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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### Titrations: Strong Acid and Strong Base

- A titration follows the change in pH that occurs as you add more and more of a solution you DO know the concentration of to a solution whose concentration you wish to know.
- Again, do the stoichiometry of the neutralization reaction using mols, then do the pH calculation using molarities.
- A titration curve is the graphical representation of the titration process, and has a characteristic “S” shape.
- The equivalence point is the point where you’ve added just enough of the delivery solution to react completely with the species in the receiving solution.
- The pH of a strong acid/strong base titration is 7.

### Titrations: Strong Acid and Strong Base

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
- Titrations: Strong Acid and Strong Base
- Scenario 1: Strong Acid- Strong Base Titration
- Question
- Part 1: No NaOH is Added
- Part 2: 10.0 mL of NaOH is Added
- Part 3: Another 10.0 mL of NaOH & 20.0 mL of NaOH are Added
- Part 4: 50.0 mL of NaOH is Added
- Part 5: 100.0 mL (Total) of NaOH is Added
- Part 6: 150.0 mL (Total) of NaOH is Added
- Part 7: 200.0 mL of NaOH is Added
- Titrations Curve for Strong Acid and Strong Base

- Intro 0:00
- Titrations: Strong Acid and Strong Base 1:11
- Definition of Titration
- Sample Problem
- Definition of Titration Curve or pH Curve
- Scenario 1: Strong Acid- Strong Base Titration 11:00
- Question
- Part 1: No NaOH is Added
- Part 2: 10.0 mL of NaOH is Added
- Part 3: Another 10.0 mL of NaOH & 20.0 mL of NaOH are Added
- Part 4: 50.0 mL of NaOH is Added
- Part 5: 100.0 mL (Total) of NaOH is Added
- Part 6: 150.0 mL (Total) of NaOH is Added
- Part 7: 200.0 mL of NaOH is Added
- Titrations Curve for Strong Acid and Strong Base

### AP Chemistry Online Prep Course

### Transcription: Titrations: Strong Acid and Strong Base

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

*We finished off, last time, talking about buffers; and today, we are going to talk about the next major topic in aqueous equilibria, and that is going to be titrations.*0004

*We are going to spend a couple of lessons on titrations--maybe three lessons, actually.*0016

*We are going to start, of course, by defining what a titration is; we have already seen it, if you remember, way back early in the course, when we talked about acid-base neutralization and things like that.*0021

*We also discussed some precipitation.*0031

*We will define what a titration is; and today's lesson is essentially going to be one big problem that we are going to dissect in detail; and we are going to follow a titration as we add a volume of a certain base of a known concentration to an acidic solution to decide what the pH is along the titration.*0034

*Very, very common procedure: it is done in labs all the time; it is done, of course, in the first-year chemistry lab curriculum.*0054

*You actually spend a fair amount of time titrating, and it is a very common technique in actual analytical laboratories.*0062

*Let's go ahead and get started and see what we can do.*0068

*OK, let's start off with a definition of titration; this definition that I am going to give you is sort of the...it's a little bit more specific.*0071

*Titration, in and of itself, is a little bit more general; and I will talk about that in a minute; but I would like to throw at least something out there, so we have a point of reference.*0079

*A titration: it is the delivery of a solution of known concentration to a solution whose concentration we wish to know...OK, actually, I will leave it at that.*0087

*It is the delivery of a solution of known concentration to a solution whose concentration we wish to know.*0136

*Now, I said this is specific, because this is actually the definition that we used earlier on in the course, when we discussed acid-base and precipitation reactions, and when we titrated one solution with another.*0141

*Now, in this particular case, what you are doing--the problem that we are about to start: we have, we are going to deliver, a solution of known concentration to a solution, also, whose concentration we actually know.*0156

*Here, we are not necessarily going to be trying to find what the concentration of the unknown solution is; we are going to try to find what the pH is at different points along the delivery of the base solution to the acid solution.*0170

*So again, a titration, generally speaking: it is just adding some solution to another solution to find out something about that solution; that is all that is going on--or to find out something about the solution at large.*0184

*It is just a fancy name for adding one solution to another one; that is the broadest definition of titration.*0198

*I probably shouldn't say too much more about it; it's best to just sort of jump on in; a lot of the terminology of science makes more sense when you actually see the science.*0204

*So, let's just do a quick problem to recall what it is that we mean, using this definition that we have; and then, we will move on to what we really came to talk about today.*0213

*Let us say: Recall--if 11.50 milliliters of a 0.26 Molar sodium hydroxide is added to 37.0 milliliters of HCl solution to bring it to equivalence...*0224

*And you remember: equivalence means...in this particular case, we are talking about acid-base equivalence; it means the amount of hydroxide exactly equals the amount of hydrogen ion; so H ^{+} + OH^{-} forms water.*0267

*You get complete neutralization; in other words, there is no OH floating around; there is no H ^{+} floating around.*0281

*That is what equivalence means; this is the stoichiometric definition: equal moles of hydroxide and hydrogen ion to form neutral water.*0286

*If 11.50 milliliters of a .26 Molar NaOH is added to 37 milliliters of a hydrochloric acid solution to bring it to equivalence, what was the concentration of the HCl solution to begin with?*0296

*Now again, the reaction that we are talking about here: we are talking about an acid-base titration, neutralization; so let's draw a quick picture here, off to the side.*0327

*I have this HCl solution, so I have a bunch of H ^{+} and Cl^{-} floating around; and to it, I add sodium hydroxide, which is essentially just adding that and that.*0336

*Well, the sodium and the chloride aren't going to do anything; they're just going to float around in solution.*0349

*What is going to happen is: the hydroxide is going to react with the hydrogen ion, and it is going to form water; this is the net reaction of an acid-base titration.*0353

*This is the net reaction of an acid plus a base.*0362

*An acid, plus a base, goes to salt plus water.*0365

*In this case, water is the net reaction; the salt is soluble; sodium chloride is just going to stay in solution--it's not going to go anywhere.*0369

*This is what we are looking at; this is a basic stoichiometry problem from the early part of the course--there should be no reason why you can't do this right now.*0377

*But, just as a reminder of what we initially meant by titration, let's go ahead and finish this problem off.*0384

*Let me see here: they are saying that 11.5 milliliters of a .26 Molar NaOH; so that means...so 11.5 milliliters is 0.01150 liters, times 0.26 moles per liter; that means 0.00299 moles of OH ^{-} were delivered, right?*0391

*Molarity times volume gives us moles, just like grams per mole, times molar mass...so when we are dealing with volumes, volume times molarity gives the number of moles.*0420

*We have that many moles of OH ^{-} delivered; well, it says that it came to equivalence; so, if we have...notice the ratio here is 1:1, so for every 1 mole of OH, it reacts with 1 mole of H^{+}.*0436

*.00299 moles of OH react with 0.00299 moles of H ^{+}; that is what this reaction tells us; and again, it is important to have a reaction that you know what you are dealing with--this is chemistry, after all.*0450

*We have that many moles of H ^{+}; OK, now we want the concentration of the HCl, which is the concentration of H^{+}, because HCl, when it dissociates, releases 1 hydrogen ion and 1 chloride.*0471

*Well, it is floating around in 37 milliliters, which is 0.037 liters, of solution; so now we just do the division: moles per liter is molarity; we end up with 0.081 Molar HCl.*0486

*That is it; this was just a simple stoichiometry problem, using the definition of molarity, moles per liter.*0505

*They wanted the molarity of the hydrochloric acid solution; well, we know what the volume is (37 milliliters); we needed to find what the number of moles is.*0513

*Well, we found the number of moles, because we know the reaction of OH ^{-} and H^{+} to form H_{2}O is 1:1.*0520

*We know what the amount, the number of moles of OH is; that comes from the volume delivered, times its molarity.*0528

*2, 3-step stoichiometry problem: this is a basic titration problem.*0535

*Now, we are going to take it one step further: now we are going to do a titration (which we said, generally, is just adding some solution of known concentration to another solution; that is the broadest thing); and we are going to follow the change in pH of this titration.*0539

*We want to be able to calculate, at any given moment...let's say if I add 36 milliliters of this Molar sodium hydroxide to this acidic solution; what is my pH going to be?*0556

*We want to be able to follow that titration; and then, at the end of this titration, we are going to draw a graph of that, and we are going to think about the graph qualitatively--what it is actually saying and what is going on along the different steps of the titration.*0569

*Let's go ahead and...actually, before we start the problem itself, let's go ahead and define what we mean by a titration curve, and then, at the end, come back to it, also.*0584

*A titration curve (or it is also called a pH curve) is a plot of the pH of a solution as a function of titrant volume delivered.*0595

*If you remember, the thing that you are actually delivering, the thing that you are adding, drop-by-drop, with a buret--that is called a titrant.*0633

*Titrant volume delivered: so it's a plot of the pH of a solution as a function of titrant volume delivered.*0641

*So basically, what you have is the volume of the titrant that you are delivering on the x-axis (or, from your perspective, it will be over here: x-axis), and the y-axis is the pH.*0647

*We will do one at the end.*0656

*OK, so let's go ahead and start our problem.*0659

*Our first scenario, which is the title of our lesson for today: so we will do scenario #1.*0662

*It is going to be a strong acid-strong base titration.*0672

*What that means is that I have either a strong acid solution to which I am adding strong base (and discovering what the pH is along each addition), or it's the other way around; I have a strong base to which I am adding a strong acid, and again, calculating the pH as we proceed along the titration.*0681

*In this case, what we are going to do is: we are going to add base to a strong acid solution: so strong acid-strong base.*0700

*We will follow the progress of titrating 50.0 milliliters of a 0.200 Molar HNO _{3} (nitric acid: strong acid, fully dissociated) with a 0.100 Molar NaOH by calculating the pH of the resulting solution after specific volumes of NaOH have been added.*0710

*So, we are going to follow the progress of titrating: we have 50 milliliters of a .2 Molar HNO _{3} solution sitting in a beaker, and to that, we are going to add various volumes of this .1 Molar sodium hydroxide--different volumes (10, 20, 30, 40, 50 milliliters).*0782

*Each time we add these volumes, we are going to stop and calculate the pH at that point; we want to follow the progress of the pH to see what happens, and that is what we are going to plot.*0797

*This is going to be a bit of a long problem; like I said, it is going to occupy most of the lesson.*0807

*It is very, very important that...I don't want you to get the idea that it is actually this complicated; it is not that each titration problem is going to take this long; it won't.*0811

*We are doing it simply so that we actually expose every aspect of a titration problem, so you understand the chemistry; and hopefully, by understanding the chemistry, you can move these problems along much faster when you actually do them.*0822

*But, since we are introducing them, we are going to take our time to do it properly.*0834

*OK, so let's go ahead and start with our initial situation: our initial situation is no NaOH added yet.*0838

*With no sodium hydroxide added yet, all we have is this 50 milliliters of a .2 Molar HNO _{3} solution.*0853

*Well, this is just a...HNO _{3} is a strong acid; it's just that we want to calculate the pH of this solution, so we do what we always do.*0860

*Let's check what our major species are in solution.*0868

*Well, our major species are H ^{+} (because it is a strong acid, fully dissociated); it is the nitrate ion; and it is the H_{2}O.*0873

*That is it; well, strong acid; water--weak acid (the K _{a} is 1x10^{-14}); strong acid--there is no K_{a}; it's through the roof--a huge K_{a}.*0882

*Therefore, this one--the hydrogen ion concentration is pretty much the hydrogen ion concentration of the nitric acid, because it's fully dissociated.*0894

*Here, the HNO _{3} is fully dissociated into H^{+} + NO_{3}^{-}; that is why we have these major species--it is floating around freely in solution.*0907

*So, a .2 Molar solution of this produces .2 Molar of that and .2 Molar of this; so the hydrogen ion concentration is 0.200 Molar; that implies that the pH of this solution is 0.70--so, we have our first pH.*0917

*At 0 milliliters of sodium hydroxide added, the pH of our solution starts at .70.*0936

*And now, we are going to follow what happens to the pH as we add more and more hydroxide; we are titrating the acid solution.*0943

*OK, so our second is, now: What is the pH going to be when 10 milliliters of sodium hydroxide are added?*0950

*There we go: OK, so now, the major species before reaction, before anything happens; we want to see what everything is before anything happens, so we decide what is going to happen.*0969

*Well, we have H ^{+} floating around; we have NO_{3}^{-} floating around; certainly we have H_{2}O; we have just added sodium hydroxide, which is a strong base.*0983

*So, these are the five things that are floating around in solution: H ^{+}, NO_{3}, H_{2}O, Na^{+}, and OH^{-}.*0993

*What is the chemistry that is going to take place?*1000

*Well, you know that anytime that you put a hydrogen ion anywhere near a hydroxide ion, they are going to react.*1002

*That is acid-base neutralization.*1012

*Hydrogen ion, plus hydroxide ion, in the same vicinity--they are going to react to form water.*1015

*Here is how it looks...I am actually going to move to the next page, because I need a little bit more room to do the mathematics; so, let me write down what happens.*1020

*We have H ^{+} + OH^{-} forming H_{2}O; that is the reaction that is going to take place.*1031

*OK, well, we have a Before; we have a Change; and we have an After situation.*1043

*This is stoichiometry, OK?--when we work in stoichiometry, we work in moles.*1053

*H ^{+} + OH^{-}...we need to...there is a certain number of moles of H^{+} reacting with a certain number of moles of OH^{-} that have been added to the solution.*1059

*Well, let's calculate what we have.*1069

*Well, what is the concentration of free ions floating around the H ^{+}?*1071

*Well, it is 50 milliliters, times 0.200 Molar; moles per liter is also the same as millimoles per milliliter.*1075

*So, because often, in titrations, we are working with small volumes, milliliters, we are also going to end up working in millimoles; so again, moles per liter (which is molarity) is equivalent to millimoles per milliliter.*1091

*When we talk about a .2 Molar solution, we are talking about .2 moles per liter; we are also talking about .2 millimoles per milliliter.*1112

*This milli-, as long as that milli- is there on top and bottom--everything is fine; no numbers are changed.*1120

*So here, the number of moles of H ^{+} is: I have 50 milliliters of the HNO_{3} solution; it is .2 Molar (millimole per milliliter); and you end up with 10.0 millimoles.*1126

*Now, what about the sodium hydroxide--what about the hydroxide ion?--how much hydroxide ion is there?*1145

*Well, we have added 10 milliliters, and the molarity is 0.10 millimole per milliliter.*1150

*So here, we end up with 1.0 millimole.*1162

*There is the water; it doesn't matter.*1168

*Well, these are going to react; anytime H and OH ^{-} come in contact with each other, it's going to react to completion--in other words, it's going to be all to the right; there is going to be nothing left over.*1171

*Well, this OH ^{-} is going to react with H^{+}; all of it is going to react.*1182

*Therefore, all of it is going to disappear, leaving you with none left.*1189

*Well, this H ^{+} is going to react with 1 millimole; 10 millimoles is going to react with 1 millimole, so that means what you are left with is 9.0 millimoles, because this H^{+}...1 millimole of it disappeared because it reacted with the OH.*1196

*And here...but the water doesn't really matter, so we are not really concerned about that; we will just go ahead and put it there, but it forms.*1218

*So now, at the end of this reaction, what we are left with is 9 millimoles of hydrogen ion, no millimoles of hydroxide ion.*1230

*Now, let's calculate the concentration of the hydrogen ion, because we have added volume.*1242

*And then, we will take the negative log of that.*1247

*So now, the hydrogen ion concentration is...well, we have 9.0 millimoles of hydrogen ion; it's floating around in 50 + 10 milliliters, right? *1250

*We had 50 milliliters of solution; we have added 10 milliliters of the hydroxide solution, so now there is a total of 60 milliliters of water, of solution floating around.*1267

*There is a new concentration; when we do this, we end up with 0.15 Molar.*1278

*And then, we take the negative log of that; it implies that the pH is 0.82; so our pH went up from 0.70 to 0.82 by the addition of 10 milliliters of hydroxide ion.*1287

*Notice how I did this: I calculated the number of moles based on the reaction that is going to take place; the reaction is going to be H ^{+} + OH^{-} forming H_{2}O; a certain amount of OH^{-} is going to vanish; a certain amount of H^{+} is going to vanish--1 millimole until it runs out--this is a limiting reactant problem right here.*1303

*Here, you have 1 millimole; here you have 10 millimoles; it's 1:1; this is the limiting reactant.*1321

*When it runs out, the reaction stops.*1326

*You are left with 9 millimoles of hydrogen ion; you divide by the total volume in that solution to get its molarity; you take the negative log to take the pH.*1328

*OK, now, we will do the third part: another 10 milliliters of sodium hydroxide is added.*1339

*OK, another 10 milliliters added to the original 10 milliliters--that means a total of 20 milliliters has been added.*1353

*Whenever we do these titration problems, we always go back to the beginning, original solution, and take the total amount of hydroxide that has been added.*1361

*So, what this means is: it is as if you have added 20 milliliters of sodium hydroxide to the original solution...has been added...OK.*1370

*Now, always go back and start from the original solution.*1388

*Here is why: the original solution, because if a mistake has been made--if a mistake was made in a previous calculation--it will not carry forward.*1402

*Therefore, if you made a mistake in this calculation over here, for the 10 milliliters, and then you just took that as your starting point for the next 10; if you made a mistake here, that mistake is going to carry forward.*1440

*But, by adding another 10 milliliters, it is the same as the fact that you have added 20 milliliters to the original solution; that means you get to start the problem all over again.*1449

*The mistake will not carry forward; so always treat the solution as if you are starting from the beginning.*1458

*OK, well, same thing: major species--you are going to have the same as before: you are going to have H ^{+}; you are going to have NO_{3}; you are going to have the H_{2}O; you are going to have Na^{+}; OH^{-}.*1464

*Just go back to the previous problem; the same thing is going to happen.*1480

*H ^{+} and OH^{-} in proximity--they will react to form water; therefore, we have, again, the same reaction.*1484

*We have H ^{+} + OH^{-} (and I am leaving enough room so that we can actually do the math) → H_{2}O.*1491

*We have a Before; we have a Change; we have an After.*1499

*Well, the original solution contained 50 milliliters, times 0.20 Molar, which gives me 10.0 millimoles.*1503

*Now, we have added 20 milliliters of hydroxide; that is 20 milliliters of hydroxide, and the molarity is 0.10 Molar, leaving us with 2.0 millimoles; the water doesn't matter.*1518

*These react; this is the limiting reactant; this runs out, -2.0, leaving 0 millimoles.*1534

*Here, this runs out; well, it doesn't run out--it's 10 minus 2, leaving 8.0 millimoles; and that doesn't matter.*1545

*So, in the solution, what we are left with is 8 millimoles of hydrogen ion.*1555

*Now, the concentration of hydrogen ion is 8 millimoles; but now, we have added 20 milliliters.*1562

*So, the total volume of the solution is now 70 milliliters.*1575

*The concentration is 8 millimoles, over 70.0 milliliters (millimoles over milliliters is the same as moles per liter); we end up with 0.11, which implies that the pH is 0.94; so it has gone up again.*1578

*With the addition of 20 milliliters of the hydroxide, now our pH is 0.94.*1601

*OK, let's keep going: situation 4: 50.0 milliliters of NaOH is added.*1606

*OK, proceeding as before...I won't do this one; I hope you will actually do it, so you can check it to make sure.*1621

*Just do the same thing that we just did with the previous two.*1627

*Proceeding as before, we get a pH equal to 1.3; so the pH is rising--OK.*1631

*Now, let's do #5: Now, let's add 100 milliliters total--I want to write "total," actually--of the NaOH.*1645

*OK, now let's do the mathematics; again, you are going to have H ^{+} + OH^{-} forming H_{2}O.*1669

*You have Before; you have Change; you have After.*1682

*You have 50 milliliters times 0.20 molarity; you have 10.0 millimoles; and now, you have 100 milliliters of the 0.10 Molar NaOH.*1687

*You end up with 10.0 millimoles; well, now you have 10 millimoles of H and 10 millimoles of OH ^{-}.*1706

*They are going to react completely and equally, because it's 1:1; this is going to be -10.0; -10.0; you are going to have no H ^{+} left in solution; you are going to have no OH^{-} in solution.*1714

*The water doesn't matter; it is going to be all water.*1726

*Now, the major species...the only thing you have in solution, floating around, now, is a whole bunch of water, a whole bunch of sodium ion (which doesn't do anything), and a whole bunch of nitrate ion (which doesn't do anything).*1730

*This is what is going to control the equilibrium, and we know that water has a neutral pH, 7.*1743

*So, we are going to end up with...well, let me just write it out explicitly: the H ^{+} equals 0, which means that we have pure water, which implies that the pH equals 7.*1750

*OK, for a strong acid-strong base reaction, this is called the equivalence point.*1770

*This is called the equivalence point; it is also called the stoichiometric point.*1783

*Now, it is when enough base has been added to react completely with the acid present.*1794

*In other words (I wanted to do that in red; let's try this again; red), 10 millimoles of base were added; in the solution were 10 millimoles of acid.*1824

*Those 10 millimoles of base will react completely with the 10 millimoles of acid, and there will be nothing of either left.*1839

*That is equivalence--that is stoichiometric point; that is why it is called "stoichiometric point"--it is about the stoichiometry.*1846

*At the stoichiometric point for a strong acid-strong base titration, the pH is going to be 7; the reason it is 7 is because there is no more H ^{+}; there is no more OH^{-}.*1853

*All that is left is pure water; that is why the pH is 7.*1863

*Now, that doesn't mean that every equivalence point is going to be pH 7, as we will discover in the next lesson, when we do a weak acid-strong base titration.*1867

*pH is not going to be 7; equivalence point is defined by the stoichiometry--it's defined by the reaction--a certain number of moles of H ^{+}, reacting with a certain number of moles of OH^{-}.*1875

*There might be one or the other left over; when both are completely and equally used up, when there is nothing of either left over--that is the equivalence point.*1888

*You have reached equivalence, an equal amount of OH ^{-} and H^{+}--very, very important; and it is easy to forget this on a test.*1897

*pH does not define equivalence point; stoichiometry defines equivalence point.*1904

*OK, so once again, it's when enough base has been added to react completely with the acid present; or, if you do it the other way, it's when enough acid has been added to react completely with the base that is present, so that no more of either is left over.*1909

*OK, so let's move on to the next one here (let's go back to blue): 6: now, 150 milliliters total of NaOH added.*1925

*Well, let's go ahead and do our reaction: H ^{+} + OH^{-} goes to H_{2}O.*1945

*We have a Before; we have a Change; we have an After.*1953

*Here, the same as before: 50 milliliters, times .2 Molar--it's the original solution; remember, we always go back to the original solution.*1957

*We have 10.0 millimoles of the H ^{+} floating around; now, the OH^{-}...we have 150 milliliters, times the 0.1 Molar; you end up with 15 millimoles of the OH^{-} (15.0); water doesn't matter.*1964

*Now, the limiting reactant is the H ^{+}; this is 10; this is 15.*1986

*That goes away, leaving no H ^{+}; well, 10 millimoles of H^{+} reacts with 15 millimoles of OH^{-}, leaving 5.0 millimoles of OH^{-}; and water doesn't matter.*1991

*Now, we take...now notice: now it's the OH ^{-} that is in excess--again, you have to keep track of what species you are dealing with.*2009

*Here, this is 0; now it's OH ^{-}; we have to calculate the OH concentration, convert that to pOH, and then take 14 minus the pOH to get the pH, because it is pH that we want.*2018

*The OH ^{-} concentration is moles per liter; well, we have 5.0 millimoles floating around; the total volume that has been added is now 50 + 150, right?*2029

*We added 150 milliliters of solution.*2042

*What we have here is 5.0 millimoles, divided by 200 milliliters; and we get 0.025 molarity.*2048

*Now, this implies that the pOH is equal to 1.6, which implies that the pH, which is 14 minus the pOH (minus 1.6), is equal to 12.4.*2060

*There you go: so we see that we have a basic solution; well, of course we have a basic solution--we don't have any hydrogen ion floating around, but we have an excess of hydroxide ion floating around.*2079

*This confirms the fact that it is basic; if you end up with a number here that is something like 6, 5, 4...anything less than 7, there is a problem, because you know that you are dealing with a basic solution.*2088

*You have an excess of hydroxide ion: let the chemistry decide what happens.*2098

*Don't get wrapped up in the math; it's just numbers--it is the chemistry that is important.*2102

*OK, and now, our final, just to round things out: 200 milliliters of NaOH is added total; well, proceeding as before--proceeding as in 6 above, which we just did, this one right here--we get a pH equal to 12.6.*2107

*So, now that we have a bunch of pHs, let's go ahead and see what the titration curve looks like for this.*2143

*So, our titration curve for a strong acid-strong base looks something like this.*2149

*Again, we said that the pH is on the y-axis, and this is the volume of OH ^{-} delivered.*2163

*Let's go ahead and put a 50-milliliter mark...yes, let's make this milliliters.*2175

*50 milliliter; 100 milliliter there; and let's do 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13; OK, so we will start at around right there; and then we go a little higher; and then we did 20; and then we did this; and then at 100 milliliters, this is 100; this is 150; we end up...1, 2, 3, 4, 5, 6, 7; it's going to end up like right about there.*2181

*And then, we have...let's see...12.6, 12.4, this 13, 12.4; it's going to be right about there.*2217

*So, you're going to end up, if you follow this along...you are going to get a titration curve that looks something like that.*2228

*This is a pH of 7.0; so notice, we started our pHs somewhere around .7, .8; we kept adding base (10 milliliters, 20 milliliters, 50 milliliters); our pH was rising, rising, rising; suddenly, when we add 100 milliliters of base, which gives just enough hydroxide to react with exactly the amount of hydrogen ion--all of a sudden, the pH flies through pH 7 and ends up over here, on the basic side.*2243

*pH 7: this is the equivalence point.*2277

*There you have it; now, notice some things about this.*2286

*Now, this pH, this titration curve, is characteristic of all titration curves; they all have this sigma shape.*2290

*This particular titration curve is specific to a strong acid-strong base; the equivalence point happens to be at a pH of 7.*2299

*This is for strong acid-strong base; it will not be like this for weak acid-strong base or a weak base-strong acid; this is for strong-strong.*2308

*So, notice, you have this slow, steady increase before this big, huge, dramatic jump.*2316

*Well, the reason is because there is a lot of hydrogen ion in here, and you are adding a little bit of hydroxide at a time.*2323

*As you add a little bit of hydroxide, the amount of hydrogen ion gets lower and lower and lower and lower, but it's still hydrogen ion in there.*2328

*You actually have an acidic solution: "acidic solution"--down here is acidic--pH below 7.*2336

*But, all of a sudden you're at equivalence, and all of a sudden it jumps up here.*2342

*This is very, very characteristic of all titration curves.*2347

*It will all end up looking like this: the difference is (and you will see it in the next lesson) that where equivalence happens will be at different pHs.*2352

*When we do a weak acid-strong base, our equivalence point is actually going to be at a pH higher than 7, which is going to be characteristic.*2360

*OK, so let's just put a couple of notes here.*2369

*The pH changes slowly until close to the equivalence point--then a sudden jump.*2376

*OK, what else?--let me see: so again, really, the important couple of things, qualitatively, to take away from this: you have this general S-shape; for a strong acid-strong base, the equivalence point (which is when you have the same amount of base added as there has been acid, or the same amount of acid as there is base in the solution, so that now you have pure water, nothing else)--you are going to have a pH of 7.*2398

*Strong acid-base: equivalence point happens at 7.*2424

*Every titration curve, every pH curve, will look like this when you are adding strong base to a strong acid.*2426

*Now, just to let you know that if you did this in reverse, if you added a strong acid to strong base (...adding strong acid to strong base...), that means you are starting off with a strong base solution; it's this same curve, except it's flipped, because now you are starting with a basic solution, and you are becoming acidic.*2435

*The graph looks exactly the same, except it is actually flipped: it begins up here, and it goes down like this.*2458

*But again, the one thing that is going to stay the same is that, at equivalence, your pH is going to be 7; and that is the real take-home lesson here.*2469

*OK, so what we have done is: we followed the progress of titrating a strong acid with a strong base, and the different volumes that we have added.*2484

*We have stopped, and we have done the stoichiometry.*2493

*We have calculated what was left over, as far as hydrogen ion or hydroxide ion; we divided by the total volume to get the concentration; and then, we took the negative log of it to give us our pH.*2496

*That is what we did: we followed the progress of the pH along the titration.*2506

*You should be able...we did one for a bunch of these different volumes; any given problem that you might have might simply ask you, "I have 50 milliliters of a .2 Molar HNO _{3} solution; what is the pH when I add 37 milliliters of NaOH?"*2511

*Then, you only have to do one of these problems.*2529

*So again, it's the same exact procedure; we have just done that problem 5 or 6 times here, so that we could follow the progress of the titration.*2533

*OK, so we'll go ahead and stop it here, as far as titrations of strong acid-strong base is concerned; I hope that that gives you a little bit of an idea of what is going on.*2542

*The next lesson, we are going to do the same thing, except we are going to titrate a weak acid with a strong base--kind of similar; kind of different.*2551

*Thank you for joining us here at Educator.com.*2559

*We'll see you next time; goodbye.*2561

1 answer

Last reply by: Professor Hovasapian

Tue Apr 12, 2016 3:30 PM

Post by Bahaa Jabbar on April 10, 2016

Hi Dr. Raffi

Would you explain this question to me:

Q/If 10 mL of 1 M NaOH is titrated with 1 M HCl to a pH of 2, what volume of HCl was added?

2 answers

Last reply by: sadia sarwar

Wed Nov 26, 2014 1:47 PM

Post by sadia sarwar on November 20, 2014

hi sir

can you please help me with the question below, I need your help badly!!

a standard pottasium carbonate solution is made by adding 1.227g of k2co3 to a 250 ml volumetric flask and filled to the mark with water. 20.00ml of aliquotes are taken and titrated against sulfuric acid, using methyl orand indicator. the average titre was 22.56 ml of sulfuric acid.

calculate the concentration of the k2co3 solution?

calculate the concentration of the sulfuric acid solution.

1 answer

Last reply by: Professor Hovasapian

Thu Mar 13, 2014 11:05 PM

Post by Angela Patrick on February 27, 2014

Hi Raffi,

A cool idea for these video series might be to add a lab section of several labs that you could walk through. Just some of the most important core lab practices in chemistry. I don't know how you would do it as I don't believe anything close to it has been done on educator. I don't know how it would be done in terms of taping and interface but thought it might be a cool idea you might pass along.

1 answer

Last reply by: Angela Patrick

Thu Feb 27, 2014 8:17 PM

Post by Christian Fischer on February 3, 2014

Hi Raffi, Great lecture. Every time we have some OH(-) and some H(+) in a solution they will always react to form H2O (right)?

So at pH = 7 we still have a concentration of 10^-7M H+ and 10^-7M OH(-). How come they don't react, so we end up with zero molar H+ and Zero Molar OH(-)?

Have a great day,

Christian.

1 answer

Last reply by: Professor Hovasapian

Fri Jan 4, 2013 6:31 PM

Post by Nigel Hessing on January 4, 2013

Hello,

Do you have a date for when your biochemistry educator videos will be coming out?

Thanks.