AP Chemistry Titration Examples & Acid-Base Indicators

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

Today, we are going to round out our discussion of titrations with some titration examples--three of them, actually--just to sort of broaden our exposure to titrations and maybe get some slight variations on the types of problems that you can run across.0004

And, we are going to close off this particular lesson with acid-base indicators.0019

Now, well, you know, let's just jump right on into the examples.0023

OK, so Example 1 (let's do this in black today; no, I guess we're doing blue; OK): OK, our first example: If 60.0 milliliters of a 0.100 Molar (I'll just put .10 Molar) hydrocyanic acid, HCn, is titrated with a 0.15 Molar sodium hydroxide, calculate the volume of sodium hydroxide solution needed and the pH at the halfway point of titration.0028

The K_a of hydrocyanic acid is 6.2x10^-10 (very, very weak acid).0114

OK, so if 60 milliliters of a .10 Molar hydrocyanic acid solution is titrated with a .15 Molar NaOH, what is the volume of sodium hydroxide solution that I need to halfway point; and, once I reach the halfway point, what is the pH of that solution?0123

This is a slight variation on what it is that we have been doing: we have been, the last couple of examples--we have been following the progress of a titration, just to get a sense of how to actually work the problems with the stoichiometry part and the equilibrium part.0142

Now, slight variation: we want you to find the equilibrium; they are specifying a specific point in the titration, the halfway point, and we know what the halfway point is--that is the point where exactly one-half of the original solution (the HCn solution) has been used up.0156

So again, this is the stoichiometry part: when we say "halfway," that means half of the HCn has reacted with the OH^- that has been added.0174

We know how to do this; this is actually a basic stoichiometry problem from early on in the course.0184

So, let us write down: At the halfway point, one-half of the HCn has reacted with the added OH (with the added OH^-).0188

In other words, what has happened is that a certain amount of OH^- has reacted with the HCn to produce cyanide ion, plus H₂O.0219

Halfway means half of what we started with has been used up, so if we started with 10 moles, 5 moles have been used up; if we happen to have started with 15 moles, 7.5 moles have been used up; so we know exactly what the stoichiometry is at this point.0228

Well, let's find out how many moles of hydrocyanic acid were actually there to begin with.0241

Well, it says 60 milliliters (let's use a red); it says 60 milliliters of a .1 Molar solution; so, we do 60.0 milliliters, times 0.10 millimole per milliliter (right?--molarity--millimole per milliliter; as long as milli- is on top and bottom, we are good).0248

We end up with 6.0 millimoles; well, if 6.0 millimoles is what we start off with, the halfway point is that we have reacted 3 millimoles.0270

Therefore, we can say 3.0 millimoles has reacted, leaving...well, 3.0 millimoles of HCn are left over, and, in the process of this reaction, if we have used up 3.0 millimoles, that means we have used up 3.0 millimoles of OH that were added; we created 3.0 millimoles of Cn^-.0281

You notice: I am using the chemistry to sort of write out what is going on--that is the whole idea.0322

If you understand the chemistry, you can follow the chain of logic, and just end up going where you want to go--provided, of course, that you have a reaction that you are working with...and this is the reaction that we are working with.0328

OK, well, this says that 3.0 moles of HCn react with 3.0 millimoles (I'm sorry, not moles, millimoles; let me erase this--so millimoles) of OH^-; it's a 1:1 ratio in this.0338

So, we have 3.0 millimoles of OH^-; and now, the molarity of the OH^- solution is .15; and sodium hydroxide is 1:1--when it dissolves, 1 sodium ion is produced; 1 hydroxide ion is produced.0365

It is 1 milliliter of the OH^- solution; we have 0.15 millimoles of OH^- (right?--.15 Molar sodium hydroxide produces .15 Molar sodium ion and .15 Molar hydroxide ion).0390

Here, we are able to calculate that 20.0 milliliters of the OH^- solution is added.0420

That is the first part of the problem; this is a basic stoichiometry problem.0432

The fact that it is at halfway point tells us what is going on: we need to find the moles of HCn; we take half of that; the 1:1 ratio means it reacted with that much OH^-, and then we use the molarity of the OH^- to find the number of milliliters of solution.0436

The first part of the problem: 20 milliliters of OH^- solution is added to bring it to the halfway point.0451

Now, we want to do the pH; OK.0457

Once there is a reaction, major species--we want to find out what reaction is going to dominate.0460

Well, we have HCn floating around still, 3 millimoles; now we have Cn^- floating around, 3 millimoles; we certainly have the sodium ion floating around--that is however many millimoles that is--it doesn't really matter--it's not going to contribute.0466

And of course, we have H₂O.0488

Among these species, what is going to dominate the equilibrium?--this right there: this is going to dominate.0491

The K_a is 6.2x10^-10; water, 1.0x10^-14; four orders of magnitude.0498

So, we write the equilibrium part; we go with: HCn is going to be in equilibrium with (well, let me make it a little bigger, because I need a little bit more room) H⁺ + Cn^-.0506

We have an Initial; we have a Change; we have an Equilibrium.0531

Well, the initial equilibrium (this is an equilibrium part--we deal in concentrations)--well, we have 3.0 millimoles of HCn; now, it's floating around in 60 milliliters + 20 milliliters of solution that we added; the total volume now is 80 milliliters.0534

That is why we need to work in molarity, in concentration; it's very important, because the volume changes when you do a titration; you are adding volume to volume: 80 milliliters.0553

You get 0.0375; there is no hydrogen ion before equilibrium; now, the cyanide ion--again, we used up 3 millimoles of HCn; we produced 3 millimoles of Cn^-; and again, it is floating around in 80 milliliters; 0.0375--interesting, isn't it, that it's the same?0564

A certain amount of this is going to disappear; a certain amount of this is going to appear; the same amount is going to appear; so we have our equilibrium concentrations of 0.0375-x; this is x; and this is .0375+x.0593

Well, our K_a is 6.2x10^-10, equals x, times 0.0375+x, all of that divided by .0375-x, which is approximately equal to x times 0.375, divided by 0.0375.0610

Again, x is pretty small; you are talking about a common ion effect; it is going to be very, very little dissociation.0635

So, these, of course, cancel; and we are left with: x (which is equal to the hydrogen ion concentration) is equal to 6.2x10^-10, which is the K_a of the acid.0641

We knew that already: at equivalence point for a weak acid titration, we know that the halfway point...the pH is just the pK_a; the hydrogen ion concentration is the actual K_a of the acid.0653

This implies that the pH of our solution is 9.21.0668

Now, I would like you to show that we could have...I'm going to do this problem again, really quickly; I'm going to do it using the Henderson-Hasselbalch equation.0675

So, the Henderson-Hasselbalch equation (this second part here, just to give you a slightly different version, in case you like that Henderson-Hasselbalch equation): we could have used the Henderson-Hasselbalch (I'll just abbreviate it as H/H).0684

It says that the pH is equal to the pK_a, plus the log of the base over the acid concentration.0704

Well, notice: the Cn^- concentration is .0375; HCn is .0375; this is 1; the logarithm of 1 is just 0.0718

Therefore, our pH equals our pK_a; it equals 9.21--the quick version.0731

You can do the ICE chart if you want to see exactly what is happening and decide how you want to work the approximation, or you can just use the Henderson-Hasselbalch equation; you are going to get the same answer.0740

So, I hope that helped.0750

OK, let's do our second example--this is kind of an interesting one.0753

Here, what we want to do is...well, I'll just write out the example, and we will see what it is that we want.0762

OK, a chemist synthesizes a monoprotic weak acid and wants to determine its K_a; this is how we do it.0770

He dissolves 2.5 millimoles of the acid in 100 milliliters of water.0806

Sorry, I know the statement of the question is a bit long, but this is the nature of chemistry; this is the nature of science--word problems: that is what it is, and sometimes, as things get more complicated, you need a lot of information.0822

So, don't let the length scare you.0833

He dissolves 2.5 millimoles of the acid in 100 milliliters of water, then titrates this solution with 0.050 Molar sodium hydroxide solution.0836

OK, now: after 23 milliliters of NaOH solution are added, the pH equals 6.10.0859

What is the K_a of the acid?0887

OK, a chemist synthesizes a monoprotic weak acid and wants to determine its K_a.0891

He dissolves 2.5 millimoles of the acid in 100 milliliters of water, then titrates the solution with a .05 Molar sodium hydroxide.0897

After he adds 23 milliliters of the sodium hydroxide solution, the pH ends up being 6.10.0905

With this information, calculate the K_a of this acid.0912

Well, stop and think about this for a second: usually, when we talk about titrations, we are asking for the pH, and we are given a K_a.0916

Notice that this is just reversed: we give you the pH at a given place in the titration, and we are asking for the K_a.0926

So again, the only thing that is going to be different is: your ICE chart is just going to end up being slightly different.0933

Other than that, you handle this the exact same way.0938

So, let's go ahead and take a look at what we are doing here.0941

Major...so this is why, when we present a problem in a certain way, when we present the theory, we are presenting it in its given frame of reference.0950

If you understand the relationships among the players in that frame of reference, you can fiddle around with them as you need to.0959

It doesn't follow that...not all titration problems are this, this, this, and this.0968

We want to give you the idea of what happens in a titration, how to handle it mathematically and from a global perspective, and then, depending on what the individual situation, you want to sort of play around with the pieces and where they go.0973

OK, so the major species upon addition of OH^- before reaction: well, we have the acid, HA; we have H₂O; we have OH^-; and we have Na⁺.0985

Well, we know what is going to happen--the OH^- is going to react with that completely--it's a strong base--it's going to pull off every hydrogen it can until it runs out; that is the whole idea.1009

We have OH^- + HA producing A^- + H₂O; we have a Before; we have a Change; and we have an After.1021

Well, before--23 milliliters of a (I'll go back to blue) .050 Molar sodium hydroxide are added, so we have 23 times 0.050; that gives 1.15 millimoles.1035

Well, how much HA do we have?--we have 2.5 millimoles; remember, this is the stoichiometry part--in stoichiometry, we work in moles, not in molarity yet.1060

We have 2.5 millimoles; there is none of this left; and the water we don't care about.1070

Well, this is the limiting reactant, so this is going to run out; minus 1.15 millimole leaves no hydroxide at the end of the reaction (oops, I have these random lines showing up again; OK).1077

0; -1.15 millimoles here leaves 1.35 millimoles; here, it's going to be +1.15 millimoles; so we have 1.15 millimoles, and water does not matter; OK.1091

Now, major species after the reaction: well, we have HA; we have A^-; we still have Na⁺, which doesn't matter; and we have H₂O, which doesn't matter.1111

These two will dominate the equilibrium; so now, let us go ahead and run our equilibrium reaction.1129

OK, we have HA in equilibrium with H⁺ + A^-; we have an Initial; we have a Change; we have an Equilibrium.1136

Now, HA: we have 1.35 millimoles; now it's floating around in...we started with 100 milliliters; we added 23 milliliters of the sodium hydroxide; so now, the total volume is 123 milliliters.1152

The molarity is 0.01098; there is none of that; we have 1.15 millimoles of the A^- floating around in 123 milliliters; therefore, it is 0.00935.1169

OK, here is what is interesting: some of this is going to dissolve; some of this is going to show up; some of this is going to show up.1189

Well, guess what: we know what x is already, because they told us that, at equilibrium, once it comes to equilibrium, its pH is 6.10.1197

Let me come down here and do a quick little calculation in red: they tell me what the x is already--it equals the hydrogen ion concentration.1210

Well, the pH equals 6.10; therefore, the hydrogen ion concentration, which is x (I'll write it again) equals 10 to the negative 6.10 (right?--when you go backwards...10 to the negative 6.10--I hope you get that).1219

The pH equals the negative log of the hydrogen ion concentration; well, if I have the pH, I just go backwards; so it becomes -pH=log of the hydrogen ion concentration; the hydrogen ion concentration is 10^-pH.1242

That is reverse; so, if you are given the pH and you want to go back, you just put a negative sign on the pH and exponentiate with a base 10.1266

You end up with: this equals 7.9x10^-7; therefore, -x equals -7.9x10^-7; this is +7.9x10^-7 (let me make this a better multiplication sign; it looks like an addition sign here); and this is 7.9x10^-7.1275

Well, we can just add these; these are all just numbers.1303

10 to the negative 7 is actually pretty small, so what you actually end up with: this minus that--you still end up with 0.01098; it's one number.1306

Here, you end up with 7.9x10^-7; that is another number.1318

And here, you end up with 0.00935; when you do these additions, well, we want the K_a.1325

The K_a is equal to (by definition) the hydrogen ion concentration, 7.19x10^-7 (is that right?--yes, to the negative 7), times the A^- concentration, which is 0.00935; and then, it is divided by the HA concentration, 0.01098.1334

And then, when we do the calculation, we get 6.7x10^-7.1363

So notice the only thing that was different here: we still did the stoichiometry; we still did the equilibrium; stoichiometry in moles, equilibrium in molarity.1371

Except here, instead of finding x given K_a, they gave us x, because they gave us the pH; from the pH, we can get the hydrogen ion concentration, so the ICE chart actually changes.1379

So now, these numbers just go into the definition for K_a, and we just do the arithmetic, and we end up with 6.7x10^-7.1391

I hope that makes sense.1401

OK, so let's go on to our final example here: this is going to be the titration of a weak base with a strong acid--the reverse of what we did last lesson--just to see what it looks like and the reactions that we are dealing with.1404

And again, it is the reactions that decide where you go next; don't just jump into the mathematics.1418

OK, Example 3: 100 milliliters of a 0.050 Molar NH₃ solution is titrated with a 0.15 Molar HCl solution.1425

So, we have 100 milliliters of this .05 Molar ammonia solution sitting in a beaker, and we have this .15 Molar hydrochloric acid sitting in a long buret on top of the beaker, and we are going to be opening the stop clock and titrating it.1460

We are going to be dropping acid into this base solution to find out whatever it is that we need to find out, which we'll find out in a second.1473

So, after 10.0 milliliters of HCl solution is added, what is the pH?1481

What is the pH of the final solution?1502

OK, so we have a weak base solution sitting in a beaker; we have the hydrochloric acid on top of it; we are getting ready to add the hydrochloric acid.1509

I add 10 milliliters of this particular acid, which is .15 Molar; where does my pH end up?1517

OK, same thing: major species, stoichiometry, equilibrium: so (you know what, let me do this in blue; I like to actually change colors here) major species upon addition of H⁺, before reaction; I need to decide what is happening.1522

I have NH₃; I have H₂O; I have added HCl (which is a strong acid; it's going to fully dissociate, so I have free protons and free chloride ion floating around); what reaction is going to dominate?1559

What is going to dominate is this reaction right here.1574

What is going to happen is: NH₃ is going to react with the H⁺, and it's going to form ammonium (NH₄⁺), plus...what else?--that's it.1577

That is it; that is the reaction that is going to take place--OK, so we have a Before; we have a Change; and we have an After--this is stoichiometry, so we're working in moles.1594

NH₃: we have 100 milliliters, times 0.05 Molar; that means we have 5.0 millimoles of ammonia floating around.1603

We added (let me circle that) 10 milliliters of a .15, so 10 milliliters (I hope you'll forgive me--I'm skipping the decimals, but you should keep track of the decimals if you have a teacher that actually cares about significant figures) times 0.15 moles per liter (molarity) gives me (in millimoles per milliliter)--I get 1.5 millimoles.1617

Well, 5 millimoles, 1.5 millimoles; these are going to react; this is the limiting reactant.1650

This is going to go away: -1.5 millimoles--that is the change there; we are going to end up with an After of 0 millimoles of H⁺.1655

This is going to be -1.5 millimoles; that is going to give us 3.5 millimoles of that; and this is going to be...the beginning is 0; we are going to add 1.5 millimoles, because as this depletes, that same amount shows up; so we get 1.5 millimoles of NH₄⁺.1665

OK, so now, species after the reaction: well, we have NH₃; we have NH₄⁺; we have the Cl^- that doesn't do anything; and we have H₂O.1688

What is going to dominate the equilibrium?--this is going to dominate the equilibrium.1710

OK, here is where it gets really, really interesting: as it turns out, I have a choice here--I can use one of two reactions.1714

I can use this one--I can use the NH₃ + H₂O is in equilibrium with NH₄⁺ + OH^-.1725

If I do that, I have to use the K_b, because this is, by definition, a base association reaction; a weak base reacts with water, pulls off the proton from water, forms the ammonium ion, and releases a hydroxide ion into solution.1738

Or, I can use the acid version of this: I can do NH₄⁺ gives up...the acid dissociation is...because this is just an acid, this is the conjugate base; this a base; this is the conjugate acid; acid-base--it just depends on a perspective.1752

It is like heads or tails; it's the same coin.1770

H⁺ + NH₃: if I use this equilibrium for this problem, I have to use the K_a--that is the only difference.1773

This is a base association reaction; I need the K_b; this is an acid dissociation reaction--I have to use the K_a.1784

Well, the relationship between these two is (you remember): K_a times K_b equals K_w, which is equal to 10^-14, always.1791

So, if you are given a K_a, you can find K_b; in this case, if you are given a K_b, you can find K_a.1802

Well, what to choose, what to choose...I think I'm just going to go ahead and, because this is a weak base, I'm going to go ahead and use the base equilibrium.1807

I'm going to use that one, just to keep things consistent.1816

So, let's go ahead and write: NH₃ + H₂O is in equilibrium with NH₄⁺ + OH^-.1821

We have an Initial; we have a Change; we have an Equilibrium.1838

This is an equilibrium part, so we have to deal in molarity.1842

Well, the NH₃...we said we had 3.5 millimoles left over in solution after reaction.1845

Now, we had 100 liters; we have added 10 milliliters to it...sorry, 100 milliliters; we have added 10; so the total volume is now 110 milliliters.1851

We have 0.0318; water doesn't matter; the NH₄ is 1.15...I'm sorry, not 1.15; that was the previous problem; 1.5 millimoles floating around in 110 milliliters--that is equal to 0.0136; there is no hydroxide left.1860

This is going to diminish a little bit; water doesn't matter; this is going to show up; this is going to show up; we end up with 0.0318-x; water doesn't matter; 0.0136+x; and x.1886

And now, K_b: K_b (which equals, in the case of ammonia, 1.8x10^-5), is equal to this, times this, divided by that; it equals x, times 0.0136+x, divided by 0.0318-x.1906

I hope that you are getting sick of seeing this over and over again; that is a good sign--like I said last time, when you are sick of seeing a problem over and over again, that means you completely understand it; that is where you want to be.1933

You want to be sick of these problems.1942

OK, equals x times 0.0136, divided by 0.0318; x, which in this case is the hydroxide ion concentration (keep track of the species that you are dealing with--don't lose your way), equals 4.2x10^-5.1945

When we take the negative log of that, we get a pOH of 4.38, which implies that the pH is 14 minus that; you get 9.62.1967

That is it--very, very nice--OK.1984

Now, let's go ahead and...so again, nothing is different here; it was just a different type of reaction; we had a weak base titrated with an acid, and so we have to just sort of look at the chemistry that takes place.1989

The first reaction that takes place is the stoichiometry, the acid actually reacting with the base; and then, the equilibrium part is a base equilibrium.2004

You could have used the acid equilibrium, just as long as you, again, just did the K_a instead of the K_b; so that is all you really have to watch out for.2017

Make sure...there is a lot going on in these problems, but it isn't a difficult lot going on; it is just the details.2027

Keep track of the details; make sure you don't lose your way; don't let the math guide the chemistry--let the chemistry guide the math.2034

OK, so I'm going to round out this lesson by discussing something called acid-base indicators.2041

Now, I'm not going to actually say everything about acid-base indicators, and the reason is because it can...the discussion of acid-base indicators has a tendency to become complicated, without any reason for it to become complicated, because it's not like that.2048

It is just somehow...during the description of it, things just...I don't know; just for me, they tend to sort of look like they slightly get out of hand.2064

What I am going to discuss is what is going on--how acid-base indicators work--so that you know what is happening with the chemistry.2071

And after that, I'm going to give you a general rule of thumb to decide which indicator to use for which titration.2078

An ultimately, that is what matters most.2084

OK, so acid-base indicator (OK, I'll just write ab indicator) is used to mark the equivalence point of a titration by changing color.2087

We have these molecules, these--most of them are--organic molecules (carbon-based), and what is really, really interesting about them is that, when they are in their acidic form (meaning when they are protonated), they are one color; and when you pull off those protons, they are another color.2125

this is actually fantastic, that we can actually use this to decide when a titration actually ends, because you remember the titration at the equivalence point; there is this big jump in pH--it goes from a low pH to a high pH, or a high pH to a low pH.2140

That change...and because it happens really quickly, you remember the pH curve--it goes like that--we can use these acid-base indicators to let us know when it happens.2154

Other than that, we would have to use a pH meter; but if we don't have a pH meter, these indicators work just great; and they really do--they work fantastic.2164

They are still used in analytical work.2171

OK, indicators are themselves weak acids, and we symbolize them as HIn.2173

So again, it is the proton that matters in acid-base chemistry; it is that thing that matters--this is just the conjugate base.2201

Some indicator is usually a very complex molecule; but again, for understanding the chemistry, we can just deal with the symbol.2210

OK, now they are one color when the proton is attached--in other words, HIn; and another color when H is unattached.2218

So, it looks something like this: it's a weak acid, so there is an equilibrium associated with it.2250

So, in the example of...let's do, actually, an example here...let's use phenolphthalein.2260

Phenolphthalein is a very common acid-base indicator in the lab; when it's in its protonated form, it's colorless.2270

It looks like water; it is colorless; this is the acid form.2280

The acid form means that it has its hydrogen attached.2285

When it is in its unattached form, in this form, it's not colorless; it's pink.2289

This is called its base form (right?--acid, conjugate base; so base is when it doesn't have its proton).2294

That is it; basically, depending on what is in solution, if all of this is in solution, you have a colorless solution; if this is floating around, you have a pink solution.2304

Now, we'll talk about how this actually works.2313

How this works: OK, so let's just start off with an acidic solution.2318

The other way around, it's the same thing, except in reverse.2334

Start with an acidic solution: in an acidic solution, here is what you have floating around in there: basically, an acidic solution means that you have a bunch of free hydrogen ions floating around.2337

OK, so I will indicate those with H⁺, H⁺, H⁺; I am not going to put the base--it's the conjugate base of a particular acid (it could be hydrochloric acid, so you would have a bunch of Cl's floating around; it could be hydrocyanic acid, so you would have a bunch of Cn^-s floating around); what I want to concentration on is the actual indicator chemistry--what is going on.2351

To this solution, I add some indicator--very little, in fact--as little as possible.2379

I'm just going to represent that with (let me use a different color here, and let me put a couple more H⁺s in here, because again, we are talking about an acidic solution; and now)...I'm going to use black for my indicator.2385

HIn; HIn; so very, very few indicator molecules; now, notice--this is an acidic solution; so, because it's an acidic solution, any indicator that is there is going to be fully protonated.2408

In other words, there are more than enough free protons to protonate the indicator, so you are not going to find any In^- in the solution.2425

This is going to be a colorless solution.2432

It is acidic; that lets me know that it's acidic, because the indicator is protonated.2437

Now, as we add OH^- when we titrate it (so we are going to take OH^-, and we are going to drop it into solution; we are titrating it with OH^-), well, what does the OH^- do?2444

The OH^- eats the H⁺.2461

I am actually going to write "eats the H⁺"; it reacts with it, in other words, because that is what hydroxide and H⁺ do--they react to form water; it reacts with it.2465

Well, so I am dropping in H⁺, so I am eating up that; I'm eating up that; I'm eating up that; I'm eating up that.2475

So now, let's take a look at what our solution...as I'm adding more and more H, I'm eating up a little bit more of that and a little bit more of that.2486

So now, where is my solution?--my solution ends up like this.2495

Now, I have HIn, and I have HIn (let me go back to blue); I have an H⁺, an H⁺; I have eaten up some H⁺ with my additional; so at some point, the OH^- you add will react with all of the H⁺ present--the free H⁺ present.2500

And, here is the important part: start (actually, I don't want to write it that way)...OK, so at some point, the OH^- you add will react with all of the H⁺ present, then move on to pull off protons from the indicator, because that is what OH's do.2538

OH^- is a very strong base; it will seek out every source of protons it can.2567

Once it eats up all of the free hydrogen in solution--once it eats this up and eats this up--now it's going to start pulling off hydrogens from the indicator itself.2572

So, the OH^- that I add...it used that up; it used that up; now, it's going to pull that H; now, it's going to pull that H; so then, it will move on to pull off protons from the indicator itself.2581

And when that happens, now what you are left with in solution is this.2598

You have created a whole bunch of water, because every time an OH^- pulls off an H⁺ or reacts, you are just creating water; now, what you have is this: In^-, In^-.2602

Let me do it in black, because I did it originally in black.2612

Let me go there; so I have In^- and In^-.2617

Now, notice: there are no more free hydrogens floating around; there is probably a little bit of excess OH^- floating around, because once it actually pulls off all of the hydrogens from the indicator, well, there is no more indicator, so there is nothing else that it can actually pull hydrogens off of.2622

Now, there is a little bit of excess OH^-; now, the solution becomes basic.2637

And now, the indicator is in its pink form, the In^- form.2643

That is when the solution turns pink; now the solution turns pink, because what you have done is: you have eaten up all of the free hydrogen ion; any excess that you add beyond that is now going to pull hydrogen ion off the indicator (which, hopefully, you have very, very little of).2649

What you are left with is indicator in its basic form; it's a different color; that change of color tells you, "Stop the titration; I have reached my equivalence point."2667

There you go; now, the idea is: we want to use (let's see, do I have another...yes, I do) very little indicator, and here is why.2676

Very little indicator, so that the last drop of hydroxide solution that we add converts all of the indicator--because we want a clear color change--it converts all of the indicator.2695

Now, one drop is approximately equal to 0.05 milliliters.2731

So, if I reach equivalence, when the color changes, that is called the end point of the titration.2740

The end point usually marks the equivalence point; but equivalence point is when the OH^- and H⁺ have exactly balanced each other out.2749

The reason I want to use a little indicator: because, once I have actually eaten up all of my hydrogen ion, those few molecules of indicator that have their hydrogens pulled off to indicate the color change--I don't want to end up adding a whole bunch of hydroxide to that.2758

If I have a whole bunch of indicator in there, I'm going to end up adding more volume of hydroxide; but by adding just that one drop, that .05 milliliter, that is very little difference to take me from one pH to another pH, and it marks off the equivalence point almost exactly where the end point of the titration is.2773

That is the whole idea; but this is all that an indicator does: it sits around in solution; under acidic conditions, it's one color; when you add a base to it, it eats up all of the free hydrogen ion; once it has eaten up all of the free hydrogen ion, any excess that you add will pull off hydrogens from the HIn, the indicator in its acidic form; it will convert it to basic form; that will be the color change.2796

Hopefully, it will come down to one drop, because one drop doesn't make much of a difference.2820

If you are adding 50 milliliters or 20 milliliters, .05 milliliter is not going to be that big of a deal.2824

OK, now, our final comment about acid-base indicators: let me rewrite the equilibrium again: H⁺ + In^-...here is a general result; and this is what is important.2832

General result: for an acid-base indicator with a given K_a (because again, it's just a weak acid; it has its own K_a), the color transition occurs over a range of pH values equal to the pK_a, plus or minus 1.2850

So, if I have a K_a of 1.0x10^-7...1.0x10^-7, let's say, is the K_a of our particular indicator (some indicator, HIn; put HIn)...well, the pK_a of this is equal to 7.2896

That means that 7, plus or minus 1--that means from a range of 6 to 8 (7-1, 7+1)--this pH range is where the color of this indicator is going to start to change from one color to another.2919

In your book, if you actually look, every single book has a list of indicators and the pH ranges at which they change, and what colors they change to.2937

It could be colorless to pink; it could be blue to yellow; it could be yellow to green...depending on the particular molecule; but, that is what is important.2946

So, the final thing here: when choosing an indicator for a given titration, we want (put a comma there) the indicator end point (end point pH--I'll say the end point pH) to be as close as possible to the equivalence point.2955

I'll just say the end point; no, I'll say pH...to the equivalence point.3015

So, when I'm choosing an indicator to use, I will, more often than not, know where my equivalence point is going to be.3020

I will know what pH it's going to be.3029

I want to find, I want to use, an indicator whose pK_a is close to that pH value.3032

OK, so let's say, for example, if I know that my equivalence...well, I know that my equivalence in a strong acid-strong base titration is going to be a pH of 7.3042

I want to find an indicator that has its midpoint somewhere around 7; in this particular case, whatever it might be--whatever the indicator--this one would be a good one (1.0x10^-7); it's close.3055

It is in the range of 6 to 8; the pH at equivalence falls between those two, so that is the real thing to do here.3067

Now, the general result: for an acid-base indicator with a given K_a, the color transition occurs over a range of pH values equal to the pK_a, plus or minus 1.3074

That is where the color transition changes.3084

When choosing an indicator for a given titration, we want the indicator end point to be as close as possible to the equivalence point.3086

I'll go ahead and write pH--to the equivalence point pH--because that is what we are dealing with; we are dealing with pH's.3095

I want the indicator end point pH (in other words, this value here, the pK_a) to be as close as possible to the equivalence point.3102

That is the whole idea.3113

OK, thank you for joining us here at Educator.com to discuss indicators and titrations.3115

We'll see you next time; goodbye.3122