AP Chemistry 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 (H₂SO₄), carbonic acid (H₂CO₃)...probably the most important of the polyprotic acids is H₃PO₄, 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 H₂CO₃, 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 H₂CO₃, it's in equilibrium, so it dissociates into H⁺ + HCO₃^-, 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 HCO₃^- 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: CO₃^2-.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

H₃PO₄ is in equilibrium with H⁺ + H₂PO₄^-, and the first acid dissociation constant is 7.5x10^-3.0200

And then, this H₂PO₄^- dissociates into H⁺ + HPO₄^2-; 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 H₂PO₄, the HPO₄, the PO₄, H₃PO₄; 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, HPO₄, the hydrogen phosphate (which is 2-), becomes H⁺ + PO₄^3-.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 H₃PO₄ solution, as well as the equilibrium concentrations of H₃PO₄, H₂PO₄^-, HPO₄^2-, and PO₄^3-.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: H₃PO₄.0409

That H₃PO₄ is just going to be floating around in solution, not very dissociated.0413

The major species in the solution: well, you have an H₃PO₄, 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: H₃PO₄ that dissociates into H⁺ + H₂PO₄^-.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 H₃PO₄ 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 x² over 4.00, just to make our math a little bit easier.0540

When we do this, we get x² (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 H₂PO₄^- concentration.0629

The H₂PO₄^- concentration is the same thing, 0.173 Molar.0634

Well, since we have x, we also have the phosphoric acid concentration (H₃PO₄), 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 HPO₄^- concentration and the PO₄^3- (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 HPO₄^2-, is the following.0713

We have the H₂PO₄ concentration; well, that is in equilibrium with H⁺, plus the HPO₄^2-, 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 HPO₄^2-.0757

And then, for the PO₄^3-, 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 H₂PO₄^-), becomes the concentration of H⁺, times the concentration of HPO₄^2-, over the concentration of H₂PO₄^- (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 HPO₄^2-, the concentration--moles per liter--is equal to the K_a2 times the H₂PO₄^- 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 H₂PO₄; 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 HPO₄^2- 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 PO₄^3- concentration; that is the final concentration--the final species whose concentration we want.0870

OK, well, now that we have found the HPO₄^2- concentration (it's this), we have an equilibrium.0880

The third equilibrium is (the third dissociation of phosphoric acid--excuse me): HPO₄^2- (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⁺ + PO₄^3-.0886

We can write a K_a for this: the third dissociation constant is equal to the H⁺ concentration, times the PO₄^3- concentration, over the HPO₄^2- concentration (which we just found).0912

We rearrange this, and you get: the PO₄^3- concentration is equal to the K_a3, times the HPO₄^2- 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 HPO₄ 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 PO₄^3- 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 H₂SO₄.1057

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

Its first dissociation is: H₂SO₄ dissociates into H⁺ + HSO₄^- (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_as; they are huge.1097

The second dissociation is: HSO₄^- 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 H₂SO₄ 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 HSO₄ going to H⁺ + SO₄^2-, 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) H₂O.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 HSO₄, 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 HSO₄ going to produce under these circumstances?1397

So, the equilibrium that we want to look at is the HSO₄^-: H⁺ + SO₄^2-.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 HSO₄ is 1.2 Molar, because it is produced in the same way that this is produced.1436

When H₂SO₄ 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 HSO₄ floating around; that is what these two are.1455

But, there is no SO₄^2- 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 HSO₄; 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 H₂SO₄ 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 H₂SO₄.1872

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 HSO₄ 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 HSO₄^-.1927

That is the equilibrium we want to look at, HSO₄^- goes to H⁺ + SO₄^2-; 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 x²; get your x term; get your...it's going to be ax²+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