AP Chemistry Complex Ions & Solubility

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

Today, we are going to round out our discussion of aqueous equilibria, and we are going to finish off by discussing complex ions and solubility.0004

We talked about solubility, and we talked about complex ions; now, we are going to talk about how we can mix the two.0013

Let's go ahead and move forward.0021

OK, so let's recall, from a couple of lessons ago: we did something called this classical qualitative analysis.0029

Recall: the classical qualitative analysis--remember, we had this mixture of ions, all of these metal ions, and we added different reagents, and we used this process of selective precipitation to precipitate out certain groups.0042

Group 1: the insoluble chlorides; Group 2 was those that are insoluble in acidic solution; those that are insoluble in basic solution; and then the carbonates; and then the alkali metals.0070

We are sort of taking this mixture, and we are sort of separating it out, based on solubility.0083

OK, well, now we are going to combine this: so we had this...I won't write out all of the ions again...mixture of ions.0088

The first thing that we did was: we added dilute hydrochloric acid; and when we did that, we precipitated out the insoluble chlorides.0107

There were three of them, if you remember: there was lead (2) chloride; there was silver chloride; and there was mercury (1) chloride, which is Hg₂Cl₂ (and the reason is because mercury (1) tends to aggregate as 2 mercury atoms together...but it's still mercury 1+).0123

And then, of course, we had the others; OK.0142

Well, here is the question: now I have these things sort of sitting at the bottom of the solution; so I filter it out, and now I have this solid clump of lead chloride, silver chloride, and mercury chloride.0147

Well, now, the next step: How do I separate these into their individual ions (lead 2, silver 1, mercury 1)--how do I do that?0163

OK, well, from a practical standpoint, we need to basically put all of these three back into solution again; so we throw away these others; we take this; and we have to find a way to dissolve these back again, so that we can run a different set of reactions to separate these individually.0175

We separated them from the group, but now we need to separate them; how do we do that?0196

OK, so again, we need to find a way to re-dissolve these so that we can do some chemistry to separate these three out.0201

Well, as it turns out (let me write this down...well, let me write out the whole thing): OK, now the question becomes: How do we separate these three--how do we separate these three?0213

OK, the answer is: we have to re-dissolve them and run some other chemistry.0233

OK, well, the "run some other chemistry" part is probably going to be taken care of when you actually do your lab work for this particular General Chemistry course, or your AP Chemistry course.0254

What we are going to concentrate on is: How do we re-dissolve them--how do we take these and put them back into solution (or at least one of these species back into solution) to separate them out?)0267

OK, as it turns out, forming complex ions with insoluble salts (that is what these are--they are insoluble salts; their K_sps are really tiny--silver chloride, lead chloride...they don't dissolve very much; they just sit there at the bottom of the beaker) makes them soluble.0279

We will concentrate on AgCl; so I'm just going to pick one of these to show you how to re-dissolve that.0329

If you have this precipitate, that is what is going to happen; anytime you put the chloride--when you have silver and chloride together in solution, they are going to bind, and they are going to drop to the bottom of the flask as a precipitate.0346

Well, in order to do some more chemistry with that, I need to find a way to dissolve silver chloride.0357

Well, silver chloride doesn't dissolve in water; how can I, therefore, dissolve it?0363

Well, here is how you do it: let's start with the actual silver chloride equilibrium.0368

AgCl, solid, is in equilibrium with Ag⁺ + Cl^-; OK, if you remember (well, you don't have to remember), the K_sp for this is 1.6x10^-10; so clearly, not very soluble at all.0377

All right, well, watch what happens: so I have this solution; I have a whole bunch of silver chloride, solid, down at the bottom; so this is AgCl, solid; and there are a few ions of silver and a few ions of chloride; there you go.0396

Now, here is what we do to actually dissolve this stuff that is sitting at the bottom.0417

We need to find a way to dissolve it to run more chemistry on it.0423

We add NH₃; well, in the last lesson, you saw what happens when you have silver solution reacting with NH₃; here is what happens.0428

The free silver, the little bit of free silver that is actually floating around in solution, the few ions--well, they actually react with the ammonia that you drop into the solution according to the complex ion equilibria.0438

They form the monoamine complex ion; that K₁, remember, was 2.1x10³.0456

Now, this one goes and reacts with another molecule of ammonia, and it forms the diamine complex ion.0465

So, when this happens (and let me just go ahead and write the K), let's see what is going on here.0480

I have some solid sodium chloride; it is in equilibrium with some silver and some chloride, based on the K_sp, so this is what is happening.0489

Mostly it's this, but a little bit of this; but the moment I drop ammonia into this solution, the ammonia starts binding to the silver--the free silver that is floating around--and it binds it up as this diamine complex ion.0497

Well, when that happens, that is what is happening; you are basically tying up this silver concentration as something else.0511

Well, when you do that, you have reduced the silver ion--the free silver ion concentration.0520

Well, if you have reduced the free silver ion concentration, Le Chatelier's Principle says this reaction will adjust itself to re-increase the silver ion concentration.0525

The only way the silver ion concentration is going to increase again is if this dissociates to release more free silver ion.0535

Yes, it's going to also release chloride, but chloride doesn't matter in this case.0543

That is what happens; when it dissociates, that is dissolving; so, every time it dissolves to produce more silver ion, up to the value of the K_sp, and then we add ammonia--that ammonia is going to react with the silver ion and pull it out of solution again.0547

If that pulls it out of solution, it is going to create an empty space for silver ion; more of the AgCl will dissolve; more of it will dissolve; more of it will dissolve, until all of it dissolves.0563

Because, what you have done is: you have locked up the silver as diamine silver complex ion.0574

It is kind of like the same thing that we did when we added acid to an insoluble hydroxide.0580

The acid reacts with the hydroxide, forming water, pulling the reaction forward, causing the insoluble salt to dissolve.0587

It is the same thing here: silver chloride is insoluble in water; therefore, to that water, I add ammonia.0594

The ammonia binds with the silver, pulling this reaction forward, dissolving the silver ion.0600

Now, I have silver ion in solution; yes, it's bound up, but it is still silver ion.0606

And now, I can do other chemistry to it or separate it out from other solutions; that is the whole idea.0612

OK, so let's actually write these reactions in order...oh, let me write down what is going on here; it's very, very important.0618

Ag⁺ reacts with NH₃ to form a complex ion (and as we know, ions in solution are soluble); the Ag⁺ concentration is reduced (that's nice; it would be good if I could spell); as compensation for that, AgCl responds (oh, I'm having a hard time spelling tonight) by dissociating to release more silver ion (in other words, it dissolves), until the K_sp value of 1.6x10^-10 is reached again.0628

Because that is what the K_sp is; the K_sp is the concentration of that, times the concentration of that; once that is reached, no more of this will dissolve.0705

But, if I add ammonia to it, it reacts with the silver, depleting the silver ion concentration; now, the K_sp doesn't match--there is chloride ion, but there is virtually no silver ion.0714

More of this dissolves to form silver ion until it reaches the K_sp value.0724

That is what is going on here.0729

OK, now let's do the chemistry; so we have AgCl in equilibrium with Ag⁺ + Cl^-.0730

The K_sp for that, we said, is 1.6x10^-10: very, very insoluble.0744

We have Ag⁺ + NH₃ going to form the monoamine complex ion.0751

Oh, in case you are wondering, ammonia, when it acts as a ligand, is called "amine"; so that is why I'm saying "amine" instead of "ammonia"; I should have mentioned that earlier.0765

K₁ equals 2.1x10³, and then the monoamine reacts with another molecule of ammonia to form the diamine complex ion.0779

Symbols, symbols, symbols...I tell you...2...no, this is 8.2, times...oh, -3; it's positive 3; I'm telling you, there are so many places to make mistakes when you do this chemistry, because there is so much symbolism.0800

That is the thing: concepts are not difficult--it's the symbolism that gets in the way, and that is what is often frustrating about science.0819

It's not too difficult, but just the symbolism makes you crazy!0825

OK, well, we can form...so this is what is going on: there is equilibrium among all of this stuff.0829

Well, we have some things on the left-hand side and the right-hand side that are actually the same; so let's add these equations for a net reaction--what is happening in solution when you take silver chloride, solid, and drop in ammonia?0839

All of these reactions are taking place; what is the net reaction?0852

Well (let's do this in blue), that is on the right; that is on the left; they cancel; that cancels with that; and we are left with the following.0856

Solid silver chloride will react with 2 moles of ammonia to form the diamine complex ion, plus chloride ion.0872

The K_net (remember what we said: we multiply these) is equal to K_sp times K₁ times K₂, and it ends up being 2.8x10^-3; so now, let's look at this.0893

Let's take a look at it and think about it.0910

We have this reaction taking place: solid silver chloride; drop in some ammonia; the ammonia reacts with silver and forms the monoamine; the monoamine reacts with another ammonia and forms the diamine.0915

Mostly this is going to be there; I can add these equations together for a total reaction, a net reaction; that means one mole of silver chloride, solid, will react with 2 moles of ammonia to form 1 mole of the diamine complex ion and a free chloride ion.0926

The total equilibrium constant for this reaction--these over this; this doesn't count--is 2.8x10^-3.0948

2.8x10^-3 is small, but it's not super small; so it's large enough, so this definitely is a classical equilibrium situation; there is going to be a fair amount of this, this, and this floating around in solution (and perhaps even some of this, depending on how much was actually there).0957

So, given that, now we can do our example.0976

Example: Calculate the solubility of AgCl in 11 Molar ammonia solution.0986

There you go: remember, solubility is the amount of a solid that dissolves.1005

We calculated the solubility of silver chloride earlier; it just has to do with the K_sp value--writing out the K_sp, doing your ICE chart.1009

Well now, we are not saying "Calculate the solubility of AgCl in pure water"; we are saying, "Calculate the solubility of AgCl in 11 Molar NH₃."1016

When that happens, we know what happens when it reacts with NH₃.1026

This happens; we add the equations--this is our net equation; this is our net equilibrium constant; that is what we use in our ICE chart.1031

That is the whole idea: the chemistry first, then the math.1040

OK, so now, let me go ahead and move to another page here, and write...yes, that is fine: so, we have AgCl + 2 NH₃ in equilibrium with Ag(NH₃)₂⁺ + chloride ion.1045

Our initial concentration of AgCl: well, this is a solid, so we don't care how much is actually there--it doesn't affect the equilibrium, so we don't really care about it.1081

How much ammonia do we have--what is the concentration of ammonia?--well, the ammonia concentration, before anything happens, is 11 Molar; that was the thing that was given.1090

There is none of this formed yet, and there is none of this formed yet.1100

This is before anything takes place.1105

The change: what kind of a change takes place?--well, a certain amount of the silver chloride is going to dissolve; this is what we are looking for--this is the x, solubility, how much dissolves.1107

Well, for every mole of this that dissolves, 2 moles of NH₃ is used up.1123

So, it's -2x; that is it; this 2 and this 2--that is where they come from.1131

Well, for every mole of this used up, one mole of the complex ion shows up, and one mole of that shows up.1137

Our equilibrium is that; that doesn't matter, because it's a solid; 11.0-2x; x; and x.1146

The K_net is the diamine concentration, times the chloride ion concentration, divided by the NH₃ concentration squared (law of mass action, right?).1164

This is not a liquid; this is solution--this is aqueous; OK--this is a mixture of water--all right.1185

Now, well, this is x; this is x; this is 11.0-2x; we know what the K_net is--we had it already; that is 2.8x10^-3, is equal to x, times x, divided by 11.0-2x squared, equals x squared over 11.0-2x squared.1192

Now, I know what you are thinking: can we actually go ahead and ignore this and use the 5% rule to decide that it is valid--so, can we make the math easier?1227

Yes, you can if you want to; but notice something here: you actually don't have to simplify here, because this is squared and this is squared.1238

This is equal to x over 11.0-2x squared; so it's the whole thing squared; so because the top is squared and the bottom is squared, I can just take the square root of both sides.1245

When I do that, I get (the square root of this, the square root of that)...I end up with 0.05291=x/11.0-2x (and let's not have any stray lines here).1266

When we multiply through, we get 0.5821-0.10582x=x; and we end up with 0.5821=1.10582x; and we get x is equal to 0.53 moles per liter.1292

There we go: we just calculated the solubility.1324

In 11 Molar ammonia solution, silver chloride: .53 moles of silver chloride will dissolve in a liter of that solution.1329

Compare this with the solubility of AgCl in pure H₂O (in pure water): 1.3x10^-5 moles per liter--very big difference.1343

1.3x10^-5 is virtually nothing; .53--that is very significant solubility--that is very, very soluble.1372

There we go: by simply taking an insoluble salt (like silver chloride or any other insoluble salt), if that insoluble salt--if the cation actually ends up forming a complex ion, we can use the fact that it forms a complex ion to dissolve that insoluble salt.1382

If we need to run some chemistry on it, in this particular case, we just need to add some ammonia to the solution, and sure enough, what is insoluble in water will dissolve in the ammonia, by an application of Le Chatelier's Principle, by the formation of a complex ion--very, very important.1402

OK, so now let's take sort of a global view of what it is that we have done.1419

Let me go back to blue.1424

We have now seen two ways to dissolve a water-insoluble ionic compound.1426

One, which we discussed a few lessons ago, is to acidify the solution.1458

If we acidify the solution (in other words, if we add hydrogen ion to the solution), the anion of the solid reacts with the added H⁺ (oops, not going to have that), thus pulling the dissociation forward (forward--to the right--dissolving).1468

I'll just write "forward."1518

We have seen two ways to dissolve a water-insoluble ionic compound; so when we have something that is actually not soluble in water--it doesn't dissolve--we need to find a way to dissolve it.1523

Well, one of the ways that we have is to add acid to that solution; when we add acid to the solution, the acid (in other words, the H⁺) reacts with the anion of the solid; and, by Le Chatelier's Principle, it pulls the dissociation forward until everything is dissolved.1533

That is the whole idea.1549

A couple of examples: the examples that we did of that were like magnesium hydroxide--remember, we had magnesium hydroxide, solid.1552

Well, the equilibrium, the K_sp for that...it's true that it is in equilibrium with magnesium and 2 hydroxide ions, but take a look at the K_sp for this: it is (oh, wait, do I even have the K_sp for this?)...oh, it looks like I forgot to write the K_sp, but if I'm not mistaken, it was somewhere on the order of 10^-12.1561

It's very, very insoluble; however, when we added H⁺ to this solution, here is what happens.1584

That OH^- ends up reacting with that H⁺ to form water.1594

When the water is formed, this is depleted; this is now reacting with that; well, since this is depleted, Le Chatelier's Principle wants to produce more hydroxide, which means...the only way to produce more hydroxide is: this has to dissolve to release free hydroxide.1602

More will dissolve; more will dissolve; more will dissolve until the whole thing is dissolved.1619

We have bypassed that, the fact that it is insoluble, by adding acid to the solution.1624

Another example would be calcium carbonate: calcium carbonate...1631

I could be wrong about the 10^-12; that could be 10^-16...but it doesn't matter; it's insoluble--that is the whole idea.1637

...goes to calcium 2+, plus carbonate 2-; well, the anion of the salt (this is solid)--small K_sp--very small--not very soluble.1645

Well, this reacts with the acid; when it reacts with the acid, it forms the bicarbonate ion; that means it is now sequestered as bicarbonate ion.1664

Well, now there is an empty spot there; in order to fill up that empty spot and reach the K_sp value again, more of this has to dissolve.1679

More dissolves; more dissolves until it is all dissolved.1688

OK, #2: The other way that we have to dissolve an insoluble ionic solid is to add a solution of a ligand that forms a stable complex ion with (I'm always putting an 'e' at the end of my "with") the cation (this time, with the positive--with the cation) of the ionic solid.1693

Our example (well, almost our example; we used silver chloride, but I'm going to write it out as silver bromide): AgBr, solid, is in equilibrium with Ag⁺ + Br^- (do I have another page?--yes, I do; good).1741

Well, the Ag⁺ reacts with 2 molecules of ammonia to form that diamine complex ion.1761

In the process of forming that diamine complex ion, now the silver is bound up as this complex ion.1776

Well, because now there is no silver ion in solution anymore, the K_sp value is not satisfied for this.1783

So, Le Chatelier's Principle: when you deplete this, the system will react by trying to raise that concentration again.1791

In order to raise the concentration of free silver ion, this has to dissolve; it has to dissociate to produce, so more dissolves.1798

You add more ammonia--it binds the silver and pulls the silver out of solution as a complex ion; more dissolves until eventually, you have dissolved all of your silver bromide.1806

And that is the whole idea; so again, when we are dealing with insoluble salts, yes, it is true--the problems that we did--we dealt with them directly; but in certain circumstances, we do have to find a way to dissolve them.1817

And now, we have two ways of dissolving them: we can acidify the solution by adding hydrogen ion (in which case, the hydrogen ion--if the hydrogen ion reacts with the anion of the insoluble salt, that acidified solution will actually dissolve that salt, where it is not otherwise soluble in water).1832

Or, if the cation of the insoluble salt forms a stable complex ion with a given ligand (it could be any ligand), if we add a solution of that ligand, by doing that, we can make that insoluble salt all of a sudden soluble, so that we can do further chemistry with it.1854

OK, that brings us to the end, actually, of our discussion of aqueous equilibria.1876

With that, I will thank you again for joining us here at Educator.com.1882

We will see you next time for a further discussion of chemistry; take good care; goodbye.1887