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

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

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### Titration Examples & Acid-Base Indicators

- An Indicator is a weak acid itself which reacts with that one drop over and above the equivalence point to let you know you’ve reached equivalence. We cannot see molecular species, so we use a change in color to signal when we’ve reached Equivalence.

### Titration Examples & Acid-Base Indicators

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

- Intro 0:00
- Examples and Indicators 0:25
- Example 1: Question
- Example 1: Solution
- Example 2: Question
- Example 2: Solution
- Example 3: Question
- Example 3: Solution
- Acid/Base Indicator Overview
- Acid/Base Indicator Example
- Acid/Base Indicator General Result
- Choosing Acid/Base Indicator

### AP Chemistry Online Prep Course

### Transcription: 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 H_{2}O.*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 H _{2}O.*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 H_{2}O; 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^{-} + H_{2}O; 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 H_{2}O, 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 NH _{3} 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 NH _{3}; I have H_{2}O; 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: NH _{3} is going to react with the H^{+}, and it's going to form ammonium (NH_{4}^{+}), 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

*NH _{3}: 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 NH _{4}^{+}.*1665

*OK, so now, species after the reaction: well, we have NH _{3}; we have NH_{4}^{+}; we have the Cl^{-} that doesn't do anything; and we have H_{2}O.*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 NH _{3} + H_{2}O is in equilibrium with NH_{4}^{+} + 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 NH _{4}^{+} 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 ^{+} + NH_{3}: 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: NH _{3} + H_{2}O is in equilibrium with NH_{4}^{+} + 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 NH _{3}...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 NH _{4} 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

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