For more information, please see full course syllabus of Biochemistry

For more information, please see full course syllabus of Biochemistry

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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

### 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: H _{3} 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 NH _{3}^{+}, 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 RNH _{3}^{+}, 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: RNH _{3}^{+}, and this is going to be in equilibrium with H^{+} + RNH_{2}.*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 RNH _{2}.*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 RNH_{2}, that is the conjugate base, over the acid form, RNH_{3}^{+}, 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 = RNH _{2}/RNH_{3}^{+}.*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 H _{2}PO_{4} and the HPO_{4}, 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: H _{2}PO_{4}^{-}, that is going to be in equilibrium with H^{+} + HPO_{4}^{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 H _{2}PO_{4}^{-} plus the concentration of HPO_{4}^{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 PO _{4}^{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 HPO _{4}^{2-}.*0823

*The hydrogen phosphate concentration, HPO _{4}^{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 HPO_{4}^{2-} over the acid, which is the H_{2}PO_{4}^{-}, 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 HPO_{4} in terms of this, so I’m going to write 0.15 - H_{2}PO_{4}^{-} over the concentration of H_{2}PO_{4}^{-}.*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 H _{2}PO_{4}^{-}.*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 - H _{2}PO_{4}^{-}/H_{2}PO_{4}^{-}.*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 HPO _{4}^{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 H _{2}A 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, H _{2}A 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 H _{2}A 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 H _{2}A 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 H _{2}A to begin with.*2152

*That is what this means: 100mL of a 0.15M solution to begin with, I have 0.015mol of H _{2}A.*2174

*Well, I know that at pH of 2.5, the pK _{A} is 2.5, so I know that my H_{2}A is actually equal to this; so, half of this H_{2}A 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 H _{2}A, 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 H _{2}A, that is left over, the little bit that is there; because that is the first ionization, H_{2}A, and it's going to produce H_{2}O + 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 H _{2}A 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 H _{2}A, 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

2 answers

Last reply by: Swati Sharma

Sat Jan 27, 2018 10:16 AM

Post by Swati Sharma on January 21 at 08:23:24 PM

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 * 10.0*10 - 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?

1 answer

Last reply by: Professor Hovasapian

Tue Sep 27, 2016 5:03 PM

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