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

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

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### Percent Dissociation: Strong & Weak Bases

- Strong Bases are salts that dissociate completely and produce Hydroxide upon dissociation.
- Weak bases react with water to abstract a Hydrogen Ion, thus producing Hydroxide Ion.
- Again, in A/B reactions, Hydrogen Ion is the only thing that moves and switches partners.

### Percent Dissociation: Strong & Weak Bases

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
- Bases 0:33
- Percent Dissociation: Strong & Weak Bases
- Example 1
- Strong Base Dissociation
- Example 2
- Weak Acid and General Reaction
- Example: NaOH → Na⁺ + OH⁻
- Strong Base and Weak Base
- Example 4
- Example 5

### AP Chemistry Online Prep Course

### Transcription: Percent Dissociation: Strong & Weak Bases

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

*Today, we are going to continue our discussion of acids and bases--a profoundly important topic.*0004

*We are going to be talking a little bit about percent dissociation, sort of a continuation of our weak acid topic from last lesson.*0009

*But mostly, we are going to be talking about strong and weak bases; and, as you are going to see, how you handle the calculations is exactly like we did for weak acids.*0017

*Anyway, let's just jump in and talk about percent dissociation, and move forward and see what we can do.*0027

*So again, the idea is to do a fair number of problems to get you comfortable with what is going on--comfortable with the chemistry--how to handle this intuitively and turn that intuition into mathematics.*0033

*OK, so weak acids: remember, we said that a weak acid is some species, like for example, let's say, hydrogen fluoride (which was an example that we did in the last lesson): that dissociates a little bit into H ^{+} and F^{-}.*0045

*Now, not too much; again, that is the whole idea behind a weak acid.*0062

*Well, there is this notion called percent dissociation; we want to know how much of the original HF actually came apart--how much of it dissociated.*0065

*Well, a percent is always the same thing; a percent is always the part over the whole, times 100.*0075

*So, by definition, our percent dissociation is equal to the amount dissociated (and the amount dissociated is always this amount, or this amount, because again, if .2 moles dissociates of this, well, .2 moles is produced of that and .2 moles is produced of that...so it's always the amount dissociated), divided by the initial amount (or the initial concentration--either way--the initial concentration); it's the part over the whole, times 100.*0080

*That is it; a percent dissociation--you are just measuring: "To what extent has this weak acid actually come apart? 10%? 5%? 3%? 2%?"*0121

*Strong acids--100% dissociation; percent dissociation for an acid, a strong acid, is 100%.*0132

*You are going to find, in general, for weak acids, anywhere from about .5% up to maybe about 5 or 6% for a weak acid.*0138

*Like, for example, your vinegar solution: you are talking about acetic acid, which is maybe 2% dissociated; it's very little.*0146

*It is actually kind of interesting--with such little dissociation, and yet, it has that really, really strong, strong acidic quality to it.*0156

*OK, as a quick example, let's think about the last example that we did when we talked about the hydrogen fluoride.*0164

*Remember, we found that the hydrogen ion concentration for the fluoride was 0.033 Molar, in the last lesson.*0173

*Well, the initial hydrogen fluoride concentration that we started with (so initial is a little 0 down at the bottom)--it was 1.5 Molar.*0183

*Therefore, our percent dissociation is equal to 0.033 over 1.5, times 100%; you ended up with 2.2 percent.*0193

*That means, when we stuck that 1.5 Molar hydrogen fluoride solution...well, when we have that 1.5 Molar hydrogen fluoride solution...only 2.2% of the hydrogen fluoride has dissociated into H ^{+} and F^{-}.*0207

*That means the other 97.8% is still hydrogen fluoride, floating around in solution--pure hydrogen fluoride, not hydrofluoric acid.*0225

*Acid is when it is this way; but again, we just sort of have become accustomed to saying "hydrofluoric acid," but it's important to distinguish.*0234

*When it is together like this, yes, it is an acid, because it is in water--but it is not dissociated; it is actually still together.*0242

*Only 2.2% of the original amount has dissociated; that is it--it's all this is.*0249

*OK, let's list a general trend, actually--a general trend with percent dissociation.*0256

*The more dilute the solution, the greater the percent dissociation.*0269

*It is a nice little thing just to sort of keep in the back of your mind: in other words, if I have, let's say, a 1.0 Molar solution of HF, it's going to be some percentage.*0287

*Well, if I have a .5 Molar solution of HF, which is more dilute (right?--lower the concentration), the percent dissociation is actually going to be higher.*0299

*If I have a 0.1 Molar HF solution, the percent dissociation is going to be even higher.*0309

*That doesn't mean that the pH is going to be lower; that doesn't mean that the concentration of the hydrogen ion is going to be higher.*0317

*As you have lower and lower concentrations, you are actually going to end up with less concentration of H ^{+}; it just means that there is a greater dissociation, that more of the original acid has actually completely come apart into free H^{+} and free conjugate base ion.*0324

*That is what that means; be very, very certain to distinguish between the two.*0341

*Lower concentration, more dilute solution, gives a greater percent dissociation.*0346

*The percent dissociation is not a measure of acidity; pH is a measure of acidity (in other words, how much hydrogen ion is actually floating around freely).*0356

*Remember, we said: when we talk about an acid, we are talking about free H ^{+} floating around in solution; more H^{+}, more acidic, more damage.*0366

*That is it; so let's use this idea of percent dissociation to actually calculate a K _{a} (an equilibrium constant).*0375

*Our first example is going to be: In a 0.100 Molar lactic acid solution, the lactic acid is 3.8% dissociated.*0384

*OK, and for those of you who are biology majors and biochemists, lactic acid is the product that is produced by your cells under anaerobic conditions (under anaerobic respiration).*0423

*When you start to get really, really fatigued, that is the lactic acid building up in your muscles, until your body can get enough oxygen going to actually start burning the sugar aerobically (meaning with oxygen), as opposed to anaerobically (without oxygen).*0435

*So, dissociated...Calculate the K _{a} for this acid.*0449

*Calculate the equilibrium constant for this acid.*0456

*OK, so let's go ahead and write down an equilibrium expression.*0466

*I'm just going to write HL for lactic acid; actually, you know what--I think I'll write HLac, is in equilibrium with H ^{+} + lactate ion.*0471

*Well, they tell me that the initial concentration is 0.100 Molar, and we have none of that; the change was -x, so it becomes x and x.*0484

*That means the equilibrium concentration is .100-x, x, and x.*0496

*Well, the K _{a} for this is equal to...well, sure enough, the same equilibrium expression: the hydrogen ion concentration, times the lactate ion concentration, times the concentration of lactic acid (or hydrogen lactate).*0502

*Now, we are looking, this time, for K _{a}; that is what we want to find, which means that we need this number, this number, and this number in order to multiply and divide--which means we need x.*0516

*Once we have x, we have this and this, and we can subtract from .1 to get this; we multiply and divide, and we get our K _{a}.*0527

*How can we find x?--well, they tell us that the percent dissociation is 3.8 (is it 3.8?)...yes, 3.8%.*0535

*What does that mean?--that means that this number, the amount that is dissociated (which is the same as this number, x), divided by the original .100, times 100, equals 3.8.*0545

*So, we have a way to find x; and now, we just go ahead and solve it.*0561

*We get: x is equal to 0.0038; there you go--nice and simple!*0565

*Now, we have our values; now, the hydrogen ion concentration (which is this one) is going to be 0.0038; the lactate ion concentration is 0.0038 (that is what this is, right?--they are the same; for every mole of this broken up, it produces a mole of this and produces a mole of that); and our HLac concentration is going to equal the original .100, minus 0.0038.*0574

*When we put all of these values in, we get: K _{a} is equal to 0.0038, times 0.0038, divided by 0.0962 (that is this one....962...).*0613

*And, when we run this number, we get 1.5x10 ^{-4}.*0639

*So again, if you understand the chemistry, you should be able to do the math.*0645

*We have this; we set up the ICE chart; we knew that we needed to find x.*0651

*But, they gave us x already--they gave us a way to find x; they told us the percent dissociation.*0656

*That means the amount that was dissociated; well, the amount dissociated is the amount produced of the H ^{+}, the amount produced of the lactate, divided by the initial amount, which was .1.*0661

*So, this over that or this over that, times 100, is 3.8; we have x; we plug it in; and we get our K _{a}.*0671

*This is a common problem; you may actually run across it in your AP exam.*0679

*OK, now let's go on to discuss bases.*0684

*All right, now, the base is the opposite of an acid.*0688

*An acid is something that has hydrogens that it wants to give away; a strong acid--it gives up all of its Hs (in other words, it comes apart completely); a weak acid is one that sort of doesn't really give up its Hs too easily--it actually holds onto them.*0693

*A base is the opposite; a base is something that actually...strong base: let's actually define a strong base.*0707

*A strong base--it fully dissociates to produce OH ^{-}--so, for example, sodium hydroxide.*0716

*Sodium hydroxide: when you take solid sodium hydroxide (it's kind of like hard rocks) and drop it in water, it dissolves; it dissociates.*0731

*It produces free sodium ion and free hydroxide ion.*0741

*It is a basic solution; it is a base, because when it dissociates, it produces OH ^{-}.*0745

*Remember, water is also a base; water breaks up into hydrogen ion and OH ^{-}; it's amphoteric--it's an acid and a base.*0751

*This is just a base.*0758

*Potassium hydroxide, another strong base: notice, I wrote the arrow in one direction--full dissociation.*0760

*K ^{+}, OH^{-}: if you had a solution of sodium or potassium hydroxide, you wouldn't find any NaOH, any KOH, in solution; it would all be this, this, this, this--free ions in solution.*0766

*Let's do a quick example.*0783

*Handle it the exact same way, except now, in reverse.*0787

*Calculate the pH of a 4.0x10 ^{-2} Molar NaOH solution.*0791

*Well, let's see: NaOH is a strong base; any of your alkali metals (sodium, potassium, all of those in the first group)--with a hydroxide, they are all strong bases; they all fully dissociate.*0808

*"Strong base" means full dissociation.*0822

*OK, so the major species in water: again, nothing new--we handle it the same way.*0829

*Take a look at what it is: it's a strong base; there is going to be full dissociation; now, let's see what is floating around in water to decide what is going to dominate the equilibrium.*0834

*The major species in water floating around: you have sodium ion; you have hydroxide ion; and you have H _{2}O.*0844

*Well, we know that H _{2}O is also a source of hydroxide ion, right?--because H_{2}O dissociates into H^{+} + OH^{-}.*0853

*But, this is 10 ^{-7}; let's write the K_{a} for this--the K_{w} is 10^{-14} (1x10^{-14}; I just ignored the 1).*0863

*Well, this is a strong base, and this is 4.0x10 ^{-2} Molar; that means 4.0x10^{-2} Molar sodium hydroxide has fully dissociated, and has produced 4.0x10^{-2} moles of hydroxide.*0876

*Well, 10 ^{-2}; 10^{-14}; a huge difference--that is 12 orders of magnitude--so this is virtually...you can ignore it.*0894

*The species that is going to dominate the chemistry is this; the sodium is not going to do anything at all--it just sits there like a spectator ion.*0905

*So, our pOH (or actually, let me)...our OH ^{-} concentration is equal to 4.0x10^{-2}, because we have full dissociation.*0913

*In other words, NaOH goes to Na ^{+} + OH^{-}, in case you are not sure what I am talking about.*0930

*It starts off with 4.0x10 ^{-2} Molar--full dissociation.*0941

*This is 00 minus 4.0x10 ^{-2}; all of it dissociates--this ends up being 0.*0946

*This ends up being 4.0x10 ^{-2}, 4.0x10^{-2}.*0954

*That is what we mean by the OH ^{-} concentration, is that; because all of it is gone away--there is no more of that left in solution.*0961

*It is that; now, let's calculate the pOH; remember, p is just a function--it means the negative log of something.*0970

*Negative log of 4.0x10 ^{-2}; we end up with...actually, you know what, I did this a little differently.*0976

*I did it this way: OH ^{-} equals that; well, what do we know about the hydroxide ion concentration and the hydrogen ion concentration in any aqueous solution?*0988

*They multiply to 10 ^{-14}.*1000

*So, OH ^{-} times H^{+} equals 10^{-14}.*1002

*We are looking for pH, not pOH; so the H ^{+} concentration is 10^{-14}, divided by 4.0x10^{-2}, equals 2.5x10^{-13}.*1013

*And then, we will take the pH, equals the negative log of that number (2.5x10 ^{-13}), and we get 12.6--basic solution.*1030

*Remember what we said: if the pH is above 7 and below 14, or just above, you get a basic solution.*1042

*This confirms the fact that this is a basic solution--the pH.*1049

*pH is the standard by which we decide.*1053

*OK, now let's talk about weak bases.*1059

*Weak base--all right, the general reaction for a weak acid, we said, was this: we said, if we had an acid, plus water, we'll go to hydronium ion (which is the same as H ^{+}), plus the conjugate base of the weak acid.*1065

*Now, the weak base general reaction is this (the general reaction; I'll just use B for base): it is: the base, plus water, goes to BH ^{+} + OH^{-}.*1095

*In other words, let me rewrite this another way: B, plus (let me write H _{2}O as HOH)--what is happening is that the base actually takes a hydrogen ion away from water.*1120

*But, it doesn't take it as a hydrogen atom; it takes just the hydrogen ion.*1135

*Hydrogen leaves its electron with the hydroxide, which is why you end up with BH ^{+} + OH^{-}.*1139

*I want to show you why this happens.*1148

*Most bases, weak bases--they have a lone pair of electrons, and we'll do an example in a minute.*1151

*H is here; this is usually not something that you are going to do until Organic Chemistry, but I want you to see what is happening, because I want it to make sense to you.*1157

*These electrons--they reach out and they actually grab the H; they rip it away from the water molecule.*1165

*When these electrons move in, these electrons move out, OK?*1173

*H cannot be attached to two simultaneously (well, it can for hydrogen bonding; but for our purposes); this base wants this, so when it takes this, it goes this way, and it kicks these electrons; they move onto the hydroxide, and what you end up with is this.*1180

*Because this is only coming as a proton, minus its electron--it's coming as an H ^{+}--now this whole species has a plus charge, and you have an OH^{-}.*1201

*This general reaction is the reaction of a weak base with water; this is very important reaction--it always happens like this.*1211

*We said that a base is something that produces hydroxide ion; well, here you go.*1221

*A base doesn't necessarily have to have hydroxide ion in it; so, for example, you know that sodium hydroxide produces hydroxide by dissociation.*1226

*Well, an example like ammonia--well, let's actually use a specific example.*1236

*NH _{3}, which has a lone pair, plus HOH (I'm going to write it as HOH, water) goes to NH_{4}^{+} + OH^{-}.*1245

*OH ^{-} is still produced in solution, but the OH^{-} doesn't come from the base itself; it is the base that reacts with water.*1259

*Water is now acting as the acid; it is giving up its H; this ammonia is acting as the base--it pulls the H off.*1268

*Now, it's NH _{4}^{+} and it's OH^{-}; so it's producing OH^{-} in a roundabout way--not by dissociation--it's doing it by breaking up the water.*1274

*This is...and there is, of course, a K _{b} associated with this.*1286

*It is an equilibrium constant expression for the reaction of a base (or something, some species) that will extract a hydrogen ion from water to produce the conjugate acid of this base and hydroxide ion.*1294

*The K _{b} is exactly what you think it is; this is liquid; this is aq; this is aq; and this is aq; it is equal to NH_{4}^{+} times OH^{-}, over NH_{3}.*1311

*That is it; this is the generic expression for a weak base; it is the base, plus the water, plus the BH ^{+}, plus OH^{-}.*1332

*It is taking the hydrogen from water, leaving the hydroxide ion.*1342

*In the process, it produces the hydroxide ion.*1346

*It produces the hydroxide ion; the solution becomes basic; that is why it is called a base.*1351

*It is also called a base because it actually takes the hydrogen ion from something (in this case, water).*1356

*It is acting like the base; water is acting like the acid.*1363

*In the weak acid equilibrium, this is the acid; this is the base; now, these are the lone pair of electrons that are taking this hydrogen and producing that as a conjugate.*1367

*Back and forth: it's just a competition between two things for the hydrogen ion.*1379

*Here, the base...it just depends on what the equilibrium is; that is all acid-base chemistry is--it's a competition between two different species for the hydrogen ion in between.*1385

*It is a tennis game between the two: which one is stronger?--the stronger one will take the hydrogen ion.*1393

*OK, so again, a base--whether weak or strong--produces OH ^{-}.*1401

*This is the take-home lesson: a weak base, strong base...they all produce OH ^{-}.*1418

*A strong base does it by dissociation; a weak base does it by hydrogen abstraction from water.*1423

*OK, so now, let's do a couple of examples here.*1430

*Oh, let me just give a couple of versions: Strong base--we already mentioned one, potassium hydroxide; it dissociates into potassium ion plus OH ^{-}; here is your OH^{-}.*1436

*A weak base--I'm going to use something called pyridine, and it is a molecule that looks like this, and it reacts with water.*1450

*Let me actually write it as HOH; you will find it very, very convenient to write water as HOH--I certainly prefer to do so, although a lot of people censure me for it--I don't know why.*1462

*It becomes N + OH ^{-}; again, here is your OH^{-}; it happens in a roundabout way.*1475

*Here it comes apart; here it takes hydrogen from the water to produce the hydroxide.*1488

*Let's do an example.*1494

*All of it will make sense as we sort of do more examples: OK.*1498

*Let's see: what can we do?*1503

*Oh, calculate the pH of a 13.0 Molar NH _{3} solution.*1508

*The K _{b} is equal to 1.8x10^{-5}.*1526

*OK, so here, because we are dealing with a weak base, we are producing hydroxide ion concentration.*1533

*The x-value that we found is actually going to be hydroxide ion concentration; we are going to have to use the relationship of hydroxide, times hydrogen ion concentration, equals 10 ^{-14} to find the hydrogen ion concentration, and then take the negative log of that.*1540

*If we just take the negative log of the hydroxide ion concentration we find for a weak base problem, we are not going to get the pH; we are going to get the pOH.*1555

*Now, you can do that; that is fine, as long as you take the pOH and subtract it from 14, and that will give you the pH; because again, we have chosen hydrogen ion concentration as the standard--pH as the standard.*1563

*Well, we can calculate the p of anything.*1576

*OK, so let's see what we have: well, let's take a look at the major species.*1580

*Again, same procedure: major species: we have NH _{3} (weak base--it's a weak base, which means that it is not dissociated--it stays mostly NH_{3}), and its K_{a} is 1.8x10^{-5}.*1586

*The other species is water, because water is also a source of hydroxide.*1607

*But, its K _{a} (which is K_{w}) is equal to 1.0x10^{-14}; -5; -14; I think I'll go with the -5.*1612

*We can ignore this one; this is going to dominate the equilibrium--this is going to dominate the solution.*1622

*The hydroxide ion in this solution is going to come mostly from the chemistry of ammonia.*1629

*So, let's write it out: NH _{3} + HOH (same equation over and over again: base plus water goes to conjugate acid plus hydroxide; "conjugate acid": conjugate acid just means stick an H on top of it--attach an H to it and put a plus sign on it) + OH^{-}; the Initial; the Change; the Equilibrium.*1635

*I know you are going to get sick of these ICE charts.*1658

*Let me see where are we (where are we, where are we, where are we); oh, there we are: 13 Molar; this is before anything happens--initial means before the system comes to equilibrium.*1662

*Before this takes place, water doesn't matter; there is no ammonium, and there is no hydroxide yet.*1672

*A certain amount of ammonia disappears; OK, now this is kind of interesting.*1680

*NH _{3} is not coming apart like an acid is, but we put a -x because NH_{3}, as ammonia, is disappearing.*1686

*What is forming is ammonium; does that make sense?*1696

*Before, when we had H, like HF, and it dissociated, we knew it was -x; and it produced x amount of H and x amount of F ^{-}.*1700

*But here, it is still -x, even though this NH _{3} is actually taking something from the water and becoming more.*1715

*It is becoming NH _{3} to NH_{4}^{+}.*1724

*But what is happening is that NH _{3} is disappearing as a species; NH_{4}^{+} is showing up; OH^{-} is showing up.*1726

*That is the idea; you have to get your mind around the physical reality of it.*1735

*This doesn't matter; x is showing up; hydroxide is showing up; our equilibrium concentration is going to be 13...actually, this is 13.0, if I am not mistaken.*1740

*Yes, I think it's three significant figures; sorry about that: 13.0, 13.0-x; that doesn't matter; let's make sure this is clear; that doesn't matter; that doesn't matter; that doesn't matter; this is +x, +x; handle it the same exact way.*1753

*So, we have: K _{b} is equal to the ammonium ion concentration, times the hydroxide ion concentration, divided by the ammonia concentration.*1773

*Let's put some numbers on this: 1.8x10 ^{-5} is equal to x, times x, divided by 13.0-x.*1788

*Well, again, because we are talking about a weak base (1.8x10 ^{-5}), x is probably going to be pretty small compared to the 13.0.*1800

*Let's just ignore it for the time being, and we will check the validity of our approximation in a minute; and just leave it as 13.0.*1812

*x squared is equal to 2.34x10 ^{-4}, and x is equal to 0.0153 Molar, which equals the hydroxide ion concentration.*1820

*Well, the pOH equals the negative log of the hydroxide ion concentration, equals negative log of 0.0153, is equal to 1.82; and pH + pOH is equal to 14; pH plus 1.82 is equal to 14, so our big fat pH is 12.18 (bigger than 7--a lot bigger than 7).*1838

*This is a basic solution; it's confirmed.*1876

*There you go; that is it--handle it the exact same way as an acid, except it's different; it's a base--it's a weak base.*1882

*Instead of a K _{a}, we have a K_{b}; nothing is different; everything is exactly the same.*1891

*Initial, Change, Equilibrium; decide what the major species are; decide which species is going to dominate the solution--which is going to produce, in this particular case, the most hydroxide ion.*1896

*It is the weak base; it's weak, but it is still stronger than the water.*1907

*OK, now let's do the percent; I'll put in quotes "dissociation."*1913

*Well, percent dissociation works for acid, because an acid is going from HA; it is actually dissociating into H ^{+} + A^{-}.*1920

*A base, like NH _{3}, is actually becoming BH^{+} + OH^{-}...so we say "percent dissociation," but what is more appropriate for a base is percent association (n other words, percent of the association of B with the H to produce H^{+}).*1929

*They still call it percent dissociation; but as long as you know that now, you are talking about a base; it is a reverse process.*1953

*So, percent "dissociation"--I actually call it percent association, because I like things to make sense--I don't like things to just drop out of the sky.*1959

*OK, so we said that it is the OH ^{-}; it's the amount of thing produced--over the initial concentration of what we had.*1969

*We have 0.0153 Molar, over 13.0 Molar, times 100, equals 0.12%.*1983

*That means, of the NH _{3}, only .12% of the ammonia actually ripped off a hydrogen from the water to become ammonium ion and produce that much hydroxide.*1998

*That is it; that is all this means.*2016

*OK, so let's see what we have here: let's try one more, and I think we'll wrap it up for weak base discussion.*2020

*Example 3: Calculate the pH of a 0.10 Molar (this time) methylamine, which is CH _{3}NH_{2}, solution; so a methylamine is just like an ammonia, except it has, instead...I've taken out one of the hydrogens and put a CH_{3} on there.*2031

*And the K _{a} for methylamine is 4.38x10^{-4}.*2070

*I would like you to see what this actually looks like, because I'm a big fan of structures.*2078

*NH _{3} looks like this, as you know, and it has a lone pair of electrons; well, methylamine--it has its 2 hydrogens (not a problem), but now, it has a CH_{3} attached to it; so, it's the same thing; it's like an ammonia; it has a lone pair there; but instead of an H, it just has that.*2082

*It behaves exactly the same way; it's a base; it pulls a hydrogen off the water to produce hydroxide.*2104

*So, let's write our major species: well, we have the CH _{3}NH_{2}, and we have our H_{2}O.*2110

*Well, 4.38x10 ^{-4} versus 1.0x10^{-14}: yes, I think we can ignore the 1.0x10^{-14} as a source of hydroxide.*2123

*Most of the hydroxide in this solution is going to come from this--a weak base, but still a stronger base than that.*2134

*So, let's write our CH _{3}NH_{2} + HOH (or you can write H_{2}O, not a problem--same equation--plus water) goes to CH_{3}NH_{3}^{+} (just add a hydrogen and stick a plus charge on it) + OH^{-}.*2140

*We have an initial; we have a change; we have an equilibrium.*2159

*0.10; nothing; 00 (before anything happens); as the system comes to equilibrium, this species disappears; this species appears; and this species appears.*2163

*At equilibrium, we are left with .100-x; this doesn't matter; that doesn't matter; this is +x; this is +x; now, we have: 4.38x10 ^{-4} is equal to this times that, divided by that.*2176

*x squared over 0.10-x approximately equals x squared over 0.10.*2196

*Now, when we do this, we end up with the following (when we do this approximation): we end up with x equal to 6.6x10 ^{-3}.*2206

*Let's check the validity of this; let's see if our approximation here, from going from here to here, is valid.*2219

*Well, 6.6x10 ^{-3} over 0.10, times 100: guess what--it actually equals 6.6%.*2224

*6.6% is too high; it's close to the 5, but it really is too high.*2237

*That means that you have a choice; you can either...you have to go back; you can't use this approximation, in other words.*2243

*You have to actually solve this whole equation as it is.*2249

*You can't eliminate the x from here; you have to solve this equation; you have to do it--either solve it as a quadratic equation (which is not a problem--it's easy enough to do; it's just numbers; you have a calculator--you can do it), or I'm going to show you a method called the method of successive approximations, which is a really, really great technique, if you don't want to use the quadratic.*2255

*It is an older technique; it still works--there are a lot of computer programs that are actually based on this method of successive approximations.*2276

*Here is how it works: OK, so now let's go back, and we said that we have 6.6x10 ^{-3}, right?*2283

*So, when we did this approximation of 4.38x10 ^{-4} equals x^{2} over 0.10-x, approximately equal to x^{2} over 0.10; we got a value of...our first value...we got 6.6x10^{-3} for x.*2295

*Let me write x=this.*2325

*x equals 6.6x10 ^{-3}; OK, we checked this 6.6x10^{-3}; we divided by the .10, and we got 6.6%; that is too high.*2329

*Instead of going back and solving this, here is what you do: you take this first value, and you put it back in for x, and you solve this equation again.*2342

*In other words, you take .10; you subtract 6.6x10 ^{-3}; and you solve the equation x^{2} over the number that you get here, when you subtract this from this.*2351

*You solve this, and you get a second value for x.*2365

*Now notice, I didn't put it here and here; that doesn't make sense.*2369

*What I am doing is: I'm going to use this first value to get closer with a second value.*2374

*I'm going to use the second value to get closer with a third value.*2380

*When any two successive values actually match each other, that means I have hit my point.*2384

*For those of you that are familiar with something called the Newton-Raphson method of solving for the roots of an equation, this is somewhat similar to that.*2389

*You are basically just sort of converging on a value, and when two successive values are equal, that means you have hit your point; you are not going to go any further.*2397

*So here, we were slightly off; I use this to put it back into this one, leaving this alone; I solve for x, and a second value I get is x=6.39x10 ^{-3}.*2407

*This is a pretty significant difference; this is fairly significant here.*2422

*And now, I check this one; well, I actually don't really--I just stick it back into this again; I take .10-6.39x10 ^{-3} in the denominator; I leave the x^{2} on top; and I solve this equation again.*2427

*My third value that I get when I solve for x: I get x=6.40x10 ^{-3}.*2443

*I stick this back in there; I do it again; I get a fourth value, x=6.40x10 ^{-3}.*2453

*Two values, one after the other, match; I can stop there.*2464

*x=6.40x10 ^{-3}.*2468

*x happens to be my hydroxide ion concentration of 6.4x10 ^{-3}.*2473

*Now, let me see: how else did I handle that?*2481

*And the pOH, which is the negative log of the 6.4x10 ^{-3}, gives me 2.19, and then pH is equal to 14-pOH; I end up with 11.8.*2483

*That is my basic solution.*2502

*So again, when you are presented with a value, and you try to approximate it by leaving this x off, and it turns out that that x value is higher than 5% of the original value that you took it from; well, you can take the value that you got and stick it in there.*2507

*Just subtract it from that value in the denominator, run this calculation again for x, and just keep doing that over and over again.*2523

*Whatever value you get, put it back in; when you get two values that are the same, that is when you stop--that is your value; you have converged on it.*2531

*You have gone from...if this is the real value, let's say you started over here; you are going bounce here; you are going to bounce here; you are going to bounce here; you are going to converge on it.*2538

*It really is nothing more than a Newton-Raphson method.*2547

*Or, you can just solve the quadratic equation.*2552

*It's up to you; personal taste--I think it's sort of nice to do whatever you feel comfortable with, because it certainly makes the act of problem-solving much more relaxing and much more enjoyable whenever you are doing something that you like to do.*2554

*OK, so I'll go ahead, and we will stop there for our discussion of weak bases.*2567

*Next time, when we meet, we are going to talk about polyprotic acids and the acidic and the acid-base properties of regular salts.*2573

*So, thank you for joining us here at Educator.com, and we'll see you next time; goodbye.*2581

1 answer

Last reply by: Professor Hovasapian

Wed Jan 27, 2016 4:07 PM

Post by Jinhai Zhang on January 16 at 01:32:41 PM

Professor:

when you mentioned the biological catabolic, the aerobic respiration means the O2 is the final electron acceptor, and anaerobic respiration is non O2 for example S2-, or HS- as a final electron acceptor. they both go through citric acid cycle, and just has a different electron acceptor in chemiosmosis stage. And lactic acid form and alcohol formation is called a respiration which is Fermentation.

0 answers

Post by Caleb Lear on April 10, 2014

Perhaps another way to think of it would be the percent dissociation of the water. You check the part of the water that's left--the OH-- over the whole that was there.

Someone correct me if there's a conceptual problem here, this just makes sense to me.

1 answer

Last reply by: Professor Hovasapian

Thu Mar 13, 2014 8:39 PM

Post by Chemutai Shiow on March 9, 2014

In Example 2, why did you compare the concentration of the NaHO solution to the equilibrium constant of water? I thought you can only compare equilibrium constants to equilibrium constants.

1 answer

Last reply by: Professor Hovasapian

Wed Dec 18, 2013 7:45 PM

Post by Burhan Akram on December 18, 2013

Hello Prof. Raffi,

Can you also make a course on Physics as well? or is it not your field?

Thank You

Burhan Akram

1 answer

Last reply by: Professor Hovasapian

Fri Oct 12, 2012 4:59 PM

Post by Luis Esposito on October 12, 2012

question. in example 1 in the previous lecture, you used 10^-7 as the water concentration whereas in example 2 in this lecture you used 10^-14. why?

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Post by chenglong piao on September 12, 2012

That's what I thought too

1 answer

Last reply by: Professor Hovasapian

Thu Jul 26, 2012 10:08 PM

Post by Weiwei Gerl on July 26, 2012

I am confused about Example 5. Ka is known, but should I use Kw/Ka to find out the Kb first since CH3NH2 is a base?