We are now ready for our third learning objective. In this learning objective we are going to try and understand how a buffer works. We know that a buffer resists change to pH. That it's job. We also know what it has in it. It has a weak acid and its conjugate base. or a weak base and its conjugate acid. So it has something weak with its conjugate. In significant quantities. And the question is how does that do what its supposed to do? How does that cause a resistance of the change of pH. Now, this learning objective is very important for you because you have gone to understand how it works, in order to do what is next which is calculate the pH of a buffer after you add and acid and add a base. If you do not understand what is going on in that you won't be able to write good reaction equations. So make sure you pay very close attention to this learning objective. you won't be able to write good reaction equations. So make sure you pay very close attention to this learning objective. We are going to start with a clear question. If we have a buffer that consist of acetic acid that is a weak acid. And we see acetic acid written right here. And it also has its conjugate base salt. So it has it conjugate base that is this portion here, in it. And we take that buffer, and lets draw a little picture of a flask, containing that buffer. So we have this buffer in there a flask, containing that buffer. So we have this buffer in there and we are going to add to it some base. Now a base has with it hydroxide ions. So we are putting hydroxide ion in there, the base. Which portion of that buffer will the OH- react with? Think about that, and answer the question and then we will move on. Which portion of that buffer will the OH- react with? Think about that, and answer the question and then we will move on. Well, hopefully you said the acid. What neutralizes a base? Well an acid does. So the reason you have an acid in the buffer is so it will neutralize any added base. So the next question is this, we are going to have that exact same buffer that contains a weak acid and its conjugate base in solution that contains a weak acid and its conjugate base in solution so we have that same flask, with that same buffer in it and this time instead of adding a base, we are going to add an acid. So we have an H+ or it could be considered as H_3O+ as added into that flask. What is going to neutralize that added acid. considered as H_3O+ as added into that flask. What is going to neutralize that added acid. Did you say the base portion of the buffer? Well if you did you are correct. The CH_3COO- which is the acetate ion will react with the added acid to neutralize it. So lets kind of examine this with our pictures here the flask in the middle is our original buffer and we see that it consists of a weak acid HA and that is represented with all of these in solution and then we have got our weak base which is the A- and that is represented with all these in solutions. So this solution contains both and that is represented with all these in solutions. So this solution contains both a weak acid and its conjugate base in significant quantities,. Lets heads towards the left. Lets add the base to this buffer. As we add the base to the buffer it is the acid that has to neutralize it. So a base is going to be reacting with an acid in this buffer. As we add one OH-. an acid in this buffer. As we add one OH-. It is going to be neutralized by one of these acids. When this happens then we see what is one the left with the addition of this base. Well, the HA that I have circled here in this flask gets converted to an A-. It is dropping the amount of HA's down. We have less HA's in this flask It is dropping the amount of HA's down. We have less HA's in this flask then we had over here in the original. But it increased the amount of A-. We see that the A-, we have 1, 2, 3, 4, 5 represented over here. Where we only had 1, 2 3, 4 represented over here. So the A- is increasing, the HA is decreasing but we contain a buffered solution. We still have some of both of these. pH might change a little bit but we have removed the strong base that we added in, we have neutralized that so it only changes a small amount. Lets erase all of this and lets go in the opposite direction. Let us start with our buffer and now we are going to add added acid, H+. Or we could think of it as H_3O+. added acid, H+. Or we could think of it as H_3O+. Adding that into our buffer. Or we could think of it as H_3O+. Adding that into our buffer. It is going in. Well which portion of this is going to neutralize the added acid? Well, the base portion will the A- will. And H+ will react with an A- and it will turn into HA. So we see that we had 1, 2, 3, 4 here. As we go to the right we now have only 3. We have dropped the amount of A- down because we have used some of it to neutralize the acid. But we increased the amount of HA. On the left we had 1, 2, 3, 4 in our original buffer. And as the reaction takes place and it reacted and took on the H+. We end up increase the amount. So we see that the HA is going up. The HA- is going down, it is still a buffer. But we used up all that added acid. And it resists change to pH. It may change some because the ratio of And it resists change to pH. It may change some because the ratio of acid and base over here is changing but it won't change a lot because we are taking that acid out of the mix. So you have a couple things to think about here. Why can't you use a strong acid So you have a couple things to think about here. Why can't you use a strong acid and its conjugate base? For example, why can't you just have HCl and sodium chloride? And is it a buffer? and sodium chloride? And is it a buffer? Well the reason you cannot to that, is because the HCl we know 100% ionizes. And you would not have an acid in your solution to neutralize your base. So you would have just a bunch of Cl- floating around in a solution. And you would not have any HCl. So you would have just a bunch of Cl- floating around in a solution. And you would not have any HCl. So that is why you need both And you would not have any HCl. So that is why you need both in a significant quantity, both the acid and the base, in significant quantity, in your solution and when you have a strong acid in your solution you do not have any of the HCl solution and when you have a strong acid in your solution you do not have any of the HCl actually floating around. Now you can't just add any old weak acid, and any old weak base together and make a buffer out of that. You might think any old weak acid, and any old weak base together and make a buffer out of that. You might think well you would have some of both present. But they would just end up reacting with each other. well you would have some of both present. But they would just end up reacting with each other. So if we took a weak acid and ammonia and let them , put them in a flask, they end up reacting with each other. They have to be conjugates because conjugates of each other do not react with each other. They sit on opposite sides of the equation from each other. OK so we learned how to calculate the pH of a buffer. In a previous learning objective. OK so we learned how to calculate the pH of a buffer. In a previous learning objective. Now we are ready to learn how to calculate the pH, after you add a strong acid or a strong base. You do not have to add a strong acid or a strong base to a buffer to see it resist change. But we are only going to ask you to calculate the pH to a buffer to see it resist change. But we are only going to ask you to calculate the pH after adding a strong acid or a strong base because it is a simpler calculation for us to work through. We remember that a buffer resists change to pH by neutralizing that strong acid. Or that strong base in solution. by neutralizing that strong acid. Or that strong base in solution. So we have to write that reaction and that is hardest part about doing these calculations. So we have to write that reaction and that is hardest part about doing these calculations. As we go through and do problems for this as students are doing them for the first time they generally can do all the calculations but they struggle with the very beginning with how do I write the reactions. So really focus on this portion of the lesson. Once we have written the reaction then we are going to follow a common procedure. And I am going to walk you through the procedure. The first this you do is to write the reaction. This is going to be a one way reaction, by that I mean it is that is going to go to completion. The reason for that, is because This is going to be a one way reaction, by that I mean it is that is going to go to completion. The reason for that, is because one of those things that we add either the acid or the base that we add is going to be strong. That strong will react with the weak of the buffer whatever we need. And the strong component pushes this reaction to completion. whatever we need. And the strong component pushes this reaction to completion. With that we are going to do something called a ICF table instead of an ICE table. F stands for final. It is a way of doing a limiting reactant problem that you F stands for final. It is a way of doing a limiting reactant problem that you probably have not seen before, but it is really handy when the ratios are one to one in your balanced equation. Which is always the case here in this chapter. When we have completed the table we are going to look at the F line. And we are going to see that we still have a buffer. If we haven't wiped out the buffering capacity we still have a buffer and as a buffer we can still determine the pH as we would any other and as a buffer we can still determine the pH as we would any other buffer solution. I general choose to use the Henderson-Hasselbalch equation because I have it memorized and I see that is saves me a lot of effort and time. So lets work this problem. We are going to calculate the pH first of just the buffer. So we see where the starting point is. We have the Henderson-Hasselbalch equation telling me pH equals pK_a. That would be the negative log of 1.8 x 10 ^-5 plus the log of the base over the acid. Well, here is the base, it is 1 molar. Here is the acid, it is also 1 molar. If both the acid and base concentration are the same, that is the log of 1, and the log of 1 is zero. So pH is just equal to the negative log the same, that is the log of 1, and the log of 1 is zero. So pH is just equal to the negative log of that K_a value, or the pK_a. pH is equal to in this case 4.74. pH is equal to in this case 4.74. And that is the pH of that buffer. Now for the more challenging part. we are going to add strong base, sodium hydroxide, to that buffer. So we had to think about that buffer and what would react with that strong base. Now the first thing I want you to do is to write this as a net ion equation. We need to get rid of the spectator ions because they just complicate things. They make it difficult to write a good equation. As a strong base it 100% breaks apart. We are not going to worry about the sodium ions. Any time I have a strong base I am going to write in my react OH- Any time I have a strong base I am going to write in my react OH- to represent that strong base. Then I consider my buffer. So look back at you notes and look at what is in that buffer and which component of that buffer would neutralize that strong base. Well it would be Acetic acid. The job of the acid in that buffer is to neutralize any added base. The job of the acid in that buffer is to neutralize any added base. Since the base is strong it is going to be a one way reaction. Now don't try to memorize the products. Realize that we have a base and we have an acid Now don't try to memorize the products. Realize that we have a base and we have an acid so we just to swap the proton. The acid will donate to the base so we just to swap the proton. The acid will donate to the base when the acid The acid will donate to the base when the acid gives it proton to the base we are left with CH_3COO- and water. So we have our reaction that like I said is the most challenging part so you might want to pause and really think about why did I write like I said is the most challenging part so you might want to pause and really think about why did I write that reaction the way i did. pause and really think about why did I write that reaction the way i did. Since this is a one-way reaction we are going to do an ICF table instead of an ICE table. And an ICF table is a limiting reactant type problem and it is really handy to use moles instead of molarity. in the equation. So lets get the moles of OH- first. The moles of OH- would be equal to the moles of NaOH because there is one hydroxide in there. To get the moles, we will take the molarity of .10 mole per liters, times the volume in liters. of .10 mole per liters, times the volume in liters. it is 1 milliliter or .001 liter. and that is going to give me 1.0 x 10 ^-4 moles. And I can place that into my table. For the moles of the acetic acid we look at the buffer solution we know that it is going to be the molarity it is not .1 molar it is 1 mole per liter it is not .1 molar it is 1 mole per liter times how many liters of that buffer. Now they tell me up here in the problem that I had 100 milliliters. So that is .1 liter. That is going to give me .1 moles and I can place that in my table. Now, this is a buffer so it also has the acetate ion present. That came from the sodium acetate. So the moles of the acetate would be the same as the moles of sodium acetate. So we take the molarity times the volume again, it is 100 milliliters or .1 liter. And that gives me .1 again, it is 100 milliliters or .1 liter. And that gives me .1 moles of the acetate ion. And I put that into the table. moles of the acetate ion. And I put that into the table. I really don't care about the water, so I scribble that line out. Now here is how you do a ICF table. You don't variables, you don't use X's. What you do, is say this reaction is going to You don't variables, you don't use X's. What you do, is say this reaction is going to go until it is finished. What you do, is say this reaction is going to go until it is finished. And it can only go until it uses up go until it is finished. And it can only go until it uses up the least of the two reactants. And it can only go until it uses up the least of the two reactants. So look at those reactant quantities there we have 1 x 10 ^-4 and .1 and we know this reaction is going to go until we use up all of the lesser one. And thats will be 1.0 x 10 ^-4. Now as in an ICE table the change line is the reaction taking place. Now as in an ICE table the change line is the reaction taking place. So we are going to use it up change line is the reaction taking place. So we are going to use it up the same quantity of this So we are going to use it up the same quantity of this and produce 1.0 x 10 ^-4 of that. That is going to consume all of the added base. That is why a buffer works it uses up and consumes the added base. That is going to consume all of the added base. That is why a buffer works it uses up and consumes the added base. But it effects the amount of the others. So if I take .1 and use up a little bit. I am going have to .0999 and really to three decimal places it is still .100 but i am going to carry extra places. And I have got .101 over here. Again according to the rules of significant figures those two number won't change at all. .101 over here. Again according to the rules of significant figures those two number won't change at all. But we will carry extra until we finish the problem. At this point we see that we still have a buffer We have some weak acid. At this point we see that we still have a buffer We have some weak acid. That is the weak acid. And we have some weak base. That is the weak base, so it is a buffer. And as we can use the Henderson-Hasselbalch equation. pH equals pK_a. We got the pK_a from that previous problem. pH equals pK_a. We got the pK_a from that previous problem. So I will go ahead and write that down plus the log of the concentration of base over the concentration of acid. Well these are not concentration, these are moles. But I am going to tell you something that I haven't told you yet. In the Henderson-Hasselbalch equation we could put moles of base over moles of acid instead of concentration of base over concentration of acid. And I am going to show you why that is here in just a minute. But let us go ahead and think of the numbers for the base, we have 0.1001 and for the acid we have 0.0999. Now we put those numbers in we are going end up not changing the pH at all. Now we put those numbers in we are going end up not changing the pH at all. Take the log of that ratio, it is so small that it doesn't change the value of the pH. Now I said I would mention why you can use moles. Lets say we couldn't use moles. Lets say we have to take this term on the right and put in concentrations instead. Well how do we go from moles to molarity. We divide by volume. So we take our moles of the base and divide by the total volume. Total volume would be 100 ml of the one. 1 milliliter of the other Total volume would be 100 ml of the one. 1 milliliter of the other converted to liters because we need moles per liter and we would have converted to liters because we need moles per liter and we would have .101 liters. Lets do that for the acid. It would be moles of acid over volume and it would be the same volume. Since I am dividing the numerator and denominator by exactly the same volume Since I am dividing the numerator and denominator by exactly the same volume I can use the ratio of moles instead the ratios of concentrations. So know that the Henderson-Hasselbalch equation you can plugs those moles in, instead of molarity. So since moles comes out of the table up here you can plugs those moles in, instead of molarity. So since moles comes out of the table up here I can plug the moles directly into the Henderson-Hasselbalch equation. and obtain a value. Now you might think well of course it didn't change. Of course it resisted change. We have a very low number of volume, 1 milliliter being added to a whole 100 milliliters of solution. And the concentration of that base is very very low compared to the concentration of the buffer which is 1 molar. And you say surely that would not have affect the pH no matter what. Whether it is a buffer or not. So the next question is to counter that argument. If I were to add the same quantity So the next question is to counter that argument. If I were to add the same quantity of sodium hydroxide to just water would it also barely change the pH? Well, lets see. pH of pure water is 7, we know that. Well, lets see. pH of pure water is 7, we know that. So maybe this would change the pH of the pure water. Well, this is just a dilution we are adding sodium hydroxide to water. So I am going to use a dilution equation. M_1 V_1 equal M_2 V_2. The molarity of the base to begin with was .1 molar and I had a milliliter of it. The molarity of the base after it has bee diluted I am trying to solve for and the total volume after dilution was 101 milliliters. So the molarity of the sodium hydroxide after this dilution is equal to So the molarity of the sodium hydroxide after this dilution is equal to 9.9 x 10 ^-4 molar. It brings that concentration down quiet a bit. What is that the concentration of? That is the concentration of sodium hydroxide or after it is dissolved sodium hydroxide or after it is dissolved and we know it breaks apart, it is the concentration of the hydroxide. or after it is dissolved and we know it breaks apart, it is the concentration of the hydroxide. So pH would be the negative log, no not pH. pOH. pOH would be the negative log of that number. And the pOH is 3. The pH would be 14 minus 3 which would be 11. So yes, a tiny amount of sodium hydroxide added to water would take the pH from 7 to 11, that is not resisting the change of pH. You add that same amount to a buffer we saw no change what so ever. For our next problem we are going to add an acid to the same buffer, so we have to figure out how to write this reaction and follow the same procedure. Whenever I came across a strong acid. Rather then writing it as HCl in my reaction I will write it as H_3O+. Because it 100% ionizes to I will write it as H_3O+. Because it 100% ionizes to produce H_3O+. It is the H_3O+ that gets neutralized by the base of the buffer. So look back at your buffer and identify the base. We have a clear question about that This is the base that is what neutralizes the acid. It is one way reaction because this that is what neutralizes the acid. It is one way reaction because this strong acid pushes it to completion and we end up producing. by a proton swap the acid the H here is being donated to the base and we are left with water as the other product. We will go an ICF table because it is an one-way reaction. We will plug moles into the table So lets start by determining the moles of H_3O+. That would be the same as the moles of of H_3O+. That would be the same as the moles of HCl that I put in there. They gave me the molarity Moles per liter of HCl time the volume in liters of HCl time the volume in liters is going to give me 0.004 moles. Or 4.0 x 10 ^-4 moles. So it is way more moles then we had sodium hydroxide maybe this will change the pH. We are going to put 4.0 x 10^-4 here. The same buffer we had before and the moles we determined before were .1 The same buffer we had before and the moles we determined before were .1 and I don't care about the water. So these are being added to the same buffer we have been working on all along. OK we are going to consume the smaller of the two reactants. That is the 4.0 x 10 ^-4. Carry that across and the change line according to the balanced equation this represents the reaction actually happening. until I use up all of the added acid. That is the reason a buffer works we are neutralizing reacting out, consuming all of that added strong acid. But that does change the amount reacting out, consuming all of that added strong acid. But that does change the amount of the buffer components so this is going to drop down .0996 this is going to be increased to 0.1004 and we are ready once again to use the Henderson-Hasselbalch equation. pH equals pK_a. Well I am using the same buffer it has got a pH a pK_a of 4.74 plus the log of the number of moles of base. So we come up to our F line and there is the base I can use the number of moles of base. Of the number of moles of acid. And guess what this still does not change it. It takes it down ever so slightly but but to two place after the decimal point which is all we can really know it but but to two place after the decimal point which is all we can really know it it still is .74 . If we had, maybe added 4 milliliters instead of 2 milliliters I think we would have start getting enough in there that the ratio of base of acid would have change enough to actually bring the pH a little bit down. Now when we are adding this acid, eventually that is what is going to happen. pH will drop down Now when we are adding this acid, eventually that is what is going to happen. pH will drop down as the acid concentration goes up, becomes a buffer and the base concentration goes down enough, we will see it drop. So that ends learning objective number three. You have learned how to calculate the pH of a buffer after you add an acid or a base to that buffer it is a challenging procedure. Not to do the work but to start off with a good reaction. You have got to remember that if you add an acid into the buffer it reacts with the base of the buffer. If you add an base to the buffer, it reacts to the acid in that buffer. You have got to write a good first reaction and then you base to the buffer, it reacts to the acid in that buffer. You have got to write a good first reaction and then you can proceed with that ICF table. You will have extra problems that you can watch and work through that are posted in this lesson as well. And once you have done that, work on those practice problems.
