AP Chemistry Buffer

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

Today, we are going to continue our discussion of buffer solutions; and after this, we are going to go even further with buffer solutions.0004

Last time, we introduced what a buffer solution was; we said that it is a solution made up of a weak acid, plus the salt of its conjugate base; or, it's a weak base plus the salt of its conjugate acid.0011

It is designed for one thing only: it is designed to resist changes in pH once you add acid or base to that buffer solution.0026

We make a buffer solution at a specified pH that we want, and we design it specifically so that, if you add acid or base to it, the pH doesn't change; or, if it changes, it changes so slightly that it is negligible; that is what a buffer is.0036

Today, we are going to talk about how buffer solutions actually work.0051

In the last lesson, you remember, we actually created a buffer solution--two of them, in fact: we did one with hydrofluoric acid and sodium fluoride, and then the last one that we did was: we did acetic acid and sodium acetate.0055

We ended up with a pH of about 4.66; so here, we are going to talk about how buffers actually do what they do--in other words, resist changes in pH.0068

Let's get started; now, let me just write down something really quickly--what it is that we actually said about buffers.0080

We said that a buffer solution is a solution at a prescribed pH that resists changes to that pH upon addition of H⁺ or OH^-.0091

Again, when we add acid, we are adding H⁺; when we add base, we are adding OH^-.0138

That is what we mean by adding acid or base, ultimately.0142

This is the acid; this is the base; that is what is important--the other ions don't matter.0145

OK, how does this happen?--that is the question: how--how does this happen?0154

OK, so let's take a picture of a buffer solution; we have a choice--we can either do acid, salt of conjugate base; or weak base, salt of conjugate acid.0161

Let's stick with what we have done, which is a weak acid, plus the salt of its conjugate base.0175

Hydrofluoric acid: so, we have some HF floating around in solution, some HF floating around in solution, some HF floating around in solution.0180

We have the salt of its conjugate base; so let's ignore the cation--let's just deal with the fluoride ion.0189

We have some F^- floating around; we have F^- floating around; and we have F^- floating around.0196

Now, it's true: we also have a little bit of H⁺ floating around--I'm going to put a circle around it; H⁺ floating around; some H⁺ floating around.0204

Now, here is the idea: how does something resist a change in pH?--well, how does pH change?0213

Hydrogen ion concentration changes--it changes one of two ways: it changes by either going up or going down.0219

If I add acid to a solution, I have added more hydrogen ion to the solution, so the hydrogen ion concentration goes up; the pH drops.0226

If I add base to the solution (I mean any normal solution; I'm not talking about a buffer right now--any normal solution), well, the base ends up sort of...the hydroxide ion concentration rises; the pOH drops; the pH actually rises.0235

That is what is going on; so, what we want to do is...resisting changes upon addition of acid or base means, if I add acid, how can I eat up that acid so it doesn't float around freely in solution and change the molarity of the hydrogen ion concentration?0254

In other words, how can I sequester it--how can I bind it--how can I lock it up, so it isn't just floating around freely, changing the concentration of the hydrogen ion?0275

If I add base, how do I bind it--how do I lock it up--how do I pull it, get rid of it, so it doesn't actually end up reacting with the hydrogen ion concentration and dropping the hydrogen ion concentration, and raising the pH?0284

That is the whole idea of a buffer; how does it do that?0299

Here is how it does that: if I add base (which is OH^-, right?): I add base; here is the reaction that takes place.0302

Let's say, all of a sudden, I drop in some base (which is OH^-); well, the OH^- is a strong base; it does one thing--it's going to seek out hydrogen ions.0316

The base that you add is going to seek out hydrogen ions; well, the only source of hydrogen ion (actually, let me erase these, because actually will just get in the way and confuse you--what is important in a buffer solution is the acid and its anion) in a buffer solution is right there--is the Hs--these Hs right here.0331

When I drop in OH here, the OH is going to pull off these Hs, and here is the reaction that is going to take place...+HF; it's going to form HOH, which is water (acid-base neutralization), plus F^-.0365

What it is going to do is: it is going to...what did we say?--we said, if we add base, we need a way to bind that base so it doesn't affect the hydrogen ion concentration.0382

Well, here is your binding right here: if you add base to a buffer solution, it reacts with the hydrogen fluoride; the hydrogen fluoride is there for that reason--it is there to react with added base...production of F^- ion and water.0395

Water is neutral; F^- doesn't really affect anything all that much in this solution; so, what we have done is: we have sequestered the hydroxide by making it react with the hydrogen on the hydrofluoric acid, to produce water and F^- ion.0409

So now, it isn't floating around freely, affecting the hydrogen ion concentration.0426

Now, if we add acid (which means adding H⁺), well, H⁺...here is what is going to happen: it is going to react with the F^- now.0434

So, if I add acid, now the F^- is there to react with it, to sequester it--to bind it so that it is not floating around freely.0462

Because F^- is a strong conjugate base, HF is a weak acid; it is going to be mostly in that direction, not mostly in that direction; that is what the equilibrium says.0476

F^- reacts with any excess acid that we add, on top of what the pH is, to produce HF; it binds the hydrogen ion, so that the hydrogen ion is not floating around freely, contributing to the concentration of free hydrogen ion.0487

Acids...the measure of the acidity of a solution is based on the concentration of hydrogen ion.0506

A buffer solution has free base floating around the F^- (that is the conjugate base of the weak acid), to react with any extra acid that I might add--to bind it, so that it isn't floating around freely.0513

It also contains the weak acid itself, to react with any base that I might add, so that the base doesn't float around freely.0526

That is the whole idea; that is all that a buffer solution does--it binds any base or acid that I throw in there, and keeps it from doing any damage.0535

It keeps it from changing anything; it's as if...remember those H⁺s that I initially erased?--it is so that these H⁺s' concentration stays reasonably the same.0543

If I add H⁺, it would go up if these F^-s were not there.0556

Because the F^-s are there, they bind the added H⁺ so they are not floating around freely.0560

This is constant.0565

OH^-, if I add that...these are there to react with the OH^- to bind it so that it doesn't pull these out of solution and reduce the hydrogen ion concentration.0567

That is what a buffer solution does, and this is how it does it.0578

OK, let's do an example.0583

OK, now, I have to tell you something about these examples that I am going to do: I'm going to do 2 examples; they are going to be reasonably detailed, and they are going to be very, very, very important.0588

If you understand (both of the examples are the same--just different species; I just thought I would do it twice, so that you get a sense of what is going on) one or both of these examples, make sure you understand them completely.0598

Take your time with these: you will understand the nature of buffer solutions completely.0612

I promise you, everything that you need to know about buffers is contained in these examples--both technically and conceptually.0616

OK, Example 1: We want you to calculate the change in pH when 0.0100 moles of solid NaOH is added to 1.0 liters of the buffer solution from the last example of the last lesson (and don't worry, I am going to actually review what the last lesson was).0623

Then, we want you to compare this to the change in pH that occurs when 0.0100 moles of NaOH is added to 1.0 liters of pure water.0690

We want you to calculate the change that occurs--the change in pH--when .0100 moles of solid sodium hydroxide is added to 1 liter of the buffer solution from the last example of the last lesson.0729

Let's review what that was: Recall, we had 0.55 molarity acetic acid and 0.45 Molar sodium acetate.0741

This was our buffer solution.0758

Remember that?--we ended up with a hydrogen ion concentration that was 2.2x10^-5, and the pH was 4.66.0767

That is our prescribed pH; we have a buffer solution; our pH is 4.66; now, to this buffered solution, we are going to add .0100 mol of sodium hydroxide, and we want to see what the final pH is.0787

That is what is going on here; OK.0799

Now, let's just take a look at where we are starting.0802

Recall: well, let's see--so we need...our HAc concentration, if you remember right, is going to be the initial 0.55; that was minus x; remember, we started with .55 Molar HAc; some of that dissolved--that was the x.0814

Well, remember, x was really, really tiny; so, for all practical purposes, this is x right here.0838

0.55, minus the 2.2x10^-5: well, that equals 0.55 Molar; this is so small that it actually doesn't have an effect on the initial concentration.0846

The concentration of HAc floating around in solution is that.0863

The concentration of Ac^-...same thing: it was 0.45+x=0.45+2.2x10^-5, and again, it's so small that this is essentially 0.45.0870

This is the concentration of acetic acid and acetate ion floating around in the buffer solution, before we add the sodium hydroxide.0891

Now, this is where we stand as a buffer before the addition (so let me write that down--this is where we stand as a buffer...)--before OH^- is added.0900

And again, we are adding sodium hydroxide, but the sodium can be ignored; it's the OH^- that you are actually adding; you are adding base.0922

OK, buffer problems consist of two parts: the first part--you have to do the stoichiometry problem; and the second--you do the equilibrium problem.0928

In solution, these happen simultaneously; but in order to do the mathematics, we have to treat them as if they happen separately.0954

OK, so let's see what is going on.0964

What is the first thing that we do?--major species.0971

Major species (let me actually...OK) upon addition of sodium hydroxide: OK, so now that I have added the sodium hydroxide, here is what I have floating around in solution.0979

I have my acetic acid, HAc; I have my acetate ion, Ac^-; I have water; I have put in sodium hydroxide, so I have (sodium hydroxide is a strong base--fully dissociates into sodium ion and hydroxide ion, so) sodium ion floating around; and I have hydroxide ion floating around.1013

OK, let's take a look at what happens here.1038

I have HAc, Ac^-, H₂O, Na, and OH^-.1041

What did we say about adding hydroxide?--when you add a base, it is going to react; the first thing that it is going to do is react completely with any source of hydrogen ions that are available to it.1045

OK, the only source of hydrogen ion available to it is that, so these two will dominate what happens immediately in the reaction, before the system actually has a chance to come to equilibrium.1058

This is the stoichiometry part of the problem.1070

It will always be like this: when you add the base or the acid, you do the stoichiometry (this reaction); then, you do the equilibrium.1073

Let's go ahead: so, OH^- is a very strong base and will react with the only source (not with the only source; there are a couple of sources, but) the primary source--with the best source of H⁺, which is HAc.1081

We said that is how a buffer works; if it is base you are adding, it's going to react with this; if it's acid you are adding, it's going to react with that; that is the whole idea behind a buffer.1119

Here is the reaction that takes place: write the reaction--this is chemistry; you need the reactions in order to see what is going on.1127

OH^- will react with HAc to produce Ac^- + HOH.1134

OK, OH^- takes this H, releasing Ac^-; HOH--that is water.1148

This is the reaction that takes place; now, we have to do the stoichiometry--this reaction takes place, and then what happens after the reaction takes place?1156

Well, we are going to have some more Ac^- floating around; we are going to have a little less HAc floating around; then, we can do the equilibrium of Ac and HAc.1168

So again, hydroxide...this is going to react with that before anything else happens.1179

So, let's do this: I'm going to rewrite it again, a little bit bigger: I'm going to write OH^- (a little more room), HAc (and you will see why in just a minute), Ac^-, + HOH.1185

OK, so we are going to do something similar to an ICE chart, except I like to use Before, Change, and After.1203

The reason I do this is: I like to do Initial, Change, and Equilibrium when I'm dealing with equilibrium situations.1211

Before, Change, and After is used (I like to use it) for the stoichiometry part of the problem, before anything happens, after this reaction has taken place, but before any equilibrium has been established.1219

That is why I don't like to use ICE, because this isn't really an equilibrium situation.1236

This is a before-and-after situation.1240

So, before anything happens, I have dropped in 0.0100 moles of OH^-.1242

Now notice, this is a stoichiometry problem (let me actually write this here: this is stoich...)--this is the stoichiometry part of the buffer problem.1252

The stoichiometry: in stoichiometry, we work in moles; in equilibrium, we work in concentrations (moles per liter)--it's very, very important that you recognize that, and you will see why in a minute.1264

In stoichiometry, we are working in moles; notice: we are not working in concentrations just yet--that is for the equilibrium part.1278

So, HAc; well, we said that we have 1 liter of this solution, this buffer solution; and that contains .55 moles per liter of the HAc.1285

Well, 1.0 liter times 0.55 moles per liter gives me 0.55 moles; it's one of the conveniences of using one liter of a buffered solution--because the molarity actually becomes the moles.1303

But, it is important--again, with stoichiometry, we work in moles; with equilibrium, we work in molarity.1320

The Ac^- concentration: well, we just said in the previous slide that the Ac concentration is 0.45 moles per liter.1328

We have one liter of this solution; therefore, its concentration is 1.0 liters, times 0.45 moles per liter; that means there are 0.45 moles of Ac floating around.1338

Water doesn't matter.1354

The change: all of the OH^- that I have added is going to react with any HAc that is available; .010 moles reacts with .55 moles of this; this is going to vanish.1357

That is what this reaction says: this OH^- is being sequestered--it's being bound up in the form of water.1378

It is going to vanish: 0.0100; this is also going to disappear--this is 1:1.1385

For every one mole of hydroxide, one mole of HAc reacts; therefore, it's going to be -0.0100.1396

This--for every one mole of hydroxide added, one mole of Ac^- shows up; therefore, this is going to be +0.0100; water doesn't matter.1405

After this reaction has taken place, we have no hydroxide left; it has been sequestered--it has been bound by this.1420

We have 0.55 minus 0.01; 0.54 moles of that left over; water doesn't matter.1431

0.45 moles; we have produced 0.01 moles; we have 0.46 moles.1443

Now, there is .54 moles of HAc floating around; there is .46 moles of Ac floating around; there is no more OH floating around--that has been bound up; it has formed water (it doesn't matter; water is water).1456

Now, we can do the equilibrium part.1470

Again, major species: now what do we have floating around in solution?1480

Well, we have HAc, a little bit less than what we had before, because it reacted with the OH^-; we have Ac^-, a little bit more than we had before, because it was produced upon reaction with the OH^-.1486

We have H₂O, and we still have the sodium ion floating around: irrelevant, irrelevant.1501

These two are going to dominate.1508

Now, the system can come to equilibrium according to: HAc, H⁺, plus Ac^-; now we can do our ICE chart; Initial, Change, Concentration--I'm sorry, Initial, Change, Equilibrium.1511

OK, ICE charts--we have to deal in molarity.1527

We said we have (I'm going to make this a little bit bigger--I'm sorry--I need some room to actually do the chemistry here, so let me go...): HAc is in equilibrium with Ac^-, plus H⁺.1530

Let me keep it consistent: I usually always write the H⁺ first.1553

H⁺ + Ac^-: Initial, Change, Equilibrium.1558

We said we had 0.54 moles of HAc left; it is floating around in 1.0 liter; so the molarity is 0.54 Molar.1564

H⁺ hasn't come to equilibrium yet; this is initial--we said we had 0.46 moles of Ac^- floating around in the 1 liter, so the concentration is 0.46 molarity.1578

This will diminish; this will augment; this will augment; at equilibrium, we have 0.54-x; x; and we have 0.46+x.1592

Now, we have our K_a, 1.8x10^-5, equals the hydrogen ion concentration x, times the Ac^- concentration, 0.46+x, divided by the HAc concentration of 0.54-x.1608

OK, and this is approximate, because x is going to be very, very small in this case.1630

We have x times 0.46, divided by 0.54.1635

When we do this, we get x is equal to 2.11x10^-5, which gives us a pH equal to 4.68.1642

There we go: now, recall: before the addition, we had a hydrogen ion concentration of 2.2x10^-5; that was a pH equal to 4.66.1658

After we added the sodium hydroxide, now our concentration is 2.11x10^-5; now, the pH is equal to 4.68.1675

The ΔpH is 0.02; it is virtually nothing.1694

This is the power of a buffer solution.1700

Notice what happened: I had a buffer solution at a prescribed pH of 4.66; I added .010 moles (it doesn't seem like a lot, but it's a fair amount) of sodium hydroxide; the pH only changed by .02.1711

In fact, the hydrogen ion concentration (if you were to take the difference of the 2.2x10^-5 and the 2.11)--the Δ of the hydrogen ion concentration is 9x10^-7--it's virtually nothing.1725

The hydrogen ion concentration wasn't affected at all.1738

That is the power of a buffer solution.1741

OK, now we are going to actually do the comparison that we said we were going to do.1744

We want to now find what the pH is going to be if I take neutral water (not a buffer solution, just regular water) and drop in .010 moles of sodium hydroxide.1749

OK, so now, let's do the comparison.1759

Now, our major species upon addition of the hydroxide (sodium hydroxide) to water: we have--well, we have water; sodium ion; hydroxide ion; and we have water; that is it.1769

Sodium hydroxide is a strong base--completely dissociated--that is floating around; that is floating around; that is floating around.1801

Something is going to dominate the concentration here; well, water can behave like a base; OH can behave like a base; the difference between these two...it's that that is going to...1807

So basically, what we are going to do is take the H⁺ concentration (remember, we said that is equal to 10^-14), over the OH^- concentration: 10^-14 over 0.0100 equals 1.0x10^-12.1820

Our pH is going to be 12; well, neutral water is pH=7; the ΔpH (it might be nice if I actually wrote properly here) is equal to 12.0-7.0; look at the difference.1842

If I add .01 moles of sodium hydroxide to pure water: .01--that is not very much!--but apparently, it is a lot--the pH changes by 5 units: it goes from pH 7 to pH 12.1878

But, in a buffer solution, I add the same amount of sodium hydroxide, and the pH only goes up by .02--virtually nothing; almost not even noticeable.1896

This is the power of a buffer solution.1906

OK, now, I know that I had a second example, but I think I'm going to hold off on that second example, and I'm going to actually start the next lesson with that second example, just to give you a chance to sort of work on this one a little bit.1909

And then, it will be something nice to start the other one with, as a review of the previous lesson; because this is very, very, very important, and I don't want to end up concentrating all of it into this one lesson.1921

So, again, go through this very, very carefully--it's profoundly important that you understand what is happening in a buffer solution.1931

If you understand the chemistry, the math should make complete sense.1937

Thank you for joining us here at Educator.com.1941

We'll see you next time; goodbye.1943