Titrations of Diprotic and Polyprotic Acids

So in this video, guys, we're gonna continue with our discussion of buffers. But now, looking at die product as well as Polly product buffers, Now we're gonna say here, Die product or Polly Product Buffer can be approached in a way similar to mono product buffers. The main difference, though, is with mono product acids. We're dealing with just one K value, but with di product buffers, we're dealing with two K values because there's too acidic hydrogen and with try product buffers, we're dealing with three K values because there's three acidic hydrogen now here, it says For mono product buffers, remember, buffer simply a weak acid. In this case we have Hi Paul Cloris Acid, which is a week oxy acid. And then we have sodium hypochlorite, which is its conjugate base. Remember, you're confident based normally has one less hydrogen and usually toe disguise the chart that's formed. They will replace that hydrogen we lost with a metal. When we have a buffer, we just use simply the Henderson household back equation. And remember what the Henderson Hasselbach equation. It's just pH equals P K, which remember, is negative log of K plus log of the conjugate base over the weak acid. Now, the units that go in here, they can be either mole arat e if we're given just polarity or they could be moles. We tend to use moles when they're giving us volume and molar ity. So, for example, if this that you had 50 miles off 500.100 Moeller hcl Oh, as the amount for your weak acid remember, polarity equals moles over leaders. So if they're giving you volume and polarity realize that if I multiply both sides by leaders here, we can see that moles equals leaders times more clarity. So you would divide the MLS here by 1000 to get leaders and then multiply them by their mole arat e to find your moles. Okay, so those are the units that we're gonna use their remember, if they're giving you just polarity, the units will be more clarity. But if they give you volume of a polarity there really asking you to find your moles, this is the basic set up. When it comes to mono product buffers. We'll continue onward with looking at die product buffers as well

So with di product buffers, remember, a die product. Acid basically has three major forms. We have the former. It has both of its acidic hydrogen. So we're gonna say that this is the acidic form. When it gives away its first acidic hydrogen, we have its intermediate form. And then when it's finally giving away its last acidic hydrogen we have it's basic form. Now, remember, if we're talking about giving away the first acidic hydrogen, that means we're going from H two a form T H a minus form A just represents non metals. We're gonna say, since we're moving that first tested a hydrogen, that means we're dealing with K one. So we're gonna say here in this example, let's say you had 0.10 Mueller Okay, carbonic acid and 0.15 Mueller of sodium bicarbonate. We can see that carbonic acid here has both of its acidic hydrogen, so it's the acidic form. This one here has one less hydrogen. So that's the conjugate base form. It only has one hydrogen left, so that's that intermediate form. In this case, we'd say that pH equals p K one. Since we're dealing with removing the first acidic hydrogen, which connects the acidic form in the intermediate form plus log of my conjugate base, which is the 0.15 Mueller sodium bicarbonate divided by the weak acid form, which is 0.10 Mueller carbonic acid. So that's the equation would use for that version. Now, if we're talking about removing the second acidic hydrogen to give us the basic form, that means we're dealing with K two. So let's say they gave us 50 m Els off 0.100 Mueller sodium bicarbonate and 30 Emmaus off 0.100 Mueller sodium carbonate. So we can see here that this form has one acidic hydrogen left and this form here has lost both acidic hydrogen. So that's the basic form in this case is were dealing with these two forms now and they're connected by K two. We'd say pH equals p k two plus log again of my conjugate base, which in this case would be the carbonate ion. Remember, we divide this by 1000 and then multiplied by the molar iti to find our moles. So they give me 10000.3 moles of my sodium carbonate divided by divide This by multiplied by the molar. It gives me 005 moles of sodium bicarbonate. So that's how we solve for the buffers, dealing with the different forms of our die product acid as we continue onward into try product buffers. It's basically the same type of idea, except we have mawr than two K values. So we're gonna have to be careful of mawr, different types of forms that exist.

Now we're looking at Poly product Buffers. So with Polly, product buffers were mainly dealing with, like, try product buffers. So we have three acidic hydrogen. So we have three K values we're gonna say, try product buffers. They have three K values, so there's four different forms that exist here. We have the former, all of it all the acidic hydrogen zehr there. So this is my acidic form. And if we're taking off that first acidic hydrogen to make this first intermediate form, we're dealing with K one here. We say that this is intermediate form one. Then if we're gonna remove that second acidic hydrogen to give us this intermediate form, that means we're gonna pass through K two to create my intermediate too. Then if we're gonna remove that last acidic hydrogen to give us this final form here, the basic form that means we're dealing with K three. So a good example of a tribe product asset is phosphoric acid, right? So we could say here that the acidic form would be this form. Then we'd say here that the first intermediate form we removed the first acidic hydrogen, so this would be this form we remove the next one. So that BHP all four to minus and then removed the last of city hydrogen to give us people for three minus. So here, let's say we have 10 Moeller phosphoric acid and 0.12 Moeller Potassium, potassium di hydrogen phosphate. Right. So that would be PK one plus log of conjugal base over weak acid. So that would be over 10 Mueller phosphoric acid. Then we're talking about these two. Okay, so basically, this first one gave us this one. And now for the next two, we could say we have 0.100 Mueller, potassium di hydrogen phosphate. And we could do 0.200 Moeller potassium mon off hydrogen, phosphate or hydrogen phosphate. So we'd say here that this will be dealing with P K two plus log of conjugate base over weak ass. Remember, the continent faces just the one with less hydrogen is one less hydrogen. And then finally, we're dealing with these two. So here, let's say we had 0.15. Mueller, can we have the phosphate hydrogen phosphate and point 14 Moeller? We're gonna say potassium phosphate so that b p k three plus log off conjugate base over weak acid. Conjugate base is the one with one less hydrogen, 15 more on the bottom. So these would be all the different, all the different types of of buffers that could be created when we're dealing with a try. Podcasts such as phosphoric acid. So again, we're customers seeing regular mono product buffers. We're just dealing with one K, but things definitely get more complicated once you move into Di product and tribe or Polly product arenas. So you have to be careful of which form are you dealing with because that determines which cave al you're gonna use within the Henderson Hasselbach equation. So as we continue with the discussion off these different types of buffers, keep in mind when you're looking at any of them, learn which form we're using to figure out which of the equations we need to use to solve for Ph.

here, it says. Calculate the pH of 100 miles of a 1000.25 Mueller carbonic acid solution when. 70 miles of 700.25 Mueller sodium hydroxide er Added here carbonic acid is die protic acid, so it has to K values. Now the challenge with these types of questions is understanding which forms of the die protic acid or Polly protic acid are involved in the question. So remember carbonic acid H two co three. This is it's acidic form when it still has both of its acidic hydrogen. When it loses its first hydrogen, it becomes H seal three minus, which is hydrogen carbonate or bicarbonate. Then, when it loses its last acidic hydrogen, it forms carbonate ion. We have to determine in this question which one of these forms are involved, and what we need to realize here is that carbonic acid, because it's K values, are less than one. It represents a weak acid, which I'll abbreviate is W. A. Any weight we should know at this point represents a strong base. Now, remember what I've said in the past. If you have a weak species reacting with a strong species, then that means you're going to have to set up in I c f chart. But again, we still have to figure out which versions are reacting here. So we're gonna say that we have to figure out basically the equivalent volume of the strong species needed to reach the equivalence points for this die protic acid. All right, so we're gonna say here M acid times V acid equals M base times V base. And all we're gonna do here is we're gonna calculate how much of the strong species do I need to add in order to get to my first equivalents point? So my acid here it's polarity is 0.25 Moeller. Its volume is 100. Emmaus. The polarity of my strong base here is 0.25 Moeller. Now, I know I give us a volume here, but we're gonna ignore that volume for right now. What I'm trying to determine is how much of the strong species in milliliters do any toe add to get to the first equivalents point of carbonic acid. So divide both sides now by 25 Mueller and you'll see here that the volume of my base needed to get to the first equivalents point is 100 m. Els. This is important, okay? Because we need 100 miles to get to the first equivalents point for carbonic acid. And we're not there were only using 70 m. Els. Okay. And so the way you look at it is this You're gonna say that we use these two forms before we reach the first equivalents point point volume. Okay. So, again, what you have to do when it comes to die product and Polly product acids when they're reacting with the strong species is determined. What is the equivalent volume needed of these strong species to get to the first equivalents point? If we haven't reached the first equivalents point volume yet, then we use these first two versions. Now, if our volume of the strong base had been greater than 100 Emmaus, then that would mean that we passed the first equivalents point. So we're gonna say here these other two forms, um, this is after we have reached Well, this is after we have passed the first equivalents point volume. That's when we use the other two forms. So, again, we needed 100 mls of strong base. We only have 70 MLS. So it's not enough to get us beyond the first equivalents point. So that means we're gonna use the first two forms of carbonic acid, Theus acidic form and its intermediate form. So we're gonna bring down H two co three, which is the weak acid form. Remember, the strong species must always be a reactant. So my bases the strong species, it reacts with the acid. That's why I drew the weak acid form with it. Because in a way, which is a strong species, we're gonna have a single arrow going forward. Remember, when we have weak and strong together, we use an I C F chart. So initial change final here. This is my acid. This is my base acids donate h plus. So the carbonic acids gonna donate an H plus to the minus. That's gonna produce water. And then we're also gonna have the n A plus, combining with the three minus left behind. Remember, in an I C F chart, we only care about three things. We only care about what is strong. We only care about the weak acid. We only care about its conjugate base. The fourth thing, we're gonna ignore it. Also remember that in an I C f chart, the units have to be in moles. Moles equals leaders, times more clarity. So divide the MLS by 1000 to get leaders and multiplied by the mole Aridjis. So when I divide the the MLS 5000 get leaders and multiply them by their polarities, I'm going to get 0. moles of carbonic acid and 0.175 moles of sodium hydroxide. We don't have any of the conjugate base, so that zero initially remember Look at the react Inside, the smaller moles will subtract from the larger moles and whatever happens to the react inside, the opposite happens to the product side. So this will be plus 0.1 point +0175 Moles bring down everything so would get 0.75 moles. Zero here 0.0 175 moles. We look to see what we have at the end of our reaction. At the end we have weak acid left and at the end we have conjugate base left. So remember, if you have weak acid and conjugate base left at the end, you have a buffer. Which means I get to use the Henderson household back equation. So Ph equals p. K. And we have to determine which PK so remember, if we're dealing with the acidic form and its intermediate form, that means they're dealing with K A one k one is what connects them together. If we're dealing with the intermediate form and the basic form, that means we're dealing with K two. All right, so in this case, we're dealing with the acidic four minutes, the intermediate form. So that means we're dealing with K one, which would mean that this is PK one plus log off conjugate base over weak acid. Alright, So PK one that will be negative log of K one, which we're told this 4.3 times 10 to the negative seven plus log off the moles of my conjugate base divided by the moles of my weak acid. If you do it correctly, you'll get Ph equals 6. Okay, so you just take the negative log of this value here, take the negative log of these guys here, and then add those two answers together. So again, dealing with buffer questions. When it's dealing with a model product, acid can be challenging. But the difficulty of the questions increases when you're dealing with di product and Polly product acids. So again, you have to make sure. At what point are these tight rations occurring? Are they occurring before the equivalents point? Are they after the first equivalence point? Are they occurring after the first equivalents point that determines which versions are being used in the question and then also determines which peek A you'd have to use in the question. So as we continue with these discussions, make sure you take note of all the little things we talked about as we approach problems like this.

Okay, calculate the pH of 75 MLS of 750.10 Mueller Phosphorus Acid Solution When ADM Elza 0.15 Mueller sodium hydroxide are added here, were given two K s. Now, before we approach this question, realize here that false force acid, which is H three p 03 looks like it will be a try product acid, where it would have three acidic hydrogen. But based on its lewis dot structure, it only actually has to que values. So phosphoric acid is not a try protic acid. It's a die product acid. And remember, when it comes to doing these types of questions, we have to analyze the information given to determine which form of the die protic acid is involved in the reaction. So we have the acidic form of phosphorus acid. When it loses its first H plus, it becomes H two peel three minus. And then finally, when it loses its last acidic hydrogen, it becomes HP all three to minus. So these are the three forms of the dive protic acid, and we have to determine which one of the of the forms are involved in this question. Now, remember, we have a strong base reacting with our weak acid here, and we set up above that. If we're dealing with these two forms, this would happen before the first equivalents volume. And then we'd use these two after the first equivalents volume. So here to determine that we say M acid times v acid equals M base times V base. All right, so we're gonna use the information of the week species, which is the acid. It has 0.10 Mueller and then it's 75 MLS. We know the molar ity of the Strong's bases. 15 Mueller. Now, remember, when we're looking for the equivalent volume, we're looking for the equivalent volume of the strong species, which in this case represents a strong base. We're gonna divide both sides now my 0.15 Mueller so V base equals 50 amounts. So that means we need 50 miles to get to the first equivalents point. And to get to the second equivalents point, we would need double that. We need 100. Immelt's. So we have 80 mls. So we've gone beyond the first equivalents point. So that means we're dealing with these two, and we're not above 100 MLS. So we're not dealing with stuff beyond the second equivalents point. Okay, So if we're dealing with stuff beyond the second equivalents point, um, that means that we'd really be looking at this form here, reacting with the base in some additional way. All right, so we know that we're in between basically the first equivalent volume in the second one. So we've gone beyond the first equivalent volume. So we're not gonna use these two. We're gonna be using these two because we've gone after the first equivalent volume, but before the second. So we're gonna have hte to peel three minus, which is the acid form. It's going to react with my strong base. So we have initial change final. So here. This is my acid form in this case or intermediate form, which is acting like an acid. This is my base. Remember, acids donate h plus, so the age plus will connect with the O. H minus of the base to give us water. What's left behind will be n a to p 03 h p Okay, so that's our structure. And we know that's the formula for for this product here because remember when this gives away an H plus, It becomes h p +03 to minus. It's gonna react with the sodium here, which is plus one. So the two comes here and the one comes here. So that's how I came up with this formula here, remember, In an I C F chart, we only care about three things we only care about. Whatever strong. We only care about the weak acid form, and we only care about its conjugate base form the fourth species we ignore. Next, we're gonna say our units have to be in moles, so divide the MLS by 1000 and multiplied by the mole Aridjis. When we do that, we get 0075 moles and 0.45 moles here. This is gonna be equal to zero because we're not talking about it all yet. And remember, the smaller moles will subtract from the larger moles. Yeah, so this is gonna be 0.30 moles. This is gonna be zero. Whatever happens on the product on the reacting side, opposite happens on the product side. So we're gonna add this many moles. So we're gonna say at the end. What do we have left? We have weak acid left. We have conjugate base left. So we have a buffer. So we use the Henderson hassle back equation. Remember, because we're dealing with these two forms here. They're connected to each other by K two, which means we're gonna use P k to within my Henderson Hasselbach equation. So p k two plus the log of conjugate base over the weak acid form. So p K two is just the negative log of K two, plus the log of the moles of the contract based over the weak acid. So when we do that, so take the negative log of this. Take the log of this expression, add those two numbers together. So here this would be 6.698 and this would be 0.176091 So when we add them together, pH equals 6.88 or so when we round. So this would represent the pH of our buffer system that we created again. Die poetic and Polly product buffers are tricky. You have to look to see which form you're dealing with when approaching any of these questions. That's the key to getting the correct answer