Our seventh learning objective is dealing with polyprotic acids and looking at titration curves of these polyprotic acids. The polyprotic means you have more then one proton that can be donated from this acid. They call that being a liable proton donatable proton. So we are looking at a titration donatable proton. So we are looking at a titration curve on the screen right now of a diprotic acid being titrated with a strong base. What we want to be able to do is recognize what titrated with a strong base. What we want to be able to do is recognize what species would be present at every point along the titration curve. We will consider the pHs at those equivalence point but we will not actually calculate the pHs during a polyprotic acid titration. As we look at this titration curve we have a diprotic acid. A diprotic acid we are goign to be looking at is carbonic acid. It is going to be reacting sodium hydroxide. is carbonic acid. It is going to be reacting sodium hydroxide. That is what is occurring during this titration curve. So we see that we will have two equivalence points. We have one here, we have one here and we have the pH changing gradual fashions and then spiking up, and then gradual fashions gradual fashions and then spiking up, and then gradual fashions and spiking up during the course of this titration curve. and then spiking up, and then gradual fashions and spiking up during the course of this titration curve. Lets begin by looking at the over all reaction of carbonic acid with sodium hydroxide. This is the overall reaction. You are going to need two sodium hydroxides for every carbonic acid moles that you have in there. If we wrote the net ionic equation we still see that 2 to 1 ratio between the OH- and the H_2CO_3. between the OH- and the H_2CO_3. But when this reaction takes place it the NaOH doesn't pull off both protons off the carbonic gas at a time. What it does is occurs in a step-wise fashion. What to I mean by that? Well, the first step will be removal of the first proton. It will be the easiest one to pull off of the H_2CO_3. So we will be removing one of these protons and forming the bicarbonate ion during this first titration curve. So we begin by adding some base. It will convert some of the acid over to its conjugate base. We will have a range where we will have a buffer. the acid over to its conjugate base. We will have a range where we will have a buffer. Because there will be both We will have a range where we will have a buffer. Because there will be both the H_2CO_3 and the HCO_3- present in solution and we will see that buffer occur. So it will be a gradual increase and eventually you will be at the equivalence point. When you are at the equivalence point you will have converted all of your base and all of you H_2CO_3 over to the bicarbonate ion. Now this bicarbonate ion is still able to donate a proton Now this bicarbonate ion is still able to donate a proton so we still expect the pH to be in a acid range, because it will behave as an acid. So lets look at the curve. We are looking at the area inside this blue box. And we are once again considering H_2CO_3 And we are once again considering H_2CO_3 and one of the OH-'s converting over to the bicarbonate ion and water. So we see this area of a flattening out sort of. That is occurring at the buffer, so we have a buffer between then That is occurring at the buffer, so we have a buffer between then bicarbonate ion and the carbonic acid here. bicarbonate ion and the carbonic acid here. We can use the Henderson-Hasselbalch equation to calculate the pHs during that range if we were asked to. But like I said I won't ask you to calculate pHs of these diprotic acids. We hit this equivalence point and when we are at the equivalence point we pHs of these diprotic acids. We hit this equivalence point and when we are at the equivalence point we see that we still have a pH in a We hit this equivalence point and when we are at the equivalence point we see that we still have a pH in a acid range because this thing can still behave as an acid. So we get to this first equivalence point and then we still behave as an acid. So we get to this first equivalence point and then we continue adding OH- and then at that point it is going to start taking it off on the second the second hydrogen. So we have carbonate ion that got converted. In the first equivalence point the OH is continually being added. In the first equivalence point the OH is continually being added. And we are going to be forming the carbonate ion as we continue on this reaction. And we are going to be forming the carbonate ion as we continue on this reaction. So the pH is going to slowly climb, where once again in a buffer range where we have both the bicarbonate, and the carbonate ion both present. So lets look at the curve. We are looking at this blue box area. In this blue box area we have the HCO_3- and the base converting over to the carbonate ion and water. We have this zone here where we have significant amount of both. After the climb it starts leveling out because we have present in the solution, both. And these are conjugates of each other and they are weak. There is that buffer range. We finally reach the point where we just have consumed all of the bicarbonate with the added hydroxide and there is only carbonate present. When there is only carbonate present, this is definitely a base so can see a pH in the base range. So at each point along the curve we see what is happening. so can see a pH in the base range. So at each point along the curve we see what is happening. If we keep on adding more and more base So at each point along the curve we see what is happening. If we keep on adding more and more base after we have reached the equivalence point we now will climb our pH as there is an increased concentration of OH-. So there is both OH- and CO_3 2 minus in the solution, it is definitely basic and it will continue to climb. So that is our learning objective number seven in which we are just looking at the titration curve in this case a diprotic acid. talking about what exists at each point along the way in the graph and what the pH at the equivalence point in general will be.
