Transcript - Lecture 22

All right.

We are in the middle of acids and bases. And we are just going to finish up the handout from Monday starting first. And we were talking about relative strengths of acids. If you have an equation such as this, there is acid-base chemistry going on in both directions.

In the forward direction, HNO3 is acting as the acid. It is giving up its hydrogen ion or proton to NH3. And when it gives it up, it is forming its conjugate NH3-. And when NH3 accepts a hydrogen atom or proton, it is forming its conjugate acid NH4+.

That is what is happening in the forward direction of this reaction. In the reverse direction of this reaction, NH4+ is acting as the acid. It is giving up a proton to NH3-. When it gives up the proton, it is forming NH3, the conjugate.

And when this accepts the hydrogen ion or proton it is forming its conjugate acid. In this direction this is the acid. And in the reverse direction this is the acid. You can predict something about where the equilibrium lies, whether it is to the right or the left.

If you have a big K or a small K for this reaction based on whether this is the stronger acid or this is the stronger acid. If this is the stronger acid the equilibrium will lie to the right. If this is a stronger acid the equilibrium will lie to the left.

And so we can take a look at this and work out which is the stronger acid and where the equilibrium lies. Here is the equilibrium constant for this reaction. Again, it would be products over reactants.

And so to find this number, we can work through it by getting information about the Ka or acid ionization constant for this acid and for this acid. We can consider the reactions separately. We can consider this acting as an acid and also this one.

Let's start here and consider the forward direction of the reaction and consider how good an acid NHO3 is. You all should be able to write reactions of acid in water or bases in water and figure out what the products are going to be of that.

If we have NH3 in water then we are going to be forming hydronium ion and the conjugate base of this conjugate acid here. And then we can write a Ka. And it is a Ka because we are talking about an acid in water.

The Ka is going to be equal to the products over the reactants. And here water is not included because it is the solvent and its concentration really isn't changing through the course of this reaction.

And we can look up an equilibrium constant for this acid and that is 20. Is that a strong acid or a weak acid? How did we define strong acid last time? It was a Ka if it is greater than one. 20 is greater than one, so this is a strong acid.

Now we can do the same thing for the reverse direction. In the reverse direction, the acid is NH4+. We can write NH4+ in water. And that would go to hydronium ions. And the conjugate of this acid, which is NH3.

We can write out a Ka expression for this. Again, it would be products over reactants. And here the Ka, which could be looked up, is 5.6 times 10 to the minus 10. Is that a strong or weak acid? That is a weak acid so the Ka is quite a bit less than one.

Already we can predict that for this overall reaction it will lie to the right. This is a much stronger acid than this one. We can prove that that is true by using these two Ka values to get the overall equilibrium constant for this reaction here.

If we took those two individual reactions that we wrote out and subtract them, we can get our overall expression for the reaction. And if we subtract them then we can come up with the K by dividing the Kas from each other.

Remember we talked about Ks, if you add together you multiply, if you subtract reactions you can divide? And so we could prove that this is true, that if we have the Ka up here and the Ka down here, it will cancel to form the equilibrium expression for the overall reaction.

And we can substitute our Kas in 20 over 5.6 times 10 to the minus 10th. And we get a K for the overall reaction of 3.6 times 10 to the 10th, which is an enormous number. And so this would lie very, very far to the right.

This is a much, much stronger acid than NH4+. If you know something, if you have acid-base on both sides of the equilibrium and you know about the strengths, you could predict where the equilibrium of an overall reaction is going to lie.

All right. This is still on the handout from Monday. We mentioned that there are five different types of acid-base problems that you are going to encounter in this unit. And there are problems of salt and water, which we are going to see today break down to a weak acid or a weak base in water problem.

It is really not a different type of problem at all. There is weak acid in water, weak base, strong acid, strong base, and there are buffer problems. And when we do titrations it will seem like that is a different kind, but it also breaks down into these categories.

One trick that I have learned from teaching this for a number of years is to get people starting to think about what type of problem it is whenever they see it and figuring out then they know how to address that problem.

Always be thinking about what type of problem is this, and so how do I go about solving it. We are going to look at number one here, a weak acid in water problem. And the weak acid that I picked to look at is vitamin C.

In this particular problem we are going to ask what the pH would be if I take this vitamin C tablet, a 50 mg tablet and put it into 100 milliliters of water. And usually when you try to swallow vitamin tablets you try not to dissolve them, but if we did then we could look at the pH.

I actually tried to do the calculation and do the actual experiment one year and discovered that these vitamin C tablets are coated really well. You cannot get them to dissolve very easily. That is because you don't want the bad taste in your mouth.

Nature's Bounty, if you don't want this table to dissolve in your mouth before you swallow it, I guaranty it won't. You have to go to extreme lengths to get this to dissolve in your mouth. We cannot actually measure the pH of this because it is not going to dissolve, the coating is pretty good, but we can do the calculations if the coating was less good what the pH of that would be.

All right. The first thing we need to know is molarity. And so you can do this calculation. You know the number of milligrams. Convert it to grams. Look up the molecular weight. Calculate how many moles that is.

And then that number of moles in your 0.1 liters or 100 milligrams. And you can get the molarity of a vitamin C tablet in 100 milliliters of water. Then you would need to write the equation for a weak acid in water problem.

And so here is our weak acid in water. It is forming hydronium ions and a conjugate base. Then you can set this up and figure out what happens as initially the change and then what the concentration would be at equilibrium.

In the beginning you just have this molarity, the 0.0284 molar solution. And you don't have any of these initially. The change would be minus X and then plus X plus X. At equilibrium 0.0284 minus X and, again, plus X.

Now to solve for X, which is what we want to solve, we want to find out pH. And pH is the minus log of the hydronium ion concentration. We need to find X to be able to find pH. And to get X, we can do that if we know the Ka for the solution, write out the expression, hydronium ion conjugate base over acid, we are leaving out the water again, that would be equal to X squared 0.0284 minus X.

Now we can try an assumption. And this will be useful to you on exams. First you can try to say what if X was really small? What if X was much, much smaller than this 0.0284? Then this term just collapses into the number.

You can make that assumption and try it and see if it works. And I will tell you what the rule will be to check that out. If we try the assumption then we have just X squared over this number. That is easier to solve.

And we can solve for X. Here X is 0.00151. And I am carrying an extra significant figure. Then we can go and check is this true that that is, in fact, much smaller than this? And the rule that we will use is 5%.

You can calculate whether the number is 5%. Take X over your initial number times 100%. In this case, it was 5.3%, which is more than 5%, and so you should use the quadratic equation to solve for X.

Most of you normally use calculators, which there is no problem in doing quadratic. But, of course, the calculators we use on the exams are not so amenable to doing this. I suggest on an exam, you always try the assumption that you can make it a simpler problem to save yourself time.

The rule is 5%. It has to be less than 5%. I will just mention that the book talks about this expression and refers to percent ionized to percent deproteinated. This equation, which is really X over the number minus the initial molarity minus X, is how much you have gone to that side of the equation, how much you have deproteinated, what the percent ionization is.

You will see those terms in the problem set and in the book. If you use the quadratic equation then you get 0.00147. And we are still carrying an extra significant figure. That is really two significant figures.

The Ka just had two significant figures that were given. If we now solve, this is X when X was the hydronium ion concentration, so pH is minus log of the hydronium ion concentration, which would be 2.83.

And since this was really two significant figures, now you have two significant figures after the decimal point. That would be the correct significant figure number for this. Remember this 5% rule to make your life easier, and also in all of these problems you want to pay attention to the significant figures.

Again, the rules for logs in significant figures are a little complicated and require some thought. All right. That is a weak acid in water type problem. And the weak base is going to be very, very similar to that.

And so today we are going to be going on and talking about weak bases. We are going to talk about salt solutions. And I will try to prove to you that salt solution problems just break down to weak acid problems and weak base problems.

Yeah? No, that is the same X. It is just a question of is it smaller than the number, so back here. It's not that it is an incredibly small number. It is just within the significant figures smaller than this number so that if you subtracted it from that number it would not really make any difference.

This one? This is the assumption that the number X is going to be enough smaller than 0.028 that if you subtracted X from that number you would still get that number. Here you are getting plus X. You are forming this amount of hydronium ion, this amount of the conjugate, and that is the same amount that is forming here that what you are taking away from your initial over here.

It is the same X. You are not really saying that this is an infinitesimally small number. You are just saying that it is smaller than this to the degree that it is not changing this very much so you don't have to actually do that subtraction.

And so the rule we use for that is 5%. Again, there are these types of problems. Now we are going to look at a weak base problem. Let's look at a weak base problem. I need better chalk than that.

All right. Weak base in water. If we have NH3 plus water, it will go to the conjugate of NH3, which is NH4+ and hydroxide ion. If you are talking about a weak base in water problem, you are going to expect to see hydronium ion on the other side.

And this is just a check that you want to make sure you are asking yourself as you are doing these problems. If you write the equation wrong, you are going to be in a lot of trouble. Weak base should yield a basic solution.

You should be getting hydroxide ions out. All right. If we look at this problem, we have an initial concentration of 0.15 molar. And we don't have any of this and we don't have any of that. The change is going to be minus X plus X and plus X.

Then at equilibrium we will have 0.05 minus X and we will have X and X. Now it is a weak base in water, so we are going to talk about a Kb. And so Kb is the base ionization constant. Kb is also going to be equal to the products because it is an equilibrium constant, products over reactants.

And we are not including water because it is the solvent. This will be equal to X squared over 0.15 minus X. And we can try this assumption again that that might be equal to X squared over 0.15. The Kb is given.

And for a lot of the problems, you will be able to look this up in your book of 1.8 times 10 to the minus 5. Again, it is a weak base. It is less than one. It has a Kb that is less than one. And so if we solve for X here, we get X as 0.00164.

And now we can check our assumption that that is OK using the 5% rule. 0.00164 over 0.15 times 100 gives you 1.1%. You are checking the assumption, and the assumption is OK. It is less than 5%, so we can use that number.

Now we want to find out the pH of this mixture at equilibrium. And we have X. X is equal to the hydroxide ion concentration. We can calculate first pOH minus log of the hydroxide ion concentration.

And so we have minus log of 0.00164 and we get 2.79. This only had two real significant figures. We were carrying an extra one. We only had 0.15. We only had two significant figures here. This is going to get two after the decimal point.

Since we have two here, we get two SF after decimal. Now we have pOH. How do we get pH? We can subtract from 14, so the pH is going to equal 14.00 minus 2.79. And it will be 11.21 that will be the pH.

And I just want to remind people, sometimes when you are in a hurry on a test and you get to this point and have this number of 2.79 and you say, oh, I am done. But that is the pOH. And, if you think about it for a minute, a weak base in water, a base in water should give rise to a basic pH.

i.e., a pH greater than 7. Below 7 is acidic. Above 7 is basic. You have to remember to convert to pH. The pH should be a basic number. Just keep it in mind, as you are doing these problems, think about what the problem was.

Oh, it is a weak base in water, does my answer make sense, to kind of double-check that you have gone the full direction. OK, so that is weak base. Now we are going to talk about salts in solution for a minute.

And salts in solution are going to break down into weak acid and weak base problems. If you have a salt, a salt can be formed by mixing an acid in a base, as shown up here, but the pH of salt water is not always going to be neutral.

If the salt contains, for example, the conjugate acid of a weak base then it contains a conjugate acid, so it will produce an acidic solution. Also salts that contain some small highly charged metals like iron 3+ will also affect the pH.

This is something that you should remember. Group 1 and Group 2 metals or metal CAD ions with charges of plus one, we have lithium, Ca2+, Ag1+. These are all going to be neutral, but other things like Fe3+ will not be.

Or, if you have a salt that contains a conjugate base of a weak acid, it is a conjugate base, so it can produce a basic solution. If you remember on Monday, we measured the pH of MIT tap water and found that it was not neutral.

It was pH 6 rather than 7, so that would suggest that there was some salt in that tap water that was giving rise to a non-neutral pH. This is also review from last time. There is this relationship between the conjugate acids and bases.

This is not in your handout, but just to remind you what we were talking about. If you have a strong acid, its conjugate is ineffective as a base. If you have an acid that is moderately weak to weak then it will have a conjugate base that is also moderately weak to very weak.

And if you have a really strong base then its conjugate will be ineffective as an acid. So, there is this relationship. Let's take a look at some examples of salts. If we have NH4Cl. If you are asked whether putting some amount of this in water will lead to an acidic solution or a basic solution or a neutral solution, this is how you would approach the problem.

First you are going to break this compound through the middle and consider two sides. This salt will be composed of NH4+ and Cl-. Then you are going to ask whether this is a conjugate acid of a weak base.

Is this a conjugate acid of a weak base? If it is a conjugate acid of a weak base then that will affect the pH of the solution. What is the conjugate of NH4+? The conjugate of this is ammonia. If you take the weak base, the conjugate, the weak base is NH3.

And if you added a proton to this you would get the conjugate acid NH4+. This is a weak base, so the answer to that question was yes. You could look it up in the table. Here is ammonia. It has a Kb that is fairly small so it is a weak base.

And we also had it on this chart. You had NH3 over here. And so you could see that it is, in fact, a weak base. And we can look up the Kb for that. And so the Kb is 1.8 times 10 to the minus 5. That is a weak base.

Because the conjugate is a weak base that means that the conjugate acid is also a weak acid. And so this should be acidic in solution. That should be acid in solution. And let's just write that out over here.

If we have NH4+ that has got to be a weak acid because its conjugate is a weak base. And if we put the weak acid in water, we should get the conjugate base back plus hydronium ion. And the presence of the hydronium ion is going to make that acidic, and so this is a weak base over here.

Its conjugate is a weak base. And we can look up a Ka number for it. 5.6 times 10 to the minus 10th. That is, in fact, a weak acid. And so here it is on this table. We can look it up. And it is less than 1 and is definitely a weak acid.

This part on this side of the compound, we have something that is going to produce an acidic solution. Now we have to ask, well, what about this side over here? And so we would ask, is this a conjugate base of a weak acid? Is Cl- a conjugate base of a weak acid? What would be the conjugate acid of Cl-? HCl.

So the conjugate is HCl. Is that weak? No. This is a strong acid. And that is one that you will encounter a bunch, so you will definitely have that memorized by the time you get through the problem.

It has a Ka over here of 10 to the 7th. If you put that in water, if you had HCl in water, it is a really strong acid, Ka 10 to the 7th. And so it would go pretty much all the way to completion. That means that this would be ineffective as a base if you have a strong acid.

Again, there are these relationships that we talked about. If something is a really strong acid up here then its conjugate is completely ineffective as a base. This then is going to be neutral. Because this is not going to change the pH anymore, it is ineffective as a base.

It is not going to make the solution basic at all. Whereas, on the other side, NH4+ is a conjugate acid of a weak base so it will produce an acid solution. The answer is that this salt in water will be acidic.

If you were given an amount of this compound, if you said 0.15 molar, for example, what is the pH, you could work that out by just doing a weak acid type problem and solving for X here and calculating pH.

This particular salt in water problem is just a weak acid in water problem. OK. Let's look at another example. Again, we are going to take this compound and break it up. And we will form NA+ and CH3COO-.

Then we ask is this a conjugate acid of a weak base? And what is the answer? Where do you think it probably came from? Yeah. It probably came from NAOH. This one, the answer here is no. And we talked about rules that all Group 1 and Group 2 are neutral.

Group 1 ions are neutral. Group 1 and Group 2 are neutral, so this in solution will be neutral. All right. Let's look at the other side and ask is this the conjugate base of a weak acid? And what would be the acid here? Yeah, acidic acid.

We have CH3COOH. And we can look to see if that is a weak acid. And we can look it up here. And so we have, get the right line, 1.76 times 10 to the minus 5. Is that a weak acid? Is that a weak acid? Yes.

If it is a conjugate base of a weak acid then in solution it is going to be basic. And we can work that out. If we wrote it out, we have CH3COO- in water is going to yield hydroxide ions and the conjugate acid over here, and the presence of this, of the hydroxide ion will make the solution basic, so this is a weak base on this side.

And so if you were asked, given an amount, you could solve the problem using the Kb of the weak base solving for OH-, finding pOH and then pH. Overall, this salt in solution will give rise to a basic solution.

You should have a pH greater than 7, so this salt and water would give rise to a basic solution. And this salt and water problem breaks down into a weak base in water problem. Let's just do the generic -- -- type of problem.

In the general case, if we have made up compound XZ. Made up compound XZ will break down into X+ and Z-. And you are going to ask is this a conjugate acid of weak base, or you are asking basically is this going to be acidic? If the answer yes -- If you have a conjugate of a weak base, which is a weak acid, you will have an acidic solution.

If the answer is no, you will have a neutral, at least in terms of the X. On the other side, for the Z-, you are asking the question is this a conjugate base of a weak acid? If it is a conjugate base of a weak acid then it is going to be basic.

It will be a weak base. And so the answer is yes, you are going to get a basic solution. If the answer is no here it will be neutral. And in these problems, in the short answer type format here, you are not going to be given a case where one side will be acidic and the other side will be basic because then you cannot give a short answer to that problem.

You would have to figure out which one is more acidic and which one was more basic. In these types of problems one side will always be neutral, or they could both be neutral, you will see examples where both of them are neutral, but you are looking at each side and asking a question about it.

That is how you go about doing these salt and water problems. All right. This leads us right into buffers. Buffers, this is another type of problem that we will have. Again, you are talking about weak acids and weak bases here.

A buffer is any solution that maintains pretty much a constant pH despite the addition of a small amount of acids and bases. Buffering is all about constant pH. An acidic buffer is just a buffer that is going to pH or stay at sort of a constant pH in the acid range, so the acid side of neutral, and it would consist of a weak acid and its conjugate base supplied as a salt.

A basic buffer will have a weak base and its conjugate acid supplied as a salt, and it is going to buffer on the basic side of neutral. That makes a lot of sense. All right. Let's look at a buffer example.

In all buffers, you are going to have an acid mixed with its conjugate in the form of a salt to get a dynamic equilibrium. And so if we have this, we have the acid and its conjugate added here. And we ask what happens if you add a strong acid to this solution? Again, what we want to have a buffer do is maintain a constant pH.

You need to have about the same amount of this guy as this guy for that to be true. If you added strong acid and you had some of this and some of this, the acid is going to react with a conjugate base and drive it in this direction using up the amount of strong acid that was added.

The added acid, hydronium ions are effectively removed and the pH stays pretty much constant. Now what happens if you add a strong base? Well, a strong base is going to react with the weak acid over here.

And it will also be consumed. And so the OH is removing a proton, it is reacting with the weak acid, and it will form more conjugate base and it will also form water. Again, it is effectively removed and the pH stays pretty much constant.

A buffer should be a source in a sink for added problems to the reaction. It is going to use up either the acid or the base that is added and keep the pH pretty much constant. Again, what is happening is that the weak acid that you use, and you can have anyone there, is going to transfer protons to this.

And that will cause the reaction to happen and use up the hydroxide ion. A conjugate base that is added as a form of a salt would then be able to accept protons from any acid that is used. So you need both the weak acid and the conjugate base added in the form of salt.

One will react if you add extra acid. One will react if you add excess base. Why wouldn't a strong acid and the conjugate base of that strong acid make a good buffer? Right. Because the conjugate would be ineffective as a base.

And so the strong acid would go to completion, will drive to the right. This is ineffective as a base so you are not going to push it back the other way. You are only going to have a good buffering system with a weak something and its conjugate that is also a weak something.

A weak acid and its conjugate weak base, a weak base and its conjugate weak acid. That will buffer because it will go both directions. Both conjugates have to work as an acid or a base. If you have a strong acid then its conjugate is ineffective as a base.

It won't work. If you have a strong base then its conjugate is ineffective as an acid and it won't work. All right. Base buffering example, the same stuff applies here. If you add a strong acid, it will react with the base.

The base will accept the protons and make more of the NH4. And it will use up the strong acid. When a strong base is added, it will react with the conjugate acid and go the other direction. And so you will form more of the conjugate base and water.

Again, the pH is staying the same. You are using up the strong base that is added. In a buffer action you have a weak base accepting protons supplied by the strong acid and you have the conjugate acid of the weak base transferring protons to the hydroxide, the strong base that is added using that up as well.

Again, it is about going both directions of the reaction. That makes something a good buffer. A buffer mixture of a weak conjugate acid and base stabilize the pH of a solution by providing a source or sink for the protons.

Either they are providing a source to neutralize the OH that is minus or a sink for the protons that are added in the form of acid so the pH stays pretty much the same. You need something that can do this dynamic equilibrium to be a good buffer.

Let me just give you, in the last couple minutes, an example of why buffering is important from a human health perspective. All right. Your blood is buffered around a neutral range, and the buffering agents are shown here.

Propionic acid and bicarbonate are effective as buffering agents. They can provide the source or sink for added stress to the system, for added acid or added base so you blood is buffered. And this is really important if your blood pH changes.

Enzymes are designed to work at particular pHs. Molecules are not necessarily stable at certain pHs. Our body would not work well if it was not appropriately buffered. Let me give you an example of a genetic defect that overwhelms the buffering capacity of the blood and leads to a condition.

This is an enzyme that you have in your body. It is in your mitochondria. And it requires vitamin B12. This is what the enzyme is involved in. It is involved in breaking down fatty acids to form energy.

Most people like this enzyme. They are very happy with this enzyme. Most people want their fat converted to energy, and so this enzyme helps out with that going on. All right. What happens if you have a genetic defect or if you don't have enough vitamin B12, but in this case they are talking about a genetic defect, this step is blocked.

Your body is making a lot of methylmalonyl-CoA and it doesn't know what to do with it. It is a huge amount, and so it actually gets excreted from the body in the form of an acid. And so this leads to a condition known as methylmalonic aciduria.

There is so much of this made that you are excreting like a gram of this acid every day. The blood is overwhelmed. It cannot handle that much acid. Its buffering capacity is not that good. All right.

What causes this problem? I will tell you that this now is the amino acid sequence of the enzyme. Enzymes are made up of multiple amino acids strung together in a polypeptide chain. And down here is the number 721, so this is a lot of amino acids.

And at the time there were these babies born with this condition, and the doctors did not know what to do to treat them and a lot of the babies died. A number of them got together and tried to sequence the gene.

Now, of course, there is the Human Genome Project so we know a lot of the sequences of the genes. But the doctors sequenced this gene and sat down with the parents and said, OK, there is a mistake right here, where that red arrow is.

It should be a glycine residue but it is something else. All right. Imagine you are a parent. You have a kid and the doctor tells you there is a mistake right here. Well, that is fantastic. What are you going to do about it? Well, at the time, the doctors had no idea what to do about it because they didn't really know what was going on here.

This answer partly came through structure determination. They found that vitamin B12 changes shape when it binds to a protein. Here is vitamin B12. It has this group attached to the cobalt when it is free.

When it is bound to the protein it looks like this. This long group is extended into the core of the protein. The protein needs a big hole to put this group. And so this is a model of what the protein look like and here is a model of vitamin B12.

We discovered that it fits like this. It is fantastic. There is a huge hole in the protein into which you can put the vitamin. How did nature create a hole? How did it make this space for the vitamin to fit and then with the vitamin bond it is active? What it did was put the smallest amino acids it could find over here.

Now, what happens in this mutant protein is that one of those small amino acids was converted into something larger in the dark blue. This was the protein that these babies had. And the doctors kept giving them more B12, more B12, more B12 and it didn't do any good at all, because no matter how much you gave it couldn't bind.

It was blocked. The binding was blocked. This suggested to doctors that instead of giving them the full vitamin B12, basically take a pair of scissors and chop off part of it and create something that looked more like this and then try to see if that would then bind to the mutated protein.

That is the kind of thing that they were looking at trying to alleviate this particular condition because the buffering of the blood is pretty good. If could you restore some activity, get rid of some of the acid, you might be able to have a more normal pH.

OK. Have a good weekend.

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