For more information, please see full course syllabus of AP Chemistry

For more information, please see full course syllabus of AP Chemistry

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### Polyprotic Acids

- Acids dissociate ONE hydrogen Ion at a time.
- Sulfuric Acid is strong in its first dissociation, weak in its second.
- Equilibrium problems involving solutions of Sulfuric Acid more dilute that 1.0 molar must use the entire expression without approximations ( no ignoring the “x”).

### Polyprotic Acids

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Polyprotic Acids 1:04
- Acids Dissociation
- Example 1
- Example 2
- Example 3

### AP Chemistry Online Prep Course

### Transcription: Polyprotic Acids

*Hello, and welcome back to Educator.com, and welcome back to AP Chemistry.*0000

*Today, we are going to continue our discussion of acid-base equilibria, and we are going to talk about polyprotic acids.*0004

*If you remember, in the first lesson that we discussed acids in, we said that polyprotic acids are simply a fancy name for acids that have more than one hydrogen that they could possibly give up.*0010

*So, for example, sulfuric acid (H _{2}SO_{4}), carbonic acid (H_{2}CO_{3})...probably the most important of the polyprotic acids is H_{3}PO_{4}, phosphoric acid.*0021

*It is how the inside of a cell actually maintains its pH; it is the buffer solution inside of the cell.*0033

*And the carbonic acid, the H _{2}CO_{3}, is how the blood maintains its pH; it is a buffer solution outside of the cell.*0040

*We are going to discuss polyprotic acids today and talk about they behave; and it is not altogether different than anything that you have seen already, as far as weak acids and strong acids.*0048

*We'll just jump right in, do a quick couple of definitions, and we'll start with the examples.*0058

*OK, so once again, the really important thing about all acids (polyprotic or otherwise--well, polyprotic especially) is that acids dissociate one hydrogen ion at a time.*0065

*So, in the case of, say, carbonic acid (let me actually write that out first: so acids dissociate one hydrogen ion at a time), which is H _{2}CO_{3}, it's in equilibrium, so it dissociates into H^{+} + HCO_{3}^{-}, and there is a K_{a} associated with this, an acid dissociation constant, 4.3x10^{-7}.*0076

*And notice, we have a little 1 here, so K _{a1}, because it gives up one ion at a time.*0112

*Now, this HCO _{3}^{-} has another H that it can give up, and it also behaves like an acid; so it now shows up on this side, and it is in equilibrium with the proton that it gave up: CO_{3}^{2-}.*0118

*And this is K _{a2}; this is the second acid dissociation constant, which is equal to 5.6x10^{-11}.*0135

*One thing you want to notice here, which is a general trend for all polyprotic acids, is that the first K _{a} is going to be a certain number; the second K_{a} is one that is very, very small: 10^{-11}...very tiny--certainly much smaller than this.*0143

*So basically, when it comes to polyprotic acids, it's really the first dissociation that counts the most.*0160

*The only exception to this case (which we will see in a little bit when we discuss sulfuric) is sulfuric acid, and the reason being that sulfuric acid is strong in its first dissociation--it behaves just like hydrochloric acid or hydrobromic acid--but in the second dissociation, it's actually a weak acid, so it's slightly different depending on the particular molarity of the solution that we are trying to find the pH of.*0168

*But, we will get to that in a minute.*0191

*Another example would be phosphoric acid, like we mentioned, and it has three dissociations, because it has three protons to give up.*0193

*H _{3}PO_{4} is in equilibrium with H^{+} + H_{2}PO_{4}^{-}, and the first acid dissociation constant is 7.5x10^{-3}.*0200

*And then, this H _{2}PO_{4}^{-} dissociates into H^{+} + HPO_{4}^{2-}; so make sure you keep track of the charges.*0214

*It's very, very important, especially with these polyprotic acids; they all sort of look alike--the H _{2}PO_{4}, the HPO_{4}, the PO_{4}, H_{3}PO_{4}; it gets very, very confusing, so definitely work very carefully, very, very slowly.*0228

*Chemistry is very symbol-heavy, and it's easy to get lost in the symbols and make silly mistakes, simply for that--as opposed to conceptual mistakes.*0241

*The second dissociation constant, the K _{a2}, is 6.2x10^{-8}.*0251

*And then, this species, HPO _{4}, the hydrogen phosphate (which is 2-), becomes H^{+} + PO_{4}^{3-}.*0258

*Now, the phosphate is finally alone, and the third dissociation constant is equal to 4.8x10 ^{-13}; so clearly, 10^{-3}; 10^{-8}; 10^{-13}: you see the pattern.*0270

*At is dissociates more, it becomes harder and harder to actually pull off that second and third hydrogen.*0285

*Let's just go ahead and do an example, and I think everything will make sense.*0291

*Example 1: We want to calculate the pH of a 4.0 Molar H _{3}PO_{4} solution, as well as the equilibrium concentrations of H_{3}PO_{4}, H_{2}PO_{4}^{-}, HPO_{4}^{2-}, and PO_{4}^{3-}.*0297

*So, not only do we want the hydrogen ion concentration (in other words, the pH, when we actually take the negative log of it), but we want to find the concentration of all of the species in solution, once you take that phosphoric acid and drop it into water to create a 4 Molar solution.*0349

*What is the concentration of each of those species, in addition to the pH?*0365

*OK, well, let's just go ahead and start.*0368

*The first thing we do is the same thing we always do: we check to see the major species in the water before anything happens.*0371

*That is the whole idea; we want to see what is in there before the situation comes to equilibrium, before the system comes to equilibrium.*0377

*We want to check out the chemistry; we want to make some decisions about what is going to dominate--what chemistry is going to dominate.*0384

*There is always going to be one that dominates.*0390

*OK, so major species: well, phosphoric acid is a weak acid; you notice, K _{a1} is 7.5x10^{-3}; that is kind of small, so it's a weak acid.*0392

*A weak acid means that it's not going to dissociate very much.*0405

*So, most of it is going to be in this form: H _{3}PO_{4}.*0409

*That H _{3}PO_{4} is just going to be floating around in solution, not very dissociated.*0413

*The major species in the solution: well, you have an H _{3}PO_{4}, and your other major species is just the water.*0418

*Well, both of these contribute hydrogen ions; well, the K _{a} of, or the K_{w} of, water (which happens to be the K_{a} of water) is 1x10^{-14}.*0425

*Well, the K _{a} of this one is 7.5x10^{-3}, 11 orders of magnitude bigger; so we can ignore water's contribution to the hydrogen ion concentration; this is the dominant species.*0435

*Because that is the dominant species, that is the equilibrium that we are going to work with.*0447

*The same thing we have always done: major species; decide which is dominant, which is going to control the chemistry; that is the one you concentrate on--ignore everything else.*0451

*So, let's go ahead and write our equilibrium expression and do our ICE chart.*0459

*Let's see...yes, that is fine.*0465

*OK, so we have: H _{3}PO_{4} that dissociates into H^{+} + H_{2}PO_{4}^{-}.*0470

*We have our initial concentration, our change, and our equilibrium.*0482

*Our initial concentration is 4.00 molarity, and this is 00, so this is before anything has had a chance to come to equilibrium.*0486

*A certain amount of this H _{3}PO_{4} is going to dissociate, and for each amount that dissociates, 1:1, 1:1 ratio--that is the amount that shows up in solution of the other species.*0495

*Therefore, our equilibrium concentration is 4.00-x; this is x; this is x; and now that we have our equilibrium concentrations in terms of x, and we know what the K _{a} is, we set it equal to each other.*0505

*7.5x10 ^{-3} is the K_{a}; it is the x, the hydrogen ion concentration, times x, the dihydrogen phosphate concentration, divided by the phosphoric acid concentration, 4.00-x.*0521

*We do our normal approximation, which we can check the validity of; so this is equal to x ^{2} over 4.00, just to make our math a little bit easier.*0540

*When we do this, we get x ^{2} (actually, you know what, I'm just going to skip this line altogether; and I'm just going to go ahead)...and you multiply by the 4; you take the square root; you end up with x is equal to 0.173.*0548

*That is equal to the hydrogen ion concentration; therefore, the pH is the negative log of the hydrogen ion concentration--you get a pH of 0.76.*0563

*That is our first part: 0.76 is our pH; let's check the validity of what we have; by checking the validity, that means...*0574

*So let's do it over here; let's say "check validity of our approximation"--I mean, we know it's good, but it's good to just sort of check it.*0583

*0.173, which is x, divided by 4.0, times 100; you end up with 4.3%.*0592

*It's close to 5, but it's still under the 5% rule, so we're still actually pretty good; it's not a problem.*0600

*A perfectly good, valid approximation: we don't need to do the quadratic equation; this is a good concentration; this is the pH.*0606

*Now let's see where we stand: we found the pH, but we also wanted to find the concentrations of all the other species in solution.*0613

*Let's see what we have; we have the H ^{+} concentration; that is equal to x, which is 0.173 Molar.*0620

*Well, the H concentration also happens to be the H _{2}PO_{4}^{-} concentration.*0629

*The H _{2}PO_{4}^{-} concentration is the same thing, 0.173 Molar.*0634

*Well, since we have x, we also have the phosphoric acid concentration (H _{3}PO_{4}), which is equal to 4.00-0173, and I'll just go ahead and...no, we'll just do 3.8 Molar (I'll just go ahead and round it up a little bit; sorry about that).*0641

*So, that is going to be 3.8 Molar; so now, the only thing that we are actually missing is: we are missing the HPO _{4}^{-} concentration and the PO_{4}^{3-} (this is 2-; see, again, I am making the same mistakes; I have symbols and charges all over the place).*0667

*So now, we need the concentration of the hydrogen phosphate, and we need the concentration of the phosphate ion.*0688

*Well, like we said, an equilibrium is established; it doesn't matter where these species come from (the phosphate, the hydrogen...); the K _{a} that we have for a given species, for a given equilibrium; if it involves the species that we want, that is good enough.*0693

*The equilibrium that we do have, that involves the HPO _{4}^{2-}, is the following.*0713

*We have the H _{2}PO_{4} concentration; well, that is in equilibrium with H^{+}, plus the HPO_{4}^{2-}, which is the species that we are looking for.*0721

*Well, we have this concentration, and we have this concentration; that is this and this; and we also have the second K _{a}, the second dissociation constant.*0734

*That is what this is here: this is the second dissociation of phosphoric acid, and the K _{a} for this one, K_{a2}, is equal to 6.2x10^{-8}.*0742

*So now, we can just basically rearrange the equilibrium expression and solve for HPO _{4}^{2-}.*0757

*And then, for the PO _{4}^{3-}, we do the same thing with K_{a3}, the third dissociation constant.*0764

*Let's go ahead and write everything out, so that we see it.*0769

*K _{a2}, based on the equation that was just written (the dissociation of the H_{2}PO_{4}^{-}), becomes the concentration of H^{+}, times the concentration of HPO_{4}^{2-}, over the concentration of H_{2}PO_{4}^{-} (oh, this is crazy--all kinds of charges and symbols going on).*0774

*Therefore, I rearrange this to solve for this species, because I have this, I have this, and I have this; it's a simple algebra problem.*0797

*We have HPO _{4}^{2-}, the concentration--moles per liter--is equal to the K_{a2} times the H_{2}PO_{4}^{-} concentration, divided by the H^{+} concentration.*0806

*We get: that is equal to (I'm going to move this over here) 6.2x10 ^{-8}, which is our K_{a}, times 0.173, which was our concentration of H_{2}PO_{4}; it also happens to be the concentration of our H^{+}.*0825

*So, these cancel, of course, leaving us with 6.2x10 ^{-8} molarity for...that is the HPO_{4}^{2-} concentration--that is one of the last things that we needed.*0849

*Notice, it's very, very, very small.*0866

*Well, now we want to know the PO _{4}^{3-} concentration; that is the final concentration--the final species whose concentration we want.*0870

*OK, well, now that we have found the HPO _{4}^{2-} concentration (it's this), we have an equilibrium.*0880

*The third equilibrium is (the third dissociation of phosphoric acid--excuse me): HPO _{4}^{2-} (oh, wow, you see what I mean--all of these charges, all of these symbols floating around; you're bound to make mistakes--you have to go very carefully) is in equilibrium--it dissociates into H^{+} + PO_{4}^{3-}.*0886

*We can write a K _{a} for this: the third dissociation constant is equal to the H^{+} concentration, times the PO_{4}^{3-} concentration, over the HPO_{4}^{2-} concentration (which we just found).*0912

*We rearrange this, and you get: the PO _{4}^{3-} concentration is equal to the K_{a3}, times the HPO_{4}^{2-} concentration, divided by the H^{+} concentration.*0929

*So, the K _{a3} is going to be (let me see: what is the K_{a3}?) 4.8x10^{-13}, times the HPO_{4} concentration, which we just found, which is 6.2x10^{-8}, over the hydrogen ion concentration, which we found in the first step: .173.*0953

*Well, we get a very, very tiny number: we get 1.7x10 ^{-19} molarity; that equals the PO_{4}^{3-} concentration.*0976

*There you have it.*0990

*So, a polyprotic acid: you handle it in the exact same way--most polyprotic acids are going to be weak acids.*0993

*The only exception is sulfuric; sulfuric is strong in its first dissociation, weak in its second dissociation; we are going to handle a problem in just a minute.*0998

*But aside from that, you handle it in exactly the same way.*1007

*And then, if you happen to need the concentrations of the other species--a further dissociation (so, phosphoric acid, dihydrogen phosphate, hydrogen phosphate, phosphate), you use the concentrations that you found, and then you use the next dissociation, and the next dissociation, with the appropriate K _{a}, to find the concentrations of all the species that you want.*1010

*About the only difficult problem here, as you saw, was just keeping track of the symbols and not losing your way in the symbology.*1035

*That is going to be the hardest problem, which, as far as I am concerned, that is the problem you always want--you don't want conceptual problems; you want mechanical problems; those are easy to deal with.*1042

*OK, so let's do another example; this time, we are going to deal with sulfuric acid, and we are going to do a couple of variations of it.*1050

*So, let's just start Example 2: We want to calculate the pH of a 1.2 Molar H _{2}SO_{4}.*1057

*OK, well, so sulfuric acid: let's write down its dissociations.*1074

*Its first dissociation is: H _{2}SO_{4} dissociates into H^{+} + HSO_{4}^{-} (the hydrogen sulfate ion, also called the bisulfate ion).*1079

*This K _{a} is large: we said it's a strong acid in its first dissociation; it's large--it doesn't even have a number.*1091

*You remember, strong acids don't have K _{a}s; they are huge.*1097

*The second dissociation is: HSO _{4}^{-} dissociates into H^{+}...*1101

*Actually, I'm not even going to give the equilibrium sign here; this is going to be a one-sided arrow pointing in one direction--full dissociation; there is no H _{2}SO_{4} left--very, very important to remember that for strong acids.*1107

*For strong acids, there is an arrow pointing to the right, and that is it; it's not in equilibrium.*1119

*We have HSO _{4} going to H^{+} + SO_{4}^{2-}, and the K_{a} (it's small, but it's actually still pretty large, relatively speaking): 1.2x10^{-2}.*1125

*So, a little larger than the other weak acids, but it still behaves as a weak acid; there is an equilibrium here; it is still less than 1.*1138

*OK, we want to calculate the pH; well, let's do what we always do--let's see what the major species are.*1145

*Be careful here: the major species in solution--you have taken sulfuric acid; you have dropped it into water; sulfuric acid is strong in its first dissociation.*1152

*Therefore, a solution of sulfuric acid is entirely composed of free hydrogen ion, free hydrogen sulfate ion, and (the other species in water is) H _{2}O.*1163

*There you go; now, again, we are calculating a pH here; once again, strong dissociation; it's a strong acid in its first dissociation; that means in species, before any equilibrium is reached, it's this and this and this.*1179

*"Strong acid" means full dissociation--it breaks up completely: there is none of this left.*1197

*It is all that in solution: this floating around, this floating around, this floating around; what is the hydrogen ion concentration?*1202

*Well, the question is: Since it's a strong acid in its first dissociation, can we treat it just like any other strong acid and just take the molarity and just take the negative log of it?*1211

*Can we just do -log of 1.2?*1220

*Because 1.2 moles per liter produced 1.2 moles per liter of H and 1.2 moles per liter of that, and this is actually pretty weak, it's probably not going to contribute too much.*1223

*However, we need to make sure; so let's ask our question: Can we just treat this like any other strong acid problem?*1235

*The answer is maybe; so, we have to check.*1263

*The reason we have to check is, we need to see if this actually does contribute anything.*1269

*It is small, but it's not so small (like on the order of 10 ^{-5} or 10^{-6}, 10^{-7}; it's 10^{-2}.*1274

*When you get to the 10 ^{-2} range, you're going to have to be careful; and we'll see in a minute what a general rule of thumb is.*1284

*A general rule of thumb is: when you have second-dissociation constants, such as 10 ^{-2}, if you are below about 1 molarity, you can't really ignore it; you have to include it.*1291

*So, not only do you have to take the 1.2 Molar, but some of this HSO _{4}, because it also dissociates, is going to produce a little bit more H^{+}; so you might have to include that; that is what we are going to check.*1302

*We are going to check to see if the second dissociation is significant--produces enough H ^{+} so that we have to include it into the 1.2.*1317

*Let's write: first of all, maybe, so we have to check; so now, mind you, our final H ^{+} concentration that we are going to take the negative log of to find the pH is going to equal, in this case, the 1.2+x.*1327

*This x is going to be any hydrogen ion that comes from this dissociation.*1345

*It's strong in this first dissociation, so I know that, before anything happens, I know that at least 1.2 moles per liter of hydrogen ion is floating around in solution.*1352

*However, I need to know if the second dissociation also produces enough hydrogen ion; so I have to add it to this.*1360

*That is the thing; we are going to compare x and 1.2 to see what the real difference is; if it's small, we can ignore it, and just treat it as a regular strong acid problem; if it's not small, we have to include it before we take the final negative log of that final hydrogen ion concentration.*1369

*I hope that makes sense.*1384

*The ICE chart is going to look slightly different; OK.*1386

*Well, let's go ahead and do our ICE chart, then: so again, we know that we have the 1.2of that, so we don't have to worry about this.*1389

*We have to worry about this equilibrium: how much H ^{+} is this HSO_{4} going to produce under these circumstances?*1397

*So, the equilibrium that we want to look at is the HSO _{4}^{-}: H^{+} + SO_{4}^{2-}.*1405

*I hope that makes sense: major species, ICE charts...you do ICE charts with weak acids; you don't do them with strong acids.*1414

*I have at least 1.2 moles per liter of H ^{+} floating around, because it's a strong acid in its first dissociation.*1421

*I need to check the second dissociation to see if it produces a significant amount of H ^{+} to add to the 1.2.*1428

*So, Initial, Change, Equilibrium: our initial concentration of HSO _{4} is 1.2 Molar, because it is produced in the same way that this is produced.*1436

*When H _{2}SO_{4} dissociates, it produces 1.2 Molar of this, 1.2 Molar of that.*1448

*Now, what is our initial H ^{+} concentration?*1453

*Here is where we have to be careful, and where the ICE chart is different: our initial H ^{+} concentration is also 1.2, because we have free H^{+} floating around and free HSO_{4} floating around; that is what these two are.*1455

*But, there is no SO _{4}^{2-} yet, before anything else happens.*1471

*In solution, before the system comes to equilibrium, this and this are the same from the first dissociation.*1475

*Now, the change is going to -x here; this is going to be +x here; +x here; we add vertically down; we get 1.2-x for the HSO _{4}; we get 1.2+x...like I said, 1.2+x is going to be our final hydrogen...and we get +x.*1482

*Now, we form our equilibrium; let's see, our K _{a} is (oops, here we go with the stray lines again; they show up in the most interesting places; OK--let me just do it off to the side here) 1.2x10^{-2}, equals 1.2+x, times x, over 1.2-x.*1504

*So now, we have to solve this equation.*1542

*OK, here is what we are going to do: we are going to presume that x is small in order to do our approximation; then, we are going to check our approximation to see if it's valid.*1544

*So, when we take x small, watch what happens to this approximation.*1556

*OK, it's approximately equal to...that means this x disappears--not this x, this x.*1561

*So, it becomes 1.2 times x, over 1.2.*1568

*Well, these cancel, and I am left with: x is equal to 1.2x10 ^{-2}, which is the same as 0.012.*1579

*I hope you'll forgive me; I actually prefer to work in decimals, as opposed to scientific notation.*1590

*I like scientific notation, but I just like the way that regular numbers look.*1594

*OK, now, we said that x is .012; now, we have to go up here: our final hydrogen ion concentration is going to be 1.2 (the original amount floating around), plus the x, plus the 0.012 that came from the second dissociation.*1599

*That equals 1.2.*1627

*It equals 1.212, but it's 1.2 to the correct number of significant figures.*1632

*So again, significant figures become a little bit of an issue; I personally don't care about significant figures all that much; a lot of people sort of criticize me for that, but you know what, it doesn't really matter; I'm more interested in concepts than I am in actual significant figures.*1640

*I think they are fine, and I think it's good to sort of use them, and in this context, yes, it sort of helps, because you notice: 1.2 plus .012--this is actually pretty small compared to this.*1652

*Since, working with significant figures, you still end up with 1.2, this says that you are actually justified in ignoring the second dissociation.*1665

*In other words, it doesn't really contribute all that much, in terms of hydrogen ion; you can just take the negative log of the 1.2.*1674

*So, here we have the 1.2; so the pH is equal to -log of 1.2, and you get a (let me see what...yes) -0.079 for a pH.*1682

*Yes, a pH can be negative; that means it's a really, really strong acid.*1699

*Now, if you ignore the significant figures, like I personally tend to do, I'll just go ahead and put that number down, too.*1703

*If you ignore the significant figures, and take the final hydrogen ion concentration to equal 1.212--if I include that .012, there is a little bit of extra hydrogen ion in there--well, you know what--your pH is going to be slightly different.*1711

*This is .084; 0.079, 0.084...you know, it's a small difference; if you are doing analytical work, it's important--for normal work, it's not altogether that important.*1737

*But you are going to end up with a pH of .08; that is really what it comes down to.*1749

*So, as it turns out here, the rule of thumb ends up being the following.*1754

*We notice that, for a solution which is about 1.2 Molar, roughly 1 Molar, the contribution of the hydrogen sulfate dissociation can be ignored.*1765

*You can treat it like a strong acid problem.*1776

*Below about 1.0 Molar, 1 Molar, .9 Molar...you can't ignore it; the second dissociation does produce a significant amount of hydrogen ion that has to be included in the concentration that came from the first dissociation.*1779

*You have to add those two; so, the first dissociation and then a little bit of the second dissociation...that is when you have your final hydrogen ion concentration.*1796

*For solutions of H _{2}SO_{4} more dilute than approximately 1 Molar, the full expression for the equilibrium constant must be used.*1805

*In other words, we can't approximate; we have to do the 1+x, times x, over 1-x; we have to solve the quadratic, which means the quadratic expression must be solved.*1842

*OK, so now, we'll do a quick example; I'm actually just going to set it up for you (I'm not actually going to solve it), just so you see what it looks like.*1864

*The Example 3 is: Calculate the pH of a 0.010 Molar H _{2}SO_{4}.*1872

*This is pretty dilute: .01 Molar--a lot less than 1.*1889

*So, we can't ignore the second dissociation; OK.*1896

*Let's take a look at what we have; we have our major species, and again, we have the H ^{+} from the first dissociation; we have the HSO_{4} from the first dissociation; and we have water.*1900

*These two are going to dominate the equilibrium; we know what this is already--this is just .01 Molar, because it's fully dissociated (strong).*1918

*The final hydrogen ion concentration is going to be 0.010+x, and x is the hydrogen ion concentration we get from the dissociation of the HSO _{4}^{-}.*1927

*That is the equilibrium we want to look at, HSO _{4}^{-} goes to H^{+} + SO_{4}^{2-}; we have Initial; we have Change; we have Equilibrium.*1943

*The initial concentration is 0.010, right?--and 0.010, strong, dissociates into this and this (first dissociation).*1955

*This is 0; the change is -x, +x, x; we get 0.010-x over here; we get 0.010+x over here; we get x over here, and we write K _{a2}, which is 1.2x10^{-2}, is equal to 0.010+x, times x (oops, let me just take out this parentheses here), and then, that is going to be over 0.010-x.*1965

*That is it; we have to solve this equation.*2003

*Multiply through; get your x ^{2}; get your x term; get your...it's going to be ax^{2}+bx+c=0.*2007

*b and c and a--they can be negative, if they need to be.*2019

*And then, you just plug it into the quadratic equation, or use your graphing utility (your TI-83, 84 calculator, your graphing calculator) to find the roots of this equation.*2022

*Now, you are going to get two roots; check to see what makes sense--one of the roots is completely not going to make sense.*2031

*The other root will make sense, and that is the whole idea.*2037

*So, your final hydrogen ion concentration (oops, let's make this a little bit...): when you get x, you are going to get x equal to some number; we'll just put a little box there.*2041

*Your final hydrogen ion concentration is going to equal the 0.010, plus this boxed number, and your pH is going to equal the negative log of your final hydrogen ion concentration.*2054

*And that is it: so, for polyprotic acids, treat them like any other equilibrium; usually, most polyprotic acids are weak acids; it is going to be the first dissociation that dominates.*2071

*You can pretty much ignore the second, third, fourth...I don't think...I don't even know if there is a...well, there is, but...*2081

*Weak acid--you can pretty much ignore the second and third dissociations.*2086

*For sulfuric acid, maybe; maybe not.*2092

*If you are talking about 1 Molar or above, you can ignore the second dissociation; if you are talking about more dilute than 1 Molar (.8, .7, .6...below that), the second dissociation is going to contribute a significant amount of hydrogen ion.*2094

*You have to take the amount that comes from the original concentration, the first dissociation; find the amount that comes from the second; add them; and then take the negative log of that.*2110

*I hope that helps; again, you see: it's pretty much all the same--slightly different when you are dealing with multiple dissociations in the case of sulfuric acid.*2119

*Thank you for joining us here at Educator.com.*2128

*We will see you next time to discuss the acid-base properties of salts.*2130

1 answer

Last reply by: Professor Hovasapian

Fri Feb 26, 2016 1:07 AM

Post by RHS STUDENT on February 12, 2016

Sir, regarding to the first example, would you plz explain why the Hydrogen ion you found when dealing when the dissociation of H3PO4 can also be used in that of H2PO4- and HPO42-? Aren't the breakdown of H2PO4- and HPO42- also produce extra H+? Thx

1 answer

Last reply by: Professor Hovasapian

Thu Mar 13, 2014 7:57 PM

Post by Alexis Yates on March 12, 2014

Wonderful lesson (as always) Professor, thank you for all of the help!

0 answers

Post by William Dawson on December 14, 2013

You are defeinitely right about decimals being better when they are not unwieldy, and signifcant figures being overrated!

1 answer

Last reply by: Professor Hovasapian

Wed May 1, 2013 11:42 PM

Post by morgan franke on May 1, 2013

I am unsure which root to use in my h final equation. Can you help me!

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Last reply by: Professor Hovasapian

Wed May 1, 2013 5:41 AM

Post by Antie Chen on April 30, 2013

Hello Raffi, I cannot understand at the time about 30:00, "For solution...", can you explain for me again?

The negative pH in example 2 is because the solution is so concentrated?

1 answer

Last reply by: Professor Hovasapian

Tue Feb 26, 2013 3:36 PM

Post by kevin vaughn on February 26, 2013

very good,wish i found this site early. would have to take my chem class for a 3rd time at university ><