We are ready to begin our second learning objective in this unit, and that is to calculate the pH of buffered solutions. We are going to learn how to do this in two different ways. First way is using an ICE table. So this is not a new concept there is nothing really new to learn expect how to apply an ICE table in this environment. how to apply an ICE table in this environment. It is a similar procedure that you learned in the last unit when when you were dealing with acid solutions or base solutions. Then we are going to learn a new equation called the Henderson-Hasselbalch equation. We are are going to learn how to calculate the pH using it. Then we are going to learn a new equation called the Henderson-Hasselbalch equation. We are are going to learn how to calculate the pH using it. Either method is fine, they both have their advantages. The advantages of taking the energy to learn the Henderson-Hasselbalch equation is that it is a whole lot simpler, shorter procedure, to calculate the pH when you utilize this equation. So the first thing we are going to do is see this buffer here, and we are going to calculate the pH of this buffer using an ICE table. So again, this is with no new method. Now what you have to be able to do is write the correct reaction, and this is where a lot of students struggle. You don't react HF and NaF with each other. There are conjugate acid/base pairs. They do not react with each other. You want to choice the, They do not react with each other. You want to choice the, in this case the, not the salt, but either the acid of the base [so in this case it is an acid] and write the reaction of that acid in water. Now here I do things a slight different and write the reaction of that acid in water. Now here I do things a slight different then what you saw in the acid/base lesson. Whenever I write a weak acid, I do not just break it apart. It is just another way of doing the same thing I add my water, because I want to be consistent I add my water, because I want to be consistent always having the water as a part of the reaction. Now the acid will donate to the water, to leave you with F- and H_3O+. to the water, to leave you with F- and H_3O+. Now Dr. salt would have written this reaction and there is nothing wrong with this reaction it is just a preference. As just breaking it apart into its ions. You can choose to do that as well. But like I said I like to have the water there the acid donates to the water and we have this exchange. I will explain why I like to include that as we proceed on down the line. So I want to do an ICE table. I see that is the concentration of HF is .10 molar. I will not include the water of HF is .10 molar. I will not include the water I come over to the right hand side and I will look up and note that the amount of NaF is .3 molar. We know that breaks apart NaF is .3 molar. We know that breaks apart 100% into Na+ and F- 100% into Na+ and F- there the concentration of F- in solution before anything happens what so ever there the concentration of F- in solution before anything happens what so ever is already .3. That is the common ion. And that is the difference between a regular acid/base problem, and a common ion problem. What is the difference? You will always have Present in solution prior to anything happening with this reaction, already some of that ion. I will put a zero under the H_3O+. because this reaction has not occurred yet. There is a teeny teeny tiny amount of H_3O+ in water There is a teeny teeny tiny amount of H_3O+ in water but such a small amount will not have to consider it. So what is going to happen? We use out change lines some of this will react and according to the coefficients we will have equal quantities of those substances produced. I carry the change line into this spot. And I have these amounts at equilibrium. Now I will write the equilibrium expression for K_a, the law of mass action. And it is F- time H_3O+ over HF, we do not include time H_3O+ over HF, we do not include the water because it is a liquid. over HF, we do not include the water because it is a liquid. We plug in what we know K_a is 3.5 times 10 ^ -4. F- is 0.30 plus X. H_3O+ is X. And HF- 0.10 minus X. Now at this point we are going to assume X is very very small compared to the .30 or the .10 . We see that X is being added to .30 here. It is being subtracted from .10 here. And if X is much much smaller then .10 and of course it is much much smaller And if X is much much smaller then .10 and of course it is much much smaller .30 as well, then we can ignore this term right here. Now do not ignore that X .30 as well, then we can ignore this term right here. Now do not ignore that X that is sitting here, we cannot get rid of it. But we can ignore the ones that are added or subtracted from that value. Then I can solve for X, it will be 3.5 times 10 ^-4 time .10 and divided .30 . This gives me a X of 1.2 times 10 ^ -4. Now what you solve for X, you look back up and say what did they want me to determine? They wanted me to determine the pH of this solution. So this is the H_3O+ concentraion. So this is the H_3O+ concentraion. I know that because I look up here and I see that H_3O+ concentraion. I know that because I look up here and I see that X is sitting right underneath that H_3O+. So I can take the negative log of 1.2 times 10 ^ -4 and that will equal the pH. And the value is 3.93. Comment on this pH, it is an acid pH and when you make a buffer with a weak acid typicality it is in the acid range. So that is nothing really new. We are seeing an ICE table, we are writing our reaction, we are solving for X. and this is very very similar to what we saw in the previous unit on acid/base equilibrium. and this is very very similar to what we saw in the previous unit on acid/base equilibrium. The only difference once again is that there was a value under the F that we are starting with, and that is the common ion scenario. So now lets learn the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation is only usual for this common ion, or buffered solution. The Henderson-Hasselbalch equation is only usual for this common ion, or buffered solution. So we have got to keep that in mind. is only usual for this common ion, or buffered solution. So we have got to keep that in mind. You have to know, that you are dealing with a buffer before you can use this equation. But if you are, then you can utilize it. Now lets talk about the pieces. First of all pK_a, pK_A is simply the negative log of the K_a value just like pH is the negative log of H plus concentration. And pOH is the negative log of the OH- concentration. pH is the negative log of H plus concentration. And pOH is the negative log of the OH- concentration. This p function is take the negative log of what you see after it. This p function is take the negative log of what you see after it. When we use this equation, it incorporates all the steps we just saw in the previous example. When we use this equation, it incorporates all the steps we just saw in the previous example. It also incorporates the assumption all the steps we just saw in the previous example. It also incorporates the assumption that your initial concentration is big enough so that X is small compared to it. that your initial concentration is big enough so that X is small compared to it. It is taking into account this fact that we utilized in that. This assumption is a good assumption generally when you have a concentration of that acid. for your buffer, or a base we will see later. That is in the .1 range or larger. If it gets much smaller it might not be a great assumption and it will just get you close to the pH, and not exactly the pH. So lets do that previous problem that we did but utilizing the Henderson-Hasselbalch equation. pH equals pK_a. but utilizing the Henderson-Hasselbalch equation. pH equals pK_a. Which would be the negative log of 3.5 x 10 ^ -4 plus the log of 3.5 x 10 ^ -4 plus the log of the concentration of the base is on top so we have to find the base, this is the base I know it is the base, because it has the F- ion. I know this is the acid, it has the acid at the beginning. So the concentration of the base goes on top I know this is the acid, it has the acid at the beginning. So the concentration of the base goes on top the concentration of the acid goes on the bottom. The negative log of 3.5 x 10 ^-4 is 3.45 and I am going to carry to carry and extra significant figure for now. It should only have to two places after the decimal point when we are talking about significant figured, but I am going to carry and extra one. It should only have to two places after the decimal point when we are talking about significant figured, but I am going to carry and extra one. Plus the logo of this when we are talking about significant figured, but I am going to carry and extra one. Plus the logo of this is 0.477 and I again I am carrying an extra figure. I will add them together and then round to two places after the decimal point and that gives me a pH of 3.93 . two places after the decimal point and that gives me a pH of 3.93 . This is the exact same value we obtained when we did the previous problem. Now you can look back are you notes and see it took a whole page of a lot of writing to determine the pH of a buffer. it took a whole page of a lot of writing to determine the pH of a buffer. When we utilized the previous method. to determine the pH of a buffer. When we utilized the previous method. So the previous method, there is nothing new it is a procedure you can follow without memorizing an equation. It is an advantage in that way. The Henderson-Hasselbalch advantage is that it is short it is sweet and you can get to your answer fairly quickly. Without a whole lot of work. Now this equation can be used even if the buffer is created from a base. So if you have a base you don't go and change the pH to pOH So if you have a base you don't go and change the pH to pOH and the pK_a to pK_b you still utilize the K_a and the pK_a to pK_b you still utilize the K_a of the acid that is in there. So lets think about a base buffer. Lets say it is Methylamine . CH_3NH_2 and its conjugate acid would be NH_3 and we have to have a salt with a positive charge. and its conjugate acid would be NH_3 and we have to have a salt with a positive charge. The salts go with it, so this would be a buffer. What you want to do is say what is the K_a of this guy. And use the pK_a of this guy in the equation. And this still make sure you put the base on top and the acid on bottom. There is no need to recreate this equation in some different form, I see student do this all the time. to recreate this equation in some different form, I see student do this all the time. To come up with, with what we would do if it were a base buffer. Use the equation as you see it. With that piece of information in mind lets say we have a buffer and its consists of this weak base ammonia and its conjugate acid salt. And we want to use a Henderson-Hasselbalch equation. We look up and find that ammonia has a pK value given to us. And we want to use that Henderson-Hasselbalch equation. pH plus pK_a plus the log of the base or the acid. My question is what pK will you use in this Henderson-Hasselbalch equation? Think about it and answer the question. If you said number 4 you were correct. A lot of students will take the negative log of this number that is up here, and say that is the answer. of this number that is up here, and say that is the answer. And that is not what you want to do. In the Henderson-Hasselbalch equation you want to put pK_a and you first have to first find K_a of the acid. The acid is NH_4+ the K_a would be 1 x 10 ^-14 The acid is NH_4+ the K_a would be 1 x 10 ^-14 divided by the K_b. And then you take the negative log of that number and that is how we get the correct answer. So now we are ready to utilize the Henderson-Hasselbalch equation. It is a weak base buffer. Contains its conjugate acid salt. To determine the pH pK_a plus the log of the base concentration over the acid concentration. So I have written in what the base and acid are. We determined on the previous slide it was 9.25 was the pK_a plus the log of the base which was.15 molar. Over the acid, which was .10 molar. And that is going to be 9.25 plus 0.176 and that is going to give me a pH of 9.42. We see that the pH is close to the pK_a of the acid of our buffer. So the pK_a of the acid is 9.25. the pK_a of the acid of our buffer. So the pK_a of the acid is 9.25. This term, this portion right here of the Henderson-Hasselbalch equation adjusts up and down around that value so if you need a little bit higher pH you make sure you have a little bit more base then you do acid. If you needed a buffer with a little bit lower pH than 9.25, then you reverse this. And you put more on the acid side and less on the base side. That is what you would do if you wanted to bring the pH down a little bit. This one has more base then acid, so it is a little higher. That is what you would do if you wanted to bring the pH down a little bit. This one has more base then acid, so it is a little higher. Then the pK_a value. So that ends our learning objective number two in which we learned to calculate the pH of a buffer, and you can utilize the ICE table, which is lengthy but requires no memorization of an equation. the ICE table, which is lengthy but requires no memorization of an equation. Or you can learn the Henderson-Hasselbalch equation and save yourself time. But know that equation is used for the buffer,don't use it for anything else. and save yourself time. But know that equation is used for the buffer,don't use it for anything else.