A **diprotic acid** is an acid that can donate two hydronium ions (H^{+}).

Diprotic Acids

Since a diprotic acid has two acidic hydrogens it would have 2 equilibrium equations.

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concept

## Diprotic Acids

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in this video, we'll take a look at die product acids. Now we're gonna stay here that die product acids and bases are compounds that can donate or accept to H plus ions. Now let's talk about die product acids. First, we're gonna say Die product acids. The general formula for them will be H two A. That's if you don't know what the dye product as it looks like. But you know what? The dye product as it so it has to acidic hydrogen. That's why it's H two and then a represents the nominal portion. Okay, now they're equations can be illustrated by when we're dealing with di product acids. That means we can donate H plus two times. The first H plus that we donate is associated with K one. Okay, so that's your acid association, Constant one. Then, if you're gonna give away the second acidic hydrogen that deals with K two, what you need to realize here about K one and K two is that K one will always be greater than K two. It's always easier to donate the first H plus and then much more difficult to donate the second hydrogen and in fact, they're usually different by several magnitudes. What I mean by that is a k A one. Let's say it Waas 1.3 times 10 to the negative too. You would expect your K two to be much smaller than that. Maybe like 8.5 times 10 to the negative five. I'm just making up these numbers, but I'm illustrating how they're different by several magnitudes here. This is to the negative to whereas this is to the negative five. So they're different by it. Magnitude of three here. Okay, so there's a big difference in strength. Now, if we're looking at our equilibrium equation, remember H two way this will represent our acid water. He will represent our base. Remember? Based on the Bronston Laurie idea of acids and bases, acids donate H plus bases except H plus. So now, my ass, it has become my conjugate base and water accepted in H plus. So now it's H +30 plus. So this is the conjugate acid. If we're talking about the equilibrium expression that means K one equals products overreact. INTs remember, we ignore solids and liquids. Water is a liquid, so it'll be ignored. So this would be a check a minus times H plus divided by H two Way. Now, after we've given away the first acidic hydrogen we have. This left this because there's still another H plus in it. It has the potential to donate that second one if it decides to donate that second H plus. Then we come down here to this equation. So here it is now it's already given away the first H plus. Now it's about to give away the second one. So again it's the acid water is gonna act as the base, so it's gonna give away an H plus. As a result, it becomes even more negative. So now it's a two minus. So this would be the conjugate base here. Water accepted an H plus. So now it's the conjugate acid we're talking about giving away its second acidic hydrogen. So now we're dealing with K two, and so this would be a to minus times H 30 plus, divided by H A minus. So these will be our to equilibrium expressions dealing with di protic acid, donating it's H plus ions, so give yourself a minute or two to absorb the information that we've just presented in terms of die product acids and then we'll move on to die product basis.

The diprotic bases would also have 2 equilibrium equations.

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concept

## Diprotic Acids

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now looking at die product basis, We're gonna say the general formula is a to minus because it has a to minus. There, it can accept to h plus is one at a time. Now for those equations, let's take a look. So here we're dealing with the first part of a die product base. It's a to minus form, so it's gonna act as the base so water will be the acid again. Remember, acids donate H plus so that a to minus, except on a H plus to become H A minus water, as a result, becomes O. H minus here. Since we're accepting the first H Plus, we're dealing with K B one are based association constant and just like a one in K two K B one will always be greater than K B two, meaning that it's always easiest to accept that first H plus than it is to accept that second H plus. So here it be H A minus times O. H minus, divided by a two minus. Then, finally, we're gonna say here that we have the possibility of this accepting the second H plus, So if it does, it gives us the second equation here. So again, water is gonna act as the acid. This will be my base. We're gonna donate an H plus. So now it becomes ht way, which is the conjugate acid, all h minus of the conjugate base. So here, up here, this could also be the conjugate acid. And this is the conjugate base again, we're accepting the H plus here for a second time. So we're dealing with baby, too. We ignore solids and liquids. So this is Htwe times of H minus, divided by H A minus. So those were the equilibrium expressions for whether you're dealing with the dye product acid form of the die basic form. Knowing those are key to helping you set up your equilibrium expressions and from there determining your P h and P o h. Now that we've seen, this will continue onward to the last remaining portions when dealing with di product acids and bases

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## Diprotic Acids

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So based on what we've seen so far, we're gonna say, based on these equations, the relationship between the different forms off died, product species are all right. So before it's given away any of its H plus values, it's going to be Htwe. So you could think of this as it's acidic form. When he gives away the first H plus, it has become H A minus. We'll call this the intermediate form and then we have it giving away its second H plus to become a to minus. So it's lost. All of its H plus is now, so this would be its basic form. These would be the three major forms of any dye product acid or base. Now, as a result of having multiple K s and KBS, we can't just simply say anymore that K Times K B equals K w. Now you have to remember that K one is connected to K B two and multiplying them together gives us our K W, which is still 1.0 times 10 to the negative 14 when the temperature is 25 degrees Celsius and then remember that a k A to is connected to k B one and they're both still connected to K. W. Now, why is that? Well, if we take a look here at the three forms, we're gonna say we're going from the full it fully acid acid form. And we're giving away our first H plus to give us our intermediate form. So the relationship between them is connected by K one. Then if we're giving away the second acidic hydrogen, that's gonna give us K two, and that's what we're looking at it this way. Now, if we look at it the opposite way going this way, this is its basic form in the very beginning, and it accepts in H Plus to become this form. If we're talking about it accepting its first H plus, we're dealing with CBI one. So notice here that K B one and K two are connected here. Then if we're talking about accepting our second H plus to give us back our acidic form, that means we're talking about CBI, too. So that's why K one and K B two are connected, and that's why K two and K B one are connected. So just remember the relationships between those acid Constance and based Constance. Now we're gonna stay here when dealing with di product acids. What? We're trying to figure out our pH. We're gonna say H two way can be treated as the mono product acid in the same way and in that case would use K one. Then, for the intermediate form, we would use K two because it give away its second acidic hydrogen to become just like the base. Then we'd say that a to minus represents the basic foreman in that case would use K B one and technically, for the intermediate form, we could use either que two or k B two. Depending on. Are we is this thing accepting an H plus? If it's accepting an H plus, then we'd use K B two. Is it giving away an H plus? In that case, we use K two, so the intermediate forms a little bit tricky. It's all based on. Is it acting as an acid, or is it acting as a base? Because, remember, here it has ah, hydrogen in front and a negative charge in the back so it can act as either an acid or base because it is anfo Terek. And because of that, we could go and use K two if it's acting, hacking acid and donating an H plus. Or we could use K B two if it's gonna accept an H plus and act like a base. So these are just some of the fundamental principles you need to keep in mind when dealing with di product acids and bases. As we go further into this concept will look at problems and we'll learn. Learn how to approach those different types of questions to figure out what our pH or P O. H will be.

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## Diprotic Acids

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so sulfurous. Acid, which is H two s 03 represents a die product acid with K one equal toe 1.6 times 10 to the negative two and K two equals 4.6 times 10 to the negative five. Here, it says, calculate the pH and concentrations off sulfurous acid by sel fight and sulfide ion. When given 0.250 Mueller sulfurous acid here they're giving us the concentration of the acid for so we're dealing with the acid form, which is a weak acid because it's K is less than one. We're dealing with K one because we're dealing with the fully acidic form of it. It hasn't given any H plus yet. It's gonna react with water, which is the liquid. Remember acids donate H plus two bases. So we're gonna get here is we're gonna get H S 03 minus since its eye on its acquis plus h +30 plus acquis. Since it's K one is less than one, it's a weak acids. We're gonna use an ice chart in order to determine our pH and also to help us determine the final concentrations off these ions. So we have initial change. Equilibrium. Remember, in a nice chart, we ignore solids and liquids. What will be ignored? Because it's a liquid. Our initial concentration of our weak acid is 0.250 Mueller. We're not given initial concentrations of the products, so they're both zero initially, remember for change, we lose react INTs in order to make products, bring down everything we're talking about the first acidic hydrogen coming off. So we're dealing with K one, so K one equals products. Overreact INTs. Now remember, we have to determine Can we ignore the minus X on the bottom or not? To do that, we do the 5% approximation method. So we take our initial concentration and we divided by our constant that we're using, which in this case is K one. If we get a total, that's greater than 500 will be able to ignore minus X the initial concentrations 0.250 Mueller The K one is 1.6 times 10 to the negative too. When we divide those that gives me 15.6 to 5, this number is not greater than 500. So I kidnapped cannot ignore the minus X within my equilibrium expression. So this will be 16 times 10 to the negative to for K one. Both my products air X at equilibrium. So that's X squared, divided by 0.250 minus x. Again, we cannot ignore the minus X here because the total we got was not greater than 500. So we're gonna do here is gonna keep the minus X and use the quadratic formula. These cancel out, we're going to distribute, distribute. So when I distribute, I'm gonna get here 4.0 times 10 to the negative three minus 1.6 times 10 to the negative two x equals X squared. Because this X has the highest exponents, it's our lead term. So we have to subtract everything over to its side or add everything over to its side. So we're gonna add 16 times 10 to the negative two x two Both sides here and they're going to subtract 4.0 times 10 to the negative three. So we're gonna stay here. X squared equals So X squared plus 1.6 times 10 to the negative two x minus 4.0 times 10 to the negative three. So this represents our equation. So this is a B and C. We're gonna use the quadratic formula. So negative B plus or minus B squared minus four A. C over to a. So that's negative. 1.6 times 10 to the negative, too, plus or minus 1.6 times 10 to the negative. Two squared minus four times one. Don't forget the negative sign for C times Negative 4.0 times 10 to the negative three divided by two times one. Now, remember, because this is plus or minus, that means we're gonna get two possible answers here. When I do everything inside of here and I take the square root of it, what I'm gonna get is X equals negative one six times 10 to the negative two plus or minus 20. to 7 divided by two. So you get two possible answers again because this is plus or minus. So when we add, that would give us 0.5575 Moeller. And if we kept the negative sign and give me or negative 07175 Mueller, both cannot be the answer on Lee. One of them is the true answer here. E in your ice chart stands for equilibrium at equilibrium. It is impossible to have a number that is negative. So if I took this second answer and I plugged it in here or here, then we would have a negative value at equilibrium, which is impossible. So that means we cannot use this X instead will use this X here, plug it into here into here and into here to figure out each one of those compounds concentrations at equilibrium. So at equilibrium, h two s 03 equals to 50 Mueller minus X. So plug in the X variable that we just found. So at equilibrium equals 0.194 to 5 Mueller. So we got that answer at Equilibrium HS. All three minus equals X, just like h 30 plus equals X. So they both equal the same thing at equilibrium. All right, so we figured out the concentration of this one end of this one. We still have to find the concentration of vessel three to minus. Plus, we still need to determine the pH. All right. We should know what the ph is at this point because we know what h 30 plus is. PH equals negative log of h 30 plus. So we're gonna take the negative log off 0. So that's 1.25 now to figure out the concentration of S three to minus is a little bit trickier, because at this point, we don't have s all three to minus anywhere within this equation. What we would say is, if by Sulfide ion decided to donate its second h plus, that could create this particular ion. So that's what we're gonna do. We're gonna come down with that. So H s 03 negative comes down here. It's gonna donate its second H plus toe water. So that's how we're gonna get s three to minus plus H +30 plus here. We won't need to set up a nice charter calculate. This is what we need to realize if we take a look at the K is between K one and K two. Look at K one K one is 10 to the negative, too. Whereas K two is 10 to 95. They're different by a lot. A magnitude of three fold. That's a big difference. So what that means is, the amount off product we're gonna make the second time around is gonna be very, very, very small. So small, in fact, that we don't need to do a nice chart. This is how he would solve for the basic form of this die protic acid. Since we're dealing with giving away the second acidic hydrogen, that means we're dealing with K two Que two equals products. Overreact INTs. So that's what we have initially. What we need to do here is we need to solve for eso three to minus. So you're gonna multiply both sides by HS 03 minus so hs three minus times ca two equals sl three to minus times H plus divide both sides here by H 30 plus. So we're gonna get here that are sulfide ion. Concentration equals H s minus times K two divided by H plus. Here's the thing We just found out in the first ice chart that at equilibrium H s 03 minus, which is the intermediate off the dye protic acid, it equals my hydro Nehemiah on concentration and equals H plus. So that the same exact number. Now, if they're the same exact number, what does that mean? That means that they will cancel out here since their same exact number. So we'll see that at the end. The concentration of the basic form of my die protic acid is just equal to K two. So the concentration of sulfide iron is just 4.6 times 10 to the negative five Mueller. So it's important that you remember this when we're dealing with a dye protic acid here to figure out the concentration of the acidic form before it's given away any h plus, we would set up the ice charting sulfur acts and plug it in here so that X would subtract from the initial concentration of my die protic acid to find the intermediate form of the die Protic acid again, we'd still just need to find X whatever that X is, that would be the concentration of the intermediate form. Sall. Fight here represents the basic form of my die protic acid, the basic form of my die protic acid. Its concentration at equilibrium is just equal to K two. So it's important that you guys remember this. And if you're still confused of what I mean by acid form, intermediate form and basic form makes you take a look at the videos before this one. Because remember, for a dive product acid has too acidic hydrogen. So this would be the acid form after gives away its first H plus, This would be its intermediate form. And then it's basic form. Is this so H two s 03 would be the acid form. H s 03 minus is the intermediate form, and eso three to minus is the basic form. When it comes comes to being given the concentration of the acid form of a die protic acid, we set up a nice chart to help us find the concentration of the first two and then just simply say that the concentration of the basic form is equal to the K two of the die proto gas ID. So knowing that is the best way to answer this type of question and again because you're K two is so much smaller than your K one. This number is gonna be so small, in fact, that it's not gonna really affect the overall concentration of H 30 plus. Okay, so you're Ph would still be 1.25 And in fact, at equilibrium, these two would give us the same exact number of this. And if you added this concentration to the amount of H 30 plus, you would still see that we have the same exact ph at the end.

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example

## Diprotic Acids

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So here we're told to determine the ph of 0.115 Mueller sodium sulfide. Now here we're told that hydro sulfuric acid, which is H two s, contains the following K values. Here. We have to make a slight correction to the values that were presented before. So these are the actual K s of hydro sulfuric acid. We see that it has two K values, so of course it's a die product acid. But which form are we dealing with when we're looking at sodium sulfide? We'll realize here that the acidic form is H two s. That's when it has both acidic hydrogen. Once it loses its first acidic hydrogen, it becomes S H s negative. Then, once it loses its second acidic hydrogen, it becomes as too negative. So this is the acidic form. This is the intermediate forms. So the form that's in the middle and then this is the basic form. The basic form is once it's lost all of its acidic hydrogen. Now, if we're looking at sodium sulfide here, sodium sulfide is made up of a plus and there's two of them, plus the sulfide ion, and you can clearly see that right there that represents the basic form of this die protic acid. And remember, when we're dealing with the fully basic form, we have to use K B one, this form here would use K one and then this intermediate form. It depends if it's acting like an acid where donates an H plus. Then we'd use K two. But then, if it's acting like a base, depending on what it's reacting with, would use K B to Okay, so just remember which version of the intermediate form that's reacting to determine which K value on to use. All right, so we're dealing with a K B one here because we're dealing with the basic form. Remember here to find K B one, we remember that K B one is connected to K two. So kay, two times K B one equals K W. K. Two. Is this value here? We don't know what K B one is. That's what we're looking for. And K W is 1.0 times 10 to the negative 14. Divide both sides now by 9.1 times 10 to the negative. Eight makes you put these in parentheses when you punch them into your calculator. Otherwise, you may get the incorrect answer. If you do it correctly, you should get CBI one equals 1. times 10 to the negative seven. We're dealing with the basic form here, so it's gonna react with water. If it's going to be the base, that means water is gonna act as the acid. Remember, acids donate h plus. So when the process will create H s negative plus O H minus. So we have initial change equilibrium. Okay. Remember, we ignore solids liquids in our ice chart. Here. The initial concentration off our base is 0.115 Mueller, these initially your zero. Now remember, we lose reactant to make products so minus X plus X plus x bring down everything plus X plus X. Now, at this point, we'd say K B one equals products. Overreact INTs. Both products are multiplying, so they'd be x squared on top, divided by 0.115 minus x on the bottom. We can check to see. Can I ignore this minus X or not? To do that, you take your initial concentration. You divided by the K value that you're using. In this case, it's bkb one. If this ratio was greater than 500 we could ignore that minus X. We refer this to this as the 5% approximation method. So my initial concentration is 0.115 Mueller we divided by K B one, which we determined was 0.1 point 0989 times 10 to the negative seven. That gives me 1.465 times 10 to the six. Definitely a value greater than 500. So I could ignore this minus X multiply both sides now by 0. and 0.115 is gonna multiply, Mike one, which is this value here. And when we do that, we're gonna get what are X squared is equal to X squared of equal to 1.26374 times 10 to the negative eight Take the square root of both sides. So X equals 1.12 times 10 to the negative four. And remember when we find X X gonna give us either h 30 plus or O H minus in our ice chart. We see that X here gives me Oh, h minus. So if I took the negative log of this number, it give me P o h. So I'm gonna take the negative log of this number When we do that, it gives me 3.95 as my answer. And remember, if you know p O h. You know ph because ph equals 14 minus p o h. So this equals 10. So that would be the ph of my solution. So remember, when it comes to a die protic acid you have tow, analyze the compound being asked, Is it a acidic form? If so, would use is K one, is it the intermediate form intermediate form? A little bit tricky if it's acting like an acid by donating in H plus tau water would use K two. If it's acting like a base, it'll except in H plus from water. So would use K B two here for the basic form we'd say here for the basic form we're dealing with K B one, and as a result of this, we'd say in this equation when we're using K B one that help us find o h minus and from there we can figure out our P O. H

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PRACTICE PROBLEMS AND ACTIVITIES (14)

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