AP Chemistry 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 H₂O.0844

Well, we know that H₂O is also a source of hydroxide ion, right?--because H₂O 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 H₂O 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

NH₃, which has a lone pair, plus HOH (I'm going to write it as HOH, water) goes to NH₄⁺ + 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 NH₄⁺ 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 NH₄⁺ times OH^-, over NH₃.1311

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 NH₃ 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 NH₃ (weak base--it's a weak base, which means that it is not dissociated--it stays mostly NH₃), 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: NH₃ + 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

NH₃ is not coming apart like an acid is, but we put a -x because NH₃, 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 NH₃ is actually taking something from the water and becoming more.1715

It is becoming NH₃ to NH₄⁺.1724

But what is happening is that NH₃ is disappearing as a species; NH₄⁺ 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 NH₃, 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 NH₃, 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 CH₃NH₂, solution; so a methylamine is just like an ammonia, except it has, instead...I've taken out one of the hydrogens and put a CH₃ 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

NH₃ 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 CH₃ 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 CH₃NH₂, and we have our H₂O.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 CH₃NH₂ + HOH (or you can write H₂O, not a problem--same equation--plus water) goes to CH₃NH₃⁺ (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 x² over 0.10-x, approximately equal to x² 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 x² 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 x² 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