted a proton from sulfuric acid which used to be sitting over here. Which means that if I wanted to draw my curved arrows to account for this transformation,
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I would start my curved arrow from one of the electron pairs on the oxygen, move those electrons towards the hydrogen and now the OH bond has to break moving electrons onto the oxygen which will result in a negative charge on the oxygen exactly what I have right over here. In this case, if I closely pay attention to where my proton is, I started with my proton on the sulfuric acid and I ended up with my proton over here attached to my carboxylic acid. So, based on our definition that means
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that the sulfuric acid here is the acid and my acetic acid, my carboxylic acid is actually the base because it is accepting the proton. That means that whatever has accepted the proton, in this case that's going to be my carboxylic acid, going to end up being the conjugate acid while what has lost the proton
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which is my sulfuric acid, going to become a conjugate base. So as I've mentioned a moment ago, the terms conjugate acid and conjugate base only have meaning within the scope of a certain reaction to certain equilibrium. Now let's look at a slightly different example. In this case, I again have a carboxylic acid reacting with a different species but now if I pay very close attention to what's going on with my proton, I used to have a proton on the carboxylic acid and now this
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proton ended up on my methoxide making metanol molecule. So that means that now the carboxylic acid on the left, that is my acid and the methoxide anion that I have reacting with that, that is my base. Because of that acid, once it loses the proton becomes the conjugate base and my base, once it gains a proton becomes my conjugate acid in this equilibrium.
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And this is a very important point here because the same species can be either acid or base depending on what it reacts with. If I wanted to complete this reaction with my curved arrows, I would start my curved arrow from the negatively charged oxygen and show that that oxygen gives electron density to my proton and then hydrogen releases those electrons to the oxygen on the carboxylic acid to give that a negative charge so instead of a hydrogen we now ended up with a negative charge on this O.
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So when you are dealing with the acid-base equilibrium, always look at where your proton was originally and where your proton ended up. The acid-base reactions like that, the Bronsted-Lowry acid-base equilibria, they're often called proton transfer because we're quite literally taking a proton from one species and moving it all the way to the other species. So we're just going to end up with a proton hopping from one molecule to another one. Whichever lost the proton is going to be
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our acid, whichever gained the proton is going to be our base. Another thing that you are probably remembering from your general chemistry course is that we often classify our acids and bases as strong and weak. While the strong acid is going to be the one that dissociates completely or nearly completely, while the weak acid is the one that dissociates poorly or partially. So what that means in practical terms, if I look at two equations, let's say I have a strong acid HBr, if this molecule
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dissociates, if that acid dissociates in a solution, we're going to make H+, and we're going to make Br-. Since this is an example of a strong acid, that means that we are pretty much going to have solution consisting of H+, and Br-, and virtually none of the HBr left. In other words, the concentration of my HBr is going to be nearly zero, while the concentrations of my H+, and my concentration of my Br- is going to be significantly greater than zero. From the equilibrium standpoint, we often
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refer to that as the equilibrium constant, so we're going to say that the equilibrium constant for this reaction going to be the products, which is H+, and Br- over the concentration of the starting material or reagent, which is going to be HBr, and since in this case the concentration of HBr is nearly zero, and concentrations of my H+, and Br- are significantly more than zero, we're going to say that that equilibrium constant kind of looking at the infinity.
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On the other hand, if we're looking at a weak acid, the one that dissociates partially, then we're going to have a somewhat different picture. So let's say I have acidic acid CH3COOH, which is a well known weaker acid. When that one dissociates, we're going to end up with CH3COO- and H+, just like that. So in this case, since this molecule does not dissociate very well, it means that the starting material or reagent, CH3COOH, is going to be at a meaningful concentration,
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it's not going to be going to zero, it's going to be quite meaningful for us. Even the concentrations of our products in this case are actually going to be relatively low. So if I were to write out the equilibrium constant for this one, it is going to be the concentration of my products, which is CH3COO- multiplied by the concentration of H+, over the concentration of the starting material, which in this case is CH3COOH, and that number is far from infinity. As a matter of fact, that number is
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roughly 1.8 10 to the negative fifth power, which is not that large, which means that in the solution my acidic acid is predominantly in an undissociated form. On top of the Ka values, which are these equilibrium constant values, we're often going to call those Ka values rather than some sort of generic equilibrium constant. So those Ka values, they are typically tabulated values and you can easily find them in the
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different tables in your textbooks and tables dealing with the acid-based chemistry. However, within the scope of organic chemistry, we're going to deal with such a wide range of numbers that the PKa values become significantly more important to us. I will remind you that the by definition PKa value equals the negative log of the Ka value, which means that if we take all of these Ka values that we have for our different molecules and take the negative log of that, we're going to have a
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significantly more meaningful numbers that are much easier to operate with. In this particular case, if I take the PKa value for my acidic acid and I were to take a negative log of 1.8 10 to the negative 5th power, I'm going to end up with a number that is roughly 4.75 or I can round it to about 4.8. Well, this is a much easier number to work with and within the scope of organic chemistry, we are typically going to be operating with the PKa value. So make sure you understand the
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difference between the Ka equilibrium or Ka values and the PKa values for different acids. When it comes to the large Ka values, that going to correspond to typically a negative PKa value, which corresponds to a strong acid.
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So if let's say I were to look up the PKa of HBr, the PKa of HBr is something like negative 9 or so, which means that the Ka value for that is going to be somewhere around 10 to the 9th power which is an astronomically large number. However, on the other hand, if we have a small Ka value, that corresponds to a positive PKa simply how the math works and that corresponds to weak acids.
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So the example of that is our acidic acid with a relatively small Ka value, 10 to the negative 5th, which corresponds to a positive PKa value of 4.75, so it's not the weakest of the acids that we are going to see but by far not the strongest one either. And since the PKa values are also commonly tabulated and that is pretty much what you are going to end up working with in your course, we can easily use those to compare the acidity of different species. So let's say we have a set of molecules,
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I have CH3OH, methanol, the Ka of that is going to be 3.2 10 to the negative 16th which translates to a PKa of about 15.5. So this row is going to be my Ka values, this row is going to be my PKa values. The next one, let's say I'm going to take the acidic acid CH3COOH, the one that I have already looked at, the Ka value for that is about 1.8 10 to the negative 5th, the PKa value is going to be roughly 4.8. Then let's say we have CH3SH which is going to be 1 10 to the negative 11th for the Ka
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value which gives me 11 in the PKa scale and finally let's take for instance CH3NH2, a simple primary amine, the Ka value for that one is going to be 1 10 to the negative 36 which is going to give me 36 as the PKa value. As we know that the larger the PKa, the weaker the acid is, that means that my methyl amine, the last molecule over here is going to be my weakest acid out of all of them and since my acidic acid has the lowest number out of all of them, 4.8, in this particular lineup
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it's going to be my strongest acid.
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If I were to compare it with some strong inorganic acids like let's say HCl, well HCl has a PKa value of somewhere around -7 which is significantly stronger than acidic acid or if I look at something like H2SO4, well that PKa value is going to be somewhere around -3 to -6 depending on which PKa table you look at because there are two common versions for that number but anyways so that still is a negative number which does mean the sulfuric acid is a pretty strong acid as well.
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So using the PKa values we can easily predict the state of our equilibrium in any reaction. So to illustrate how I would use that let's say I have the following reaction. I have for instance pH NH3 plus a protonated aniline reacting with CH3OH methanol giving me the products going to be aniline by itself pH NH2 and I'm going to have protonated CH3OH2 plus protonated methanol like that. So in order to estimate where the equilibrium lies in this case, whether the equilibrium
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favors my starting material or my products, I would need to know the PKa values for my acid on the left side and my conjugate acid on the right side. The acid on the left side is going to be this protonated aniline and the acid on my right side is going to be protonated alcohol. If I look up those values in my PKa table, I'm going to see 4.6 on the left side and roughly negative 2 on the right side. So the one on the right negative 2 is a stronger acid while the one on the left this 4.6 that
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is a much weaker acid. And equilibrium always favors the weaker acid. So in this case, our equilibrium going to favor the starting materials. Next we are going to use our PKa values to estimate the actual equilibrium constant for the acid-base reaction. So let's say I have a reaction between acidic acids so CH3COOH reacting with CH3OH and in this equilibrium I'm going to end up with CH3COO- and I will have CH3OH2 plus like that. So the acid on my left side is going to
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be the acidic acid so that is my acid. The conjugate acid on my right side is going to be the protonated species so that is my conjugate acid. And in order to estimate where the equilibrium is, or where equilibrium lies I should say I need to know the PKa values for those species. The one on the left we have already seen that one before, the PKa here is going to be roughly 4.8 and the PKa of the one on the right we have already seen that one today as well, that one is going to
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be around negative 2. So in order to estimate our equilibrium constant we are going to use a simple equation.
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Equilibrium constant for the reaction is going to be 10 to the power of the PKa value of my conjugate acid minus PKa value of my actual acid. So it is the right PKa minus left PKa. So in this case my equilibrium constant is going to be 10 to the power of negative 2 minus 4.8 which is going to give me 10 to the power of negative 6.8. Now since my equilibrium constant is the large negative value I am going to say that this is highly unfavorable which means our reaction favors the
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starting materials or my reagents rather than products. So I can show that also with a big arrow pointing towards my starting materials as well and of course we can already see that or we knew that that's going to be the outcome because we know that 4.8 being much higher number than negative 2 which means that this is the weaker acid and we know that the equilibrium always favors the formation of the weaker acids. Now in order to be able to easily operate with all of those
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numbers you need to become very familiar with your PKa table. Typically instructors will give you a PKa table on the exam and of course you will have access to the PKa table during the homework as well. However the PKa table that your instructor gives you is typically the one that your instructor is going to be using on the exam as well. So it's always a good idea to make a copy of that, stick that under the cover of your chemistry binder or attach that to the front page of your
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notebook or something like that so you have always access to the same PKa table. Having familiar with your PKa table will mean that you can find the corresponding PKa values for different pieces much faster than you are not going to spend too much time on that on the exam because a few seconds here, a few seconds there, they add up and during the exam when every second counts you cannot afford to waste those seconds. So make sure that you become very familiar with your PKa
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table and know how to use it well. I will do another video where I explain how exactly you use your PKa table and how the PKa table becomes your best friend within the scope of your course so to make sure that you get notification about that video and you don't miss it, subscribe to the channel, hit the like button and I will see you in the next video.
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