we've taken a look at mono protic acid, dye protic acid and now we're taking a look at polyp Roddick acids. This means the inclusion of additional equivalence points. And even more calculations involved with tight rations. So as always you look for the equivalent volume of our tightrope being used. Here are tight is a strong base. We're going to say here, polarity of my acid times volume of my acid equals polarity of my base times the equivalent volume of my strong base here will plug in the values. And when we divide both sides by the 0.100 moller of the strong base will see that the first equivalent volume here will be 50 mls. Since we're dealing with phosphoric acid, that means we have three equivalents points involved. So we're gonna have three equivalent volumes needed to get to the second equivalence point, we need an additional 50 mls. So that would be 100 mls. And to get to the final and third equivalence point, we would need another 50 mls, so that would be 100 and 50 mls. So those are the volumes needed for each equivalence point. Now, before any strong strong base has been added, we essentially just have a weak acid. Therefore we can set up an ice chart in order to determine our equilibrium expression. And since we're dealing with removing the first acidic hydrogen from phosphoric acid, that means we're dealing with KA- one. So the acid donates an H plus to water to produce H two P +04 minus which is di hydrogen phosphate plus hydro ni um ion. Remember we'd have initial concentration of our acid but no initial concentrations of our products because they haven't yet formed. We're losing reactant in order to make products bring down everything helps us to come up with our expression. Now remember we could do our 5% approximation method to help us determine if we can ignore the minus X or not. When it comes to phosphoric acid. It has three K. A. Values. So we have K. One K. Two and K. Three. So K one is 7.5 times 10 to the minus three K. To 6.2 times 10 to the negative eight. And then we have 4.8 times 10 to the negative 13. If we use the initial concentration of .100 moller and divided it by the K. A. That we're using in this example which is 7.5 times 10 to the minus three. We would not get a value greater than 500. Therefore we have to keep the minus X in our expression and perform the quadratic formula. Now using the quadratic formula, we find out that X. Which equals R. H plus concentration was going to be equal to approximately 0.239 molar. And by taking the negative log of that concentration, we'd find out that our ph is approximately 1.62. So at this point we haven't even commenced tit rations yet. We haven't added any strong base. This is our initial ph based on just the concentration of phosphoric acid. Once we start adding our strong base to this solution, we should expect an increase in our ph, so we'll click over to the next video and see what happens once we start adding our strong base and a O. H two are weak acid solution.

So before we reach the first equivalence point between our weak acid and our strong base, we will have the formation of a buffer. So at some point in our calculations, we're gonna have to use the Henderson Hasselbach equation. Remember to get to the first equivalence point, we need 50 mL of our tight trend and A. O. H. Here we only have 30 mls. So we would show the story geometric relationship between my weak acid and my strong base by writing out a balanced equation. Now, within this, when we're filling out our chart, we will be dealing with either milly moles or moles. Here. I decided to just divide the middle leaders by 1000 to get leaders and multiply them by their polarities. So we'd have the initial moles of my weak acid and my strong base, our conjugate base, we would have nothing initially should be zero. Now, remember we look at the react inside the smaller mole total would subtract from the larger mold total. At the end we'd have nothing left of the strong base. We'd have some of our weak acid remaining through the law of conservation of mass. We would produce some conjugate base. We'd have weak acid, we'd have conjugate base and therefore we would have a buffer because of this. We can employ the Henderson Hasselbach equation. Now realize since we haven't reached the first equivalence point, we're basically talking about how K. One is attached to the removal of the first acidic hydrogen from phosphoric acid. So Henderson's Hasselbach equation will be P. H equals P. K. A one plus log of conjugate base over weak acid. So that would just be negative log of r k A one, which is 7.5 times 10 to the negative three plus log of my conjugate base 30.300 moles over my weak acid. And then when we plug that in, that would give me 2.30 as my ph up to this point, So realize that we've begun our tight rations by adding some strong base and we see that our ph has increased. As a result, click over to the next video and see what happens when we get to our first equivalence point.

we've now reached the first equivalence point within articulation. So at this point we have equal moles of our weak acid and strong base. Because they're equal in terms of moles, they will completely destroy one another. So at the end we have zero of each. But remember we have the conservation of mass, we're gonna form some of the we're gonna form some conjugate base. So at the end we have some of it remaining. In terms of calculating ph at this point we would discover what the formal concentration is of our conjugate base. So we'd say formal concentration which is F equals the initial concentration of my acid, which is 0.0.100 moller times the volume of the acid, Which is 50 MLS divided by the volume of the solution. So 50 mls of the acid and 50 mls of the strong base. That would give me a concentration of .050 moller. What we would then do is we plug in all the values that we need to figure out the concentration of H plus ions. Remember this here is the conjugate base but it represents the first intermediate form of my polyp protic acid. And this equation we should be familiar with because we used it when talking about calculations dealing with polyp Roddick acids. So K one we said was 7.5 times 10 to the minus three and K two. We said was 6.2 times 10 to the negative eight. Those will be the values we plug in for K one and K two, respectively. K W is 1.0 times 10 to the negative 14. When you plug all those values in, we find a concentration of H plus equal to 2.1 times 10 to the negative five molar. Since we know what the concentration of H plus is, we could find out what ph is which is just simply the negative log of H plus. So that will give us 4.70 for the ph at the first equivalence point. Remember we're dealing with the polyp product species. So we still have equivalent point number two and number three to go through as well as all the tight rations points in between and after them. So at this point we've seen that adding strong base is showing us an increase in our ph over time. So we'll continue onward with our discussion of adding even more strong base and seeing what else can the ph do as we're adding more of it.

at this point we've passed the first equivalence point. Now we have the titillation of 50 mls of 500.100 moller phosphoric acid with 70 ml of 700.100 moller N A O H. Now remember we needed 50 mls of n A O H. To get to our first equivalence point. So we are 20 ml beyond that point here, at the first equivalence point, we had only left di hydrogen phosphate and the moles of it that we had left were these moles here we have an excess of 20 mls, so it'll be 20 mls times the concentration of N A O. H, converting that into molds gives us this value here, hydrogen phosphate, which is our conjugate base hasn't been formed at all. So that's why initially it's zero moles. As usual, we subtract the smaller moles from the larger moles as a result of this will have a portion of weak acid remaining and we have the generation of some conjugate base because we're talking about the removal of the second H plus from our polyp protic acid, we're dealing with K two. So as a result of this, we're dealing with PK two within the Henderson Hasselbach equation by inputting the value so we have P H. Here equals negative log of K two, which is 6.2 times 10 to the negative eight plus log of my conjugate base in moles over weak acid That gives me a ph of 7.03. Now that we've seen this, we can take a look at what happens at the second equivalence point. So click over to the next video and see what happens once we get to the second equivalence point.

So at the second equivalence point, all of our di hydrogen phosphate has been totally neutralized by R N. A. O. H. So both of these will be gone. It's okay though because remember we have the conservation of mass. So whatever we lost here in the form of di hydrogen phosphate gets formed here in the form of hydrogen phosphate. So this is the conjugate base that we're looking at here. We find out it's formal concentration. We'd use the initial concentration of my weak acid, which is 0.100 moller, multiply it by the volume of the acid divided by the volume of the solution. So that's 50 mls plus 100 mls which is 1 50 mls. So that give me 500.333 molar. Then by inputting values for K two K three. The formal concentration as well as K W. We can find out the concentration of H plus remember this formula we've seen before when dealing with polyp Roddick assets. This was the formula to find the concentration of my intermediate to from a polyp Roddick species here. When you plug in the values that we need, we will get an H plus concentration of 2.20 times 10 to the negative 10 moller. And when by having that we can determine what our ph is. We just take the negative log of that value. And that gives me a ph of 9.66. We can see that our ph of course is increasing as we're adding more and more strong base click over to the next video and see what happens after we've passed the second equivalence point.

So we've passed the second equivalence point realized here that we had di hydrogen phosphate reacting with N A. O. H. To give us hydrogen phosphate. We are beyond the second equivalence point, which means we'll have an excess of N A. O H. This has been totally destroyed so it's no longer around. We would form some of this conjugate base here. But remember strong species have a bigger impact on the overall ph than congee gets would. So we'll focus on determining the concentration of our strong base here. We'd save. The concentration of our strong base is equal to the initial concentration of the strong base, Which is .100 moller times the volume of access. So we needed 100 mls to get to the second equivalents point, we are 15 mls over that. So that is our excess divided by our total volume, which is 50 mls plus 1 15 mls, which is 1 65 mls. This gives us a concentration of .009091 Moeller. All we do now is we determine what the P O. H. Is by taking the negative log of that number. So that gives me 2.04 as the POH. Ph would just be 14 - that value. So at this point we'd have a ph of 11.96. Once we've passed the second equivalence point. Now we'll quickly approach the third equivalence point and see what our ph would register at that point