OK we are ready to start titrations between an acid and a base where one of them will be weak. We will never do it where both are weak, but where one of them is weak and one is strong and so we are going to begin here with a weak acid and a strong base titration. Acetic acid is first substance mentioned here, and this is a weak acid. It is not on our strong acid list, it is a weak acid. And sodium hydroxide is a strong base. So we will begin by writing the reaction between the weak and the strong base. I will never include the sodium ion, it is just a spectator ion and we will write the net ion equation for this reaction. This is biggest challenge for students it to write a good reaction and know how to work the problem from there if you have a good starting point it is not a problem. So we write our acid we write OH- for our base and we do a proton swap. The acid donates the H+ to the OH- and produces the water, and we are left with the conjugate base of acetic acid. Now once this equilibrium occurs, then we have a point where all of the acid has reacted with all of the base, then we have produced this conjugate base in solution. This conjugate base, which is the acetic ion will undergo an equilibrium in water. Now this reaction of it in water is in equilibirum so we put the equilibrium arrows, again it is just a proton swap. The water is behaving as the acid and it will donate one of its protons to the anion here to produce acetic acid and hydroxide. So at the equivalence point we have this weak base that has been produced. And what do we mean be equivalence point? Well we have just consumed all of the acid and all of the base. We have produced this conjugate base, the acetate ion. It is a weak base. It will re-establish an equilibrium with its conjugate acid in water. Because of this reaction here, the solution is basic at equilibrium. OK so we are going to look at the titration curve of a weak acid / strong base titration. Now I have changed my weak acid to formic acid here, HCOOH but it is a weak acid again. I am titrating it with a strong base. So we have got ourselves a little flask and in this little flask there will be the acid. And it will be acetic. And we see that at the very beginning here where we are just looking at the pH of this weak acid. That we have an acid pH. And if we wanted to calculate the pH of that acid we would work a problem very similar to what we did in the last unit. How to calculate the pH of a weak acid. So that is at the very beginning. Then we have our burette. And in our burette, with a little stop-cock here we put the sodium hydroxide. And we start adding a little bit of sodium hydroxide. And we will put a few drops in. OK a few drops in here of our sodium hydroxide, and lets consider what happens. Look back up here at the reaction. If you add a little bit of the hydroxide to a bunch, a lot of the acid. Then the little bit of that base is going to get consumed and go away, and we won't have any of that base. And we will have converted some of this HCOOH- and produced some of the HCOOH. So this will be consumed, and we won't have any. And we will still have some of this. And we will have some of that. And what do we have? Well, we will have some of both. Well we have a weak acid and its conjugate base in solution. When you have a weak acid and a conjugate base in solution you are going to have a buffer. So when you are prior to this equivalence point, you are going to be in a buffer range. You note here that is kind of jumps up and sort levels off before it rises again and we are resisting change of pH during that range there. Well, eventually we will get to the point where, we have added, exactly the same amount of base to consume all of the acid. We will have converted all of the acid over to the formate ion. Well, what do we have there? Well, at the equivalence point. We have a weak base. This is the weak base present right there. And so it should have a pH above 7, which it does. So we have just consumed all of the acid and all of the base that was added into the flask. And we have a conjugate base and a weak acid, and we would calculate the pH at that point based upon that weak base. Now if we keep on adding hydroxide beyond that point we will produce, we won't have anything for that to be neutralized by. So at that point we will have a strong base and a weak base in solution. So we would calculate based off this strong base because it would really dominate how to calculate the pH. So that is what a titration curve looks like when you are doing a weak acid and a strong base. So now we are ready to consider an acid being strong a base being weak. This is still apart of learning objective 6 where we doing a titration where one of them is weak and one of them is strong. We are going to consider this reaction here HCl and Ammonia and do a titration between these two substances. When I write this reaction I will always write the strong acid as H_3O+. Not as HCl. I want to get rid of the spectator ion, which causes confusion when you are trying to write the reactions. So anytime we do problems with acids and base I will write H_O+ to be my strong acid. Lets just think why that would be? HCl is in water and when it is in water it will 100% ionize to give you H_3O+ and Cl-. Well, Cl- is just a spectator ion. So whatever the concentration of HCl is in there that is concentration of H_3O+ and that is what I will write when I write my reaction. So I have a strong acid and a weak base. And I do not have to memorize the products I just think about the acid donating to the base. So the H+ that is on this acid, will be donated to the ammonia H_3O+ becomes water and ammonia becomes ammonium NH_4+. Now when we are at the equivalence point and we have NH_4+ formed and we have just consumed all of the acid and all of the base. This NH_4+ will then need to undergo equilibrium with water. So lets have you think through what NH_4+ should do. First of all, what is NH_4+ stop and answer. If you said it was an acid, you are correct. It is positive charge and a H+ that it can donate. So if it is an acid when we to think about the reaction that acid will do in water. And you have to be able to write that reaction. We write NH_4+. We place it in water. We know we will have to use equilibrium arrows because this is a weak acid. We will do the proton swap, we won't memorize products. We will say the acid will donate to the base. And if the acid donates it is going to form it conjugate base NH_3 and the water will accept that proton and become H_3O+. So this is an acid, it behaves as an acid it will give us a pH less then 7 at the equivalence point. So lets look at this titration curve. We beging with our flask and in my flask is the ammonia, the weak base. So it has a high pH. There is the pH.and in my flask is the ammonia, the weak base. So it has a high pH. There is the pH. Depending on what the concentration of that base is it will have some pH. We start adding the HCl. So lets write the reaction that is going on again. We have our NH3, our H_3O+. We are pushing this to completion and we have NH_4+ and water. So we are adding from out burette the HCl. It is going into the flask and the first think that is going to happen is some of your ammonia is going to be converted into to ammonium. We are going to have a ridge in here once again that is a buffer range. This is our buffer range. Now something that I didn't mention on the previous acid base titration, where one is weak and one is strong is what happens when you are halfway to your equivalence point. When you are halfway to your equivalence point the pH is going to be equal to the pK_a of the acid, the weak acid in solution. Well, the weak acid in my solution is the ammonium ion. So the pH is going to be the pK_a of ammonium. Lets think about why halfway. When you are halfway to the equivalence point you will have converted half of your ammonia over to ammonium. And you will have equal amount of both the ammonia and the ammonium. And if we are a buffer we know that the pH equals pK_a plus the log of the concentration of the base over the concentration of the acid. When you are halfway to the equivalence point the amount of base and the amount of acid are equal therefore the log of 1 is 0 and that gives you pH equal to pK_a. So that is where you will be always at halfway to the equivalence point. When you are at the equivalence point however you have formed only ammonium. So the pH would be below 7.you have formed only ammonium. In this problem, the pH is 5.26 at the equivalence point. It is acidic, because ammonium is acidic and you would calculate the pH by doing a problem based upon a weak acid in water. So we are going to consider a problem here where we are going to calculate the pH when 10 ml of sodium hydroxide is added to 30 ml of 15 molar and that is formic acid. You can't do these problems unless you can write a good reaction. Now I want you to come up with the net ionic reaction between the acid and the base. Stop and think about those substances and choose which one would be the net ionic equation. Well, hopefully you picked number three. But if you did not lets talk about it, because it is the most important step. A lot of students can do the calculations just fine but they don't know how to start the problem.step. We consider the sodium hydroxide and we have to think about what that is. It is a strong base. I know it is a strong base because it has that OH- a metal with OH- in there. I get rid of the sodium because it is a spectator ion and I write OH-. I think about HCOOH. And I see that COO portion, I know it is a weak acid. It is an Organic weak acid and it is going to donate that H at the end. So this is a weak acid. Now you don't memorize the products you just think acid and base will swap the H+. The H+ is this H. Not the one on the left. It is going to donate to the OH-. If the OH- takes on an H+ it is going to make water. And if the acid donates it is left with its conjugate base ion right there. So we did an proton swap and formed those products. So now we ready to do the calculation. We start we our reaction OH- plus HCOOH. We have our products and since it is a one-way reaction I am going to do an ICF table. With an ICF table I am going to figure out the moles because that is what you put into an ICF table is moles. The moles of hydroxide. Well since there is one hydroxide in sodium hydroxide it would be the same as the moles of sodium hydroxide. This is going to be molarity of the sodium hydroxide times the volume of the sodium hydroxide in liters and that is going to give me .0010 moles. And that is what we would place here. For the acid again we will take the moles. To get the moles of the acid, we will take the molarity of the acid times the volume of the acid it is 30 ml or .03 liters. And that is going to produce .0045 moles of the acid. So we will put it here. We don't have any of the formate ion yet. And I do not care about the water in my table. An ICF table I consume until I consume the smallest quantity. So I am going to consume .0010. And produce those quantities. So that will have consumed all of the strong base. I will still have this quantity of weak acid and I will have quantity of its conjugate base. Well that is a buffer. And as a buffer we will use the Henderson-Hasselbalch equation. pH equal pK_a plus the log, and since the table is in moles, we will just use the number of moles of base over the number moles of acid rather then going ahead and converting them to concentration. The pK_a is the negative log of the K_a which is given there at the top of the screen. And we have to consider which one is the base. This is the base, we put it on top. This is the acid, so we put it one bottom. And this will give me 3.77 minus 0.54 and the pH would be 3.23. So we are before the equivalence point, we are in a buffer that is buffer based upon an acid and its conjugate base so it should be acidic and that is the pH. So now we are going to continue on. We are going to go to the equivalence point.So now we are going to continue on. We are going to go to the equivalence point. To be honest, calculating the pH at the equivalence point of an acid base titration is the most difficult. It is not impossible. I wouldn't be teaching it, if it was impossible but you have a lot of work to do. And you might not notice that you are at the equivalence point sometimes until you are working through the problem. This problem is stating right up front that we are at the equivalence point. So lets see why this is challenging. Well we begin the problem as we did as before. We have the same reaction, it is a strong base and a weak acid. One-way reaction to produce the conjugate base and water. We do our ICF table. Now we don't know how much we have in there yet. but we do know how much acid is in there. And this is the same amount that we had before .0045 . So that is not changing. Now we are continuing to add base until we reach the equivalence point. By definition the equivalence point is zero. Is going to be the same moles of the base as the acid we have put in the same exact number there. So our line looks like this so we have added enough base to consume all of the acid. And so we going to subtract both of these numbers. Because are the same and produce on this side and that will have consumed the reactants, and I only have a weak base. We have a weak base in solution. And we have only a weak base in solution. So anytime you are at the equivalence point this is what it is going to look like. You are going to have zeros between these two because at the equivalence point you will just consumed all of your weak acid with a strong base in this case. And you will have only a weak conjugate sitting over on this side. And so this is a weak base, we need to write the weak base reaction. So we place that weak base, knowing it is in water, we react it with water. We do equilibrium arrows because it is weak. And we proton swap we take the H off the water and put it on the base. The base is a proton acceptor. The water is left behind as an OH-. Since this is at equilibrium we don't do an ICF table, but we do an ICE table. And we have to put in, molarities into this table. Where this was moles. So we have to do a little work in order to figure out the molarity. What we have to know is the total volume of that solution. Well we know we have 30 ml of this solution. What we don't know is how many milliliters of the base we had to use in order to get to the equivalence point. So, keeping in mind we are trying to calculate the molarity of the base. We know how many moles we have. We have to divide by the volume in liters. We know we have 30 ml or .30 liters of the acid. So to figure out the liters of the base we are going to have to consider the fact that we added this many moles of base and we know the molarity of the base. So if we know moles and molarity we can get the volume of the base we added. And I will do that down here at the bottom. The moles of the base was .0045 that we had to add to our flask to reach our equivalence point. What I want to know is the volume. Well the molarity of the base was .1 mole per liter. That is going to give me 0.045 liters. And I can put that here. So I now I know the total I know the moles of the base in there and I know the total volume that is in there. And I can divide those numbers and I will get a molarity equal to 0.060 molar. And that is will go in this table. I don't care about the water. Before the equivalence point this is 0. And Hydroxide we will put 0.Although we know that in water its 1 x 10 ^-7 but it is small enough to consider it a 0. I will consume all of this. Produce x's on this side. Give me 0.60 - x. X and X [student don't have trouble filling in that table]. You have a lot of practice doing an ICE table. But at this point we are ready now to set it equal to our K expression. Well we have to keep in mind that what weset it equal to our K expression. Well we have to keep in mind that what we got sitting right here is a base. This is a base reaction so we are going to need K_b. And K_b is going to be the HCOOH concentration times the OH- concentration over the HCOO- concentration. Well, K_b, we weren't given, but we were give K_a.over the HCOO- concentration. Well, K_b, we weren't given, but we were give K_a. And K_b would be is 1.0 x 10 ^ -14 over the 1.7 x 10 ^-4. And that is going to be equal to x squared [x^2] over 0.060 - x. Now I recommend you stop the video and work through and solve for x. And then resume it once you have an X.the video and work through and solve for x. And then resume it once you have an X. Well did you get an x of 1.9 x 10 ^ -6 ? That is what you should have obtained. Once you have determined that you would think I have determined X what does X represent? Well here is X. And X represents the OH- concentration. So if I took the negative log of that value that would give me the pOH. The pOH equals 5.73 and how do we get the pH? Well, the pH equals 14 minus the 5.73. And that is going to be a pH of 8.27. Now lets stop and think if this makes sense. Should the pH at the equivalence point be a basic pH? Well certainly, because we produced a base. That is the substance in this solution when we have reached the equivalence point. So having a pH above 7 is what we should see. Here is a calculation in which we have a strong acid and a weak base. Now I have not given you the formula for that weak base it is actually HC_3NH_2 but if you do not know the formula for your base you can either look it up or use a B to represent a weak base. Now we are going to calculate the pH to the equivalence point between those, so we are going to write a good reaction. We have H_3O+ as our strong acid and here is my weak base and it is a one-way reaction because the strong forces it to completion. The acid donates to the base and turns the base into it conjugate BH+. And the acid becomes water. We will do an ICF table but before we do this, and we are going to do this at the equivalence point. Lets think about what the pH at the equivalence point should be. At the equivalence point we will just have consumed all of the acid and all of the base. And we would have converted it to BH+ and BH+ is what? It is an acid. And as an acid it should have a pH below 7. So we will keep that in mind and make sure we have a pH below 7 when we are finished. So we need to put into our ICF table what? Well, moles. So we have to put moles in. Now they did not give me a volume to use, so we can choose any volume. The pH at the equivalence point is the same whether you a big vat of it, or a small quantity of it. So I am just going to choose a quantity. I am going to choose that I am going to have in my flask, 10 ml of that acid. Now if we are at the equivalence point we are going to have equal moles not equal volume, not equal molarities but equal moles. So we will worry about the volume of the base that we need to add here in just a minute. But lets take the 10 ml of the strong acid. So to obtain the moles of H_3O+. It would be the same as the moles of that strong acid. And that would be the molarity times the volume, I and I choose 10 ml. And we will have 0.0010 moles of the strong acid. And we can place that into our table. Now that we can know that if we are at the equivalence point we have the same moles of that substance. For the equivalence point we have none of that and the water doesn't matter. We consume all of the reactants and produce that quantity of the BH+. So we always have at the equivalence point zeros on the left side. and we have 0.010 here of this weak acid. When we are at this point in our titration you might want to go back and look at the procedure to follow and consider what we have to do next. If it is a weak acid only then we have to follow this ICF table with an ICE table. We take the weak acid, it is placed in water it re-establishes an equilibrium with its conjugate base. And when this donates to the water it will produce H_3O+. This is an ICE table and we have to put molarity in there. So we have to figure out the molarity of the BH+. Well,to get the molarity of BH+ we need the moles of BH, which is sitting right here. 0.0010 moles divided by the total volume. Well, I choose to have 10 ml, and I need to write it as liters which is 0.10 liters of the acid. And that was my choice. And I will have to add how many liters of the base were needed in order to get to the equivalence point. Well if we look at our numbers here I have half of the concentration of the acid, that I have base. So I am only going to need half of that base. So I am only going to need 5 ml. But lets say, that didn't come to mind. But you wanted to use the procedure we used before. We look in our table and we see how much base there is. We have 0.0010 moles of the base. We look in our table and we see how much base there is. We have 0.0010 moles of the base. And we say I don't want moles I want volume, and the molarity of base is .2 moles per liter. And that tells me it is going to be 0.0050 liters, or 5 ml. I place that there, so now I can know the molarity of that solution. So the concentration is 0.067 molar. Now we can place that into our table down here at the bottom because we need that value in order to fill in our ICE table. 0.067 I don't care about that, these are zero. Minus x, plus X, plus X 0.067 - x, x, x. And now this is a what? Is this an acid or a base? It is an acid, so we are going to need the K_a is equal to the B times the H_3O+ divided the BH+ we plug in what we know. We don't know the K_a, but we know K_b, so we use 1 x 10 ^-14 divided by the K_b 4.4 x 10 ^-4 and that is going to be X squared [x^2] over 0.067 - x.and that is going to be X squared [x^2] over 0.067 - x. And you will need to solve for X, so pause the video put in your numbers and solve for X. Did you come up with an X equal to 1.2 x 10 ^-6? If so, you did well. That X is the ICE table, that X is the H_3O+ concentration. So we take the negative log of the value of x, we will get the pH. And the pH equal 5.91. So there is our answer. We check, is that below 7? Certainly it is it should have been so that is a reasonable value. For our pH. And jsut look at this work. Whenever you are at the equivalence point. You will do and ICF table to figure out the moles of the conjugate left in solution and it will always be followed by at ICE table so this is the lengthiest part of a titration curve, is when you are at the equivalence point. So I have this problem in here to show you are not always calculating the pH of a titration. In some former class of chemistry you may have learned about titrations and this is a more typical use of a titration. In this problem we are calculate the molar mass of an acid. Now any time I want to know the molar mass what I need to know about that acid is the grams per mole so if in my sample i knew those two things and divided them into each other I would know the molar mass. Well, the numerator is easy enough to get because they tell me about this unknown monoprotic acid. I have 0.1276 grams of that acid. The rest of the titration, we are going to use in order to figure out the moles of this unknown monoprotic acid. So lets get a picture in our mind of what is happening here. We have got a flask, and in this flask, we are going to place this unknown monoprotic acid, and I am going to call it HA. We are also adding water. So I have a solution it tells me I put 25 ml of the water into that flask. So that is my volume here is 25 ml. Then we are going to titrate, so we have a burette. In our burette we have the sodium hydroxide. And it tells me we are going to have a certain concentration of the sodium hydroxide and that it is going to require 18.4 milliliters of that sodium hydroxide to be added to this solution, in order to get to the equivalence point. So again we are not asked about pH. For a reaction of a monoprotic acid with a strong base. We can write this. The acid would donate to the OH- and leave me with A- plus water. And we are going to map out what we are going to do here. The known or the standard solution for this titration is the sodium hydroxide. And the concentration of the sodium hydroxide that we see right up here, would be the same as the concentration of the hydroxide because there is one hydroxide in there. So if I took the molarity given to me, and the volume given to me, I would be able to obtain the moles of the hydroxide. This balance equation tells me that if I know the moles of the hydroxide, then I can get the moles of the HA it is a one to one ratio and that is what I really need because to get the molar mass I need the grams of HA, which I already know. And I need the moles of HA. So lets do the work. The molarity of the sodium hydroxides 0.0633 mole per liter. That is the molarity. I multiply by the volume in liters. This is going to give me, so we see the liters canceling, that will give me the moles of hydroxide. I don't want the moles of hydroxide, I want the moles of the acid. And the balanced equation tells me it is a one to one molar ratio. So the value for this is going to be 1.16 x 10 ^ -3 and that would be moles of HA. So I have found this because I wanted to know the molar mass of my acid, so I will place the 1.16 x 10-3 moles here and that will give for me the molar mass of 110 grams per mole. So we we can use a titration to determine the molarity or the molar mass of an unknown. Very often it is used to determine the molarity of an unknown. And I just putting this problem in here to help you remember that. We don't always use a titration, and calculate a pH we are just learning about the effect of pH or how the pH change the course of a titration for the most of this learning objective. So this is the end of our learning objective 6 we are doing titrations between acids and bases where one is weak and one is strong. This is very challenging for students it is really hard because there is so many variations. And you want to follow one model and one model does not work. So I have produced a whole series of worked example problems for you to watch. You need to be continually thinking about why did she write the reaction she wrote how do I know that is what I should write? And if you can do that, and get it started and examine what you have in the F line and decide on your next course of action you will be able to that working through these problems. So work through them, watch me work through them keep practicing, until you feel comfortable with writing the reaction filling in your tables and obtaining your answers.