AP Chemistry Common Ion Effect & Buffers

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

Today, we are going to close out with a couple of comments--close out the acid-base discussion with a couple of comments on some oxides--the covalent oxides and the non-covalent oxides.0004

We are just going to briefly touch on them--not too much, because I really want to get into the common ion effect and buffer solutions.0017

It is going to be a very, very, very important application of acid-base chemistry, probably the most important application of acid-base chemistry.0023

But, I didn't want to leave out this tail-end of sort of straight, normal, run-of-the-mill, routine acid-base stuff, and I wanted to talk about some of the oxides, because the chemistry does come up, and it's a little strange how certain oxides produce acidic and basic solutions.0032

If you remember, last time we talked about salts; and when they dissolve (the ones that are soluble), either the anion happens to be the conjugate base of a weak acid, or the cation happens to be the conjugate acid of a weak base.0050

Well, these things react with water, and they actually produce acidic or basic solutions.0064

Well, kind of the same thing happens with oxides; so let's take a quick look, and we'll just write down what they are.0069

We won't get too much into the chemistry of it, but I am going to demonstrate, actually, how it happens, because I want you to be able to see that this doesn't just fall out of the sky--that it is just a movement of electrons and things actually do make sense.0077

It isn't just about memorizing reactions; it's about understanding why things react the way that they react.0090

OK, so let's write down...covalent oxides (and you will see what we mean in just a minute) will produce acidic solutions in water.0095

In other words, if I take a covalent oxide, if I drop it in water, it is actually going to form an acidic solution.0120

Here are some examples: SO₃: SO₃ is a covalent oxide--"covalent" because you are talking about a nonmetal-nonmetal bond.0125

Sulfur and oxygen are both nonmetals; it's a covalent bond.0134

Plus H₂O: it ends up forming H₂SO₄, and when H₂SO₄ forms, it's a strong acid in its first dissociation, so it dissociates into H⁺ + HSO₄^-.0138

That is why it is an acidic solution: you drop SO₃ into water; it forms H₂SO₄; H₂SO₄ dissociates, producing hydrogen ion; therefore, you have an acidic solution.0153

SO₂: if you bubble that into water, you end up with H₂SO₃, which is sulfurous acid, which is a weak acid.0164

That is why we have a double arrow here, showing equilibrium; but again, the dissociation is the same: HSO₃^-.0175

CO₂: if you bubble CO₂ into water at high pressure, you end up with carbonic acid.0182

Carbonic acid is a weak acid, and there is an equilibrium with that and the hydrogen carbonate ion.0193

So, in each case, you see that H⁺ is being produced down the line, downstream of the actual reaction that is taking place.0199

And then, let's just do one final one; let's do NO₂ + H₂O: it produces (it's actually going to be 2 NO₂) nitric acid, and it also produces nitrous acid.0207

You will find both species in solution; nitric acid is strong, so it dissociates into H⁺ + nitrate; and here, we are going to have an equilibrium with H⁺ + nitrite.0221

There you go: covalent oxide--"covalent," meaning that the thing bonded to the oxygen--they are both nonmetals, basically; the thing bonded to oxygen is a nonmetal, so sulfur...any nonmetal.0235

These oxides, when dissolved in water, react with water to form acidic solutions.0249

Now, ionic oxides, on the other hand: ionic oxides produce basic solutions in water.0255

And you know, ionic means metal-nonmetal; so ionic oxide is basically a metal with oxygen; so, an example would be calcium oxide.0275

With calcium oxide, when it reacts with water, what you end up producing is calcium hydroxide.0284

Potassium oxide (K₂O) plus water: you end up producing potassium hydroxide.0294

Potassium hydroxide is a strong hydroxide, and it dissociates into 2 hydroxides, plus 2 potassium ions.0304

Calcium hydroxide is moderately soluble; actually, it's not very soluble at all, but it is slightly soluble; therefore, it dissociates into 2 OH^-, plus calcium 2+.0313

We write it as an equilibrium, because in fact, most of it is actually over here.0326

This is mostly a solid; this is potassium hydroxide; this is an aqueous solution; so this just means that it is not very soluble, but enough of it does dissolve to produce a hydroxide ion, and hydroxide ion--that is why we have a basic solution.0330

So again, an oxide--a covalent oxide--produces acidic solutions when reacting with water; an ionic oxide (basically, oxygen plus a metal, so any ionic bond between oxygen and a metal)--when those react with water, they produce basic solutions.0349

OK, now I'm going to quickly discuss how these are actually formed--just going to give some quick examples.0370

This is definitely not something that you have to know; however, I want you to see it, because I think it is important to see it; it is important to get a sense of the chemistry and to feel comfortable with it.0375

Those of you who go on to study organic chemistry: this is going to be a huge part of what you do--this sort of what we call arrow-pushing...movement of electrons.0385

Let's do the CO₂ example: the CO₂, plus H₂O, goes to H₂CO₃, which dissociates into H⁺ + HCO₃^-.0394

Here is what actually happens: CO₂ is a linear molecule; each oxygen is double-bonded to the carbon.0409

Water, as you know, or as you should know, maybe (and if you don't know, that is not a problem--we are actually going to get to it, because you remember: in this particular AP Chemistry course, we actually skipped over the bonding--I wanted to get to this stuff--equilibrium and acids and base--and I will actually return to the bonding)--a water molecule consists of oxygen single-bonded to hydrogens, and it has a couple of lone pairs.0415

Well, as it turns out, these lone pairs--what they do is: they attack the carbon, and they push the electrons onto one of the oxygens--one pair of electrons that are part of the double bond here.0443

What you end up getting is something that looks like this: O (now that oxygen has an extra electron), and then you have the oxygen, this and this, and 1 lone pair.0457

Well, because you have three things bonded to an oxygen, that is actually an extra bond; so now, this is carrying a positive charge.0469

You remember, charge has to balance on both sides of an arrow; here, this is neutral--everything is neutral here--net.0476

Here, this is minus; this has to be plus; that is how this works.0483

You will learn more about that later; I am just trying to get you to see what, exactly, goes on.0487

What happens next is: there is a transfer of a proton, and I am going to represent it like this: these negative charges actually take this hydrogen, and they push the electrons onto this oxygen, and what you get is OH, OH; there, you have your H₂CO₃.0492

Now, this ends up going into equilibrium with...that is how it happens.0514

Water attacks the carbon dioxide; there is a shift; there is a transfer of a proton over to oxygen; and what you end up with is carbonic acid.0522

Carbonic acid dissociates to produce H⁺.0529

I just wanted you to see that things don't just fall out of the sky; it actually makes sense what happens--this is reasonable.0533

OK, now let's do a calcium oxide example.0541

We have calcium oxide, plus H₂O, going to calcium hydroxide.0545

Here is how it happens: calcium...this is an ionic bond--this is 2+; this is O₂^-; well, O₂^-...here is what happens: H, again, with the oxygen; these electrons over here...this is fully...there are 8 electrons around here.0554

They actually take one of these; they push the electrons onto here; what you end up with is, now, calcium 2+ plus an OH^- plus another OH^-.0570

This O₂^- takes an H⁺ and becomes OH^-, right?--2 minus, 1 plus, is minus 1; it's an OH.0583

That leaves an OH, so that is the other OH^-.0589

Now, these form your calcium hydroxide, which is mostly insoluble--it's actually kind of a solid--it's pretty solid; but it is moderately (you know what, I'm not going to have these stray lines running around all over here, so let me do it this way--let me go slow)...0594

Calcium hydroxide is in equilibrium with calcium 2+, plus 2 hydroxide ion, and there is your hydroxide ion to form a basic solution.0615

I just wanted you to see it: covalent oxides form acidic solutions; ionic oxides form basic solutions--standard chemistry.0627

It will show up on the AP Chemistry exam, in terms of actual reactions; so they might say something like "aluminum oxide," "iron oxide--does it form an acidic or a basic solution?"0639

Well, iron oxide is an ionic oxide; therefore, it forms a basic solution.0649

If you have sulfur trioxide, does that form an acidic or a basic solution when dropped in water?0655

Sulfur trioxide is a covalent oxide: when you put it into water, it forms an acidic solution.0661

Qualitatively, it does show up on the AP exam; but I wanted you to see why--what actually happens chemically.0667

I want you to get a sense that these things--it's not magic; it's just atoms that are slamming into each other, and electrons are moving around; that is all that is happening--very intuitive stuff, really.0674

OK, now we are going to move on to a profoundly, profoundly important concept called the common ion effect.0685

The common ion effect, in its application, is used to talk about buffer solutions; so let's talk about the common ion effect--let's get a little bit of the math and chemistry under our belts.0693

And then, we will talk at length about buffer solutions; there are going to be, actually, two or three lessons strictly devoted to buffer solutions, because they are profoundly important.0705

Your blood is a buffer solution; it is buffered by the carbonic acid buffer system, the carbonate buffer system.0715

It is used to maintain the pH of the blood somewhere between about 7.2 and 7.4.0719

The phosphate buffer system inside the cell--that maintains the pH at a certain value inside the cell.0727

If pH goes up or down, terrible things start to happen; there is actually a very, very narrow range, as far as healthy body function, when it comes to blood pH and intracellular pH.0735

OK, so let's start with a definition.0748

Definition: When an equilibrium already exists in a solution, the shift that occurs when you add an ion (a little metathesis here) already involved in the equilibrium is called the common ion effect.0752

It is just an application of Le Chatelier's Principle, which you will see in just a moment.0829

Let's say this again: When an equilibrium already exists in a solution, the shift that occurs when you add an ion already involved in the equilibrium is called the common ion effect.0833

So, for example, if we have an equilibrium that exists in hydrofluoric acid solution, H⁺ + F^-, we know that hydrofluoric acid is a weak acid--there is a partial equilibrium.0844

It is not completely dissociated, but enough of it is dissociated to produce F^- and H⁺.0858

Now, the question is: What happens to the above equilibrium when some KF is added (when some potassium fluoride is added)?0864

Well, you know that potassium fluoride is a soluble salt, which means that potassium fluoride will completely dissociate into free K⁺ + F^-.0883

OK, you have this equilibrium right here (let's do it in red); you have an equilibrium that has been established, and all of a sudden, to the solution, you decide to add some potassium fluoride.0895

Well, the potassium fluoride dissolves; now, there is fluoride ion floating around, but there is already fluoride ion floating around.0904

Here is what we mean by an ion that is already involved in the equilibrium.0910

There is already fluoride involved in this equilibrium; now, you have added a salt that produces more of this ion; which way is it going to shift this equilibrium?0915

Well, Le Chatelier's Principle says: now that we have added more of this F^- (which is over here), the system is going to shift in the direction that opposes the change.0924

You have added F^- over to the right, the product side...well, you have just added F^-; now, the equilibrium is going to shift in a direction which reduces the concentration of F^-, which means it's going to shift that way.0936

It is going to use up F^-; it is going to produce HF.0949

So literally, what is happening, though--what is actually happening is: this F^- is going to end up suppressing the actual dissociation of this HF.0953

So, you can either establish this equilibrium, and then add the F^-, and then push the equilibrium back that way; or you can add them simultaneously, and the existence of this F^- floating around will keep this from actually moving forward.0962

It will keep the HF from dissociating.0977

There are two ways of looking at it, because again: you can establish the equilibrium, then add the common ion; or, you can just throw everything in and then let everything come to equilibrium on its own.0979

But, what is happening is Le Chatelier's Principle: you have an ion involved in an equilibrium, and then you have the presence of that same ion, and it is going to affect this equilibrium.0988

That is all that is happening--a common ion effect.0998

Two things: HF and KF--they share a common ion, the F^-; this is the common ion effect.1002

OK, let's just do an example, and I think it will start to make sense.1009

Example 1: Calculate the hydrogen ion concentration of a 1.2 Molar HF solution (something we have done before); then, calculate the hydrogen ion concentration of a solution that is 1.2 Molar hydrogen fluoride and 1.2 Molar sodium fluoride.1016

So, calculate the hydrogen ion concentration (or the pH, if you want to take the negative log of it)--calculate the H⁺ concentration--of a 1.2 Molar HF solution; then, calculate the H of a solution that is 1.2 Molar HF and 1.2 Molar sodium fluoride.1066

Let me rewrite this HF here; OK.1084

Let's do A first; so let's check our major species--it is what we always do first: we decide on what is going to dominate the chemistry.1087

Our major species are HF (because it is a weak acid, most of it is going to be in the HF form, not dissociated), and we have H₂O.1094

Well, we know from experience that the HF is going to dominate.1104

OK, so now, let's go ahead and do our equilibrium; our equilibrium is going to be: HF goes to H⁺ + F^-.1107

We have an Initial; we have a Change; we have an Equilibrium.1122

1.2 Molar, nothing, nothing; this is going to be -x; this is going to be +x; this is going to be +x.1126

Let's make sure we don't get these stray lines.1137

This is going to be 1.2-x; this is going to be +x; this is going to be +x.1140

I'm going to do it over here on the right-hand side, because I want it to be on the same page.1148

We get 7.2x10^-4, which is the K_a, which is equal to x squared, over 1.2-x; it's this times that over that; however, we are going to do an approximation, because we know that x is probably pretty small, compared to 1.2.1152

We can just do x squared, over 1.2; we end up with: x is equal to 2.9x10^-2 Molar.1172

This is the concentration of hydrogen ion, right here; so, the hydrogen ion concentration is 2.9x10^-2 molarity.1184

That is the first part; OK, so now we are going to do the second part.1193

Now, we are going to calculate the hydrogen ion concentration with the HF and the sodium fluoride in the same solution.1196

So, we have decided to just dump them both into the same solution, let everything come to equilibrium, and see what the hydrogen ion concentration is.1203

Now, the major species: well, now the major species are the following: we have the HF, which is the weak acid; we certainly have the H₂O; we also dropped in the sodium fluoride.1210

Sodium fluoride is a fully soluble salt; therefore, what is floating around in solution is Na⁺, and you have F^-.1224

These are the major species floating around in solution: HF, H₂O, Na⁺, F^-.1232

What is going to dominate here--what equilibrium is going to be established?1237

Well, the equilibrium that is going to be established is the same equilibrium as before.1241

This and this are going to come to some sort of an equilibrium; remember, the equilibrium constant--where these things come from doesn't matter; there is an equilibrium that exists between the hydrofluoric acid, the hydrogen ion, and the fluoride ion.1248

The only difference is: now the ICE chart looks different; and you will see what I mean in just a minute.1268

ICE: the initial concentration of hydrofluoric acid is 1.2; the initial concentration of hydrogen ion is 0; the initial concentration of fluoride is also 1.2--remember, we said 1.2 Molar sodium fluoride.1273

We dropped in some fluoride ion, so this is not 0 anymore.1289

Some of this is going to dissolve; some of this is going to show up--whatever shows up +x is also here, because this is a 1:1 ratio.1294

Our equilibrium concentration becomes 1.2-x in HF; it becomes x in H⁺; and it becomes 1.2+x in F^-.1303

Now, we do 7.2x10^-4, which is the K_a, equals the hydrogen ion concentration, times the fluoride ion concentration, which is 1.2+x, divided by the HF concentration, which is 1.2-x.1314

Now, again, x is probably going to be pretty small, so we are probably OK in doing our approximation.1331

You drop the approximation from this x, from this value and this value; you don't drop that x.1341

Over 1.2; well, when you solve this, you end up getting: x, which is equal to hydrogen ion concentration, is equal to 7.2x10^-4.1348

Notice: before, we had 2.9x10^-2; now, we have 7.2x10^-4; that is two orders of magnitude smaller.1364

Let's actually check the percent dissociation on this.1377

OK, for A, the percent dissociation is 2.9x10^-2 over 1.2, times 100; that equals 2.4%.1383

That means 2.4% of the original hydrofluoric acid dissociated; this is when there was no sodium fluoride in that.1399

But, by the presence of sodium fluoride, the common ion of a fluoride ion--now, the percent dissociation is 7.2x10^-4, over 1.2, times 100.1406

Now, we get 0.06%; the presence of the fluoride ion from the sodium fluoride that I put in there actually kept the hydrofluoric acid, that was in there also, from dissociating fully.1425

Le Chatelier's Principle pushed the equilibrium back to the left: only .06% of it dissolved--very, very small amount--virtually nothing, essentially--is what is going on here.1439

This is the common ion effect; OK, so once again, when you have either an equilibrium already established...it's good to think about it that way--an equilibrium that is established; and then, if you throw in some species where one of the ions of the species that you throw in is actually common to the original equilibrium, it's going to shift that equilibrium one way or the other.1451

This is how you actually deal with it--the same exact way--but notice the ICE chart: the ICE chart in the original was 1.200; there was no fluoride ion yet.1476

Remember, the Initial is just hydrofluoric acid before anything happens; when it comes to equilibrium, then we come here, like that.1487

But now, because you have the hydrofluoric acid and you have the fluoride ion--now the ICE chart looks different; and of course, it's going to give you different values.1497

That is the common ion effect.1507

OK, so now, we are going to introduce buffers.1509

Buffer solutions: so, let's see--let's go ahead and give a definition: A buffer solution (these are also called buffered solutions, B-U-F-F-E-R-E-D--I just prefer calling it a buffer solution--it doesn't really matter; it's the same thing--you are going to see them both ways) is a solution at a given pH, at a prescribed pH, that resists changes to that pH when further acid or base is added to the solution.1515

So, let's stop and think about this for a second; it's really, really important to understand definitions.1582

Definitions are profoundly important; if you understand definitions, it is the beginning of all of science.1590

We have to define what things are; a lot of the problem with understanding certain concepts has to do with the fact that many kids don't have a full grasp of the definitions involved.1597

That is why they are always sort of feeling like they are not quite on a solid footing.1609

We need to understand what we are dealing with first, and then we can deal with it.1613

A buffer solution is a solution at a given pH; so we have a solution at a given pH, and it is designed to resist changes to that pH when I add acid or base to that solution.1618

That is what the blood is; the blood is a buffer solution.1631

When you exercise, you produce CO₂; CO₂ tends to cause the blood to become acidic.1634

Well, the blood can't just drop in pH; if it does that, then everything will break loose, as far as biochemistry is concerned.1639

The blood needs to find a way to make sure that that added CO₂ is offset somehow, so that the pH is maintained within a very, very narrow range, and vice versa: there might be something that causes the blood to become a little bit more basic than it should be.1647

There needs to be something that will regulate, that will offset, that change, in order to maintain it at a stable pH--a very, very narrow range.1662

That is what the blood is--it's just a buffer solution; nothing more.1670

OK, so let's see what we have: it is (well, I know I said this before, but I might as well just write it out)--a buffer solution is--just an application of the common ion effect.1674

You will see it in a minute; it's just an application of the common ion effect.1695

OK, a buffer--let's actually define what a buffer is, physically.1708

A buffer is a solution containing a weak acid and a salt of its conjugate base, or it is a solution of a weak base and a salt of its conjugate acid.1715

In other words, we can form a buffer solution two ways: we can take a solution and put a weak acid in there, and put a salt of the conjugate base of that weak acid; or, we can put a weak base in there, and then we can add the salt--any salt--of its conjugate acid.1764

When I say "any salt"--a soluble salt; basically, we want something that dissolves.1785

It doesn't have to, but 99% of the time, you want it to dissolve.1789

So again, when we say "salt," we are talking about a soluble salt; that sort of goes without saying, but let me say it anyway.1794

OK, so here are some examples of some buffers, so this makes sense.1801

Examples of some buffers: Just think about your definitions--examples of some buffers: hydrogen fluoride and sodium fluoride.1806

HF is the weak acid; well, F^- is the conjugate base (I'll just put c.b.--oh, I'll put c.base).1826

Well, sodium fluoride is a salt of its conjugate base; that is it--it's that simple.1846

I take a weak acid; its conjugate base is--well, you know what a conjugate base is--just pull off one of the Hs, and you have the conjugate base; it's always the same, because acids dissociate one hydrogen at a time.1858

If you have an acid, take off one hydrogen; what you are left with is the conjugate base.1869

F^- is the conjugate base; a salt of the conjugate base--it just means that anion, plus some cation: in this case, sodium.1874

So, a solution of hydrofluoric acid (sodium fluoride) is a buffer solution; it resists changes in pH.1883

We will see how in just a minute.1891

OK, let's do another one: let's do acetic acid and potassium acetate.1894

Let's do acetic acid...so this is one example of a buffer; let me do the other one in blue; so acetic acid...1906

Here is the nice thing about it: you don't even need to know what acetic acid looks like--you don't even need to know the structure.1916

You know that, when you are dealing with some weak acid, it's just something plus some hydrogen ion that comes off.1920

That is the nice thing about it: it's the chemistry that matters, not necessarily the identity of the species.1926

Acetic acid and, let's say, potassium acetate: OK, so acetic acid is HAc (Ac is a short form of "acetic acid"--it is actually C₂H₃O₂; now, you can see why we use the Ac).1933

Potassium acetate is just KAc; so again, Ac^- is a shorthand for the acetate anion.1955

HAc is the weak acid (remember, 1.8x10^-5--that is the K_a); well, Ac^- is the conjugate base; again, conjugate base--just take off the H.1970

KAc is the salt of the conjugate base.1989

That is it; there is your buffer solution: you take a little bit of acetic acid; you take a little bit of potassium acetate; you drop them into the same beaker; you just stir it around; you have yourself a buffer solution--a solution which actually resists changes in pH.2000

When you add acid or base to that solution, the pH tends to stay the same; I mean, you can't add a lot of acid or base--we are talking about a little bit--and we'll get into the quantitative aspects in a while.2014

But, it is actually kind of amazing that you have a solution that actually just stays very, very...pretty much at the same pH that it was before any acid or base was added.2025

OK, let's do one more example; this time, we'll do a weak base, plus the salt of its conjugate acid.2033

Let's use CH₃NH₂.2041

This one I'll do...I think I'll do this one in black; I haven't used black ink in a while.2046

CH₃NH₂; this is methylamine; it's a weak base; and CH₃NH₃Cl (methylamine chloride, methanaminium chloride, or methylammonium chloride).2050

OK, now, the (oops, here we go with the lines again; OK) CH₃NH₂ (remember, there is that little lone pair here; that is the weak base) is the weak base.2075

Well, its conjugate acid, CH₃NH₃⁺...remember, if you have a base, the conjugate acid--just add a proton to it.2098

That is it: if you have an acid, and you need the conjugate base, take a proton away; if you have a base, and you need the conjugate acid, add a proton.2107

We added a proton; we added a charge; it's the conjugate acid.2114

And then, of course, the CH₃NH₃ (I'm thinking of NH₄ here)...NH₃Cl is the salt of its conjugate acid; that is it.2123

This is a buffer solution made from a weak base and the salt of its conjugate acid.2143

More often than not, just in terms of general laboratory work, really, most of the time, I'm guessing that you are probably going to see a weak acid and the salt of its conjugate base.2149

But again, it is perfectly reasonable, depending on what pH you want; that is the whole point.2162

You are going to choose an acid and a salt, or a base and a salt, in order to arrange a particular pH that you want; that is the whole idea behind a buffer solution.2168

You choose the pH that you want, and you choose the species, and you adjust the numbers so that you get the pH that you want.2183

That is what makes these so wonderful; you control the pH, beforehand.2190

OK, now let us do an example: OK, a buffer solution contains 0.55 molarity acetic acid, HAc, the K_a of which equals 1.8x10^-5.2195

And, it contains 0.45 molarity sodium acetate.2230

NaAc: calculate the pH of the solution--of this buffer.2235

Well, guess what--we already did this problem; we just did it a minute ago, when we did the hydrofluoric acid and the sodium fluoride.2249

We called it the common ion effect; well, this is it--we have a common ion, the acetate ion.2254

We have acetic acid and sodium acetate; we have a common ion situation; we have a buffer solution.2261

Let's see what is going on here: what is the first thing that we do?--we check the major species in solution, so that we can decide what the major chemistry is going to be.2267

There is always going to be something---one or two species that are going to dominate.2276

We have our acetic acid; acetic acid is a weak acid; this K_a tells us so.2280

That means, mostly, it is going to be HAc.2285

It is not going to have dissociated completely.2289

We have H₂O; we drop all of this in water.2292

We have sodium acetate; sodium acetate is a fully soluble salt; it is floating around as free sodium and free Ac^-.2296

Of the solutions in equilibrium, the dominant equilibrium is going to involve this species and that species.2305

Here is what the equilibrium is going to be: HAc in equilibrium with H⁺ + Ac^-.2312

Now, we will do an initial, a change, and an equilibrium.2320

Our initial HAc concentration is 0.55 Molar--0.55 molarity.2324

0 here; our initial acetate concentration is 0.45 Molar; sodium acetate--fully soluble; it dissolves completely , right?2334

Just to show you, NaAc dissolves completely to Na⁺ + Ac^-; that is what this is, right here.2346

Therefore, .45 moles per liter produces .45 moles per liter of acetate: 0.45 molarity.2354

Well, some of this is going to dissociate; some of this is going to show up; some of this is going to show up.2363

0.55-x; this is x; this is 0.45+x; now, we set it equal to the K_a.2369

Our K_a is 1.8x10^-5; it is equal to the hydrogen ion concentration, x, times the acetate concentration (which is 0.45+x), divided by the HAc concentration, which is 0.55-x.2378

Again, they are just numbers; this is probably going to be pretty small; x is pretty small, compared to the .55 and the .45.2397

It is a reasonable approximation to do x, times 0.45, divided by 0.55.2405

When we solve for this, we get a hydrogen ion concentration of 2.2x10^-5, which implies that the pH is equal to 4.66.2415

There we go: you have a buffer solution; you have a weak acid, acetic acid, and you have a salt of its conjugate base.2429

The conjugate base of acetic acid is acetate; a salt containing that is sodium acetate; when you drop all of these in solution, the things that are floating around are the acid (not the free ions; the acid itself is weak--that is what this K_asays: it's not dissociated), water, free sodium ion, free acetate ion.2444

There is an equilibrium that exists among the acetic acid and the acetate and the hydrogen ion.2466

It has a K_a; there is a constant associated with that equilibrium: it is equal to 1.8x10^-5.2474

We can use all of these numbers to establish (to find) what the hydrogen ion concentration is, under these circumstances.2479

We end up with 2.2x10^-5 moles of hydrogen ion in one liter of this solution--moles per liter--which gives us a pH of 4.66.2486

OK, so, for this lesson, we are going to stop here; next time, we will continue our discussion of buffers, and we will talk about how they actually work--what is going on with a buffer solution when you start to take this buffer solution and start adding acid or base to it?2498

That is the next step.2513

Thank you for joining us here at Educator.com; we'll see you next time--goodbye.2515