We are ready to begin our next round of learning objectives dealing with acid and base titration. within these acid/base titrations we are going to learn to calculate the pH At various points along the titration. Now if you will think back to along the titration. Now if you will think back to what you have learning in the past you have learned about acids and bases titrations, most likely. You might want to go back and review what you have learned. But one of the things you learned to do with acid and base titrations is you are given an acid where you know the concentration and you are trying to determine the concentration of an unknown base. You know the concentration of a base and you are trying to determine the pH or the concentration of an unknown acid. And its a stoichiometry problem where you are covering from the molarity and the volume of the known or the standard solution getting into moles of that substance and determining the moles of the other substance. Within these units that we are talking about here Within these units that we are talking about here and the next couple of learning objectives we are going to look at how pH changing during the course of this. We are going to look at three types of titration. We are going to begin in this lesson with strong acids and strong base titrations. This will be learning objective number 5. In learning objective number 6 we are going to have one of those be weak. Either the acid will be weak or the base will be weak. be weak. Either the acid will be weak or the base will be weak. We will never do calculations in which both the acid and the base are weak. There are very many equilibria involved and it gets very complicated. So we are going to do make sure that at least and it gets very complicated. So we are going to do make sure that at least one of those two will be strong. So lets being, our learning objective number 5 will be calculating the pH of a strong acid and strong base titration and any point along the titration, from the beginning to the, what we call the equivalence point and beyond. What happens when a strong acid and a strong base react? We we can write the reaction as follows. Here is an example of a strong acid and a strong base. The cation of the base and the anion of the acid will form the salt. And the H of the acid and the OH of the base will produce the water. We are going to consider just the net ionic equation. It would be H+ of the acid and OH- of the base combining to form water. Or we could and I am going to back this up, we could write this typically will write this with H_3O+ for the strong acid with H_3O+ for the strong acid and OH- for the strong base. And then the reaction is just a little bit different we have two waters. Because remember when a strong acid dissolves, you could either say Because remember when a strong acid dissolves, you could either say it 100% ionizes to produce those ions. Or you could say, it is to produce those ions. Or you could say, it is reacting with water to 100% give you the anion NaH_3O+ so either way, and I typically choose writing strong acids as H+. Now this is a one-way reaction. And with a one-way reaction we are going to use what is called an ICF table and we saw those tables we when were dealing with adding and acid to a buffer or a base to a buffer. We know that it is a one-way push it to completion type of reaction. So lets look at how pH is changing during the course of a titration prior to actually getting into the calculations. Here is a schematic of the reaction taking place. We have in the flask a strong acid. Here is a schematic of the reaction taking place. We have in the flask a strong acid. That is our .1 molar HCl. And lets say we have 25 ml of it. We are going to start adding base to this and I have got a sodium hydroxide in the portion I am interested in is the OH- of that base. And here is our reaction H+ and OH- forming water. I am going to plot, and kind of determine pH at every point along this titration as the base is being added. At the very beginning before I have added any base what so ever We have got the pH of this strong acid. Well, if it is .1 molar We have got the pH of this strong acid. Well, if it is .1 molar and we take the negative log of .1 we will have a pH of 1. So we have a red dot there, if you look at the graph at one, the beginning. Now we are add a little bit the base, and we are going start adding that base Now we are add a little bit the base, and we are going start adding that base into the flask. When a little bit, look at the base up here into the flask. When a little bit, look at the base up here when a little bit of the base is added it will react with some of the acid. And the base will go away, the base will be turned into water. But there will be a lot of acid left. But not as much as we started with. So the concentration of that acid is going to be dropping a little bit. of that acid is going to be dropping a little bit. during the course of this titration. Well, if that concentration is dropping a little bit the pH is going to come up a little bit so we see a slight rise of the pH. Well, we keep adding and we keep adding and it going to rise and rise and rise. But it is not changing drastically just slowly increasing as the H+ concentration is dropping a little at a time as we continue to add a little bit base and add a little bit of base. But there will come a time where the amount of acid and the amount of base are exact equal the amount of acid and the amount of base are exact equal and we have added just enough OH- to react with all of the H+. At that point we have reached what is called the equivalence point. There number of moles of base is going to be stoichiometrically equal the number of moles of acid. So there are a certain number of moles of acid in here right, we had a certain number of moles and we kept on adding until all of those right, we had a certain number of moles and we kept on adding until all of those we finally converted to acid and we had the picture and all the acid had been converted to water, and we have the right on the right. When the titration is between a strong acid and a strong base When the titration is between a strong acid and a strong base you will have converted all the H+ and OH- over to water. And the pH of water is 7. and OH- over to water. And the pH of water is 7. So always the pH is going to be 7 at the equivalence point when they are both strong. But we can keep on adding we can go beyond the equivalence point and what happens when we continue to add base? If we were take this system and add some more base to there, suddenly we got base present. We already have consumed all of the acid. base present. We already have consumed all of the acid. There is no acid left to neutralize. so the pH is going to shoot up. It is going to shoot up very rapidly and then just gradually increase as we add more and more base to the flask. So we are going to be able to calculate the pH at any point. So lets go back to this procedure. This is a similar procedure at any point. So lets go back to this procedure. This is a similar procedure that we saw when we added an acid or base to a buffer. The first thing we will have is a one way reaction in between the acid and the base. We will use an ICF table and put moles into the table which is a way of doing a limiting reactant problem. We will examine that final line and determine our next course of action. At this point there could be three possibilities of what could happen. It there remains a conjugate base pair then we would use the Henderson-Hasselbalch equation. This would never happen when we are in this portion of our lesson, because we are dealing with strong acid and strong base. There will never be a weak thing present and its conjugate. But if there is a titration and it is weak then this would occur prior to the equivalence point and we will come back to that later. If there remains only a weak acid or a weak base present this occurs when we are at the equivalence point if one of them is weak. Again, in this portion of our lesson this will not happen. What will happen in this portion of the lesson is there may remains some strong acid, or some strong base, and we put the word strong here as well. some strong acid, or some strong base, and we put the word strong here as well. And when that happens you would calculate the pH directly from the molarity of that strong acid or strong base that is left over. With this procedure we are going to hold and ready these two portions there are not going to happen when they are both strong. But we will see them later when we are in learning objective number 6. OK we are going to determine the pH beginning with a small amount of sodium hydroxide being added to the acid. So we are only adding 5 ml. Begin by writing the reaction. The strong acid I use H_3O+ to represent the strong acid. I let OH- represent the strong base. I will form 2H_2O. I will do an ICF table according to my instructions. I will form 2H_2O. I will do an ICF table according to my instructions. One way go to completion. Now I need the moles of H_3O+ and the moles of OH-. Now I need the moles of H_3O+ and the moles of OH-. Lets begin with the moles of H_3O+ coming from the acid. The molarity of HCl coming from the acid. The molarity of HCl is .1. If I multiple by the volume which is .025 liters that will produce from me .0025 moles of H_3O+. and I can put that into the table. For the moles of base I see that the sodium hydroxide has one OH in it. So the moles of OH would be the molarity I see that the sodium hydroxide has one OH in it. So the moles of OH would be the molarity of the sodium hydroxide times its volume, again in liters 0.0050 liters. That is going to give me 0.005 moles of base. And I can plug that into my table. I don't care about the water, I am just going to put a line through that. I am going to consume the smaller of the too. That is the base. I am going to consume the smaller of the too. That is the base. I would have, of course, produce some of it. I had a lot to begin with a few more moles that is still going to be a lot there and I will be left with that is still going to be a lot there and I will be left with none of the base. So at this point, I have H_3O+ in solution, acidic solution, and it is a strong acid. So I can obtain the pH by taking the negative log of the H_3O+ concentration. by taking the negative log of the H_3O+ concentration. Well, this is not the H_3O+ concentration. I plugged everything in this table using moles. So this is the moles of H_3O+. So if I want to know the concentration of H_3O+ I have to divide by the volume. So if I want to know the concentration of H_3O+ I have to divide by the volume. And I want to divide by the total volume. So I need to take the negative log of the 0.002 So I need to take the negative log of the 0.002 moles and I have 25 ml of the acid 5 ml of the base so that is .030 5 ml of the base so that is .030 liter, 30 ml, or .030 liters. When I divide those numbers out it is equal to 0.067. That is the concentration when I take the negative log of that That is the concentration when I take the negative log of that value is 1.18. So that is the pH of the solution. So it should be acidic. We are before the equivalence point we neutralize all the base We are before the equivalence point we neutralize all the base and we have converted the base to water but there is still a lot of extra acid in there. Now lets get to the equivalence point. We are going to add 25 ml of the base to the 25 ml of the acid. Since there are both at the same molarity, we know we are at the equivalence point. But lets imagine that you didn't realize that. And you start redoing the problem But lets imagine that you didn't realize that. And you start redoing the problem blindly as we have been before. Eventually you are going to recognize you are at the equivalence point. We would take the molarity and the volume of the acid. It is the same molarity of the same volume of acid we had before. So we are going to place that number .0025 here. So we are going to place that number .0025 here. And if wanted to figure out the moles of hydroxide we do the same thing. We take the molarity times the volume in liters and we see that is it .0025 moles. So at this point you might recognize that hey we are at the equivalence point. But lets say you didn't at this point. This is what I see some students doing. They just keep on working through the problem and they have a zero This is what I see some students doing. They just keep on working through the problem and they have a zero here and a zero here. They just keep on working through the problem and they have a zero here and a zero here. And then they scratch their heads and say I have nothing to base my calculation on. Keep in mind that if you neutralize all your acid and base you will only have water and what is the pH of water? It is 7. So at some point you are going to recognize you are at the equivalence point. And whenever you are at the equivalence point between a strong acid and a strong base the pH is 7. strong acid and a strong base the pH is 7. OK now I am going to have you stop the video after a bit and try to work through this problem. stop the video after a bit and try to work through this problem. We have got a titration between a strong acid and a strong base. The same one. We have got a titration between a strong acid and a strong base. The same one. But now we have gone beyond the equivalence point and added excess sodium hydroxide. So I want to see if you can calculate the pH of this solution. Well, if you came up with 11.96, very well done. That is the pH of that solution. If you struggled with that I am going to go through how to work and get that 11.96. You might not need to see this in which case you can of course skip ahead. So we have our reaction that we have been working with. H_3O+ plus OH- producing 2 waters. We have our ICF table the amount of acid hasn't changed in any one of these problems. the amount of acid hasn't changed in any one of these problems. But now the amount of in any one of these problems. But now the amount of base has certainly changed. 30 ml of .1 molar base has certainly changed. 30 ml of .1 molar is 0.0030 . So we have more of the base then we have of the acid. So know when we consume smaller amount it is the So know when we consume smaller amount it is the acid that going away. We are left with 0.005 moles of the base. Since it is a base that is left over we can first calculate the pOH. That would be equal to the negative log of the hydroxide concentration. Well this is just moles so we have of the hydroxide concentration. Well this is just moles so we have to divide by the total volume. The total volume would be 25 ml of the acid The total volume would be 25 ml of the acid and 30 ml of the base thats 55 ml or .055 liters. That would be the negative log of 0.009. That gives me a pOH of 2.04. Of course the pH is simply 14 minus that 2.04. That is how we get the 11.96. is simply 14 minus that 2.04. That is how we get the 11.96. Now hopefully when you saw this problem, you realized it had to be basic and you eliminated these two answers right off the bat. But you could not distinguish between the other two. Without first doing this process. So that is the end of learning objective 5 in which we determined the pH when the acid and the base are strong. we determined the pH when the acid and the base are strong. We have a procedure to follow. We do an ICF table there will always be either at the a procedure to follow. We do an ICF table there will always be either at the equivalence point in which case the pH is 7 no calculation is needed. Or you have got more acid strong acid in there, or more strong base in there and you can calculate of either the acid or the base using the negative log of that concentration.
