AP Chemistry Complex Ion Equilibria

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

Today, we are going to start talking about complex ion equilibria.0004

We are going to start off with some definitions, like we normally do, and then we are just going to sort of launch into an example, and again, use the example to discuss the chemistry that is going on.0008

The entire lesson is essentially going to consist of one long example.0018

Well, again, the example itself is not long; but sort of going through it step by step, making sure we understand what is going on--that is what is going to take the time.0023

So, let's go ahead and write some definitions down, and see what we can do.0030

Complex ion: well, a complex ion is a charged species with a metal center, and surrounded by species called ligands.0035

And now, I am going to define what a ligand is (or a lih-gund--most people will actually say "lih-gund"--I just happen to be a little unusual--I just like to pronounce the i long, so I say "lie-gand," but it doesn't really matter how you pronounce it, as long as you know what is going on).0097

OK, so I'm going to give a few examples of ligands, and then, I'm going to actually sort of define it, and then we will talk about what is going on and what one actually looks like.0101

OK, so a ligand example: the ligand examples that we are going to be concerned with are things like H₂O; and I'm going to actually put the lone pair of electrons on them, because...well, you will understand why in just a minute.0112

NH₃ has one lone pair; Cl^- has several lone pair; and (let's just put one more for good measure...let me actually show the triple bond here) the cyanide.0129

So, the water molecule; the ammonia molecule; the chloride ion; and the cyanide ion--they are what we call ligands, and here is what they do.0147

They basically arrange themselves around a central metal atom (either two of them, four of them, three of them, five of them, six of them...usually no more than six), and they form actual bonds with the metal that happens to be an ion, because most metals, especially the transition metals, have a whole bunch of empty orbitals.0156

Remember, they have the s orbital free; they have the d orbitals free; they have the f orbitals free; so, a whole bunch of ligands can actually fill those empty orbitals; and we will talk about it in a minute, but let me give you a quick definition of what a ligand is.0182

So again, these are just fancy words for things for which the chemistry...you already know what is going to happen.0195

A ligand is a species with a lone pair it can donate to form a covalent bond with a metal.0203

This is just another way of saying that it's a Lewis base.0238

A Lewis base--a ligand--is basically just a species that has a lone pair (this, this, this, this) that it can donate to form a covalent bond with a metal.0244

Now, let's talk about the difference between something called a normal covalent bond and a coordinate covalent bond; they are both covalent bonds, and they behave exactly the same way; it's just a question of where the electrons come from.0255

You remember: if we take something like carbon and oxygen (I'm going to represent the electron on the carbon with just a dot, and the electron that comes from oxygen as an x), this is...a bond is made up of two electrons that are shared...this is a normal covalent bond; we just call it a covalent bond.0267

What makes it normal is the fact that each species brings an electron to the table.0288

Well, in a coordinate covalent bond (like, for example, aluminum)--if there is an aluminum ion, and if a water molecule were to attach itself to it, this right here--this lone pair (there are two lone pairs on water)--one of those lone pairs actually forms a covalent bond with the aluminum center.0294

It is a normal covalent bond, like any other covalent bond, except both electrons come from one species (in this case, the water).0319

It could be the ammonia; it could be the chloride; it could be the Cn (let me put the electrons up on here, by the way).0327

That is the only difference; so, when this arrangement happens--when this Lewis base, when a ligand, actually arranges itself and bonds to a metal, but both of the electrons come from the ligand itself, and if the whole species happens to be charged, it's called a complex ion.0332

That is it; there is nothing more than that--it is exactly like anything else that you have dealt with; we call it complex simply because there are a lot of individual species involved in the whole ion itself.0351

It still behaves like an ion with a given charge; it is just a very big ion; that is it.0363

We went from monoatomic ions, like chloride; and then we went to polyatomic ions, like sulfate, chromate, dichromate, chlorate...things like that; and now, we're moved on to complex ions.0370

They are still just ions, and they form ionic compounds with things of opposite charge.0382

OK, and the last thing we are going to define is something called the coordination number.0387

The coordination number, again, is just a fancy word for the number of ligands that are around the metal center--the number of ligands bonded to the metal center.0395

That is it; if you have two ligands attached to an aluminum, well, the coordination number is 2; if you have four ligands attached to platinum, then you have a coordination number of 4.0415

That is it; so, 6, 4, and 2 are the common coordination numbers, but others do exist (3, 5, even 7 in some cases).0425

OK, so we have this thing called a complex ion; it's a charged species with a metal center, and it is surrounded by these things called ligands.0448

Well, ligands are just species that have a lone pair that it can donate to form a covalent bond with a metal--something like water or ammonia or chloride or cyanide--there is a whole list of ligands; in fact, many molecules can actually be ligands.0456

Those of you that go on to study chemistry (in particular, inorganic chemistry and organometallic chemistry)--it is going to be pretty much entirely complex ion chemistry; it's basically metal centers that have a bunch of ligands attached to them.0470

OK, now, if you recall, from a couple of lessons ago, we discussed this thing called...remember, we talked about how salts--if you drop a particular salt in a solution, it can react with the water to actually form an acid or a basic solution, and one of the species that we ended up dropping into solution, that actually produces an acidic solution, was this thing called hexaaqua aluminum.0487

Remember, it looked like this; recall, it was an aluminum ion, and it was surrounded by six water molecules; so we write (H₂O)₆; and the whole species had a +1 charge.0517

This is called hexaaqua aluminum; don't worry about the name--you will worry about the name later on, when we talk about coordination chemistry specifically.0535

Well, it looked like this: it is a metal center; it is aluminum; there is a water molecule bonded to the top...and again, remember, a single line is a covalent bond.0542

Now, I know, in this particular course, what I did is: I actually skipped the notion of bonding, and I decided to go straight into kinetics and equilibrium.0555

So, if any of this is a little strange, just know that a single line, a single bond, means a pair of electrons; that is all it means.0563

We will be going back and discussing it, but I wanted to save it for a little bit later, because it tends to move a little bit faster, and I wanted to spend more time with some of the equilibrium stuff.0572

That is why I went over it--a non-traditional sequence, but I think it's a good sequence.0581

OK, so I have...and I'll explain what these dashes and wedges mean in just a minute...H₂ (actually, you can probably imagine what they mean already); OH₂; so, I have an aluminum ion, and (actually, this is 3+, not 1+; silver would be the 1+), which is 3+ charge; and if I drop this aluminum ion in some water, what happens is: 6 water molecules attach themselves, literally bond, to the aluminum.0589

They bond according to this scheme; they don't just bond randomly--they actually take up a geometric figure around the aluminum.0627

There is one on top, one on bottom; these dashes mean that it is facing back--this is a 3-dimensional representation on a 2-dimensional flat surface.0634

When we see dashes, that means it is behind the plane of the paper; these wedges mean it's coming forward.0644

So basically, what you have is this sort of bipyramidal structure, and this is a complex ion; and the whole thing has a charge of...so, that is a complex ion.0651

That is all it is: just a metal center surrounded by ligands.0666

OK, now: fortunately for us, we are not going to be concerned with the structures and the naming of it just yet; we will, later on in the course, if and when we discuss coordination compounds.0669

We are going to be concerned with the equilibrium.0679

We will concern ourselves, here, with equilibria.0683

But, I did want you to see what it looks like, just so you have a picture of what is going on.0703

So, here is the most important thing about complex ion equilibria: Ligands attach (I will also write "detach," but we are mostly going to be concerned with attachment) one at a time.0707

For example, this OH: the OHs aren't just going to converge on the water; it is actually going to go through a single-step process--one water molecule is going to attach, then another, then another, then another...and for each of these, it's a reaction.0725

It is: an aluminum, plus a water molecule, goes to Al(H₂O)₁; Al(H₂O)₁, plus a water molecule, goes to Al(H₂O)₂; and each one of those, actually, is a reaction.0739

For any reaction, we can write a constant for it; so let's go ahead and...well, let me finish writing this.0751

Ligands attach or detach one at a time to the metal center.0759

Now, for example, let's use silver: if I have some silver ion, and if I drop that silver ion solution in some ammonia (or the other way around--if I drop some ammonia in some silver ion solution), here is what happens.0769

I'll go ahead and put the lone pair there; what happens is that the ammonia bonds to the silver, and it forms this species, right here.0789

This is an ammonia, bonded to this covalently; it's a coordinate covalent bond (both electrons came from ammonia); and the whole species is positive charge 1.0805

Well, there is an equilibrium constant that is involved here; as it turns out, when this happens, at some point, the solution comes to equilibrium--there is going to be some of this, some of this, and some of this floating around in solution.0814

But, take a look at how big the equilibrium constant is: 2.1x10³: this is called a formation constant.0828

I'll write it over here: a formation constant.0841

It is a formation constant because this complex ion is forming.0846

Take a good look at this: 2.1x10³: we are accustomed to seeing acid and base constants like 10^-3, 10^-8, 10^-7; and when we just did solubility products, we were seeing even smaller constants; we were seeing things like 10^-16, 10^-29...0850

Well, a small constant means that the equilibrium is not very far to the right; here, this is a very big constant, which means that this equilibrium is to the right.0868

What that means is that, if you drop ammonia in a silver solution or vice versa, basically, what is going to happen is: this whole reaction is going to end up going entirely to completion--meaning you are going to find mostly this species.0877

That is what constants tell you: they tell you how far to the left or to the right an equilibrium is.0889

It is true, it is an equilibrium: you are always going to find a little bit of this, a little bit of that; but when we say "little," we mean little, because this equilibrium constant tells us it's going to be mostly this.0895

It is going to be virtually no silver.0904

OK, well, here is what is interesting: this is not the only equilibrium that is going to exist; once this forms this species--like we said, ligands attach one at a time.0907

So, this species actually ends up reacting with another ligand--with another ammonia molecule.0917

NH₃⁺ + NH₃, another ammonia molecule...and it forms the complex ion Ag(NH₃)₂⁺.0924

Now, this has a coordination number of 2; so, 1 molecule of ammonia attaches itself to silver to form this; but then, another molecule of ammonia attaches itself to this complex that was formed to form this.0937

And it actually stops there; it is stable when silver ion, in ammonia, has 2 ammonias attached to it.0949

Well, there is a second formation constant; there is a formation constant associated with this one, and this is 8.2x10³; you notice, it's higher than this, so when I drop ammonia in silver or vice versa, it forms this.0957

But, as soon as this is formed, it grabs onto another ammonia, and it forms this; so really, what you have in solution at the end--if you add, actually, enough ammonia (which under normal circumstances is the case; any time we have a solution of a metal ion, the ligand that we add is always added in excess, specifically to drive the reaction forward), you are going to have mostly this.0975

Yes, you are going to have a little bit of this--a little bit of ammonia, a little bit of silver--but mostly, it's going to be that.0998

That is all that is taking place here; we have the formation of something; we can form a constant, because there is an equilibrium; we have measured the constant--we call it the formation constant, just like anything else (we have an acid dissociation constant, a base association constant, a solubility product constant; for complex ions, it's called a formation constant).1003

Usually, we have the subscripts--these 1, 2, 3--to mark down the first ligand, second ligand, third ligand, and the constant associated with them coming together with the metal center.1024

OK, now let's go ahead and jump into a particular problem.1036

In a solution of Ag⁺ and NH₃, all 4 species exist in an equilibrium.1045

And the four species are silver, ammonia, this thing (which is actually called a monoamine silver (1)), and also, this thing (this is called diamine silver (1)).1072

And again, you will learn about the naming scheme later on.1104

There is an equilibrium; so, all four of these species are in some kind of an equilibrium; however, again, when we say "equilibrium," well, again--we look at these numbers to let us know what chemistry is going on; we don't just presume that it is an equilibrium like any other equilibrium.1108

We want to use every bit of information at our disposal; these are really, really big formation constants--that means that, pretty much, it is going to be mostly this species.1122

There is going to be very little of this and this; the ammonia, in practical situations--we put in so much ammonia that it is virtually constant.1131

You will see what we mean when we do an example.1141

I just wanted you to get an idea of what is going on; we have these formation constants; we have an equilibrium equation that we can write; so now, let's go ahead and see what we can do.1143

What we want to do is: we want to calculate the concentrations of each of the species involved in a complex ion equilibrium.1154

And, in the previous case, we have 4 of those species.1187

We are going to use the same techniques that we did before, except--again, we are going to let the chemistry decide how we proceed.1195

Complex ion equilibria is a fantastic, fantastic place to practice sort of pulling back from the problem and making sure you let the chemistry decide how the math is produced.1202

OK, so we'll write: Example (I won't put Example 1, because this is going to be the example of this problem): Consider mixing 100 milliliters of a 1.9 Molar ammonia solution with 100.0 milliliters of a 1.0x10^-3 Molar silver nitrate solution.1215

Let me just draw this out, so we have a picture of what is going on.1263

I had a silver nitrate solution; that means I had silver ions and nitrate ions floating around--silver nitrate is fully soluble salt.1268

And then, I mix it with a solution of ammonia (solution--as I said, that means that is not pure ammonia--that means that there are some water molecules floating around; I just mixed water and ammonia).1278

So, you have a 1.9 Molar ammonia solution and a 10^-3 Molar silver nitrate solution; when you mix them, 100 milliliters of each, now your total volume is 200 milliliters.1290

You have to keep track of that; we are working in concentrations, so you have to keep track of the fact that now the volume, when you mix them together, is now 200 milliliters (not 100 milliliters anymore).1300

The concentration of the ions is going to change.1309

OK, well, let's see: one thing we want you to notice here--it is going to be a very, very important thing to notice, and it is characteristic of complex ion equilibria, and actually in the lab (and not just the analytical stuff that we do on paper).1313

The concentration of ligand (in this case, the NH₃) is a lot bigger than the concentration of the metal ion.1331

This is characteristic of complex ion problems.1349

The reason it is characteristic is, it tends to simplify the math.1364

It means that we can actually make approximations, where otherwise we might not be able to make approximations; because, as you are going to see in a minute, some complex ion equilibria can tend to get a little involved, a little complicated, because there is a whole bunch of species sort of floating around and doing this and doing that.1369

So, those of you that go on to chemistry and take, in your junior year or your senior year...and of course, in analytical chemistry, you will do more subtle versions of this, where perhaps, the concentration of ligand will not be all that much, and the math becomes a little bit more complex.1386

But for our purposes, and for most purposes in the laboratory, in fact, the ligand concentration is usually going to be a lot higher than the metal ion concentration, as it is here.1402

You have 1.9 Molar of the ligand and 10^-3 Molar of the metal ion--a lot bigger.1412

OK, so let's take a look at our major species before reaction in the solution once we mix them: again, we want to be able to decide what chemistry is going to take place.1419

Major species before any reaction takes place: so, we have silver ion; we have nitrate ion; we have ammonia; and we have water.1433

Well, there are a couple of things that we can do: we have been--you know from your experience in acid-base chemistry that, any time you have ammonia and water...well, what you have is a weak base situation.1445

One possibility is that the ammonia will react with the H₂O in an acid-base equilibrium, in a base association reaction, to form ammonium plus OH^-.1456

This has a K_b of 1.8x10^-5; let me write that down: K_b equals 1.8x10^-5.1478

OK, well, that is one possibility--these two might react; well, the other possibility is the silver and the ammonia forming a complex ion--that is the thing--both reactions will take place; the question is which one will dominate, because we want to concentrate on one reaction.1485

Well, we just said that the K₁ (if you will look on the previous slide) and the K₂ for the formation constant of that silver-ammonia complex ion is 2x10³, 8.2x10³.1503

Well, that versus 1.8x10^-5...what that tells us is that so little of the ammonia is going to react with the water to form hydroxide ion that we are talking about 1.9 Molar ammonia solution.1523

It is going to be virtually negligible.1541

The real chemistry in this solution...although this will take place, it is really going to be these two--not to mention the fact that that is what we are actually interested in.1543

It is possible to actually run the reaction and sort of deal with this base association reaction, if that is what the problem asked for; but the problem isn't asking for that, and it is also not really much of an issue, because there is so much ammonia, and the base association...the equilibrium constant is so small that it is negligible.1554

We are going to be concerned with Ag⁺ and NH₃.1573

So, let's go ahead and write out our reactions again: so we have silver ion, plus ammonia, going to form this, and it has a K₁ equal to 2.1x10³.1577

And then, this monoamine silver (1) species is going to react with another molecule of ammonia to form diamine silver (1), NH₃, and the K₂ for that is equal to 8.2x10³.1600

OK, let's take a look at this: we have: silver and ammonia are going to react to form this species (and a very, very big equilibrium constant--formation constant--which means that mostly, it's going to be this--all of the silver is going to be used up).1624

Ammonia concentration--because it's so high, it is probably not even going to be affected.1641

10^-3 Molar; 1.9 Molar; so, you have to keep track of all of these things and think about the chemistry here.1647

And then, when this forms (because it's going to form completely), all of a sudden, it is going to react with another molecule of ammonia, and it is going to end up forming this.1654

So basically, the formation constant for this one is 8.2x10³; again, a really big formation constant.1663

So, these numbers tell us that, in solution, mostly, we are going to have this.1672

Basically, these two reactions are going to go to completion--meaning they are going to go so far to the right that, for all practical purposes, this whole idea of an equilibrium isn't even really viable, at least as far as doing the problem is concerned...as doing the math is concerned.1679

We are just sort of concerned with what is going on globally--what is going to take place.1694

Well, because that is the case, here is what we can do...because we know we can actually add these equations for a total equilibrium situation, and that is going to be the following.1699

If I add these equations, well...remember from back when we were doing Hess's Law: anytime you add equations, what shows up on the left and right...they actually cancel.1713

Let me do this in red: this goes away and this goes away, so my total reaction, my net reaction, is going to be something like this.1722

It is going to be: one silver ion, plus two NH₃ molecules, to form one diamine silver (1) complex ion, (NH₃)₂.1731

This is the overall reaction; it takes place in two steps.1746

That is important; but the overall reaction is this--because our numbers are so high, we can just deal with this reaction, and that is the idea; so that is what we want to deal with.1749

Let me rewrite this; I'm going to write: Ag⁺ + 2 NH₃ is in equilibrium with Ag(NH₃)₂⁺.1760

OK, so now, let's deal with our Initial, our Change, and our Equilibrium situations.1777

Well, what is our initial concentration of silver ion?1784

Well, we had 100 milliliters, times a 10^-3 millimole per milliner--that is the molarity; when I multiply those two numbers together, I am going to get an answer in moles.1788

But now, I'm floating around in 100 + 100 milliliters, right?1804

We said that the total volume is now 200 milliliters.1808

So, this molarity is 5.0x10^-4 before any reaction takes place.1813

Well, the ammonia is...well, we had 100 milliliters, times a 1.9 millimole per milliliter; and we end up with 0.95 millimoles.1820

Notice: it's a lot bigger than this; and of course, there is no complex ion formed yet.1834

Well, what is the change going to be?--well, this reaction...by the way, if you also remember--I don't know if you do or not--any time you add two equations together, the final equation you get...the constant for that, I'll put K_net, is equal to the product of the two.1839

You don't add these together to get the constant for this reaction; you actually multiply this times that.1860

Remember, when we were doing equilibrium, that is how it works.1868

When you add two equations to get a net equation, and each of those equations that you added has an equilibrium constant, the final constant of the net (in other words, the net constant, the constant of the net equation) is equal to the product of the two, not the sum of the two.1871

It is not like ΔHs; you don't add the equilibrium constants--you multiply the equilibrium constants.1888

OK, so clearly it's going to be very, very high; so, this is going to react with that; these tell me that the reaction is fully far forward; so pretty much, this entire thing is going to disappear.1895

10^-4.1909

Well, this is going to be minus 2, times 5.0, times 10^-4; and the reason there is a 2 here...because there is a 2 here; for every mole of this that is used up, 2 moles of NH₃ are required.1911

So, if 5.0x10^-4 moles disappear (or moles per liter disappear), -2 times that also disappears from here.1939

What shows up here is 5.0x10^-4; so our equilibrium concentrations are this, minus that, is 0 (not quite 0, but it's 0 for our purposes; we will actually calculate what this concentration is--it is going to be pretty near 0, but for the fact that the reaction is based on these numbers, we can proceed and write the ICE chart this way.1948

This .95, minus two times that...well, guess what: it actually stays .95.1974

Let me write this...to two significant figures, it stays .95; so there is virtually no change.1980

This is 5.0x10^-4; OK, so now that we have our equilibrium concentrations, we can now calculate the species.1986

OK, well, so we know what one of these is: we know that the Ag...actually, I'll save this one for last; it will probably make more sense if I do it that way.2000

We want to know...we said we want the concentrations of all four of those species.2014

We want the concentration of Ag⁺, the concentration of NH₃, the concentration of the monoamine complex ion, and we want the concentration of the diamine complex ion.2021

Let's go ahead and deal with this one: well, we have an equation: Ag⁺, plus...actually, no: let's use...let's see, what do we have at our disposal?2036

Let's write the two equilibria: Ag⁺ + NH₃, AgNH₃⁺; and we also have Ag(NH₃)⁺ + NH₃ in equilibrium with Ag(NH₃)₂⁺...2056

This is a single; wow, now you see what we mean by "this gets a little complicated"; there is a whole bunch of symbols going on--you have to be very, very careful with what you are writing down and be very, very clear about what is going on.2082

Working slowly is the best thing to do; don't feel like you have to be in a hurry here.2093

OK, so let me see: we just calculated some equilibrium concentrations on our ICE chart, and we had this concentration; we had that concentration; we had this one, which was 0; we are looking for this one; and the only equation here (I missed the K₂=8.2x10³)...so I'm going to use that one to calculate that concentration.2098

So, K₁ is equal to...well, it's equal to the product of Ag(NH₃)₂⁺ concentration, over the AgNH₃⁺ concentration, times the concentration of ammonia.2128

That is equal to 8.2x10³; and now, you notice: we have...it's equal to...we have this one--that was the 5.0x10^-4; we have this one; we are looking for this one.2155

AgNH₃⁺, times 0.95 from our ICE chart; these numbers--we had the K₂; we have that one; we have that one; now, we solve for this; we get that this concentration (now I'm getting confused; OK)...AgNH₃⁺ concentration is equal to 6.4x10^-8 molarity.2177

We put that ICE chart together, and we used the second of the equilibria, because it involves that species; and we had a value for this and this from our ICE chart.2213

So now, we just calculated the concentration of that monoamine complex ion; and notice how small it is: 6.4x10^-8.2224

Very, very tiny: it's virtually 0, and that is the whole idea.2233

This tells us that it is really, really far forward.2238

OK, let's go ahead and calculate, now, the silver ion concentration.2241

Let me write that K₁ is equal to 2.1x10³; now, K₁ is this, over this, times that.2246

So, K₁ equals AgNH₃⁺ concentration, divided by the Ag⁺ concentration, times the NH₃ concentration.2258

It is 2.1x10³, equals the AgNH₃⁺ concentration (we just did that--that is 6.4x10^-8), divided by the Ag⁺ (which is what we want), times NH₃ (which is 0.95).2278

And, when we do this calculation, we get that the silver ion concentration is equal to 3.2x10^-11.2300

OK, and of course, our final, which (let me do it on the next page) we actually get from the ICE chart...our diamine complex, (NH₃)₂⁺ concentration, is equal to 5.0x10^-4.2312

So, here is what we have; oh, and of course the NH₃ concentration, which is virtually unchanged, stayed at 0.95.2337

Again, there was so much of it there that it was virtually unchanged.2346

OK, so let's take a look at what we have: let me write it this way again: Ag(NH₃)₂⁺ is a lot bigger than AgNH₃⁺, a lot bigger than Ag⁺.2351

Our first equilibrium...well, both of the constants let us know that, really, what was going to be the largest silver-containing species in solution was actually going to be this one.2375

This one is 5.0x10^-4 molarity.2390

This one ended up being 6.4x10^-8; now, you might be saying to yourself, "OK, 6.4x10^-8 is a small number, but 5.0x10^-4 is also a small number."2396

It's true; both of them are small numbers, absolutely; however, relatively, 6.4x10^-8, compared to 5.0x10^-4...this is 4 orders of magnitude bigger; it's huge--ten thousand times bigger; that is the whole idea.2411

So, when we are comparing these, we are not looking at absolute numbers; don't let these numbers fool you absolutely; it's relatively speaking that we are saying.2428

This, of course...we get 3.2x10^-11; and we got these values by using the two equilibria separately.2438

Here is what we did (and this is the really important part): we took a look at the two separate formation reactions, with their formation constants; and we said, "These formation constants are really large."2448

Because they are so large, basically, we can put the equations together; because, because they are both so large, the final product in solution is going to be the last product (in this case, the diamine complex ion).2463

We added the two equations, and we did our ICE chart based on that combined equation, the net equation.2477

We used the concentrations that we got from the ICE chart, under the E category, the equilibrium category, to go back and deal with the individual formations and the individual formation constants, to find this concentration and that concentration.2484

And notice how tiny these are compared to this; this is really tiny compared to this; it confirms the fact that there is virtually none of this, compared to this.2501

This is very tiny compared to both this and this, which means that there is virtually nil; that means all of the silver, pretty much, reacted to form this, and then almost all of this reacted to form this.2512

That is the whole idea; and they reacted with the ammonia--the ammonia concentration was so high that it didn't even matter.2528

You notice: the final ammonia concentration didn't even change; that is the whole idea behind using a very, very high ligand concentration and a very, very low metal ion concentration--because you can push the reaction so far forward and not have to worry about the ammonia concentration.2536

OK, I have to admit: complex ion equilibria is very, very strange.2555

We didn't exactly handle the ICE chart exactly the way that we did with solubility and with acid-base and other equilibria problems, so if there is still some confusion, I would definitely recommend going through this one more time and just thinking about the chemistry as I'm talking about it.2562

And again, let the chemistry decide what makes sense.2579

The only way that any of this--any of the numbers that you do get--will make sense is if the chemistry makes sense.2583

You constantly have to ask yourselves, "Does this fit with what I am guessing?"2589

Yes, it fits; this is tiny; this is tiny; relatively speaking, this is huge.2595

It makes sense, based on the amounts of material that you actually put in.2600

OK, so our next lesson is actually going to be a continuation of complex ion equilibria.2604

We are actually going to discuss complex ions and solubility.2610

We'll go ahead and close this lesson out; thank you for joining us here at Educator.com.2613

We'll see you next time; goodbye.2617