Raffi Hovasapian

Example Problems with Acids, Bases & Buffers

Slide Duration:

Table of Contents

Section 1: Preliminaries on Aqueous Chemistry

Aqueous Solutions & Concentration

39m 57s

Intro

0:00

Aqueous Solutions and Concentration

0:46

Definition of Solution

1:28

Example: Sugar Dissolved in Water

2:19

Example: Salt Dissolved in Water

3:04

A Solute Does Not Have to Be a Solid

3:37

A Solvent Does Not Have to Be a Liquid

5:02

Covalent Compounds

6:55

Ionic Compounds

7:39

Example: Table Sugar

9:12

Example: MgCl₂

10:40

Expressing Concentration: Molarity

13:42

Example 1

14:47

Example 1: Question

14:50

Example 1: Solution

15:40

Another Way to Express Concentration

22:01

Example 2

24:00

Example 2: Question

24:01

Example 2: Solution

24:49

Some Other Ways of Expressing Concentration

27:52

Example 3

29:30

Example 3: Question

29:31

Example 3: Solution

31:02

Dilution & Osmotic Pressure

38m 53s

Intro

0:00

Dilution

0:45

Definition of Dilution

0:46

Example 1: Question

2:08

Example 1: Basic Dilution Equation

4:20

Example 1: Solution

5:31

Example 2: Alternative Approach

12:05

Osmotic Pressure

14:34

Colligative Properties

15:02

Recall: Covalent Compounds and Soluble Ionic Compounds

17:24

Properties of Pure Water

19:42

Addition of a Solute

21:56

Osmotic Pressure: Conceptual Example

24:00

Equation for Osmotic Pressure

29:30

Example of 'i'

31:38

Example 3

32:50

Biochemistry Example Problems with Acids, Bases & Buffers

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Section 1: Preliminaries on Aqueous Chemistry: Lecture 6 | 44:19 min

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	0 answers Post by Michael De La Rosa on November 15, 2021 Why was Example 4 worded as strangely as it was? What is the purpose of wording the question this way? What is this kind of question preparing me for? Is this how chemists speak? Couldn't the quest been worded as . . . "A 100 mL solution of an unknown compound is at a pH of 2.5. 85.0 mL of NaOH is then added to this solution. The unkown compound has two ionizable groups, the first (pKa1) is at a value of 2.5. What is the second's (pKa2)?" In my re-wording, I have included all of the information from the original question. I also made sure to not include any more information. All I've done was separate the reaction chemistry from the known facts of the unknown compound and ordered the events chronologically.
	1 answer Last reply by: Professor Hovasapian Fri Jan 4, 2019 4:36 AM Post by Anthony Villarama on January 2, 2019 Professor I am so impressed with your teaching knowledge. I was blown away how good you are. The way you solved the problems 2,3, and 4 are so amazing. I am going to watch your physical chemistry soon. My only wish is I hope you will also teach Analytical Chemistry that includes the classical and instrumentation. I am going to keep track all of your videos. You deserved the highest pay than any other teachers in the world.
	2 answers Last reply by: Swati Sharma Sat Jan 27, 2018 10:16 AM Post by Swati Sharma on January 21, 2018 Dear Dr Raffi I am little confused in understanding the questions. For example in class we did this question : Calculate the PH of 1L solution containing 0.1 M of Formic acid and 0.1 M sodium formate. So in class our professor told us not to convert them into ,moles and so we directly used 0.1 M concentration into the ICE table. But if the question says for example Calculate the PH of 2L solution containing 10ml of 5M of acetic acid and 10ml of 1M of sodium acetate then we do convert the concentrations such as 5 moles/1lites * 10.010 - liters 1/2litres and we get Molar concentrations. So in your last question the question said 100.0ml of 0.15 M unknown solution so instead of converting 0.15 M into moles I directly used 0.15M and divided by 2 and got 0.075 M and I plugged into ICE tables and I got the same answer. So if I am not wrong if the question phrases such as Calculate the PH of 1L solution containing X M of acid and X M of base is different from question such as Calculate the PH of 1l solution containing Xml of acid and Xml of base. Please could you explain me if I am right. Respectfully Swati
	1 answer Last reply by: Temitope Olasusi Thu Nov 17, 2016 2:40 PM Post by Temitope Olasusi on November 17, 2016 How do you solve for the antilog?
	6 answers Last reply by: Professor Hovasapian Sat Dec 21, 2019 7:15 AM Post by pierre shaouni on September 15, 2016 at 17:24 how did u get the answer from 2.18= ((o.15-h2po4/h2po4)) to be 0.047. i am confused on what was skipped
	0 answers Post by Professor Hovasapian on February 14, 2014 Hi Tejinder. I hope you're well. My apologies for the delayed response. I'm presuming you meant that the only ACIDS you have available have pKas of 8.0 and 6.9 respectively? In this case, it really depends on whether the Buffer is going to be absorbing Acid or Base during its function as a buffer. If absorbing base, then you'll have more buffering capacity if you choose 8.0. If absorbing Acid, then 6.9 is better. Either one is a fine choice though, if there is not going to be too much movement. I hope that helps. Let me know. Best wishes. Raffi
	1 answer Last reply by: Professor Hovasapian Fri Feb 14, 2014 1:57 AM Post by Tejinder kaur on February 11, 2014 Hi Professor, I have homework question and I am little confuse on what one to pick. You want a make a buffer with a pH 7.5, but only buffers you have available have pka of 8.0 and 6.9. Which one, if any, would be better choice and why? I have picked 6.9 because the buffering region include from 5.9 to 7.9 is close to the pH 7.5. Can you please help me? Thanks
	4 answers Last reply by: Aaron Wasielewski Mon Feb 3, 2014 12:35 PM Post by Aaron Wasielewski on January 28, 2014 Hi professor, I am just hitting a bit of a snag here with example 2, and perhaps it should be obvious to me, since it is simple algebra, but at the moment it is not. How are you getting the molarity of 0.047 for the dihydrogen phosphate concentration around 16:21? Maybe after I take a break and try it again, it will jump out, like "AH-HA!", but right now it just isn't. Thank you so much for your help!
	3 answers Last reply by: Professor Hovasapian Tue Jan 28, 2014 3:03 AM Post by Madeleine Hackstetter on December 15, 2013 Hi Professor, could you please explain the exact algebra you did in Example 2: Total Phosphate Concentration to get the 0.047 M. It's around 16:27 minutes. I tried doing the question on my own and keep getting an incorrect M value but I'm not sure where my algebra is going wrong. That would be a great help, Thank you

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Example Problems with Acids, Bases & Buffers

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
Example 1 1:21

Example 1: Properties of Glycine
Example 1: Part A
Example 1: Part B

Example 2 9:02

Example 2: Question
Example 2: Total Phosphate Concentration
Example 2: Final Solution

Example 3 19:34

Example 3: Question
Example 3: pH Before
Example 3: pH After
Example 3: New pH

Example 4 30:00

Example 4: Question
Example 4: Equilibria
Example 4: 1st Reaction
Example 4: 2nd Reaction
Example 4: Final Solution

Biochemistry Online Course

Section 1: Preliminaries on Aqueous Chemistry
	Aqueous Solutions & Concentration	39:57
	Dilution & Osmotic Pressure	38:53
	More on Osmosis	29:01
	Acids & Bases	39:11
	Titrations and Buffers	41:33
	Example Problems with Acids, Bases & Buffers	44:19
	Hydrolysis & Condensation Reactions	18:45
Section 2: Amino Acids & Proteins: Primary Structure
	Amino Acids	38:19
	Amino Acids, Continued	27:14
	Acid/Base Behavior of Amino Acids	48:28
	Peptides & Proteins	45:18
	Amino Acid Sequencing of a Peptide Chain	42:47
	Sequencing Larger Peptides & Proteins	1:02:33
	Peptide Synthesis (Merrifield Process)	49:12
	More Examples with Amino Acids & Peptides	54:31
Section 3: Proteins: Secondary, Tertiary, and Quaternary Structure
	Alpha Helix & Beta Conformation	50:52
Section 4: Proteins: Function
	Protein Function I: Ligand Binding & Myoglobin	51:36
	Protein Function II: Hemoglobin	1:03:36
	Protein Function III: More on Hemoglobin	1:07:16
Section 5: Enzymes
	Enzymes I	41:38
	Enzymes II	44:02
	Enzymes III: Kinetics	56:40
	Enzymes IV: Lineweaver-Burk Plots	20:37
	Enzymes V: Enzyme Inhibition	51:37
	Enzymes VI: Regulatory Enzymes	51:23
	Enzymes VII: Km & Kcat	54:49
Section 6: Carbohydrates
	Monosaccharides	1:17:46
	Hexose Derivatives & Reducing Sugars	37:06
	Disaccharides	43:32
	Polysaccharides	39:25
	Polysaccharides, Part 2	44:15
	Glycoconjugates	44:23
	More Example Problems with Carbohydrates	40:22
Section 7: Lipids
	Fatty Acids & Triacylglycerols	54:55
	Membrane Lipids	38:51
	Membrane Lipids, Part 2	38:20
	The Biologically Active Lipids	48:36
Section 8: Energy & Biological Systems (Bioenergetics)
	Thermodynamics, Free Energy & Equilibrium	45:51
	More on Thermodynamics & Free Energy	37:06
	ATP & Other High-Energy Compounds	44:32
	Phosphoryl Group Transfers	30:08
	Oxidation-Reduction Reactions	49:46
	More On Oxidation-Reduction Reactions	56:34
	Example Problems For Bioenergetics	42:12
Section 9: Glycolysis and Gluconeogenesis
	Overview of Glycolysis I	43:32
	Glycolysis II	1:01:47
	Glycolysis III	59:17
	Glycolysis IV	39:47
	Gluconeogenesis I	41:34
	Gluconeogenesis II	34:18
	The Pentose Phosphate Pathway	42:52
Section 10: The Citric Acid Cycle (Krebs Cycle)
	Citric Acid Cycle I	36:10
	Citric Acid Cycle II	49:20
	Citric Acid Cycle III	44:11
Section 11: Catabolism of Fatty Acids
	Fatty Acid Catabolism I	48:11
	Fatty Acid Catabolism II	45:58
	Fatty Acid Catabolism III	33:18
Section 12: Catabolism of Amino Acids and the Urea Cycle
	Overview & The Aminotransferase Reaction	40:59
	Glutamine & Alanine: The Urea Cycle I	39:18
	Glutamine & Alanine: The Urea Cycle II	36:21
	Amino Acid Catabolism	47:58
Section 13: Oxidative Phosphorylation and ATP Synthesis
	Oxidative Phosphorylation I	41:11
	Oxidative Phosphorylation II	36:27

Transcription: Example Problems with Acids, Bases & Buffers

Hello and welcome back to Educator.com and welcome back to Biochemistry.0000

On the last lesson, we talked about titrations and buffers, and I said that we were going to spend this particular lesson just doing example problems mostly with buffers, because that is the real important thing in biochemistry.0003

Let's just jump in and get started.0017

OK.0021

As with all problems in the sciences, it's really, really important, I mean, we have certain equations that we deal with, for example in this particular set of problems that we are going to be doing, the Henderson-Hasselbalch equation is going to be the important one, but it's not just about plugging numbers in.0024

As you'll see in a minute, these problems, they can come across as reasonably complicated- they are not.0040

It's very, very important that you understand the chemistry behind it.0046

If you stop and take a look and ask yourself what reaction is taking place, your intuition will guide you.0049

We want the chemistry to guide the mathematics, not the other way around.0056

If you don't exactly understand the chemistry, then you're sort of going to be limited in the number of problems that you'll be able to do.0061

You'll only be able to do the simple ones, but this is biochemistry, and there tends to be a lot of things going on.0067

They are not difficult problems; it's just a question of, again, understanding the chemistry, so let's see what we can do.0074

The first example- let's go ahead and do it in black here.0082

Example 1: glycine and amino acid, those are the constituents of proteins, and we'll be talking about that very, very soon.0088

I'll go ahead and draw out the structure here: H₃ and there is a plus charge, there is a C, there is a C, H, H, H.0106

OK.0119

Glycine is often used to prepare buffer solutions in biochemistry.0120

OK.0142

Let's go ahead and concern ourselves with the amino group.0144

Let's concern ourselves with the amino group.0150

So, you notice in this particular thing, you have an OH here, so this is 1 hydrogen that can be pulled off, that is one ionizable group; but notice here, this is NH₃⁺, there is another hydrogen here, so we are going to concern ourselves with the amino group.0161

Let me go ahead and do this in blue.0174

I'm just going to call this whole rest of the thing "R", and I'm just going to write RNH₃⁺, something like that.0180

That is R amino acid glycine- the protonated form of that.0188

So, let's go ahead and write our reaction that is going to take place.0194

Again, we want to be able to understand what's happening as far as a reaction is concerned: RNH₃⁺, and this is going to be in equilibrium with H⁺ + RNH₂.0196

This is the acid, and this is the conjugate base.0212

That is our little bit of a buffer system here.0216

You are going to have a little bit of this and a little bit of that.0219

OK.0221

Now, let's go ahead and ask some questions about this.0222

I’ll give you a bit of information here, the pK_A for this, for the amino group, is 9.6.0227

So, nice, simple first question: What is the buffering region or what is the buffer region of the amino group for glycine?0234

Well, we said that the buffer region is pK_A + or - 1 unit.0256

The pK_A is 9.6, so we're looking at about 8.6 to about 10.6.0263

If we're going to run an experiment that is going to require a pH in that range, glycine buffer is a good buffer.0270

There you go.0279

OK.0281

Now, let's see what we can do.0283

Now, let's get to some quantitative stuff.0284

OK.0287

In a 0.15M glycine solution at a pH equal to 9.2, what percent of glycine is in its protonated form?0288

OK.0323

Let me go ahead and go to red.0325

The protonated form of glycine is this one; that is the one that is protonated.0327

It's the acid; it's the one that has the hydrogen ion to donate.0330

This is the unprotonated form right here, the RNH₂.0333

They want to know, in a solution that is set to a 9.2 pH, what percentage of the glycine is in this form?0337

Again, you've got a buffer solution, some of it is going be this, some of it is going to be that, what's the percentage?0348

OK.0352

A percentage is the part over the whole.0354

Let's go ahead and see what we can do here.0356

Let's go ahead and use Henderson-Hasselbalch equation, which is perfect: pH = pK_A + log of RNH₂, that is the conjugate base, over the acid form, RNH₃⁺, over that.0359

Well, they said that the pH is 9.2, so, that is going to be the left hand side of our equality.0380

The pK_A is 9.6 + the log- I'm just going to write base over acid so that I don't have to keep writing over and over again.0386

OK.0397

Well, let me see.0399

Let me go to the next page to actually solve this and rewrite it.0400

I've got 9.2 = 9.6 + the log of the base over the acid, and now, I'm going to go ahead, and I'm going to solve for this ratio- the base over the acid.0403

Because they are asking for percentage, they are asking for a fraction, I'm just going to go ahead and leave it as that; because this base over the acid, that is our percentage.0418

So, when I solve this 9.2 - 9.6 = -0.4, right?0427

This is going to be -0.4 = log of b/a.0433

Now, I'm going to take the antilog, I'm going to raise both side, I'm going to exponentiate with the base 10; and what I end up with is the following.0440

I end up getting that the, well, this is going to be 0.398 = RNH₂/RNH₃⁺.0449

OK.0469

Now, what does this mean?0470

This ratio here is telling me the ratio of base, unprotonated to protonated, is 0.398.0472

This is a part over the whole.0480

That means that 39.8% is unprotonated.0482

That is what a percent is- it's just the part over the whole.0493

In this particular case, how much of that is unprotonated?0496

Well, they didn't ask for the unprotonated, they asked for the protonated, so now, I just subtract this from 100 and what you get is: 60.1% is protonated.0500

In this particular case, I used the Henderson-Hasselbalch equation.0512

I used the ratio itself; I didn't actually solve for the numerator or denominator.0518

I used the ratio itself, because the problem asked for a percentage, and a percentage is a part over the whole.0522

I hope that made sense.0530

OK.0531

Let's do something a little bit more complex here- example number 2.0534

Let me go back to blue here.0540

Example number 2: how much in grams, actually - so, let me just write how many grams - how many grams of sodium dihydrogen phosphate and how many grams of disodium hydrogen phosphate are needed to prepare 1.0L of a pH = 7.20 buffer, with the condition that the total phosphate concentration must be 0.15M?0547

OK.0637

When you read this question, you're probably thinking to yourself "Oh my God, how the heck am I going to solve this? There's a lot going on here?".0639

OK.0645

This is where it's really, really important to take your time, relax, don't think that you have to look at this question and just automatically know what to do.0646

You want to think about this, think about the chemistry, think about what's going on; and see if we can interpret what it is this thing is actually saying here.0655

Here we're trying to prepare a buffer, 1L of that buffer- that is nice, 1L, that is good.0666

We want the buffer to be half a pH of the final pH of 7.2.0672

Well, I'm going to be adding some sodium dihydrogen phosphate and some disodium hydrogen phosphate, so I know that my buffer here is going to be the H₂PO₄ and the HPO₄, the phosphate buffer.0677

It looks like this experiment is trying to mimic what is going on inside of a cell.0689

So, I'm going to add a little bit of dihydrogen phosphate, a little bit of hydrogen phosphate.0695

I need to know how many grams I need to add to 1L in order to get myself to a pH of 7.2, but I have a condition: I need the total phosphate concentration to be 0.15M.0700

OK.0714

Let's just go ahead and write down some equations, and see where we can go.0715

And again, sometimes you just have to start with what you do know, and hopefully, something will fall out.0719

I'm going to write down the equation- my buffer equation: H₂PO₄^-, that is going to be in equilibrium with H⁺ + HPO₄^2-, right?0724

This is my acid; this is my conjugate base.0736

This is my proton donor; this is my proton acceptor.0739

Well, I need the total phosphate concentration.0743

I need it to equal 0.15M.0753

What that means is the following: that means the total concentration of H₂PO₄^- plus the concentration of HPO₄^2- -, the sum of those, that is what that means.0757

A lot of these problems in biochemistry, the difficulty is not going to be the chemistry; it's going to be interpreting what it is that it's going to ask.0774

So, questions can be asked in several different ways.0780

One of the things that you will see in biochemistry, in books, in journals, in things like that, when they talk about the total phosphate concentration or the total carbonate concentration, they are talking about all the different species.0783

In this case, the phosphates that we are dealing with, it isn’t PO₄^3--; the only phosphate species we have is the dihydrogen phosphate and the hydrogen phosphate, so the total phosphate means the sum of those two, this and this.0795

They have to add up to 0.15.0809

That is what that part means in the question.0811

OK.0813

Let me go ahead and do...I'm going to do something here.0816

I'm actually going to move one of these over, and I'm going to solve for HPO₄^2-.0823

The hydrogen phosphate concentration, HPO₄^2- - and again, be very, very careful, there's a lot of symbols going on here- 0.15 minus the concentration of the dihydrogen phosphate.0829

So, that's that.0844

Now, I can go ahead and write down the Henderson-Hasselbalch equation, and I should be able to work this out.0847

Let's go ahead and write down pH = pK_A plus the logarithm of the base, which is HPO₄^2- over the acid, which is the H₂PO₄^-, right?0853

OK.0877

We want the pH to be 7.2, so we have that number.0878

We have the pK_A of this buffer system; we can just look it up.0883

This is actually the second ionization of phosphoric acid, so the pK_A happens to be 6.86 plus the log; and now, I've already expressed HPO₄ in terms of this, so I’m going to write 0.15 - H₂PO₄^- over the concentration of H₂PO₄^-.0885

Notice what I did, is I ended up replacing two different variable here by one variable by knowing that the total phosphate concentration was 0.15, so now, I just have H₂PO₄^-.0913

Well, I have a perfectly good equation here.0925

This is my variable, the concentration of the dihydrogen phosphate.0927

I'll just go ahead and solve for that.0930

I end up with the following: 2.1 - I'll go ahead and work it out to a reasonable degree - so, I have 2.188 is equal to, once I actually move this over, take the antilog, I'm left with 2.188 = 0.15 - H₂PO₄^-/H₂PO₄^-.0933

OK.0965

And so, I'm going to go ahead and let you do the algebra here, move this over, this is just simple algebra.0966

I end up with a dihydrogen phosphate concentration equal to 0.047M.0970

So, I know that that's the concentration of the dihydrogen phosphate I need, in order to get a 7.2 buffer.0981

OK.0989

Well, since I know the dihydrogen phosphate, now, I can just do the 0.15 minus this.0990

I also know the hydrogen phosphate, so HPO₄^2- concentration, that equals 0.15 - actually, let me do that one in red just to keep them separate - 0.15 - 0.047 and that is going to equal 0.103M.0996

There is our concentrations that we need, but we want it grams, we didn't want concentration in moles per liter.1023

So, let's go ahead, and now deal with the sodium dihydrogen phosphate.1030

OK.1038

I have 0.047mol/L, and the molar mass of the sodium dihydrogen phosphate is 120g/mol, and it was going to be in 1L of solution.1040

This is why I said "I'm really happy that it is 1L, it's a very, very nice number - 1.".1057

Well, liter cancels liter, mole cancels mole.1062

I'm left with grams and my answer there is going to be 5.64g of sodium dihydrogen phosphate- that is my first answer.1065

Now, I want to do my disodium hydrogen phosphate.1080

Well, this one is 0.103mol/L, and its molar mass is 142g/mol, and of course, we are creating 1L of solution; so again, mole cancels mole, liter cancels liter - oops, sorry about these little stray lines here, they tend to show up - and here, we get 14.63g.1088

There we go.1119

In order for the total phosphate concentration to be 0.15M, using the phosphate buffer system, dihydrogen phosphate and hydrogen phosphate at a pH of 7.2, I need to add 5.64g of sodium dihydrogen phosphate to 1L of solution, and I need to add 14.63g of disodium hydrogen phosphate, and I will prepare this 7.2 pH buffer.1120

That is it.1150

I'm just using the Henderson-Hasselbalch equation and trying my best to extract as much information as I can and play around with it.1151

I hope that makes sense.1159

It's a little long, but there's nothing strange going on, and certainly nothing strange mathematically, it's just simple algebra.1160

OK.1166

Let's move on to another example here.1168

Let's see.1175

Example number 3: what is the change in pH when 6.0mL - actually you know what, let me write, I don't want to separate my unit from my number - when 6.0mL of a 0.5M hydrochloric acid is added to 1.0L of a lactic acid buffer solution containing 0.03mol of lactic acid and 0.05mol of lactate.1176

And in this particular case, the pK_A of lactic acid is 3.86.1256

OK.1267

So, what's the change in the pH when 6mL of a 0.5M hydrochloric acid is added to 1L of a lactic acid buffer solution that contains 0.03mol of lactic acid and 0.05mol of lactate?1268

OK.1281

It seems kind of complicated, a lot of numbers floating around.1283

What's nice about this, I noticed that they gave us the moles of the lactic acid and the moles of the lactate.1287

That is really, really good.1293

Let's write our equation, first of fall, so we understand what's going on.1296

We are looking at a lactic acid solution, so, I'm just going to write that as HLac, and that is in equilibrium with H⁺ + Lac^-.1299

OK.1311

That is our equation.1312

This is our acid, the lactic acid, and this is our conjugate base, so the R buffer solution contains a little bit of this and a little bit of that.1316

And, what's nice is they actually told us how much- it's 1L of solution.1321

That is fantastic, that works out well.1325

They want a change in pH, so we need the pH before, we need the pH after.1329

Let's go ahead and calculate the pH before the addition.1334

OK, and again, we're going to use the Henderson-Hasselbalch.1345

Again, pH = pK_A plus the logarithm of the base, which is the lactate, over...you know what, I hope you'll forgive me, I'm going to stop writing these brackets, I think of at this point it should be pretty clear we're dealing with concentrations, moles per liter, so, I'm just going to go ahead and write Lac^- over HLac without the brackets, but again, we're talking about concentration- moles per liter.1347

Well, the pK_A is 3.86 - that is great - and plus the logarithm of the lactate concentration is 0.05mol/1.0L, and the acid concentration was 0.03mol, so this is 0.03mol/1.0L.1375

Here is what's really nice, when the volume doesn't change because they are both in the same container, so for all practical purposes, we can just work with moles.1401

We don't have to worry about concentration.1409

In other words, even if I were to add, let's say here I'm going to add 6mL, so now, there is an extra 6mL, but the volume is the same.1411

You are dividing by the total volume, so the volume and volume actually cancel, so we can just work with moles.1420

That is the nice thing about the Henderson-Hasselbalch equation.1426

The volume actually cancels, but I want you to see that you are actually dealing with concentration, a certain number of moles per 1L.1428

OK.1437

So, when we do this, we end up with 3.86 - and by all means, please check my arithmetic here, I'm notorious for arithmetic errors - plus 0.223, and we get an initial pH of 4.08.1439

That is our initial pH before the hydrochloric acid was added.1456

OK.1460

Now, let's go ahead and calculate the pH afterward.1461

OK.1464

Now, for the pH after the addition of the hydrochloric acid.1465

So, here is where it's really, really important- we have to stop and ask ourselves what is it that's happening.1475

This is a buffer solution, so when we add hydrochloric acid, we're adding H⁺.1479

Well, we have to stop and ask ourselves what the chemistry, what's taking place?1485

So, the H⁺ is going to react with either the lactic acid or the lactate in order to be used up.1489

Well, it's going to react with the lactate.1497

The reaction upon addition of the H⁺, we add the hydrochloric acid, but the chloride doesn’t matter, it's the H⁺ that matters.1501

And, the H⁺ is as follows: we have H⁺ is going to react with the lactate to produce lactic acid.1516

Now, we have to find out how much of this lactate is actually converted to lactic acid, we calculate our mole ratios, so that we can find our new pH.1531

We're going to working with moles, so we're going to have it before the addition of the hydrochloric acid.1540

We're going to have the change that takes place, and we're going to have the afterward.1546

Well, before, our lactate concentration is 0.05mol, and our lactic acid concentration is 0.03mol, now, our H⁺ concentration happens to be 0.003mol, and you're probably wondering where I came up with that.1551

Well, I'll show you where I came up with that.1574

OK, remember we said we're adding 6mL of a 0.5M hydrochloric acid solution, so 6mL is 0.006L x 0.5mol/L.1577

0.006 x 0.5 gives me 0.003mol of HCL were added, which means 0.003mol of H⁺.1592

This is before any reaction takes place.1606

Now, this reaction goes to completion.1608

All of the hydrogen ion is eaten up, that is what a buffer does- leaving no free hydrogen ion.1613

Lactate 0.003 reacts with the lactate leaving - it's the lactate that is converted, so that is going to deplete - that is going to end up leaving me with 0.047mol of the lactate, and this is going to be plus 0.003, because now, lactic acid is being produced.1621

It is being produced that much, so we end up with 0.033mol of lactic acid.1646

So, the buffer solution stands at a certain pH, I add some hydrogen ion.1655

Some hydrogen ion converts some of the lactate to lactic acid.1660

I need to find out how much is converted.1664

I need these numbers - the final mole amounts.1665

Now, I can run my Henderson-Hasselbalch equation to get my new pH.1669

OK.1673

Let me do it over here; so, pH equals - again, I'm just going to keep writing this equation over and over again: pK_a plus the logarithm of the lactate over the lactic acid, it equals 3.86 plus the logarithm of, well, lactate is 0.047, and again, this is moles, but again, because the volume is the same, even though I've added 6mL, so now, I have 1L + 6mL, so I have 1006mL, the volume is the same.1677

So, even if I divide by the volume here to get molarity, because these are concentrations, the volumes cancels, so I can just work with moles, the volume is irrelevant...over 0.033mol.1730

And when I run this, I end up with 3 - I don't need to write out everything - I end up with the pH of 4.01.1745

My initial pH was 4.08; my final pH was 4.01, so my delta pH, my change in pH is -0.07.1758

Yes, as you see, this is a fantastic buffer.1771

I've added 6mL of a 0.5M hydrochloric acid solution.1774

That is actually a lot of hydrochloric acid, and yet the pH has gone down just a little bit, barely anything to be noticeable.1779

This is a perfectly functioning buffer solution.1786

OK.1790

Now, let's go ahead and do one last example here.1793

Do this one in blue.1802

Example 4, now, OK.1806

An unknown compound has 2 ionizable groups.1813

In other words, there are 2 sets of hydrogen ions that it can lose, that can be taken away.1826

OK.1833

It's an unknown compound.1834

Now, the pK_A1 is equal to 2.5. - that much we do know, the first ionizable group.1836

Now, when 85.0mL of a 0.1M sodium hydroxide is added to 100.0mL of a 0.15M solution of this unknown compound, which is already at a pH = 2.5, its pH jumps to 6.82.1845

My question to you is: what is K_A2 or pK_A2?1911

See here, did I give you K_A...yes that is fine.1925

What is the pK_A2?1927

OK.1931

We have an unknown compound that has two ionizable groups.1932

We happen to know that the first ionizable group has a pK_A of 2.5, now, when we add 85mL of a 0.1M sodium hydroxide to this 100mL of a 1.5M solution of this unknown compound, which is already at a pH of 2.5, its pH jumps to 6.82.1936

What's the pK_A2?1955

OK, let's see what we have going here.1958

I'm going to go ahead and assign a...let me do, that is fine, I'll go ahead and keep it as blue.1960

I need a symbol for this, so let H₂A be the unknown compound.1971

OK.1984

The equilibria are as follows.1985

In this particular problem, we definitely need to keep track of the chemistry.1989

The chemistry is what's important.1993

It will decide what the math looks like.1994

The equilibria are as follows: we have the 1st dissociation, H₂A goes to H⁺ + HA^-; and we know that this pK_A, we know that one, that is equal to 2.5.1999

And then of course, we have the second ionization: HA^- in equilibrium with H⁺ + A^-.2020

This is the pK_A that we seek.2028

So, we have number 1- we seek that.2031

OK.2036

Well, notice what we have.2040

They are telling me that the pH of this solution - let me go to red here - is already 2.5.2043

Well, as it turns out, the pK_A of the first ionizable group is 2.5.2049

This is very, very convenient.2055

So, at pH equal to 2.5, the H₂A concentration is equal to the HA^- concentration, right?2057

This is half equivalence.2076

When the pH equals the pK_A, then what you have is, that and that have the same concentration.2077

OK.2086

Well, let's see what else we can do with that little bit of information.2087

So, we have a 0.15M solution of this unknown compound, and it also gives us the volume.2093

Oh good, OK.2102

So, we have 0.100L x 0.15mol/L, so when I multiply this out, I end up with 0.015mol of H₂A to begin with.2106

You know what, I have a lot of lines floating around, so I'm going to actually do this on the next page; I hope you don't mind.2134

I want you to be able to see the math and not have it be...OK...let me actually do it down here, and see if it works a little bit better.2142

I have 100mL, 0.1L x 0.15mol/L, that gives me 0.015mol of H₂A to begin with.2152

That is what this means: 100mL of a 0.15M solution to begin with, I have 0.015mol of H₂A.2174

Well, I know that at pH of 2.5, the pK_A is 2.5, so I know that my H₂A is actually equal to this; so, half of this H₂A originally, has been converted to its conjugate base.2182

Therefore, if I have 0.015 to start off with, if I take half of the 0.015, that means I have 0.0075mol of H₂A, and I have 0.0075mol of HA^-.2198

That is before anything happens.2224

The pH of 2.5 happens to match the pK_A; I know that that is half equivalence.2227

Therefore, I know that acid and the conjugate base concentrations are equal.2231

I started off with 0.015mol; I've converted half of it.2234

That means now, that is what I have.2238

OK.2241

Now, let's move forward from here.2242

Well, let's go ahead and concentrate; let's see how much hydroxide ion was added.2245

Hydroxide ion, they said we added 85mL of a 0.1M, so 0.085L x 0.1mol/L; liter and liter cancels, leaving me with 0.0085mol of OH^- were added.2252

OK.2276

Well, this is a buffer solution; now, I'm adding a base.2277

When I add a base, it's going to react with the acid.2280

Here is what's going to happen.2285

Again, you have to reason out the chemistry; that is the only way this problem is going to be solved.2287

Well, the first reaction is going to be, the hydroxide that is added is going to react with the H₂A, that is left over, the little bit that is there; because that is the first ionization, H₂A, and it's going to produce H₂O + HA^-.2298

We're going to have a before; we're going to have a change, and we're going to have an after.2319

Well, before, our concentration is 0.0075, 0.0085 of the hydroxide, the water doesn't matter, and I have 0.0075 of that.2324

Well, 0.0075, 0.0085, this is the limiting reactant, so it's going to run out first.2340

So, the change is that, there's going to be no H₂A left over, 0.0075, there's going to be 0.0010mol of hydroxide left over.2347

Water doesn't matter; here it's going to be converted, so it's going to be +0.0075.2364

This H₂A, all of it is going to be converted to HA, so now I have 0.015mol of the HA^-.2372

OK, but notice, I still have some excess hydroxide.2384

Hydroxide is going to go after any other hydrogen ion it can find.2387

So, there's a second reaction that takes place here.2393

Now, this hydroxide is going to react with this.2398

So, the second reaction that takes place is: the leftover hydroxide is going to react with the HA^-, and it's going to convert it to water + A^-.2401

Well, I have a before, I have a change, I have an after.2416

Before, I start with 0.0010mol of that HA, I have 0.015mol of that, water, of course, does not matter; and I have none of this.2421

Well, in this case, 0.0010 is the limiting reactant, so that is going to drop to 0 - 0.0010.2437

This is going to be converted, that is why this is a minus, so we end up with 0.014mol of HA.2447

Now, in this particular reaction, it's the HA that is the acid donor, and this is the conjugate base.2457

Up here, the HA is the conjugate base, and this is the acid.2465

OK.2469

So now, we have the 0, and then we have +0.0010, so this is going to be 0.001.2470

Now, we have a certain amount of A^-- that much.2480

We have a certain amount of HA- that is that much.2483

Now, we can go ahead and do our thing.2486

Let me go ahead and we're going to write out again the Henderson-Hasselbalch equation: pH= pK_A + the log of the A^2- over the HA^-.2492

I apologize, I may have left the 2-; it lost a second proton, so now, it has a 2-charge on there.2513

Well, they said the pH jumps to 6.82.2518

Well, we're looking for the pK_A2, that is what we're looking for, plus the logarithm of the...so now, the A^2-.2525

We said we had 0.001mol, right?2535

And this time, I'm actually going to put it there.2542

So you see, I had 100mL of solution to start, I added 85mL, so now, my total volume is 0.185L - and you'll see in a minute, it actually cancels out, but I wanted you to see that it's there, I don't have to put it there, I can just work with the moles, but I do want you to see it - over 0.014mol, over 0.185L, volume, so those go away.2544

They don't really matter.2571

So, what I'm left with is 6.82 is equal to the pK_A2 that I seek, plus a -1.15.2573

That means that my pK_A2 is equal to 7.97 if I had done my arithmetic correct.2586

There you go.2595

I had a solution whose pH happens to match the first ionization- the pK_A1 of an unknown compound.2597

I added a certain amount of hydroxide to this buffer solution, and now, the pH jumped up to 6.82.2605

Well, I used the fact that, basically, what you have is a titration, you have a buffer; however, I need to keep track of what's going on.2613

I need to find out the number of moles, how much of one species is being converted to another, how much hydroxide is left over.2621

Because there was some hydroxide left over upon the first reaction, the leftover hydroxide was going to react with the next species that has hydrogen ions to give up, until all of the hydroxide is used up, is eaten up, is sequestered; and then what you're left with, was of course, this, giving us a way to actually find the pK_A.2629

OK.2654

Thank you so much for joining us here at Educator.com.2656

We'll see you next time, bye-bye.2658

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