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Lecture Comments (22)

2 answers

Last reply by: Magic Fu
Fri Feb 24, 2017 1:33 AM

Post by Magic Fu on February 21, 2017

Hi, Professor Hovasapian, can you go over want is the ending point of the Titration Curve and where could we find our endpoint?

0 answers

Post by Sun HaoHui on February 9, 2017

Hi, Professor Hovasapian, I have two quick questions. So when we deal with titration problems, do we always start off by writing out the stoichiometry reactions chart, then Equilibrium? And why?
Is there a big difference between solving buffer and titration problems?

2 answers

Last reply by: Professor Hovasapian
Sat May 2, 2015 1:24 AM

Post by Maya Balaji on May 1, 2015

Hello Mr. Hovasapian! Amazing lecture as always! quick question before my AP test this Monday.. How do I decide which species are dominating/ which species really deals with the chemistry? Thank you professor, your videos have helped me tremendously!

2 answers

Last reply by: sadia sarwar
Wed Feb 4, 2015 12:27 AM

Post by sadia sarwar on January 30, 2015

hello sir
is there any lecture on back titration??

1 answer

Last reply by: Professor Hovasapian
Tue Jun 3, 2014 7:17 PM

Post by husian alturki on June 1, 2014

Hello Raffi
thanks for the lovely videos they are helping me a lot. but i had one thing that keep confusing me
when you list the major species and compare it to water to see which one is dominating do you compare the ka of the acid to 1x10^-14 or to only the H^+ concentration which is 1x10^-7
so on example 1 lets assume the ka was 1.8x10^-10 in this case would i use the weak acid or water?

1 answer

Last reply by: Professor Hovasapian
Tue Feb 4, 2014 1:17 AM

Post by Christian Fischer on February 3, 2014

Hi Raffi. What a great lecture as always!! I have a quick question regarding the chemistry taking place during this weak acid titration reaction. I'd really appreciate if you get time to answer it one day.

Is this correctly understood: Before we add NaOH we start off with a concentration of [H+]=1,34*10^-3 molar (at equilibrium). This means the solution only contains 1,34*10^-3M*50ml = 0.067 mmoles of H+ (at equilibrium) When we add 10,0ml NaOH we are adding 10ml*0,10M = 1mmoles of OH(-). Now we have more moles of OH(-) than H(+) and All the OH must react with H(+). Does this mean that
a) OH(-) attacks HA (the weak acid) directly?  OR  
b) Does OH(-) attack the 0.067 moles of H(+) in solution directly? If it attacks H(+) in solution directly there is not enought H(+) to meet the demand so does the equilibrium reaction between the weak acid in water HA + H2O = H(+) + A(-) shift to the right to produce more H+ when it senses OH- eating up the H(+)?

I want to thank you for explaining so well why pH can be different than 7 when when the same amount of strong base and weak acid have reacted 100%. I had a hard time understanding that until now! I'm very grateful :)  

Keep up the great work,

0 answers

Post by Edwin Wong on May 19, 2013

The pH should be 8.74 not 8.72 at 30:18.

1 answer

Last reply by: Professor Hovasapian
Sat May 4, 2013 4:01 AM

Post by morgan franke on May 3, 2013

when working on part 1 should you make the concentration of HAc = .1m/.5L = .2 because you added 50 ml of a .1 molar HAc? If not, please help me understand why sometimes the amount of solution is import in the ICE table and others it isn't.

1 answer

Last reply by: Professor Hovasapian
Mon Apr 29, 2013 1:41 AM

Post by ANURAAG PRAKASH KAMLE on April 29, 2013

Sir what are the other two zones of the Titration Curve called? the ones after the Buffering zone.

1 answer

Last reply by: Professor Hovasapian
Fri Mar 22, 2013 4:10 PM

Post by Kathryn Cosgrove on March 22, 2013

After figuring out the major species, how do you decided which species need to be the reactants and which are the products?

0 answers

Post by ARLENE ROOME on May 23, 2012

not sure how you got the PH = 5.35 when 50 ml was added

Related Articles:

Titrations: Weak Acid and Strong Base

  • At half-Equivalence the pH is equal to pKa, because the concentrations of Acid and conjugate base are equal, thus making the Log term in the Hen-Has equation zero.
  • pH at Equivalence for a Weak acid/strong base titration is greater than 7.
  • pH at Equivalence for a weak base/strong acid titration is less than 7.

Titrations: Weak 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: Weak Acid and Strong Base 0:43
    • Question
    • Part 1: No NaOH is Added
    • Part 2: 10.0 mL of NaOH is Added
    • Part 3: 25.0 mL of NaOH is Added
    • Part 4: 40.0 mL of NaOH is Added
    • Part 5: 50.0 mL (Total) of NaOH is Added
    • Part 6: 60.0 mL (Total) of NaOH is Added
    • Part 7: 75.0 mL (Total) of NaOH is Added
    • Titration Curve

Transcription: Titrations: Weak Acid and Strong Base

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

Last time, we talked about titrations; we started talking about them, and we did a strong acid-strong base titration.0004

Today, we are going to take the next step, and we are going to talk about a weak acid-strong base titration.0010

What I mean by that is: we are going to have a weak acid solution, and we are going to be adding a little bit of strong base to that weak acid solution, and we are going to be following the progress of the pH of that solution.0016

We are going to do the same thing: we are going to have a bunch of pHs, and we are going to take those pHs, and we are going to construct a titration curve, and it is going to look a lot like what it did last time--same general shape--except the equivalence point is going to be different.0027

Let's go ahead and get started and see what we can do.0040

We want to consider...let us consider the titration of (oops, let's make the 50 a little bit more clear than this) 50.0 milliliters of a 0.10 Molar acetic acid solution (which, if you remember, we abbreviated as HAc, because I don't like writing out all of the C's, O's, and H' just hurts my eyes--it's too many atoms floating around), and we are going to titrate it with a 0.10 Molar NaOH.0046

Our titrant is the same as before: it is a strong base, sodium hydroxide, but this time we are titrating a weak acid, not a strong acid.0093

Now, because we are dealing with a weak acid, you can pretty much guarantee that the Ka is going to show up somewhere.0100

The Ka of acetic acid is 1.8x10-5; OK, so let's go ahead and get started.0108

The first thing we want to do is: we want to calculate the pH of a solution before any NaOH is added--so, no NaOH added yet.0116

Well, you know how to do this; this is a typical weak acid problem: there is no NaOH that had been added; you basically have 50 milliliters; you have a solution that is .1 Molar acetic acid; how do we calculate the pH?0130

Well, we are going to do an ICE chart; so, this is a typical weak acid problem, which I guess is probably pretty good--we'll do a little bit of quick review here.0142

Let's see what our major species are in solution; you will definitely want to get into the habit of doing this as a conditioned response.0157

Any kind of aqueous chemistry--you need to find out what is actually in that solution, to decide what chemistry is going to take place.0165

Our major species are going to be HAc (our acetic acid, which is a weak acid, which is why it shows up as the non-dissociated form), and we have H2O; that is it--those are the only things floating around in solution.0173

Well, between the two, 1.0x10-14 is the Ka of water; HAc has a Ka of 1.8x10-5; it's going to dominate the equilibrium.0186

The equilibrium that is going to take place here, as you drop it into solution, what is going to happen?--what is going to happen is the following.0199

The HAc, standard acid dissociation equilibrium: H+ + Ac-; it's always the same.0206

The acid loses a proton to its conjugate base--always the same.0215

We have an initial; we have a change; we have an equilibrium; when we are dealing with equilibrium issues, remember, ICE is for equilibrium; we deal in concentrations.0221

So, the concentration of HAc is 0.10 Molar; there is no this; and there is none of that.0230

Some of this is going to disappear; some of this is going to show up; and this is going to show up, the amount that disappeared, right?0238

We get 0.10-x; we have x; and we have x; OK.0245

Let's go ahead and do (I'm actually going to do it over here, off to the side, I think): the Ka is 1.8x10-5; it is equal to the hydrogen ion concentration, which is x, times the acetate concentration, which is x, all over 0.10-x, which is the HAc concentration.0252

Well, this is approximately equal to x2 over 0.10, because x is going to be pretty small; it's going to be pretty small because it's very little dissociation.0274

You can check it with the 5% rule if you want, but you can take my word for it.0284

We get x equals the H+ concentration, is going to equal 1.34x10 to the negative 1, looks like 3, which implies that our pH is going to be 2.87.0288

We are going to start off with this pH of 2.87, before any base has been added to the solution.0309

OK, so now, our next step is going to be: Add 10 milliliters of the NaOH; OK.0315

Now that we have added 10 milliliters of the NaOH, we do what we always do; what are the major species in solution, and what reaction is going to take place?0329

So, the major species are...well, HAc is in solution; H2O is in solution; you have dropped in sodium hydroxide, which is a strong base--it dissociates completely.0336

You are going to have free sodium, and you are going to have free hydroxide.0350

OK, anytime you have hydroxide in a solution, it is going to do one thing: it is going to seek out a source of protons that it can pull away to become water--that is it--that is all OH does.0354

Well, in this case, it has two sources of protons: OH- can pull H from H2O (which doesn't really change much, because it turns into H2O, but it leaves an OH- behind); so this is going to pull from that--this is the reaction that is always going to take place, always--weak acid-strong base.0364

So, the reaction that takes place is the following: OH- is going to react with the HAc, and it is going to produce H2O; well, A-, because this is going to pull off that H, plus H2O.0382

Now, we have to do the stoichiometry first: when you add this base, it is going to react with this until this is used up; any OH- that is added is going to react to completion--notice this single arrow; that means all of it reacts until there is no OH- left.0405

And then, when there is HAc left over, then we do the equilibrium problem.0426

You remember: this is no different than a buffer solution; in fact, each one of these steps that we are going to do is just a buffer problem, just like the last couple of lessons, a few lessons ago.0430

That is all that is happening here.0441

OK, so we have a Before; we have a Change; and we have an After.0444

Well, how much OH- did we actually end up adding?0450

Well, we added 10 milliliters, and the molarity is .10 Molar; OK, let me actually just write something here: this is the (let me do this in red) stoichiometry.0455

OK, so for a weak acid-strong base titration, we do the stoichiometry first; then, we do the equilibrium.0470

We didn't have to this for the strong acid-strong base titration, because it was just stoichiometry--there is no equilibrium to come to.0477

But, this is a weak acid, so there is still an equilibrium to deal with; here, we have to do the stoichiometry first, and then, we do the equilibrium.0484

When we do stoichiometry, we deal in moles.0492

So, this is going to be...I have added 10 10 milliliters, times 0.10 molarity; that gives me 1 millimole of OH-.0495

Well, the HAc...I have 50 milliliters of a .1, so 50 milliliters, times 0.1 molarity; that gives me 5.0 millimoles.0514

Well, here: 1 millimole, 5 millimoles--this is a 1:1 ratio; this is the limiting reactant, so this is going to disappear.0529

1 millimole: that is going to leave 0 OH- (oops, we don't want these stray lines floating around); the change is going to be 1 millimole of HAc is going to be reacted with; 1 millimole of H+ is going to be pulled off, leaving 4.0 millimoles of HAc.0537

We are going to produce...let me see, we actually had 0 of that to start, and we are going to produce 1.0 millimole of that.0563

So now, at the end of our reaction, we have 1 millimole of acetate (oops, I'm so used to using...let me go back to blue here...this is acetate, not just a generic acid) floating around in solution; we have 4 millimoles of HAc floating around in solution; so now, let's check our major species to see where we are after the reaction.0575

Major species after the reaction: so, we have some HAc floating around, 4 millimoles, it looks like; we have Ac- floating around, 1 millimole; we have H2O, and we have Na+.0616

The OH- has been used up; what is going to dominate the reaction now is that.0633

We have a weak acid, and we have its conjugate base; there is going to be an equilibrium established here; so now, we do the equilibrium problem.0640

When we work in equilibrium, we work in molarity.0652

This is going to be: HAc is going to be in equilibrium with H+ + Ac-; and this is just an ICE chart, and we do this over and over and over again; that is all we are doing.0656

So now, let me do this one in blue.0674

So now, we deal in molarities; so we have 4 millimoles (let me write this a little farther to the left--I need a little bit more room, because I'm going to make this right); so let me write this as I, as C, as E.0677

I have 4 millimoles floating around, right?--4 millimoles of this in solution, that I just calculated from the stoichiometry.0696

It is floating around in 60 milliliters, because it's the 50, plus the 10 milliliters of solution that I added.0705

60 milliliters; so this is 0.067.0711

There is none of this; the Ac-, well...we had 1 millimole, and it is floating around in 60 milliliters; so we have 0.0167.0717

This is going to diminish by x; this is going to augment by x; augment by x; you end up with 0.067-x; this is going to be just x; and this is going to be 0.0167+x.0730

And now, we do the Ka, which is 1.8x10-5, is equal to x, times 0.0167+x, over 0.067-x; we can...the x is going to be pretty small, so let's just go ahead and write this as x, times 0.0167, divided by 0.067; and when we do this, we get a hydrogen ion concentration of 7.22x10-5, which implies that we have a pH equal to 4.14.0754

So again, we have added some hydroxide to this acid solution; we have to do the stoichiometry first--the base that we have added, the hydroxide, is going to react with the weak acid--it's going to pull protons off until the base runs out, until that limiting reactant actually runs out.0803

And then, you are going to have some acetate floating around, as well as some acetic acid floating around; so you have to...and then you go back, and because you have changed the volume of the solution, you calculate the molarity, and you calculate the pH.0821

OK, so let's do our next one, which is going to be 25 milliliters.0837

So now, add 25.0 milliliters total; and again, we always go back--we added 10; we are adding another 15; the total that we are adding is 25--we always go back to the beginning, in case we made a mistake, so it doesn't carry over.0846

NaOH added; OK, so major species before...let me write "major species before reaction": we have HAc; we have H2; we have Na+, and we have OH-.0867

This is going to react with that, and it is going to react according to the following.0895

It is going to be OH- + HAc; it is going to form Ac- + H2O.0899

Well, we have a Before; we have a Change; we have an After; now, we have added 25 milliliters of sodium hydroxide.0910

It is 25 milliliters, times 0.10 Molar, equals 2.5 millimoles; and here, we have our 50 milliliters, times a 0.10 Molar; this is Molar; equals 5.0 millimoles.0919

Well, this is 2.5; this is 5; this is going to run out first.0941

This is going to be -2.5 millimoles, leaving us with no hydroxide; it is going to react with 2.5 millimoles of that, leaving us with 2.5 millimoles of this; and it is going to...this is 0; this doesn't matter...this is going to add 2.5: because this is diminishing by 2.5, 2.5 Ac- is showing up.0945

This is going to be 2.5 millimoles; OK, so now what are our major species? (By the way, this was the stoichiometry part).0973

I keep forgetting to write that, but of course, hopefully you know that by now.0986

So now, we will go ahead and do the...actually, let me write down what the major species are first.0991

Let me go back to blue.0998

Major species after (this was before reaction; now, after reaction): I have, now, HAc floating around; I have Ac- floating around; I have H2O floating around; and I have Na+.1000

All of the OH- has been used up; that is what that number is.1018

This is going to dominate; so we are going to do our equilibrium; we are going to have (let me do this in red) HAc in equilibrium with Ac- + H+ (actually, let me write these in the order that I am accustomed to writing them in--otherwise, I think it's going to be confusing)...H+ + Ac-; Initial, Change, Equilibrium.1023

Well, HAc--we have 2.5 millimoles; and now, it's floating around in 50+25: it's floating around in 75 milliliters.1053

That concentration is 0.0333; there is none of that; the Ac-...I have 2.5 millimoles floating around in a total volume of 75 milliliters; it equals 0.033.1065

This is going to diminish by x; this is going to augment by x; this is going to augment by x; 0.033-x; this is x; 0.033+x.1090

Now, I do my Ka, which is 1.8x10-5; it is equal to x, times 0.033+x, divided by 0.033-x, which is approximately equal to x times 0.033, divided 0.033; those cancel; x equals the hydrogen ion concentration, of course, which equals 1.8x10-5, which implies that the pH is equal to 4.74.1105

OK, this is...let me see, where are we?--so we just did that; we calculated pH; OK.1145

This is a really, really important part; notice what happened, if you flip to the back of the last slide: we ended up with half of the original acetic acid being used up.1156

We started off with 5 millimoles; we used up half of it--we reacted half of it, leaving 2.5 millimoles.1171

Now, we also ended up producing 2.5 millimoles of acetate; so now, this is confirmed right here: the concentrations are the same, because they are in the same volume.1179

When the concentrations are the same, the pH of the solution happens to equal the pKa.1191

So, let me write this as: At this point, one-half of the HAc has reacted with the added OH-, producing equal amounts of HAc and Ac-.1198

Well, what does that remind you of, when you have equal amounts of a weak acid and its conjugate base?--you have the ideal buffer.1242

We have the ideal buffer.1252

In the process of titrating a weak acid, you end up creating a buffer solution, and you will see what the buffer solution looks like when you see the titration curve.1258

It is actually going to look like it's going to be very little change in the pH.1266

That is going to increase very, very slowly before that sudden jump.1270

OK, so a buffer...which is why pH equals pKa, because, if you remember, pH in the Henderson-Hasselbalch equation, plus the log of the base concentration over the acid concentration...well, here the base concentration and the acid concentration are the same, so this is 1.1275

Well, the logarithm of 1 is 0; what you are left with is pH=pKa.1299

So, at the halfway point of the titration of a weak acid with a strong base, what you have created is the perfect buffer.1304

That is it; we just want you to know that.1312

OK, now, #4: now we have 40 milliliters of NaOH added.1314

Well, proceeding as before, just like we did the last few that we did (you should probably check this for yourself): proceeding as before, we get a pH equal to 5.35.1326

All right, now let's do 50.0 milliliters total of NaOH added; so this one, we are definitely going to do.1344

Let's check our major species (this one we are going to treat just like we did before): we have HAc; we have H2O; we have OH-; and we have Na+.1361

Well, OH- is going to react with HAc, and you are going to get the following.1375

I am not going to do all of the calculations here, but they are just like before, so you should recognize where my numbers are coming from; they are not just dropping out of the sky.1383

The OH- is going to react with the HAc; it is going to produce acetate, and it is going to produce water, because this is an acid-base neutralization.1392

We have a Before; we have a Change; we have an After.1404

Well, 50 milliliters of a .1 Molar NaOH produces 5.0 millimoles1408

Well, 50 milliliters of a .1 Molar acetic acid solution produces 5.0 millimoles; there is none of this, and there is none of that.1417

Equal amounts here--we have equal amounts, so that means they are both going to vanish--they are going to react with each other.1428

OH- is going to take this, produce that, produce water; so it is going to be -5.0, -5.0, +5.0, water doesn't matter; at the end, we have no hydroxide floating around; we have no acetic acid floating around; we have all acetate ion floating around.1435

Here is where we are now; so now, the major species in solution after reaction--major species after reaction: you end up with, well--we have Ac-, H2, and Na+.1459

There is no hydroxide left--that is what this says.1486

There is no acetic acid (there is no HAc) left--that is what this says.1490

All I am left with is the acetate, the water, and that.1494

OK, now we do our equilibrium problem.1499

However, our equilibrium problem is going to be slightly different now, and this is what you need to recognize; and this is why it's important to know the chemistry and watch what happens.1502

Notice what you have in solution: we have acetate; we have water; and we have sodium ion--sodium ion isn't going to do anything; water isn't going to do anything.1511

But this is the conjugate base of a weak acid; so what it is going to do: it is actually going to react with the water to create hydroxide ion, producing a basic solution.1520

But now...well, let me just write that out: so again, you don't have any HAc; so because there is no HAc, there is no acid dissociation equilibrium to write down; now, what you are going to have, because you only have a base and water, and this is the conjugate base of a weak acid (therefore, it's a relatively strong base): it is going to pull hydrogen ions off of the only source it has, which is water.1534

So now, the equilibrium that you are going to do is going to be a weak base equilibrium.1562

It is going to be the equilibrium of a base reacting with water to produce hydroxide ion.1567

It is going to be the base association equilibrium, and you remember this one--it looks like this: A base, plus water (I'll write it as HOH) is going to be in equilibrium with HAc, plus OH-.1575

It is going to end up producing hydroxide ion, because now this is the only thing in solution.1593

The reaction took place; nothing was left over except the Ac-; now the Ac-, when we allow the system to come to equilibrium (which we are doing right now)--it is going to follow this reaction.1598

It's really, really important that you recognize this.1609

Initial, Change, Equilibrium: well, the initial concentration of Ac- is going to be 5.0 millimoles, divided by the total volume, which is now 100 milliliters; you have added 50 milliliters of solution to 50 milliliters of solution; this is 100 milliliters.1612

You are going to end up with a 0.050; water doesn't matter; there is none of this, and there is none of this.1634

Well, some of this is going to diminish; water doesn't matter; some of this is going to show up, and some of this is going to show up.1643

Here we have 0.050-x; this is x; this is x; OK.1653

Now, here is where it's interesting: we must use the Kb, not Ka.1661

The Ka represents the acid dissociation constant, HAc going to H+ + Ac-.1679

The Kb represents the base association constant: Ac- + H2O goes to HAc + OH-.1688

Remember, now it is acting as a base; there is no acid around to get in equilibrium with--now it's completely Ac-, so we have to use the Kb, not the Ka.1700

OK, now, we say Kb is equal to the concentration of OH-, times the HAc, over the Ac-.1713

Well, we know the Ka of the acetic acid; well, how do we get Kb?1730

Remember, Kb is equal to Kw, over Ka.1737

It is equal to 1.0x10-14, divided by 1.8x10-5.1742

When we list Ka values for weak acids, we list Ka values; we don't list Kb values for their conjugate bases; but we can calculate Kb values for the conjugate bases if the situation comes up (which it does).1750

In this case, the Kb ends up being 5.6x10-10.1763

We have 5.6x10-10 is equal to x times x, over 0.050-x, which is approximately equal to x squared over 0.050.1770

Well, we get that x, which equals the hydroxide ion concentration, is equal to 5.3x10-6; that means that the pOH is equal to 5.26, which means that the pH is 14 minus the 5.26; we get a pH of 8.72.1788

There we go; now, here is the important part: the pH at equivalence is not 7.1810

This is equivalence; the amount of hydroxide that we added reacted with all of the acetic acid that was present, that gave up its proton; the OH- pulled off the H+ and produced water, leaving just acetate ion.1828

Equal amounts: 5 millimole reacted with 5 millimole; but the pH of the solution that we calculated is 8.72.1845

It is a basic solution; and the reason that it is basic is because the base that was left over, the Ac-, ended up reacting with water in a base association to produce hydroxide.1852

So, we saw that in a strong acid-strong base titration, the pH at equivalence was 7; with a weak acid-strong base titration, at equivalence, the pH is bigger than 7--in this case, 8.72.1863

The number itself is relevant only as far as this problem is concerned, because it is acetic acid; but other weak acids--the idea is: you are going to get...that equivalence of pH is going to be greater than 7, not 7.1878

It is the stoichiometry that defines equivalence, not the pH.1891

OK, now we will keep going, adding some more: 60 milliliters total of NaOH is added; OK.1896

Major species: we have HAc, OH-...1919

I know this is getting a little tedious, but believe me; when you come to a point where you are sick of a problem, that means you understand it so well that you never have to think about it again; so getting sick of a problem is good.1923

If it is becoming tedious, that is a good sign; that means that you know what you are doing.1934

60 milliliters total of sodium hydroxide is added; our major species are the (let me write them again on this page; I think it's a good idea)...major species: now, we have again HAc, OH-; we have H2O.1938

We run our stoichiometry again; our stoichiometry is going to be the OH- is going to react with the HAc (I might put a little more room here); it is going to produce acetate ion, plus water.1959

We have a Before; we have a Change; we have an After.1981

Here, again, 60 milliliters times .1 moles per liter; we have 6.0 millimoles; here we have the 5.0 millimoles (hope you know where that came from; that is the 50 times .1); 0; water doesn't matter.1986

Here, because now this is the limiting reactant, this is going to be -5.0 millimole, -5.0 millimole; this is going to be +5.0 millimole; water doesn't matter.2005

We are left with 1.0 millimole of OH- in solution; no HAc in solution; and 5.0 millimoles of Ac in solution.2020

Let's check our major species after reaction--we have: OH- in solution; we have H2O; we have Ac-; and we have Na+.2032

Na+ is not going to do anything; that is not going to do anything; ah, we have two bases--we have OH-, and we have Ac-.2050

Between the two, just for your information, hydroxide is a very strong base.2060

We saw, from the previous calculation that we did, that the hydroxide ion concentration was 5.3x10-6.2068

This was with absolutely no hydroxide in solution, other than just the hydroxide that shows up from this reacting with water.2076

Here, we actually have a whole millimole of hydroxide ion; so, for all practical purposes, any contribution that this Ac- could make to the hydroxide ion concentration is completely negligible.2084

This is the only thing that matters--this is the only thing that matters.2097

We are not even worried about that.2103

So, our hydroxide ion concentration (let me write this) is going to be 1 millimole, divided by the total volume, which is now 50 milliliters + 60 milliliters, so 110 milliliters.2105

It is going to give us 9.1x10-3 molarity.2123

That implies that the pOH of this solution is 2.04, which implies that the pH of this solution is 11.96 (very, very, very basic).2130

OK, and now, our final, just to round things out: 75 milliliters total NaOH added; proceeding as we did right here, we end up with pH=12.30.2144

OK, now, we want to take a look at our titration curve to get a pictorial view of all of these calculations that we did.2167

Titration curve; it's going to look like this: once again, this is the volume of OH- titrant added, and it's a unit of milliliters; this is going to be the pH.2181

Let's go to 25; let's go to 50; let's do 1, 2, 3, 4, 5, 6, 7, 8, 9; let's mark off our 7; 10, 11, 12, 13, 14; we are going to start off somewhere in the neighborhood of about 3, and at 50 milliliters, we are going to be at .009.2201

I just want to mark off this equivalence point; so it is going to look something like this.2229

There we go; so again, you notice that it looks the same as it did for the strong acid-strong base, but the numbers are different.2240

So, this right here--this is the equivalence point; and again, the equivalence point is where enough hydroxide has been added to react precisely with the amount of acid available; total neutralization; pure water.2249

However, it is not pure water, because, in this particular case, we dealt with a weak acid.2265

A weak acid has a strong conjugate base; so, once that Ac- shows up, it is going to start reacting with the water present to produce a little bit more hydroxide, which is why our pH at equivalence is roughly around 9--at 8.72.2271

So again, for a weak acid-strong base titration, at equivalence, the pH is above 7; it is a basic solution.2289

Or, think of strong base-weak acid; they are not balanced out; it is going to be more basic (strong base); that is another way of thinking about it.2299

OK, a couple of other things to consider here: remember, at 25 milliliters, that was the halfway point to the titration; that was when we reacted half of the acetic acid with a certain amount...and what we produced was equal concentrations of acetic acid and acetate ion.2306

In other words, we created the perfect buffer.2327

This region right here--this is called the buffering zone: basically, you have--in solution, you have a weak acid, plus its conjugate base, floating around simultaneously; that is a buffer--that is the definition of a buffer.2329

This zone is called the buffering zone; and you can tell that it's a buffering zone, because you notice you are adding all of this base, but there is very, very little change in pH.2352

As you get close to the equivalence point, it jumps up.2360

So, once again, the take-home lesson for this particular graph, qualitatively, is that equivalence is defined by stoichiometry, not pH.2364

So, on a test question (and you are going to get this question; it always shows up): acetic acid is titrated with sodium hydroxide; the pH at equivalence is 7--true or false?2386

The answer is false, because what you have is a weak acid titrated with a strong base; the pH is bigger than 7.2405

Equivalence is defined by stoichiometry; equal amounts of base and acid have reacted to produce neutralization.2411

It is not defined by the pH; the pH is only 7 for a strong acid-strong base titration.2420

OK, so there you have it: so what we have done is: we have taken a weak acid; we have followed the progress of titrating that weak acid by additional volumes of sodium hydroxide.2428

We have showed you how to actually calculate the pH of that solution during this buffering zone, at equivalence, and after equivalence.2440

So, for a weak acid-strong base, you have three different regions: you have this region; you have this region; and you have this region--and all of those were dealt with, as far as solving this titration problem.2450

We essentially did 7 problems in one big, long problem here; a typical problem that you will see is something like: 30 milliliters of a .2 Molar acetic acid solution is titrated with 17 milliliters of a .5 Molar sodium hydroxide; what is the pH of that solution?"2466

Well, you are dealing with a weak acid and a strong base; you do the stoichiometry first; after the stoichiometry is taken care of, you check the major species, and then you do the equilibrium part, depending on where you are--depending on what has been used up and what is actually floating around in solution.2485

I hope this has helped.2502

Next time, we will talk a little bit more about titrations, and we will begin to talk about acid-base indicators--how we actually know, in a laboratory, when we have reached the equivalence point--as opposed to doing it analytically (in other words, calculations).2505

Thank you for joining us here at

We'll see you next time; goodbye.2522