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Lecture Comments (11)

1 answer

Last reply by: Professor Hovasapian
Tue Apr 12, 2016 3:30 PM

Post by Bahaa Jabbar on April 10 at 02:15:00 AM

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.

Related Articles:

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 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

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 H2O 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 HNO3 (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 HNO3 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 HNO3 solution.0853

Well, this is just a...HNO3 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 H2O.0873

That is it; well, strong acid; water--weak acid (the Ka is 1x10-14); strong acid--there is no Ka; it's through the roof--a huge Ka.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 HNO3 is fully dissociated into H+ + NO3-; 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 NO3- floating around; certainly we have H2O; 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+, NO3, H2O, 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 H2O; 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 HNO3 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 H2O; 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 NO3; you are going to have the H2O; 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) → H2O.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 H2O.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 H2O.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 HNO3 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