Diprotic Acids and Bases

so die product acids and bases are compounds that could donate or accept two H plus ions. Now for dia protic acid, we're going to see that their basic formula, their generic formula is H two A. And their equations are illustrated by the following two examples. So here when we're dealing with the acid in its fully propagated form, meaning it has both of its acidic hydrogen, it's going to be the acid water is going to be our base. Remember following Bronston Laurie theory, the acid donates an H plus over to the base as a result of this, we create H A minus and H 30. Plus as our products. Since we're talking about removing the first acidic hydrogen from our diet protic acid, that means we're dealing with K. A. one and just like all other K values that equals products overreact ints. So our equilibrium expression would come down to H a minus times H 30. Plus Divided by H two A. Now hear this asset this uh form that we just created which is our conjugate base form because it lost an H. Plus it can continue to lose its second H plus and give us our second equation. So this H a minus continues onward where it still acts as an acid. Another water molecule acts as the base and just like before the acid donates an H. Plus to the base. Now we create a two minus plus H +30. Plus since we're talking about removing the second acidic hydrogen from our dia protic acid. Now it is K. Two that we're dealing with here, we're going to say that this is a two minus times H +30 plus divided by H a minus. Again, we ignore water because in an equilibrium expression we ignore solids and liquids. So these are the two basic equations dealing with the dia protic acid. And before we move on to the dia protic bases, just remember in terms of these equilibrium constants, K one and K two K one is always significantly larger than K. A. To that just means that it's always easiest to remove the first acidic hydrogen from from an acid and it's much more difficult to remove the next one. So remember K one will always be a value much greater than K. A. To click onto the next video and continue our discussion with dia product bases.

now that we've seen the dia protic acid form of a compound. We can examine its die product base form as well. We're gonna say fur die product bases. They can be represented by a two minus the two minus, representing the fact that it can gain two H plus ions here. The equations can be illustrated by the two examples given below. In this form we have our base form, it hasn't accepted an H. Plus yet. In this first equation, it's the base water is the acid water will donate an H. Plus to it, thereby creating a che minus by donating H plus the water itself becomes O. H. Minus. And because we're talking about accepting the first H plus ion, that means we're dealing with KB. One. Now K. B. one is an equilibrium constant and like all other equilibrium constants, it's equal to products overreact ints. And remember we have to ignore solids and liquids. So here would be H A minus times O. H minus Divided by a 2 -. Now the H A minus that we've created here can continue onward because it's still negative, it can still accept another H. Plus. So we're gonna say here we bring it down. It's still gonna act as the base. It's gonna react with the second water molecule which will act as the acid. Again the water will donate an H plus away to the H A minus, creating H. Two. A. Water again loses an H plus O. Becomes O. H minus. Since we're talking about accepting the second acidic hydrogen. That means we're dealing with K. B. Two. just like the other K. B. It's equal to products. Overreact ints and again, we ignore liquids and solids. So here this would be a church to a times O H minus, divided by H a minus. So at this point we've seen the dye protic acid form, the dye, basic form, die product based form. We've seen K one, K two, K B one and K B two. But how exactly are they connected to one another? By examining the next section, you'll be able to see the connections that we have with these four equilibrium constants as well as the different forms of our Dia product compound. So click on the next video and see how we relate them to one another.

So at this point we've seen the dye protic acid as well as the democratic base in action. And we've seen how we deal with four different equilibrium constants. Now, how exactly do we relate all these things with one another? Well, we're gonna say based on these equations, the relationship between the different forms of diabetic species are we can start out with the fully predominated or acidic form of our diabetic species. So that would be a church to a in its asset form. It only has one choice. It can choose to donate an H plus away by giving away that first H plus. That gives us a church a minus. And since we're talking about losing the first acidic H plus, that means we're dealing with K. A one now, H A minus. Which represents our intermediate form, can also decide to donate an H plus away giving us a two minus. Since we're talking about removing the second acidic hydrogen, that means we're dealing with K. A. Two. But remember these die product species, we can also look look at them in terms of a base. So here for this basic species here, we can say that it accepts its first H plus to become a church a minus. By accepting that first acidic hydrogen. That means we're dealing with KB one then the intermediate H a minus can decide to accept another H plus to become a church to a By accepting that second acidic hydrogen. That means we're dealing with K. B two. Now notice we have here, K K one and K B two, overlapping one another. And what do they do? They relate the acid form to the intermediate form, then we also have a on K. Two and K. B one overlapping one another. And they relate the basic form with the intermediate form. Because of this, we can establish relationships between K and K. B. So, because of the overlapping, we can say here that K one times K. B two gives us K. W. And we can also say because of the overlapping K two times KB one equals Kw. Remember K W represents the dissociation constant for water, which is 1.0 times 10 to the negative 14 when the temperature is 25 degrees Celsius. Remember all equilibrium constants are temperature dependent if we change the temperature, K W would change if they did that. They would tell you what the new K W would be. Now, because of this, we can establish some guidelines when dealing with one of these forms of our diet product species. So, if we take a look, we're gonna say here, if we're dealing with htwe, this is the fully propagated acidic form treated as you would a mono product acid. So we'd say here that we would use K one when dealing with it in order to find ph we'll skip the intermediate form for now and come back to it. H two minus has no H plus on it. So it only has one choice to accept an H plus and act like a base because we're accepting our first H plus. That means that it's going to act as a basic use KB one. Now the intermediate form is tricky because the intermediate form here can do one of two things. It could either donate another H plus and become The basic form and in that case would use K. A. two or it could decide to gain another H plus to become the acid form and therefore use KB two. So for this one it could use K two or K. B two. So it really depends on the species itself. Now realize this intermediate form has a hydrogen so it can act as an acid and a base because it's a negative charge there if you can act as an acid or base. Remember you're called an Amfa Terek species. Now, ample Terex species. We have acidic ones and basic ones. The acidic ones would donate H Plus and therefore use K two. So here we have hydrogen sulfate, we'd have hydrogen sulfite or by sulfate and by cell fight and then we have here di hydrogen phosphate. So these are your most common amfa Terek acidic compounds that would use your K two. And technically here this one fits into poly product because it has more than one. Um we'll talk about that later on in our polity product assets video then we have our basic ones. So these basic ones here would use KB two. So who fits in here? We have hydrogen carbonate, also called bicarbonate. We have hydrogen sulfide and then we have hydrogen phosphate. Again, this falls into the arena arena of polyp product species. So for right now, don't worry about these two here. Just realized later on when we talk about um philatelic species that they themselves represent acidic Amfa Terek and basic and eric species. So again, remember when it comes to the intermediate form, it could be K two or K B two, depending on if it wants to act as an acid or as a base respectively. And in terms of this topic, when we're talking about di product species, remember the first two that I listed for acidic and the first toy listed for basic. Those are the um categories they fall into and determine if they use K two or K B. To hopefully guys were able to follow along in terms of this now that we've gotten out of the way all the theory and ideas behind dive product species will continue onward with calculations in terms of finding ph and concentrations of these different types of compounds

here we're told that sulfurous acid which is H. Two S. 03 represents a dia protic acid with a K. One equal to 1.6 times 10 to the negative two and K. Two itself equals 4.6 times 10 to the negative five. Now it says calculate the ph and concentrations of sulfurous acid, hydrogen sulfide, also known as by sulfite and sulfide ion when given 50.200 moller of sulfurous acid. Now, for sulfurous acid, we're dealing with the acidic form or the fully protected form of the compound. And remember when we're dealing with the acidic form of the compound, we use K. One to help us determine what the ph will be. Now our dia protic acid is H. Two S. 03. It will react with water. Now remember the acid is going to donate an H plus to the water. It's gonna become a chess. So three Remember charged species are acquis when in solution and water itself becomes H30. Plus we have initial change equilibrium to represent the ice chart that we've set up remember in an ice chart, we ignore solids and liquids. So water which is a liquid will be ignored. Now, initially we have 0.200 moller of my die protic acid. The products initially are zero. We lose reactant to make products bring down everything At this point again, we're dealing with removing the first acidic hydrogen from our compound. So we're dealing with K. one. K one equals products overreacting. So it's H. S. 03 minus times H 30. Plus divided by H Two S. 03. We're gonna plug in the values that we know. So we have 1.6 times 10 to the negative two equals at equilibrium. Both my products are equal to X. And since they're multiplying each other that's X squared Divided by .200 minus X. At this point we have to determine can we ignore the minus X. Or not to do that? We do the 5% approximation method. Now with this 5% approximation method we take the initial concentration of our diaper tick species here and we divided by the K. value that we're using. In this case K. one. If the ratio is greater than 500 then I can ignore that minus X. And avoid the quadratic formula. So our initial concentration is 0.200 moller Divided by R. K. one which is 1.6 times 10 to the -2. When we do that we get back 12.5. So we have a value here that is not greater than 500. So we cannot ignore the -1. And we're gonna have to do the quadratic formula. So we're going to multiply both sides here by 0.200 -1. Here we're going to distribute, distribute. So here that's gonna give me 0.32 minus 1.6 times 10 to the negative two X equals X squared here. This X squared has the highest power. So it's gonna be our lead term which means everything gets moved over to its side. So we're going to add 1.6 times 10 to the negative two X. To both sides. And subtract .0032 from both sides. When we do that we're gonna get the expression X two or equation X two X squared plus 1.6 times 10 to the negative two X -1032. So this is my A my B. And my C. My quadratic formula will be negative B plus or minus square root of B squared minus four. A. C over two. A plug in the value. So negative 1.6 times 10 to the negative two plus or minus 1.6 times 10 to the -2 squared -4 times one. Don't forget the negative sign of c. so times negative .0032 Divided by two times 1. So here when we solve for everything in here and then take the square root of that, what we'll get at this point is X equals negative 1.6 times 10 to the negative two Plus or -114263 divided by two. Realize here that we have plus or minus, which means we're gonna get two possible X variable answers. So x equals .049131. If we added or X equals negative 0.65132. If we use the minus instead, how do we know which one to use? Well at equilibrium, you cannot have a negative concentration and it's not possible. So, if we use that negative acts and plugged it into here or into here for equilibrium that give us a negative value. So that would mean that that is not the correct concentration to use. So this one does not work. So my ex will be this number here. Always say now at this point is at equilibrium H two S. 03 Equals .200 - X. So, plug in what we just found in for X. And what we get for equilibrium amount for sulfurous acid at equilibrium is .151 Moeller at equilibrium hydrogen sulfide or by cell fight equals H 30 plus because they're both equal acts, Which again we said is equal to .049 131 Moeller. And because we know what H 30 plus is, we know what P. H. Is, because P. H equals negative log of H 30. Plus. So here that gives me 1.31 as my ph Once I plug that in. Now, up to this point, what have we determined? Well, we figured out what the ph three is of my dia protic acid. We determine what the final equilibrium amount of sulfurous acid is. And at this point hydrogen sulfide or by sulfide. What we're still missing is the sulfide ion itself though, if we look at our equation, we can see that cell fight is nowhere in it. So how do we determine sulfide amount? We'll realize here that this hydrogen sulfide or by sulfide still has one acidic hydrogen on it. So it could react with the second mode of water to help produce that sulfide ion. So here we have it H. S. L three minus plus a second water molecule. It donates an H plus to that second water molecule to give us S. +032 minus Water becomes H 30. Plus. Now we have initial change equilibrium again. All right. So what do we have at equilibrium for Sophia assassin? We had X. Which we determined was this amount. That's how much it had at equilibrium. Now it's reacting with another mole of water. So, what it has initially is that same amount here, we don't have anything of this yet. Initially for the last ice chart, H 30. Plus an equilibrium was equal to X. Was which was this number as well. So .049131. This would be minus X plus X plus X 0.49131 minus X plus X 0.4913 plus X. Here we're dealing with the second K. A. two. So here that's because we're removing the second acidic hydrogen and then this equals S. 032 minus times H 30 Plus divided by H. S. 0. 3 -. Alright, so now this is what's going on when it comes to acidity. The first acidic hydrogen is always easiest to remove the next ones are always much more difficult. That's that's why you see a big decrease in the K. Values for K. One which is the most acidic hydrogen. It's the negative two. To remove the second hydrogen is incredibly difficult. That's why the K two is so much smaller. The smaller K. Is the less acidic the compound, it is three magnitudes smaller. So small. In fact that that means that H. S. L. Three minus, it would be incredibly difficult to remove that H plus from it. Very very very little of it will come off. Which would mean that here this minus X. Is so small that it's not gonna affect the final equilibrium amount of H. S. 03 minus. And then this plus X. Here, same thing. It's gonna be such a small number that it's not gonna affect the overall final value of H +30. Plus. So because of that we can ignore these exes here, the plus sex and the minus X. Here, we cannot ignore this X. Because there's no number accompanying it. If we ignore that X then we won't have any value at all for S. 032 minus. So we ignored the minus X. We ignore the plus X. What does that mean? Well what that means now is K. Two equals S. 032 minus. We still don't know because we ignored these X. Variables. That means that H. 30 plus equals 300.49131 molar. And same thing with H. S. L. 3 -. It also equals that same exact number. And what does that mean? That means that they cancel each other out because of the same thing. So that means that K two is the concentration of S. 032 minus. So sl 32 minus. Its concentration will equal 4.6 times 10 to the negative five molar. So that's its final answer. Hopefully guys had enough space to write everything. I know I wrote a lot of different things, but that's the approach we would take. So for a recap when they give us the acidic form, the the acidic form, the fully propagated die protic acid species. To figure out the ph all we do is we take the K. One of the dia protic acid. Set up a nice chart, sulfur X. That will give us our H 30 plus concentration, which we then can find ph for the first acidic hydrogen that we're removing. We're using K. One that helps us to determine the concentration of our intermediate and H 30 plus concentration when dealing with the acidic form of a dia protic acid species realize here that the basic form when they ask us to find it, which is this one. All you have to ever say is that it's going to be equal to K A two. So again, this is the approach you do. When you're given the acid form of a dye product species, do these steps, you'll always be able to find each one of these ions and compounds their concentrations as well as the overall ph So keep in mind some of the methods we used here as we continue our exploration of dia product species.

here it states determine the ph of a 0.80 Mueller sodium sulfide solution. Now, here we're told that hydro sulfuric acid which is H. Two S contains K. One equal to 1.0 times 10 to the negative seven and K two equal to 9.1 times 10 to the negative eight. Now realize here that we're dealing with sodium sulfide, that's the compound we have to work with. The only reason we're giving information on hydro sulfuric acid is so that we can use its K. One and K. Two values realize here that sodium sulfide has had all of its hydrogen is removed. So sodium sulfide represents the basic form of our diaper tick acid here, it would represent to sodium and then the sulfide ion, the sulfide ion represents our basic form of our die protic acid. We can ignore the sodium because it's just a spectator ions. So we're gonna bring down the sulfide ion that we have. It's gonna react with water here, it's the base. So water is going to be the acid, the acid will donate an H plus over to the base to give us a church S minus plus O. H minus. We have initial change equilibrium. Remember in a nice chart, we ignore solids and liquids. So we're gonna ignore the liquid water. Our initial concentration is .080 moller. These initially are zero. We're gonna lose our reactant. So minus X. To help build up our products plus X plus X. Bring down everything 0.80 minus X plus X plus X. Now realize here, we're talking about accepting the first acidic hydrogen by the sulfide ion because we're accepting the first acidic hydrogen. That means we're dealing with KB one. But here, what we have we have K one and K two. What's the connection that KB one has? Well remember earlier we said that K B one and K two are connected together. And when you multiply them together, that gives us K. W. Alright, so We just have to isolate KB one. So KB one equals KW Divided by K two. So 1.0 times 10 to the -14, divided by 9.1 times 10 to the -8. When we do that, that's gonna give me my KB one value as 1.1 times 10 to the negative seven. So that's our KB one that will be utilizing. So we're gonna come over here. So KB one just like other equilibrium constants equals products. Overreact ints. Now we're gonna input the values we know for each. So 1.1 times 10 to the negative seven equals X squared divided by 70.80 minus x. We have to check to see can I ignore that minus X or not? By doing a 5% approximation method. So we're gonna take the initial concentration that we're dealing with divided by the equilibrium constant that we're utilizing, which is K B one. And look to see if that ratio is greater than 500. So here's our initial concentration Divided by the KB one that we're using. So that ratio is definitely greater than 500. So that means that this minus X is not significant enough to do any real change. So we ignore it. Alright. So now we just have to solve for X right now. So we're gonna multiply both sides by zero, So here X squared equals 8.8 times 10 to the negative. Nine. Take the square root of both sides here. So that gives me x equals 9.4 times 10 to the -5. Now, anytime you find X, it's gonna give you either H 30 plus or minus this X. That we isolated helps to give us a way to minus because we know the concentration of a hydroxide ion. We can determine P. O. H. Because P. O. H is just a negative log of hydroxide ion. So that equals 4.03. And because we know that P. H. equals 14 -4.03. So that's going to be 9.97 as our p. So the key to solving these types of dye product species questions is to determine are you dealing with the acid form, Are you dealing with the acidic form? Are you dealing with the intermediate form in the two examples on this page, we've seen what to do when asked to determine the ph of the acid form and the basic form. Later on, we'll see what do we do when dealing with the intermediate form of a dia product species.

if K one equals 4.46 times 10 to the negative seven and K two equals 4.69 times 10 to negative 11 for carbonic acid. What is the ph for a 0.15 molar solution of sodium bicarbonate. Now realize here that they're giving us information on carbonic acid but they're asking us to determine the ph of sodium bicarbonate. Our task is to determine which form of a dia product species do we have with sodium bicarbonate? Realize here that we have carbonic acid is the acid form where it has both of its acidic hydrogen. And remember when it loses its first acidic hydrogen, it becomes H. C. 03 minus so becomes bicarbonate and then it can lose its last acidic hydrogen to become carbonate ion with sodium bicarbonate which is made up of H and A plus and then carbonic bicarbonate. Here, we can see that we're dealing with the intermediate form of our die protic species. Now, because we're dealing with the intermediate form, our task of determining ph is really quite easy as long as we remember this particular formula. So, to figure out the H plus concentration of the intermediate for di product species, we say that H plus, its concentration is equal to the square root of K. One times K. A. Two times the formal concentration which is this 20.15 moller plus K one times K. W. Which the dissociation constant for water Divided by K. one plus the formal concentration of our species. Alright, so all we do here is we plug in our numbers. So K one is 4.46 times 10 to the negative seven times K two which is 4.69 times 10 to the negative 11. Our formal concentration of 110.15 moller Plus K. A one Times KW which is 1.0 times 10 to the -14 divided by 4.46 times 10 to the -7 plus .415 Polarity. Now when we plug all that into our calculator, what we're gonna get is the square root of 2.09471 times 10 to the -17. Take the square root of that gives us for the concentration of H plus as four six times 10 to the -9 Molar. So this represents the concentration of H plus and realize if we know the concentration of H plus then we know what the ph is because P H equals negative log of H. Plus. So we're gonna take the negative log of that number. And that gives me a ph of 8.34. Now this number makes sense because remember for the intermediate, these ample terrace species which can act as acids or bases because of the presence of the hydrogen and negative charge. Remember the basic filtering species which you said in earlier videos are bicarbonate, your hydrogen sulfide ion and hydrogen fast feet ion. So it makes sense that bicarbonate has a ph greater than seven because it represents one of the basic Alfa terek ions. So just remember when dealing with uh die product species, especially the intermediate form, all we need to do is utilize this formula to determine the concentration of H. Plus from that we can determine pH and if we want, we can determine hydroxide ion concentration and P. O. H. So just keep in mind the formula and you'll be able to solve questions just like this one.