For more information, please see full course syllabus of AP Chemistry

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### pH of Weak Acid Solutions

- Identify the major species in solution BEFORE any equilibrium is reached. Then use an ICE chart to find Eq. concentrations.
- ICE charts must use molarities.

### pH of Weak Acid Solutions

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
- pH of Weak Acid Solutions 1:12
- pH of Weak Acid Solutions
- Example 1
- Example 2
- Example 3
- Example 4

### AP Chemistry Online Prep Course

### Transcription: pH of Weak Acid Solutions

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

*In our last discussion, we introduced the basic notions of acids and bases; actually, we didn't talk about bases so much--we will actually talk about those mostly in the next lesson.*0004

*We introduced the notion of an acid, and we said an acid is something that has hydrogen ions that it can give up.*0014

*Often, what happens is: you will take a particular acid, like hydrogen chloride, drop it in water, and that will dissociate, and it will separate into hydrogen ions and chloride ions.*0020

*Or, it may not separate at all, which is part of the topic we are going to be talking about today: weak acids.*0030

*We talked a little bit about the pH scales, the relationship between the hydrogen ion concentration and the hydroxide ion concentration in water.*0035

*We said that each one of them is 10 ^{-7}, to come up with a total of 10^{-14}.*0045

*Right now, in a minute, I am actually going to write down some of the basic things we discussed, and then we are just going to launch right into a series of problems on how to handle the pHs of weak base solutions.*0050

*A weak base--I'm sorry, a weak acid--is something that doesn't dissociate completely; so let's just jump in and get started and see what we can work out.*0064

*OK, so, last time, we said that, remember, we have water, which acts as both an acid and a base.*0073

*H _{2}O--it dissociates into H^{+} plus OH^{-}.*0080

*We said that the equilibrium constant for this--because this is liquid water, this is aqueous, this is aqueous--so our K _{a} for this, which we ended up calling K_{w} for water, was just the hydrogen ion concentration, times the hydroxide ion concentration.*0087

*It always equals 1.0x10 ^{-14} (let me make that 14 a little bit more clear), at 25 degrees Celsius.*0106

*So, the K _{eq}, the K_{a}...these things change at temperature, but normally, we are talking about 25 degrees Celsius.*0119

*This means that any aqueous solution, no matter what is in it, the hydrogen ion concentration times the hydroxide ion concentration is always going to equal 1.0x10 ^{-14}.*0125

*That is very, very convenient, because, no matter what one of these is, we can always find the other--and that is the whole idea.*0137

*Today, we are going to talk about weak acids; and weak acids are ones that actually don't dissociate completely.*0145

*Let's talk about a strong acid first: hydrogen chloride--when we drop this is water, the dissociation reaction for this is H ^{+} + Cl^{-}; it's a strong acid--strong means it completely dissociates in water, just like sodium chloride (salt) or anything else that is just completely soluble.*0153

*Well, as it turns out, a weak acid doesn't do that; a weak acid behaves like this: like we said last time, HA--it will react a little bit with water, and it will actually establish a little bit of an equilibrium.*0174

*What you will end up getting is something like H _{3}O^{+}, plus the conjugate base of this weak acid.*0187

*Another way of writing this, which I personally prefer, is just the standard dissociation, without mentioning the water: H ^{+} + A^{-}.*0195

*Now, we said there is a K _{a} associated with this; and the K_{a} is equal to the H^{+} concentration.*0205

*And again, sometimes I will use the square brackets; sometimes I won't, simply to save myself a little bit of time.*0210

*But, these are all concentrations in moles per liter.*0216

*A ^{-} over HA: now, most of the time, for the weak acids, these K_{a}s are going to be very, very small.*0219

*They are going to be things on the order of -4, -5, -8, -10, even -12 and -15.*0227

*What that means: because a K _{a} is small, a small equilibrium constant means that the reaction has not moved very far forward.*0235

*That means most of it is here; so, if I took, let's say, some hydrofluoric acid (which is not really hydrofluoric acid--it's hydrogen fluoride, and I dropped it in water to create a solution of hydrofluoric acid), it doesn't actually dissociate too much.*0244

*Yes, it does produce a little bit of this and this; it does come apart a little bit; but it is a weak acid, which means that most of the hydrofluoric acid will be floating around as HA, as the entire species, in aqueous solution.*0260

*You still can't see the H _{a} in there; it is still a homogeneous solution; but it is solvated, so it isn't a hydrogen...*0274

*This is why, when we talk about acids, acids are always that particular compound, found in water.*0281

*We just automatically presume that we are talking about an aqueous solution.*0289

*As it turns out, hydrogen fluoride, when you make it, is not an acid--it's just hydrogen fluoride.*0293

*Things become acids when you drop them in water; but, because we always find them in water, because we are always talking about aqueous chemistry, we just often refer to them as acids.*0298

*A weak acid is one that doesn't dissociate very much.*0307

*Now, we need to find a way to actually find the pH of these solutions; and that is pretty much what we are going to be doing today, in this lesson; we are just going to be doing several problems.*0311

*Some of them are exactly like the ones before--maybe slightly different; but the idea is to develop a sense of turning what we have discussed (as far as the basic definitions of pH acid)--now we want to quantify it--now we want to use some of the math behind it.*0319

*What you are going to discover is: these problems that we are going to do are exactly like the equilibrium problems that we did in the previous chapter.*0336

*This whole ICE chart--the Initial, the Change, the Equilibrium concentration--that is going to happen over and over and over and over again.*0343

*You are going to be doing a ton of these; and I am specifically concentrating on doing a number of these problems, because it is in the process of doing these problems that we actually get a sense of what is going on with these acid-base chemistries.*0350

*There is going to be a lot of commentary, a lot of talk.*0362

*This is where most of the deep chemistry is happening; and, if you understand this stuff, the AP exam will be an absolute piece of cake, I promise you.*0365

*This is where most of the problem takes place, and we want to make sure we really understand it, because it's beautiful chemistry; it's applicable chemistry; and...well, let's just do it.*0375

*OK, so we are going to start off, actually, with a strong acid, just to get a sense.*0385

*The first example is going to be: Calculate the pH of a 0.15 Molar nitric acid solution (nitric acid is HNO _{3}), and the pH of a 1.5x10^{-10} Molar HCl solution.*0389

*OK, so let's do the first part: let's calculate the pH of a .15 Molar nitric acid solution.*0420

*All right, when we do these acid-base problems (strong acid, weak acid...later on when we work with weak base, strong base; and later on when we work with buffers), it is always the same thing.*0426

*We want to develop a nice, systematic approach, because, as the species in the solution get more complicated, it's all that much more important to keep track of what is going on.*0437

*We are going to use these problems to develop a sense of the chemistry; we are going to let the chemistry decide how the math works.*0446

*You need to understand the chemistry; if you don't understand the chemistry, the math is irrelevant.*0451

*It will mean nothing to you; but, if you understand the chemistry, the math is a piece of cake.*0456

*In fact, it is pure instinct; there is nothing here, mathematically, that you don't know or can't follow--there is nothing mysterious happening.*0460

*We just want to make sure you understand what is happening.*0467

*OK, so first of all, we notice that it is a strong acid; a strong acid means full dissociation--that means this is going to happen.*0470

*That means HNO _{3} is going to completely dissociate into H^{+} ion plus NO_{3}^{-}; that means H^{+} is going to be floating around freely, NO_{3} is going to be floating around freely, and there is going to be nothing that is left.*0479

*That is why there is an arrow pointing one way--this is not an equilibrium.*0491

*Strong acid--full dissociation.*0494

*So now, what we ask ourselves-- always the first question we ask ourselves: What is the major species--list the major species in the solution.*0497

*Well, the major species in solution are...this will decide what the chemistry is, because there is always going to be something that is going to dominate the chemistry in the solution.*0504

*Well, major species: we have...of course, we have H _{2}O--that is our solvent; we have H^{+} floating around; and we have NO_{3}^{-} floating around.*0517

*That is it; well, they are asking for the pH--what is pH?--remember, we said that pH is just a different way of dealing with the hydrogen ion concentration.*0527

*It is the negative log of the hydrogen ion concentration, in moles per liter.*0537

*Well, look at the major species we have in solution: we have H _{2}O; we have H^{+} and NO_{3}.*0543

*There are two sources of hydrogen ion: one of the sources is the dissociation of the nitric acid itself; well, the other source is the dissociation of water (remember, water also comes apart and produces a little bit of this).*0548

*But, because this is such a strong acid that all of these hydrogen ions...actually, what it is going to end up doing is: you know that this is (under normal conditions) just 10 ^{-7} molarity--the hydrogen ion concentration.*0564

*It is 7 orders of magnitude smaller than this; it is virtually...you could completely neglect it; as a matter of fact, because this dissociates (remember Le Chatelier's Principle?), all of the hydrogen ion that is floating around from the dissociation of this strong acid actually pushes this equilibrium this way.*0579

*So, there is even less hydrogen ion coming from water itself.*0598

*For all practical purposes, we can ignore that; the dominant species here is this; our problem is already done.*0602

*It says we have .15 Molar HNO _{3}; well, strong acid...0.15 moles per liter HNO_{3} produces .15 Molar H^{+} and .15 Molar NO_{3}^{-}, because it's 1:1, 1:1; that is the ratio.*0611

*This is easy; the pH is just minus the logarithm of 0.15; that is it, because the concentration of H ^{+} in a .15 Molar HNO_{3} solution is .15 moles per liter of H^{+}.*0626

*That is the free thing floating around; and we get 0.82.*0642

*That is it--nice and simple: strong acid--all you have to do is use the molarity of that acid.*0648

*Just take the negative log of it; you automatically have the pH, because you have full dissociation.*0655

*Now, let's do the second part, B: Now, we have a 1.5x10 to the negative (11 or 10?--what did we say here)...1.5x10 ^{-10} Molar HCl solution.*0660

*OK, again, let's see what the major species are in solution.*0676

*Well, again, we have HCl; it's a strong acid; it breaks up into H ^{+} + Cl^{-}; well, 1.5x10^{-10} Molar HCl is full dissociation.*0680

*That implies that it is 1.5x10 ^{-10} Molar H^{+}.*0696

*Well, think about this for a minute (and again, this is why you don't just want to jump into the math--you are going to get the answer wrong if you just all of a sudden take the negative log of this, like we did the first one): look at this number.*0701

*1.5x10 ^{-10}: that is a really tiny number; compared to the auto-ionization of water, water is 1.0x10^{-7}; 1.0x10^{-7} is three orders of magnitude bigger than 1.5x10^{-10}.*0712

*So, virtually, for all practical purposes, this hydrogen ion concentration that comes from this hydrochloric acid, which is a strong acid, is completely negligible.*0730

*The reason it is negligible is because the concentration of the H is so little.*0741

*Yes, it's a strong acid, so our instinct tells us, "Well, if it's a strong acid, then it must be the dominant force in the solution"; it's not.*0746

*There is so little of it that it doesn't have any effect; so for all practical purposes, the H ^{+} concentration here is what it would be in water at 25 degrees Celsius--which is 10^{-7}.*0753

*Well, the negative log of 10 ^{-7} is 7.*0768

*So, our pH is 7 in this particular case.*0773

*We took a look at the chemistry; we took a look at the numbers involved in the chemistry--the concentrations; that helps us decide what is going to happen next--what math to use.*0777

*Here, .15 is a pretty high concentration; 1.5x10 ^{-10}: that is three orders of magnitude less than the normal concentration of just standard water--the hydrogen ion concentration in standard water.*0787

*We can ignore it, despite the fact that it is a strong acid.*0801

*Don't let names fool you; let the chemistry tell you what is happening--pull back from the problem and think about what is going on.*0804

*Science and chemistry--well, all of science, not just chemistry...there is a lot happening; you have to be able to keep track of multiple things.*0811

*It is not just about solving the problem; solving the problem is easy if you know what is happening.*0820

*If you don't know what is happening, solving the problem will always be a nightmare; you will always be a victim--you will always be sort of a slave to algorithmic procedures.*0826

*And once something--once a problem deviates from that algorithmic procedure, you are going to be lost.*0835

*You don't want to be lost; you want to be able to think--always think; that is the idea.*0840

*OK, so now, let's go on to weak bases; and this is where the mathematics and the situation actually gets interesting.*0844

*We are going to present--we are going to do the same thing: we are going to present a nice, systematic approach for all of these problems.*0852

*Weak acid, weak base; strong acid, strong base; buffer solution; the same steps are the habit we want to get into.*0857

*OK, so let's do Example 2: Calculate the pH of a 1.5 Molar hydrofluoric acid solution.*0865

*So again, we could say hydrogen fluoride solution; it's the same thing--it just means you have taken hydrogen fluoride, and you have dropped it into some water, and you have created a solution.*0888

*Hydrogen fluoride is your solute; water is your solvent; now, it becomes an acid.*0897

*Even though we speak about HF being hydrofluoric acid, it's only an acid when it has been dropped in water; it's very important to remember that.*0902

*OK, the K _{a} for hydrofluoric acid (because it is a weak acid, it has a K_{a}; remember, strong acids don't have K_{a}--the equilibrium lies far to the right, so the denominator is virtually 0) is 7.2x10^{-4}.*0911

*That is a kind of a small number: 7.2x10 ^{-4}; well, what does a small K_{a} mean?*0927

*A small K _{a}--it means that there is very little dissociation--that is the whole idea.*0935

*You need to know what these things mean; they are not just numbers and parentheses and concentrations; it means something, physically.*0941

*It means there is very little dissociation.*0948

*Very little dissociation--that means that HF...the dissociation of HF into H ^{+} + F^{-}...that means mostly, if we looked at this, it's going to be mostly hydrogen fluoride, not hydrogen ion and fluoride ion.*0952

*That is what the small K _{a} means.*0966

*OK, in other words, it is mostly here; the equilibrium is mostly here; in solution, it's still just HF floating around, not H ^{+} and F^{-} floating around freely; so, we will say mostly here.*0969

*Well, the next step is...let's list our major species to see what is going on.*0981

*Let me actually list the major species here: our major species in this solution are going to be...again, because there is very little dissociation, it's a weak acid, you are going to find mostly HF, and you are going to find H _{2}O.*0986

*Both of these are weak acids: HF, as we said, has a K _{a} of 7.2x10^{-4}; well, H_{2}O has a K_{a}, which we call K_{w}; it is 1.0x10^{-14}.*1001

*Well, between 10 ^{-4} and 10^{-14}, this is 10 orders of magnitude bigger; it's huge--this is virtually nonexistent compared to this.*1018

*Because this is so much bigger, we can ignore any contribution that H _{2}O does in its dissociation.*1027

*Its contribution of hydrogen ion is completely negligible--not even measurable, to be honest with you.*1035

*This is going to dominate the equilibrium in the solution.*1040

*So now, we want to see: at equilibrium, when everything has stopped and settled down, what is the pH of that solution?*1043

*Well, here is how we do it: We write down the equilibrium expression, HF goes to H ^{+} + F^{-}; we have I, we have C, and we have E (our initial concentration, change in concentration, and equilibrium concentration).*1050

*Our initial concentration was 1.5 Molar; there is, before the system has come to equilibrium, 0 and 0; so again, imagine, when you are doing these problems--you are taking something; you are dropping it in water; and it is before anything happens.*1065

*Before it starts to come apart, that is the initial conditions.*1084

*Now, HF is going to dissociate a little bit, so that means a little bit of it is going to disappear; and for every amount that disappears, that is how much H and F show up, right?*1088

*1:1, 1:1...when 1 molecule of this breaks up, it produces 1 ion of H ^{+} and 1 ion of F^{-}.*1100

*That is why we have -x, +x, +x, when they put +'s here.*1108

*And now, our equilibrium concentration is just adding these: 1.5-x; when everything has come to a stop, that is how much we have of the HF; that is how much we have of the H; and that is how much we have of the F ^{-}.*1113

*Now, we can write our expression.*1126

*K _{a}, which is equal to 7.2x10^{-4}, is equal to...well, it's equal to the H^{+} concentration, times the F^{-} concentration, over the HF concentration, at equilibrium.*1128

*That equals x, times x, divided by 1.5-x.*1145

*Now, let me rewrite this over here; so let's...part of one equation: 7.2x10 ^{-4}; now, what we want to do...again, we want the pH.*1155

*We want to take hydrogen ion concentration and take the negative log of it; so we are looking for x.*1166

*We want to solve this equation for x.*1172

*Now notice, we are going to end up with a quadratic equation here; there are a couple of ways that we can handle this.*1175

*We can go ahead and (because you have graphing calculators, presumably--most of you) you can go ahead and use your graphing calculator to find the roots of this quadratic equation.*1180

*Or, you could solve the quadratic equation--turn this into a quadratic--multiply through and turn it into a quadratic--use the -b, plus or minus radical b ^{2} minus 4ac over 2a to solve for x.*1189

*Or, we can actually simplify the procedure to make this a little better for us.*1201

*And again, the simplification is based on standing back and taking a look at the chemistry.*1205

*This is a weak acid; there is going to be very little dissociation here.*1210

*Even though there is an equilibrium, there is going to be very little, actually, of the H ^{+} concentration--so little, in fact, that compared to 1.5 Molar--compared to this, this is going to be so tiny that the 1.5 probably won't even notice that it is gone.*1214

*So, we can simplify this expression by writing it as: 7.2x10 ^{-4} equals x^{2} over 1.5.*1230

*Now, when we find the value of x, we need to check to see if our assumption, if our approximation, is valid; and we will show you how to do that in just a minute.*1245

*Let's go ahead and just solve this first, and then we'll check to see if it's valid.*1253

*When we multiply through, we get x ^{2} = 1.08x10^{-3}, and we get x=0.033 moles per liter.*1257

*Well, this is the hydrogen ion concentration, because that was x in our ICE chart.*1270

*Now, we need to check if our approximation--our eliminating x from the denominator here--was valid.*1275

*Here is how you do it: most K _{a}s that are listed in tables have an error of about plus or minus 5%.*1280

*Now, this value, .033, 0.033--if it is 5% or less than the original concentration that you subtracted from...1.5 times 100; you want to calculate a percentage; if this is 5% or less of this number, then our approximation is valid, meaning the difference is not really going to be noticeable--that your numbers are actually pretty good.*1289

*If it is above about 5 or 6 percent, that means you can't make this approximation--you have to solve the quadratic equation.*1322

*So again, we are taking a look at the chemistry, and we are letting the K _{a} tell us that we have a weak acid--very little dissociation.*1329

*So, this x, compared to 1.5, is probably going to be very, very tiny--so much so that we are probably not even going to notice a difference: 1.5, 1.49--it's going to be so small.*1337

*Well, if it falls into the 5% rule, then it's valid; if it doesn't, then it's not valid; so the first thing to do is go ahead and simplify, and check to see if this is the case.*1349

*In this particular case, we get 2.2%; well, 2.2% is definitely less than 5%, which means our approximation is valid, and we can use this value of x; it's perfectly good.*1359

*So, now that we have that, we have our H ^{+} concentration of .033, our pH equals -log of 0.033, is equal to 1.48; there you go.*1372

*Now, I personally (just a little aside) don't care for pH; I think once you have found the hydrogen ion concentration, you are talking about concentration.*1387

*Taking a number and fiddling with it so that it has a more attractive number, like 1.48--I personally don't think 1.48 is more attractive or easier to deal with than .033, or 3.3x10 ^{-2}.*1396

*It's a perfectly valid number; here, we know we're talking about concentration; pH--it's a little weird, because again, the lower the pH, the more acidic.*1409

*If you remember, you have a pH scale that runs from 0 to 14; 7 is neutral--that is the pH of plain water.*1419

*Below 7, we are talking about an acidic solution; above 7, you are talking about a basic solution.*1427

*The lower the number (6, 5, 4, 3, 2, 1)--that is more acidic; what that means is that there is a greater concentration of hydrogen ion.*1434

*It's a little strange: a lower number means more powerful; but again, this has just become part and parcel of the standard chemical practice in the industry, across the board.*1444

*But again, as long as you understand what is happening, that is what is important.*1456

*OK, so let's do another one; this time, let's calculate...let's see; this is going to be Example 3.*1461

*Let's calculate the pH of a 0.15 (this time, 0.15) Molar solution of HClO, hydrogen hypochloride--otherwise known as hypochlorous acid.*1472

*The K _{a} for this equals 3.5x10^{-8}.*1501

*The first you want to notice is: look at the K _{a}; this is a very, very small number--that means we are talking about a very weak acid, very little dissociation.*1507

*When you take hydrogen hypochloride and drop it into water, it's going to stay hydrogen hypochloride; it's not going to come apart into hydrogen ion and hypochloride ion too much.*1515

*A little bit, it will; but not too much.*1526

*The fact that we have a number that we can measure means there is some dissociation; but again, this says that it is a weak dissociation.*1529

*OK, so now let's see which equilibrium is actually going to dominate in this.*1536

*We have two sources of hydrogen ion: we have the HClO, which breaks up into H ^{+} + ClO^{-}, potentially; and the K_{a} for this, as we just said, is 3.5x10^{-8}.*1542

*Then, we have H _{2}O; that is the major species in the water; that is H^{+} + OH^{-}, but again, the K_{w} for this is 1.0x10 to the (oops, not 14) negative 14.*1558

*Well, 10 to the -8--even though it is a small number, it is a lot bigger than 10 to the -14; 6 orders of magnitude bigger.*1577

*The dominant reaction here, the dominant equilibrium, is this one.*1584

*This is virtually not even noticeable.*1589

*So, let's go ahead and do the problem.*1596

*Since the HClO equilibrium is going to dominate, we are going to write HClO, and I just am one of those people that likes to write everything--so, H ^{+} + ClO^{-} (I know that it's a little...I'm just writing it over here, but I just like to make sure that everything is clear).*1598

*This is initial, change, equilibrium (and let's erase some of these stray lines here); so, our initial concentration of HClO, before the system has come to equilibrium--when we just drop it in before it has come to equilibrium--this is 0.15 Molar.*1619

*There is no hydrogen ion, and there is no hypochloride ion.*1635

*Well, a certain amount of it dissociates; and it produces, for every mole that (we're not going to have that; we want this to be clean; +x) dissociates, it produces a mole of H ^{+}; it produces a mole of ClO^{-}.*1639

*Equilibrium concentration is 0.15-x; this is +x; this is +x; and now, we can go ahead and write our equilibrium.*1658

*Our K _{a} is our hydrogen ion concentration, times our hypochloride ion concentration, over our hydrogen hypochloride.*1671

*When we do this, in terms of numbers, we get: 3.5x10 ^{-8}, which is our K_{a}; it is equal to x times x, over 0.15-x.*1681

*And again, we want to simplify, so this is going to be approximately equal to x squared, over 0.15.*1697

*Then, we'll check to see if that approximation is actually valid.*1705

*When we solve this, we get x ^{2} is equal to 5.25x10^{-9}, and we get that the x, which is equal to the hydrogen ion concentration, is equal to 7.24x10^{-5}.*1709

*Now, let's check the validity; so I'll write "check the validity of our approximation"--in other words, what we did here.*1728

*We'll take 7.24x10 ^{-5}, and we'll divide it by the initial concentration, 0.15; so, when we do that and multiply by 100, it equals (let me drop it down a little bit here) 0.048%.*1735

*Wow, look at that; that means only 0.048% is actually dissociated; that is a very, very, very weak acid--virtually none of it has come apart.*1753

*This is clearly below 5%; our approximation is good; so we can use this number, and when we take the negative log of 7.24x10 ^{-5}, we end up with (let's see what we ended up with here) 4.14.*1765

*4.14--and this our pH, and this is our hydrogen ion concentration.*1788

*Again, my preference is for the actual concentration; pH is just a number.*1794

*OK, that is good; so hopefully, you are getting a sense of what it is that we are doing; we are looking at K _{a}; we're choosing the major species; we're deciding which of those major species is going to dominate in the solution.*1800

*We know what chemistry is going to dominate; usually there is one species, one equilibrium, that dominates--everything else can be ignored.*1812

*And then, we just write out our ICE chart: Initial concentration, the Change equilibrium, and then we set it equal to the equilibrium constant, and we just deal with the math at that point; it's a simple algebra problem.*1820

*OK, now let's do a problem where we have a mixture of weak acids.*1831

*Same exact thing: it's going to be more species in water, but one of them is going to dominate; so let's do this.*1834

*Example 4: Calculate the pH of a solution that contains 1.2 moles per liter of hydrogen cyanide, or hydrocyanic acid, whose K _{a} is 6.2x10^{-10}--very, very weak--and 4.0 Molar HNO_{2} (hydrogen nitrite or nitrous acid), whose K_{a} is equal to 4.0x10^{-4}.*1843

*Also, calculate the concentration of the cyanide ion at equilibrium.*1902

*OK, so we want to find the pH; so we have a mixture--we have this certain amount of water; we drop in some 4-Molar hydrogen nitrite, or nitrous acid; we drop in 1.2-Molar hydrocyanic acid; we want to know what the pH of the solution is, and we want to know what the concentration of the free cyanide ion is in that solution.*1925

*OK, let's list our major species with their particular equilibriums.*1949

*Well, let me go back to blue here; so, major species in water.*1955

*That is what we are doing; we always want to do the major species to decide what chemistry is going to dominate.*1963

*Well, HCn is a weak acid; that means it is mostly HCn--it hasn't dissociated.*1967

*A strong acid dissociates--weak acids, not very much.*1973

*We also have HNO _{2} floating around in that, and we have H_{2}O.*1977

*Well, the HCn equilibrium (H ^{+} + Cn^{-})--we said that the K_{a} is equal to 6.2x10^{-10}.*1983

*The (oops, wow, look at that; that is a crazy line; OK) HNO _{2} equilibrium (NO_{2}^{-})--the K_{a} for this one is equal to 4.0x10^{-4}.*1996

*And of course, we have the (oh, here we go again; wow)...last but not least, we have the H _{2}O equilibrium, which is also another source of hydrogen ion, plus OH^{-}; the K_{w} equals 1.0x10^{-14}.*2022

*So now, let's compare: 10 ^{-10}; 10^{-4}; 10^{-14}.*2042

*We can virtually ignore the 10 ^{-14}--it's too tiny.*2047

*Between 10 ^{-4} and 10^{-10}, this one--the HNO_{2}; this is going to dominate the chemistry in this solution.*2050

*Because that is going to dominate, we can ignore any contribution of hydrogen ion from hydrogen cyanide, and we can ignore any contribution of H ^{+} from water.*2058

*And again, we are doing this because there are three sources of hydrogen ion: some can come from the HCn; some can come from HNO _{2}; some can come from H_{2}O.*2069

*But, the K _{a} of the HCn and the H_{2}O are so tiny that they are negligible, so this is what controls the chemistry of the solution; that is the chemistry we concentrate on.*2078

*Therefore, let's go with HNO _{2}: H^{+} + NO_{2}^{-}; now, you are going to do the same thing that you did before.*2089

*You are going to make a little ICE chart: Initial, Change, Equilibrium; you are going to do your simplification; you are going to check it; everything is going to be fine; you are going to end up with a pH of...*2103

*I hope you actually run through and do this, based on the previous two examples.*2113

*So, for this one, you are going to end up with a pH of 1.40.*2118

*Now, the question is (the second part): How do we find the cyanide ion concentration?*2126

*OK, let's write down the cyanide equilibrium; we need to find the cyanide ion concentration, so we need to work with the cyanide equilibrium.*2132

*HCn goes to H ^{+} + Cn^{-}.*2141

*K _{a} equals 6.2x10^{-10}.*2147

*Now, let's stop and think about what this means.*2152

*This is saying that, in a given solution, where you have cyanide ion, hydrogen ion, and HCn (hydrogen cyanide) floating around in solution--these three species floating around in some concentration each--the equilibrium concentration (in other words, the concentration of this, times the concentration of that, divided by the concentration of this) equals 6.2x10 ^{-10}.*2156

*Remember, that is what an equilibrium expression is: equilibrium positions can change, but the constant stays the same--the relationship among these three--that is what equals this.*2184

*It doesn't matter where these H ^{+} come from; this is talking about: at any given moment, if you have cyanide, hydrogen ion, and hydrocyanic acid in solution, the concentration of this times that, divided by the concentration of this, always equals that.*2196

*This gives us a way to find it; it doesn't matter where these H ^{+} come from--they can come from HCn, or they can come from any other source, any other source.*2215

*In this case, the primary source of the hydrogen ion is the nitrous acid.*2226

*It is the dominant species in the water; it is the one that is going to give up its hydrogen ion and suppress the others.*2232

*So, at equilibrium, there is a certain concentration of hydrogen ion, and that concentration of hydrogen ion was the 1.4--well, the pH was 1.4; the 1.4 came from the negative log of the hydrogen ion concentration.*2239

*Now, let's go ahead and work this out.*2257

*This says...so we know that the K _{a} is equal to H^{+} concentration, times Cn^{-} concentration, over HCn concentration.*2263

*That equals 6.2x10 ^{-8}; this is the equilibrium expression.*2278

*Well, let's see: let's go ahead...so at equilibrium, we have...not 6.2x10 ^{-8}; we have 6.2x10^{-10}; yes, that is 10.*2285

*That means 6.2x10 ^{-10}; now, we go back on our ICE chart, and we read off this, this, and this.*2300

*We ended up with: our equilibrium concentration is 0.04.*2310

*That was the hydrogen ion concentration that gave us the pH of 1.4.*2321

*We have the Cn ^{-} concentration, and, at equilibrium, the HCn concentration was 1.2-x.*2325

*OK, again, we are just dealing with a basic math problem, but as it turns out, we have a little bit of a problem; we have this x here...smaller than normal...suppresses the HCn dissociation...let's see here.*2335

*Let me see, how do I want to go ahead and explain this to make it most reasonable?*2358

*We have that; we have that; we have the cyanide concentration; OK.*2364

*Let me erase something here.*2369

*OK, so let me rewrite this here: HCn, H ^{+} + Cn^{-}.*2378

*Now, at equilibrium, we started off with 1.2 Molar of the HCn, right?*2389

*This was 1.2 Molar.*2404

*Now, we have the H ^{+} concentration, which is 0.04; this is the equilibrium concentration.*2407

*This is what we want--this Cn ^{-} concentration.*2412

*Now, it's true that, of the HCn, some of it will dissociate; so the fact of the matter is, the equilibrium concentration is going to be 1.2-x, but again, remember what we said: the dominant equilibrium here was the nitrous acid.*2417

*It is so dominant that it is actually going to suppress this dissociation.*2437

*Anything that HCn would actually produce, any H ^{+}, is virtually nonexistent, because not only was it small to begin with, but because there is so much of the H^{+} from the nitrous acid, it's actually going to push this even more that way, so there is going to be even less.*2442

*So, for all practical purposes, we don't even have to worry about any HCn dissociating.*2458

*Therefore, our final equilibrium will end up actually being: 6.2x10 ^{-10} is equal to 0.04 (which was the hydrogen ion concentration we found from the HNO_{2} dissociation), times the Cn^{-} concentration, over the HCn concentration (which was 1.2 Molar).*2464

*And then, when we solve for this, we end up getting a Cn ^{-} concentration (and this time, I will use the square brackets): 1.86x10^{-8} Molar.*2490

*As you can see, 1.86x10 ^{-8} Molar is so tiny--that means virtually...there is virtually no cyanide ion floating around.*2504

*That is based on this equilibrium.*2515

*So again, if you have a mixture of acids, one of those acids will dominate the equilibrium.*2518

*If you are asked for the concentration of a species from the other acid, the one that is not dominant, you can still use...you just write out its equilibrium, and it just means that--in this particular case, we wanted cyanide; well, the equilibrium expression for HCn says that cyanide concentration, times H ^{+}, divided by this, equals this.*2524

*Well, we have had this from what we just calculated from the dominant species; we have this, which is the original concentration.*2546

*And then, all we have to do is solve for this, because we already have this.*2555

*1, 2, 3, 4; we have 1, 2, 3 of them; we solve for the fourth.*2559

*There you go; I hope that made sense.*2564

*OK, let's see: we have taken a look at some weak acid problems; we have listed the major species; we take a look at the major species to decide which one is dominant.*2569

*And usually, in this particular case (for weak acids), we just see which one has the higher K _{a} value.*2582

*Whichever one has the higher K _{a} value, that equilibrium, that acid, will dominate the chemistry of the solution.*2588

*We write its equation; we list its initial concentration, the change in the concentration, and the equilibrium concentration; and then, we put the equilibrium concentrations, in terms of x, into the equilibrium expression, and we solve for x.*2595

*More often than not, they are going to ask for calculating the hydrogen ion concentration or, actually, the pH.*2610

*More often than not, they will just straight-out ask for the pH.*2616

*So, thank you for joining us here at Educator.com and AP Chemistry for weak acids.*2619

*Our next topic we are going to be discussing is going to be percent dissociation; we are going to spend a fair amount of time on weak bases.*2624

*Take care; goodbye.*2631

1 answer

Last reply by: Magic Fu

Mon Apr 17, 2017 9:49 AM

Post by Magic Fu on April 17, 2017

Hi,

Since HCl , HBr and HI are all considered as strong acids then why HF is classified as a weak acid?

Thanks

1 answer

Last reply by: Professor Hovasapian

Tue Feb 9, 2016 3:23 AM

Post by RHS STUDENT on February 8, 2016

Sir, is using the ICE chart legit in the AP exam. And also do I have to plug in the unit of some constant when asking short/long questions? Thanks

3 answers

Last reply by: Professor Hovasapian

Sat Jan 10, 2015 7:23 PM

Post by Stephen Donovan on January 9, 2015

In example 4, wouldn't the two non-dominant equations buffer the dominant one? Or is that not possible due to a lack of the cyanide and nitrite ions?

1 answer

Last reply by: Professor Hovasapian

Thu Jan 8, 2015 2:08 AM

Post by Delaney Kranz on January 7, 2015

How can you tell when it is a weak acid or a strong acid when you are only given the molarity?

1 answer

Last reply by: Professor Hovasapian

Thu Feb 27, 2014 7:25 PM

Post by Samiha Bushra on February 26, 2014

I just wanted to thank you. I think I aced my quiz because of you. This is coming from a girl that failed all three of her last quizes and had a not so great background on chemistry because I took it in summer school.

2 answers

Last reply by: William Dawson

Tue Dec 10, 2013 9:09 PM

Post by William Dawson on December 8, 2013

You do a wonderful job explaining this material. It helps tremendously to point out all the easy traps and pitfalls that typically derail students when attempting to organize these problems so that they can be comnuted. In general, however, it is better to make each problem a separate example , unless they are somehow dependent on each other. Making 1 example into 3 problems is no good. My chem prof does this and I get furious at him for it!

1 answer

Last reply by: Professor Hovasapian

Mon May 13, 2013 7:56 PM

Post by KyungYeop Kim on May 13, 2013

You said the Ka,3.5x10^8 , is too small and it means weak dissociation. How do you determine whether Ka is small or big? is there a certain number based on which to decide?

Thanks always.

15 answers

Last reply by: Professor Hovasapian

Fri May 3, 2013 9:24 PM

Post by Louise Barrea on April 22, 2013

Hi, I know this is probably not in the right topic, but there is a question I can't solve. I have to calculate the pH of a solution NH4CH3COO at 0,01 M with pK(NH4+)=9,25 and pK(CH3COOH)=4,75.

When I did the major species, I found: NH4+ CH3COO- H20

I tried to determine the major reaction looking at the pK, I found: CH3COO- + H20 = CH3COOH + OH

then I calculated pOH and did pH= 14-pOH

I still can't find the right answer. What did i do wrong?

Thanks

1 answer

Last reply by: Ali Hashemi

Sun Dec 9, 2012 8:53 PM

Post by Ali Hashemi on December 9, 2012

What if the Ka values were closer in example 4, could you figure out the equilibrium concentrations for both reactions and add them together then figure out pH using the sum of the Hydrogen ion concentrations?

0 answers

Post by nazgol farrokhseyr on May 2, 2012

you are the best.You made chemistry much easier for me.Thank you so much