Polyprotic species involve the movement of 3 or more acidic hydrogens.

Polyprotic Acids and Bases

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Polyprotic Acids

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So at this point we've mastered di protic acids and it's onwards towards polyp Roddick acids. Now because of the knowledge gained from looking at di protic acid, we can now apply them to our understanding of polyp Roddick acids here, it says for polyp protic acid which will represent by H three A. Their equations can be illustrated by the examples given below. So with H three A. We're dealing with the fully acidic form of our polyp protocol. In this case try protic acid here, it's gonna give away its first acidic hydrogen to water, thereby creating a che minus and H 30 plus. Because we're talking about removing the first acidic hydrogen, we're dealing with K one here. It's equal to products overreacting so it's H two a minus Times H 30 plus Divided by H three A. This compound here still has acidic hydrogen that can still donate. So it continues onwards and gives another H plus to a second water molecule, creating H A two minus with H +30. Plus. Again, we're dealing with K two Now because we're removing the second acidic hydrogen. So this is a church A two minus times H +30 plus Divided by H two A -. Then finally this species that's created can give away another H plus to a third water molecule to produce a three minus plus H +30 plus. So our equation at the end all be a three minus times H +30 plus over H A two minus now, as we said earlier with dia product species. Remember removing the first acidic hydrogen is the easiest with each sequential hydrogen. Afterward, it becomes harder and harder to remove. So this weak acid will make a small amount of these products because that's the easiest H plus to remove. But then making these products will have very little being formed and with the third phase even less so so less and less products are being formed because it becomes harder and harder to remove the next acidic H plus. So just keep in mind some of the fundamentals when looking at a tri product or polyp protic acid form here, now click onto the next video and see how we tackle polyp product bases.

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Polyprotic Bases

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So we've looked at the polyp protic acid form. Now let's look at the poly product based form here we're dealing with a three minus. So we have the potential of absorbing or um accepting three H plus ions. So here we're dealing with the basic form where it has no acidic hydrogen is on it. So here it's going to just accept an H plus from water, thereby creating a che two minus. By doting in H plus away, water itself becomes O. H minus now because we're talking about accepting the first acidic hydrogen. We're dealing with KB one. So he would be h. A 2 -2 times O. H minus, Divided by a three here. We're not gonna deal with H two minus. So with H A two minus um water is now going to donate a second H plus over to it to make it into H two A minus and O H minus. So here we have H two a minus times O. H minus divided by H A two minus. Finally this one that we created can react with a third water molecule to produce these new products times O. H minus Divided by H two A -. So realize here by examining the polyp protic acid form and the polyp product based form. We've gone through a lot of different variations of our polyp product species. So we've gone through the fully pro donated version. These acid form which could have lost an H plus to become a church to a minus that could have continued to lose another H plus to become a church A two minus, and then finally, we have a three minus after it's lost all of its H Plus. This understanding will help us realize the connections between the different forms of these compounds and the different K and K. P values that are associated with them. Click onto the next video to look at the final portion when talking about polyp product species.

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Polyprotic Dissociation Constants

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So now that we've looked at the polyp protic acid form as well as its base form. Let's connect them all together. Now here, as we said that we have the fully pro donated version here, we have our intermediate one. After it's lost its first acidic hydrogen. Then it's lost its second to give us intermediate too. And then we have our basic form here, realized to remove the first acidic hydrogen means that we have K. A. One involved to remove the second is K. Two. And finally to remove the third and last hydrogen is K three. Going the opposite way, adding the first, acidic hydrogen would be KB one Adding the second would be KB two and adding the third would be KB three. So just like die product species, we have overlapping of equilibrium constants. We can see that K one and K B three overlap. So that's why K one times K B three equals kW. Then we have K two times KB two equals kW because they overlap here and then finally we have K three times KB one gives us K. W because they overlap. Finally here. Now when dealing with polyp protic acid, we're gonna say here that when we have the fully protein ated version H three, A or it has all three of its acidic hydrogen, we're gonna treat it as though we would a mono protic acid and use K one when we're dealing with a three minus where it's lost all of its hydrogen. And all we can do is accept the first one. We're gonna treat it as though it's a base and use KB one. Then finally, if we're dealing with intermediate one where it's lost its first acidic hydrogen. We use this equation here to determine the concentration of H plus. So here it's K one times K two times the initial concentration or formal concentration plus K one times K. W divided by K one plus your formal or initial concentration. Recall that this is the same equation we used for the intermediate of a dia product species. But now we're talking about it in terms of the first intermediate of a polyp product or try product species. And then finally here we have our second intermediate form. And since we're talking about removing the second hydrogen to get to this form, we deal with K two times K three times the formal or initial concentration of the species, plus K two times K W Divided by K two plus initial concentration or formal concentration. So remember when we're looking at a polyp product or try product species, we have four possible outcomes when we're dealing with the fully acid form, the fully basic form. And then it's two intermediates. The intermediate forms are easier because we just have to recall the formulas, plug in the values and find our concentrations of H plus with the acid form, we'd have to set up an ice chart and use K one with the base form, we'd have to set up an ice chart and use KB one. So just recall all of these different avenues we can take in order to determine our ph or H plus concentration for each one of these compounds.

Polyprotic Acids and Bases Calculations

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Polyprotic Acid Calculations

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So here it says, calculate the equilibrium concentration of phosphoric acid, di hydrogen, phosphate, hydrogen, phosphate, phosphate, and the hydro name ion for 0.35 Mueller of phosphoric acid. All right, So here we're dealing with the tri pro decor polyp protic acid. And that's highlighted by the fact that we have three K values given to us here in this try product or polyp protic acid. We have all three acidic hydrogen. Therefore we know that it's gonna start off as an acid. It will react with water because it's in its acidic form. We know that it's going to donate in H plus to water By doing this. It becomes a church to p. and water becomes H 30. Plus here we're gonna say initial change equilibrium. We know that we're gonna ignore solids and liquids. The initial concentration of this week as it is .35 moller these initially are zero. We're losing the reactant so that we can build up our products, bring down everything. Since we're talking about removing the first acidic hydrogen or donating it to water. We know that we're dealing with K one here, that's equal to our products over our reactant Times H 30 plus, divided by our reactant. Now we plug in the valleys we know for each term. So K one is 7.2 times 10 to the -3. Both. Both of these are X. So that's X squared on top, divided by 0.35 minus X. On the bottom we do the 5% approximation method here. So initial concentration of 50.35 divided by the K one value. We're using when we do that we you should see that you don't get a value greater than 500 because of that. You'll have to keep the minus X here and perform the quadratic formula. So we're just multiplying both sides by .35 -1. So distribute, distribute. When we do that we're gonna get 0.252 minus 7.2 times 10 to the minus three X equals X squared. Our lead term is the X squared since it has the highest power so everything has to be moved over to its side. When we do that we're going to get X squared plus 7.2 times 10 to the minus three X minus 30.252. This represents my A my B and my C. So using the quadratic formula, we're going to have negative .72 times 10 to the -3 plus or minus 7.2 times 10 to the -3 squared -4 times one times negative .0025.2 Divided by two times 1 Here, that's gonna give us negative 7.2 times 10 to the -3 plus or minus .100657 divided by two. Now remember because it's plus or minus, that means we're gonna get two different variables answers for X. So x can equal .0467 moller. Or if you do the negative approach x equals negative .0539 moller realize here that an equilibrium you cannot have a negative value. It's not possible. So that means that this negative X gets dropped out. So that X represents our correct answer. Now we're gonna say at equilibrium H. Three P. 04 equals 40.35 minus X. So plug in the value we just found for X. So that equals .3033 molar at equilibrium H two p. 0. four equals H. 30. Plus because they're both equal X. So their concentrations are .0467 moller. Alright, so so far we've found the concentration for our phosphoric acid are di hydrogen phosphate as well as our hydro nia. My on we still have to figure out for our other compounds. We still have to figure out what our hydrogen phosphate as well as our phosphate ions will be. Now we need to realize at this point, I've created this first intermediate because it still possesses acidic hydrogen, it can continue to donate H plus two other water molecules and that's what's gonna happen. So we're gonna write a new equation now that di hydrogen phosphate will react with a another water molecule. It'll donate an H plus to that new water molecule and therefore become H P. +04 minus plus H +30 plus being created. So we have again, initial change equilibrium. Again, we ignore water because it's a liquid Now, initially H P 04 minus. This is the first time we're seeing this. So initially it's zero at equilibrium. In the previous ice chart, we found out that both your di hydrogen phosphate and your hydro Nehemiah equals X. Which equals this number. So they still equal that number. And now we're creating a new ice chart. So their initial amounts are that number. Now, we're gonna say again, we're still losing reactant here in order to make these products bring down everything. Now, we're gonna say here at this point, since we're dealing with the second acidic hydrogen being removed to create our second intermediate, we're dealing with K two equals products. Products overreact ints coming back up here. Let's look at what K two is, so K. One is 7.2 times 10 to the negative three. K two is 6.3 times 10 to the negative eight. Look at the differences in the powers negative 32 negative eight. Remember I've said before in the past that the first acidic hydrogen is the easiest and each hydrogen afterward becomes harder and harder to remove. Which is why we see a big decrease in our K values. They're different by magnitude of five. What does that mean? Well, that means that H two P. 04 minus although it acts as an acid and donates H plus because it's K two is so small. Very very little product is formed which means it's gonna lose very little of X. Which means that this X. Is so small that it's not gonna really affect the final amount for H. Two P. 04 minus. And therefore it can be ignored. Same thing can be said for H. 30. Plus this plus sex. Because sex is such a small number, it's not gonna really increase the concentration of H. +30. Plus. So it can be ignored this X. We have to keep around because we're looking for the concentration of H. P. H. P. 042 minus. And that's all we have for there. So we need that X. There. So going back again, K. A. Two is 6.3 Times 10 to the -8, plug in what we know. So hydrogen phosphate equals X. Hydro ni um ion is .0467. And then same thing with di hydrogen phosphate, it's also equal to .0467. So what do we see here? We see that these two numbers are the same. So they cancel out. So that means that at equilibrium the concentration of our second intermediate hydrogen phosphate equals X. Which equals K. Two. So it equals 6.3 times 10 to the negative eight Moeller. Alright so we've found basically all of our species except for the last remaining one which is our phosphate ion. So this compound still has on it. Another H. Plus. So it could react with a third mole of water. So we're gonna write yet another equation. So we have now H. P. 042 minus. It's gonna react with our third and final water because we don't have any more H. Pluses to donate after this. So it donates an H. Plus to become P. +043 minus Water becomes H. 30. Plus since we're running out of room. Guys, let me take myself out so we can finish this last portion. Alright, so how do we work this out? We have initial change, equilibrium water is a liquid so it's ignored. This is the first time we're seeing the phosphate ion here. Right? So it's the first time we've seen it. So we know that initially it should be zero because it's the first time it's appearing. Now we found out what our hydrogen phosphate is in the second ice chart, we found out it was equal to X. So this number is what it's starting off with in this new ice chart. Then hydro name ion really hasn't changed because this is what we found out it was equal to in the first ice chart. And again we said that X. Is such a small number, it doesn't really affect its final amount. So it's still this number of .0467. We're still losing reactant to try to make products. So we have that. Alright now we're gonna say we're dealing with K three Which equals p. 0. 4 3 Times H. 30. Plus Divided by h. p. 0. 4 2 Now. This K. three. Let's go up here and see what K. three is. So K three here is uh 4.2 times 10 to the negative 13. It is even smaller than K. One and K two much smaller. Which means that very little product at this point is being formed which means that the minus X. And plus X are even that much smaller and have even less of an impact on my final equilibrium amounts. So what does that mean? So like in the second ice chart that means we can ignore this uh minus X. And we can ignore this plus X. Because they're so small. They're not gonna affect those equilibrium amounts at the end for those two compounds. So we're just gonna plug in what we know. So we saw it was 4.2 times 10 to the negative 13 for K. Three. This is equal to X. H30 plus is .0467 And H Hydrogen phosphate is equal to 6.3 times 10 to the -8. All we have to do at this point is solve for X. And that will represent the equilibrium amount of phosphate ion. Alright, so Multiply both sides by 6.3 times 10 to the -8. So that's gonna give me 2.646 times 10 to the negative 20 equals X times 200.467, Divide both sides now by .0467. And we'll have our X. Variable at the end. So X. Here which represents the concentration of phosphate ion. Because that equilibrium equals acts Equals 5.67 times 10 to the -19 Moller. And what you should realize is, what do you see happening? Well, this was our last acidic compound. It has the smallest K. So it produces a small smallest amount of products, which explains why the phosphate concentration is so small. K. A two is bigger. So we produce slightly more product, which is why the concentration of hydrogen phosphate is equal to this number. And then K. One where we're taking off the first acidic hydrogen is the biggest one of them all. Which is why we have much larger concentrations at the end. So we've determined the concentrations of each one of the compounds asked of us, we determined phosphoric acid, di hydrogen phosphate, hydrogen, phosphate, phosphate and hydro nia. My on this is the approach we have to take. Now they didn't ask us to find ph but we know that H 30 plus is equal to this number here of of 300.467 moller. So all you need to do is take the negative log of that number and you'd have your ph if they asked us to determine the concentration of O. H minus, how could we figure out the concentration of O. H minus? Well, we know what H 30 plus is. So by default we know what O H minus is because remember they're connected by the formula H 30 plus times O H minus equals KW. And remember KW equals 1.0 times 10 to the -14 at 25°C. So just remember the approaches we took for determining the all the concentrations of each species produced when a try product or polyp protic acid is dissociating in solution we lose the first acidic hydrogen to help create our first intermediate. We lose the next acidic hydrogen which deals with K. To help make the second intermediate and then we lose the last acidic hydrogen which deals with K three. To help us find the basic form. Take this same approach whenever asked a question like this and remember go back and take a look at our die product species to see how the two are related because there are similarities between the two. This one just requires an additional step because we're dealing with one extra acidic hydrogen

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Polyprotic Acid Calculations 2

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here. It says determine the ph of 0.250 molar sodium hydrogen phosphate. They tell us that phosphoric acid contains K one K two and K three. With those respective values. Realize here that they're giving us information on phosphoric acid and we're only doing that because we're gonna have to utilize um one or more of those K values here. We're dealing with sodium hydrogen phosphate. Remember for phosphoric acid, it's a tri product or polyp protic acid. So it's acidic form is H three p 04. When it loses that first acidic hydrogen it becomes H two P 04 minus. When it loses that next one it becomes HP 04 minus two minus and then finally becomes P 043 minus. In this particular question, we're dealing with sodium hydrogen phosphate, which is H P 042 minus that's involved. So we're dealing with intermediate. To remember with the intermediates, we don't have to use any types of ice charts or anything. All we have to recall is the formula to help us determine the concentration of hydro knee um ion for this intermediate two were saying hydrogen ion concentration is equal to the square root of K two times K three times the formal or initial concentration plus K two times K W. The association constant of water divided by K. A two plus formal concentration or initial concentration. All we have to do now is plug these numbers in and we'll find our hydrogen ion concentration. So let's plug these values in 6.2 times 10 to the negative 84.2 times 10 to the negative 13 .250 moller Plus 6.2 times 10 to the -8. 1.0 times 10 to the -14 divided by 6.2 times 10 to the negative eight plus 80.250 moller. When we work out everything within there, it's going to be the square root of 2.852 times 10 to the negative 20. When we take the square root of that H plus equals 1.69 times 10 to the negative 10. Which means that ph is just simply the negative log of that concentration of H plus. So negative log of 1.69 Times 10 to the -10 which gives me 9.77 as the ph for sodium hydrogen phosphate. So, just remember when looking at these poly products species, you have to identify what particular form are you dealing with? Are you dealing with the fully predominated predominated acidic form? Are you dealing with the deep, deep resonated basic form? Are you dealing with one of the intermediate forms? Knowing which form you're dealing with, helps you determine which path to take to figure out what your ph of the solution will be at the end. Now that you've seen this example. Go on to example two and see if you can identify what form we're dealing with. And as a result, what path to take. In order to determine the ph.

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Polyprotic Acid Calculations 1

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So here it states determine the ph of .150 Mueller citric acid. Here, we're dealing with the fully propagated form of citric acid, it has all three of its acidic hydrogen. We know it has three acidic hydrogen because it possesses three K values. Now because we're dealing with its first acidic form, that means we're gonna deal with K. One. The acid itself will react with water because it is the acid. That means that water will be the base, so it will donate an H. Plus to the water becoming a church to C. Six H 507 negative water gains that H plus to become a +30 plus. So we know we have initial change equilibrium. Remember we ignore solids and liquids. So the initial concentration is .150 moller these products initially haven't formed yet, so there's zero, we lose reactant in order to make product. So we bring down everything. Now at this point we'd say that our K one equals products overreact ints. So X squared divided by 0.150 minus X. We can do our 5% approximation method to see if we can ignore that minus X and therefore ignore the quadratic formula. So we do initial concentration divided by the constant that we're using, which in this case is K. One, as long as the ratio is greater than 500. We can ignore the minus X. Now if we took that initial concentration and divided by R. K value, we would get approximately 202.7, which is not greater than 500. So we have to keep the minus X. Here in our expression and do the quadratic formula since you're gonna need room guys let me take myself out of the image. Alright so we're gonna plug in our K. One which is 7.4 times 10 to the negative four X squared over 0.150 minus X. So we're going to multiply both sides by .150 -1. So this gets distributed distributed So we'll have 1.11 times 10 to the negative four -7.4 times 10 to the -4 x equals x squared. Because this X. Here has the largest power it's the lead term. So everything has to be moved over to its side. So add 7.4 times 10 to -4 X. to both sides. Subtract 1.11 times 10 to the negative four from both sides. So X squared plus 7.4 times 10 to the negative four X minus 1.11 times 10 to the negative four. So this is my a my B. And my C. Remember your quadratic formula is negative B plus or minus B squared minus four A. C over two. A. So let's input those values that we know. So negative 7.4 times 10 to the negative four plus or minus 7.4 times 10 to the negative four squared -4 times one times negative 1.11 times 10 to the -4 Divided by two times 1. Remember it's plus or minus. So we're gonna get two values. So actual equal negative 7.4 times 10 to the negative four Plus. So when I take the square root of everything within their it gives me .0-1084. Or x equals negative divided by two. Or it equals negative 7.4 times 10 to the negative four minus 40.0 to 1084 divided by two. So my X. Will equal .010172 Molar or negative .010 912. Now remember this? These X values that we find represent these X values here in our ice chart. And remember at equilibrium you cannot have a negative concentration. It's not possible. It's like having negative time. So we cannot use this X. This is the X. That we would use That X. is equal to H.. Plus. So ph remember is negative log of H. 30. Plus or H plus. So plugging that concentration we just found. So that equals 1.99 as my ph So remember when it comes to polyp Roddick acids or die protic acid, it's always imperative to know which form you're dealing with knowing the form will dictate which path we take in order to find our ph