AP Chemistry Solubility Equilibria, Part III

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

Today, we are going to close out our discussion of solubility equilibria, before we move on to complex ion equilibria.0003

Today, we are going to continue our discussion of precipitation with a slight twist.0011

If you remember the last example of the last lesson, we actually used the K_sp and concentrations to decide whether a precipitate will form.0017

Well, that is nice--to be able to know whether a precipitate will form--but oftentimes, we want numbers.0027

Once a precipitate forms, that means a certain amount of the free ion has been used up; after the precipitate forms and after the system has come to equilibrium again and settled down, what is going to be the final concentration of the ions?0033

That is going to be our next step with solubility equilibria.0046

Let's go ahead and just jump in with an example, and I think everything will make sense as we go through it.0050

Example 1: OK, a solution is prepared by mixing 120.0 milliliters of a 0.040 Molar lead (2) nitrate and 180.0 milliliters of a 0.08 Molar sodium iodide.0058

OK, calculate the equilibrium concentrations of Pb²⁺ and I^2-.0105

In this case, this problem specifically gave you what ions you are looking for; so, that is what is going to precipitate, if anything does.0126

They don't necessarily have to do that; oftentimes, they will just give you the species (lead nitrate, sodium iodide); you have to figure out which is going to precipitate (in this case, it's not going to be sodium nitrate; it's going to be lead iodide), either from your experience, or from a solubility chart, or from a table of K_sps.0134

There is no K_sp value for sodium nitrate, because it's fully soluble.0152

There is a K_sp entry for lead iodide; that tells you that that is going to be the insoluble salt.0155

OK, so let's start here: I²⁺--I don't think so--I think that's I^-; yes, OK.0161

Step 1: check to see if anything actually precipitates--we don't know if it will; check to see if anything precipitates--which is the same as the last example.0173

Well, let's just sort of do this from the beginning: in solution, before reaction.0193

Let me write: upon mixing, but before reaction, our major species are: well, we have lead ion; we have nitrate ion; we have sodium ion; and we have iodide ion.0204

OK, lead nitrate--soluble; sodium iodide--soluble; so we switch partners: sodium nitrate--soluble; lead iodide--generally insoluble.0234

So, lead and iodide are what we are going to concern ourselves with.0248

We want to check to see what the lead concentration is, and the iodide concentration; multiply them together; if it's bigger than the K_sp, there will be a precipitate; we'll move on to the next step.0252

OK, so let's go ahead and write the equilibrium reaction, because this is chemistry.0261

PbI₂ goes to Pb²⁺ + 2 I^-; and let's write the K_sp expression, also: the K_sp equals Pb²⁺ concentration, times the I^- squared (because we have 2I--law of mass action).0267

OK, now let's go ahead and calculate the Pb²⁺ concentration, initially, before any reaction takes place.0286

We have 120 milliliters of a 0.040 molarity (which is millimoles per milliliter), and the total volume is going to be 120.0 + 180.0 (right?--we're mixing 180 milliliters plus 120 milliliters); total volume is 300 milliliters.0296

We end up with 0.016 Molar lead ion concentration.0323

Well, the I^- concentration to begin with is 180 milliliters, times 0.08 millimole per milliliter; and again, we are sitting inside of a 300 milliliter total volume now.0330

This one is going to be 0.048 molarity; OK.0351

Let's go ahead and move on to the next step: so Q--our Q is going to equal 0.016, times 0.048 squared; and, if I did my arithmetic correctly (which I often do not--I hope you will check it for me): 3.7x10^-5.0359

OK, 3.7x10^-5 is definitely bigger than the K_sp for lead iodide, which happens to equal 1.4x10^-8 (right?--negative 5 is bigger than negative 8; 10^-5 is bigger than 10^-8).0381

So, yes: PbI₂ will precipitate.0401

But, we didn't ask just for the precipitation; after the precipitation, we want to know "What is the concentration of lead ion and iodide ion floating around in solution freely?"0411

Well, some of those are going to be used up; let's see what is happening pictorially.0422

OK, so we have this solution; before anything happens, we have Pb²⁺, Pb²⁺, I^-; I'll do a couple more: Pb²⁺, I^-; and I have Pb²⁺, I^-.0426

I mix these solutions; the total volume right now is 300 milliliters; I have lead floating around and iodide floating around; I have not included the nitrate and the sodium, because they are unimportant.0442

Well, precipitate is going to form; that means this and this are going to connect and fall to the bottom; this and this are going to connect and fall to the bottom; this and this are going to connect and fall to the bottom.0453

And eventually, what you are going to have is: you are going to have a bunch of lead iodide, sitting at the bottom as a solid (so I'll write PbI₂, solid)--sitting there; and you have a little bit of lead left over, and a little bit of iodide left over.0465

Well, these are going to react according to the reaction of lead and iodide coming together.0483

But you notice: the volumes and the molarities of the initial concentrations of the lead nitrate and the iodide were different; so they are not in an equal ratio.0494

So, there is going to be one of them that is going to run out before the other; if they didn't, then there would be no lead and iodide left over.0508

But, because you are always going to have more of one ion than the other, you are going to have some of each ion left over.0517

So, let's go ahead and see what goes on here; so now, let me--yes, so let me write down specifically what we want.0525

What we want is (well, not is; are, because we want both): the final concentrations of lead 2+ and iodide -.0535

In other words, we want this and this after equilibrium is reached.0563

What is going to happen: like we said, this is going to react with that; this is going to react with that; this is going to react with that; it is going to keep reacting.0574

At some point, there is going to be more of one than the other; so really, what happens is: then, what you end up with is an equilibrium situation.0582

Let me erase this one; I just wanted you to see what happens--so when these react, they fall down to become solids, and what you are left with...really, what you are going to be left with is one ion or the other, depending on which you had more of and the stoichiometry (which I'll write the equation for in just a minute).0594

Once you actually form the solid, now the solid actually dissolves a little bit more, according to the normal dissolving of the solid, to reach a certain equilibrium point, where there is a certain amount of lead floating around, a certain amount of iodide floating around, and a certain amount of solid in there.0613

The first thing that happens is stoichiometry, the reaction; and then, the system comes to equilibrium.0635

We are going to have a stoichiometry part, and we are going to have an equilibrium part--just like for our buffer solutions.0641

OK, so Step 2: let's do the stoichiometry.0647

The stoichiometry: OK, let's see--so the stoichiometry: let's write: Before the reaction (in other words, here--before the reaction actually takes place), we have lead; we have iodide; we have nitrate; and we have sodium.0659

These don't matter; these are the ones that matter--they are going to react and fall down to the bottom, because they are the ones that are insoluble.0685

The reaction that we are interested in is lead 2+ (let me write these bigger), plus 2 iodides, to form lead iodide, solid.0694

There is a Before; there is a Change; and there is an Equilibrium.0718

I mix the solutions; I have already sort of worked out that a precipitate is going to form--that is actually sort of a separate problem; so now that I know that a precipitate is going to form, the real part of this problem is the stoichiometry and the equilibrium.0724

As they react...so the initial concentration of lead, before anything happens--I have to calculate it; well, that is going to be the 120 milliliters, times the 0.04 molarity; so that gives me 4.8 millimoles.0738

And remember, this is stoichiometry; when we work in stoichiometry, we work in moles; when we work the equilibrium, we work in molarity.0757

Well, how about the iodide concentration?--well, the iodide concentration is the 180 milliliters, times the 0.08 Molar, which gives me 14.4 millimoles.0765

OK, now (you know what, I probably should have done this on the next page, because there is a...well, that is OK) these are going to react; there are 4.8 millimoles of this; there are 14.4 millimoles of this.0781

For every mole of this, 2 moles of iodide react; therefore, this is going to actually end up being the limiting reactant here; so, as this depletes, there is going to be none of this left over.0798

Well, that means 4.8 millimoles is going to vanish of the lead; so, let me go back up here, and let me tell you what is going to happen.0814

All of the lead is going to react; it is going to leave no lead ion in this solution.0827

Well, 14.4 millimoles of iodide to start off with--well, 4.8, minus 2 times 4.8 (because for every mole of lead, 2 moles of iodide have to be used up)...so this is going to be minus 9.6.0835

14.4 minus 9.6 gives us 4.8.0857

That means 4.8 millimoles of iodide are now floating around after reaction.0864

This doesn't matter; this doesn't matter; and, of course, this doesn't matter, because again, it's solid, so it doesn't have any effect on the equilibrium.0872

So, (this is not Equilibrium--I'm sorry--this is After; Before, Change, and After the reaction; before equilibrium is reached) the lead and the iodide react; all of the lead is used up; there is a little bit of iodide left over (4.8 millimoles left over).0883

And now, we are going to do the equilibrium problem.0902

Step 3...actually, before I do the equilibrium problem, I want to make sure you understand something that we did earlier--why it is that I ended up multiplying that 4.8 times 2.0907

Here is why I did that--so let me write the equation one more time: it is Pb²⁺ + 2 I^- goes to PbI₂.0921

We said that we had 4.8 millimoles of this, and we had 14.4 millimoles of this.0932

Well, the stoichiometry is this: 4.8 millimoles of lead 2+...well, the mole ratio between lead and this is 1:2, right?--so it is 1 millimole of lead 2+ to 2 millimoles of I^-.0943

2 times the 4.8 equals the 9.6; that is where the 9.6 came from.0969

That means 9.6 millimoles of iodide are going to be used up for every 4.8 millimoles of lead ion that are going to be used up.0975

We started off with 14.4 millimoles of iodide; we use up 9.6; that leaves us with 4.8 millimoles of iodide floating around in solution after reaction.0983

Now, we can start the equilibrium part, Step 3.0996

Step 3 The equilibrium: OK, let's go...there we go--equilibrium--now, the major species: after the reaction and before the equilibrium, the major species are iodide ion; nitrate ion; sodium ion.1000

These don't do anything; that is the only thing floating around in solution.1031

Now, the system is going to come to equilibrium; some of the solid down at the bottom of that flask is going to start to dissolve again, and there is going to reach an equilibrium point, and the ion concentrations are going to multiply to the K_sp.1036

The equilibrium part looks like this: PbI₂, solid, is in equilibrium with Pb²⁺ + 2 I^-.1050

There is an Initial; there is a Change; there is an Equilibrium.1066

OK, the initial amount of lead iodide doesn't even matter; it's solid, so it doesn't affect the equilibrium.1070

Well, we said that all of the lead was used up; it hasn't come to equilibrium yet, so before it reaches equilibrium, there is none of that.1077

Now, how much iodide ion is there?--well, we said we are left with 4.8 millimoles of the iodide, and now the total volume is 300 milliliters...so 0.016 molarity of iodide.1085

Well, a certain amount of this is going to dissolve; a certain amount of this is going to show up; and twice that amount is going to show up for every one unit--one unit of lead is produced, two units of iodide is produced.1104

Our equilibrium concentrations there become irrelevant; x; 0.016+2x; here we go--now we can run our calculation: the K_sp for lead iodide is 1.4x10^-8; it is equal to x, times 0.016+2x (the equilibrium concentrations of the lead and the iodide--that is squared).1120

We are going to do our approximation again by assuming that this is tiny, so we are going to eliminate it from consideration.1151

x times 0.016 squared...we end up with x equal to 5.5x10^-5 molarity; that is equal to x--that was the lead ion concentration at the end of our...once everything is done.1158

Well, we said that the equilibrium concentration of iodide is 0.016+2x, equals 0.016, plus 2 times 5.5x10^-5; it's so tiny, it still equals 0.016 Molar; that is equal to the iodide concentration.1181

There you have it: a long process, but nothing that we haven't done a hundred times already.1207

Let the chemistry tell you what is going on: you have the solution of ions; you are going to mix them; you need to find out if something precipitates.1215

Well, when something precipitates, those free ions are going to be used up; they are going to be eaten up; they are going to bind together, and they are going to form solid.1225

So, the free ion concentration is going to diminish; one of those ions is going to vanish because it's a limiting reactant; once the stoichiometry is done, now the system comes to equilibrium--some of the solid that is formed is going to start to establish an equilibrium.1231

It is going to be an equilibrium based on the K_sp, so we run the equilibrium problem.1246

We do it, and we come up with final concentrations of 5.5x10^-5, and 0.016 for the iodide.1251

OK, now let's go ahead and discuss our final, or our...not our ultimate, but our penultimate...our almost-final idea: something called selective precipitation.1261

Selective precipitation: OK, we'll just start with an example.1276

The idea is this: if you happen to have two metal ions (copper, lead...), are the K_sps so different that you can actually use an anion that you know will precipitate both, but will precipitate one before it precipitates the other?1284

That is the idea--one of the things that we do in basic analytical chemistry is: you have a mixture of ions; you need a way to separate them out.1302

Well, you can separate them out because of different K_sp values; certain things will fall out as a solid, will precipitate, before others do.1312

Let's just do a problem--Example: A solution contains (oops, I'm getting sloppy with my writing, aren't I?) 1.1x10^-4 Molar copper (1) ion, and 2.1x10^-3 Molar lead (2) ion.1320

If iodide is slowly added to this solution, will PbI₂ or CuI precipitate first?1359

OK, and also: at what concentrations of iodide will each salt start to precipitate?1389

OK, so let's see what is going on, pictorially, really quickly: we have this solution, and we have some lead ion in there; we have some copper (1) ion in there; 2+; copper (1) ion; we are going to be adding iodide ion.1421

Well, we know that copper iodide is insoluble; we know that lead iodide is insoluble.1436

Well, they have different K_sps; so ideally, I should be able to precipitate one before the other, based on the difference in K_sp, because again: once the concentration of iodide (let's say for copper) is such that it equals the K_sp, it is at that point that a precipitate starts to form, because a precipitate means it's in equilibrium, now, with its solid; it has reached its saturation point.1441

There is no more of the iodide and copper that can be in solution together; the rest of it is just going to fall out as solid copper iodide (and the same for lead).1465

OK, so let's go ahead and write the K_sps for these: K_sp of PbI₂ (just as a reference) is 1.4x10^-8, and the K_sp of CuI is equal to 5.3x10^-13.1474

Now, you might think to yourself, "Well, wait a minute; why can't we just compare? We know that one of them is going to precipitate first--probably this one, simply because it has a higher K_sp."1495

Well, the problem is: PbI₂ and CuI...you can't compare them directly, and the reason is because they don't produce the same number of ions upon dissociation--so you have to run the mathematics.1505

If they actually produce the same number of ions, then yes, you can use the K_sp (when we discussed relative solubilities, you remember, two lessons back...), but here, you have to run the mathematics.1517

Let's do PbI₂: PbI₂ is going to be in equilibrium with Pb²⁺ + 2 I^-.1528

The K_sp expression equals Pb²⁺, times I^- squared; well, we know the K_sp is 1.4x10^-8, and we know the lead concentration--it says that the lead concentration is 2.1x10^-3 (right?--it gave us the concentration in the problem).1541

So, 2.1x10^-3...so it's a simple question of algebra to find out the iodide ion concentration.1563

In this case, the iodide concentration is 2.6x10^-3 Molar; that means, when the iodide concentration hits 2.6x10^-3 Molar...that means, based on the amount of lead that was already in there, it will match the K_sp; and at that point, that is when the precipitate will start to fall out--not until then.1574

If I have 2.6x10^-4, 10^-5, anything less than this, there will be no precipitate.1597

That is the idea behind the K_sp; the K_sp tells you when the precipitate will form--when it's in equilibrium with the solid.1604

OK, now let's go ahead and do the copper iodide: copper iodide is going to be in equilibrium with copper +, plus iodide.1612

Be very, very careful; you have copper + and copper 2+; make sure you don't mix them up--they are actually different ions, different compounds, different solubilities.1622

OK, so the K_sp for this is equal to (well, I'm going to use that number in a minute; let me just write it out in the expression) Cu⁺, times I^-, each to the first power; so we have 5.3x10^-13, equals...1630

Well, they gave us the copper ion concentration to begin with; what was floating around in solution was 1.1x10^-4.1652

And then, just multiply by the iodide concentration; the final iodide concentration equals 4.8x10^-9 molarity.1661

OK, here we go: here are the two numbers that we need to compare now.1673

As we add iodide to the solution that has lead floating around and copper floating around, these calculations, based on K_sp, and the K_sp expression, tell me that, when the concentration of iodide that I had added equals 4.8x10^-9 molarity, copper iodide will start to form--will start to precipitate as a solid.1677

If I keep adding iodide...keep adding iodide...when the iodide concentration reaches 2.6x10^-3 molarity, at that point, the lead iodide will start to precipitate.1700

So, in this particular case, CuI, copper iodide, will precipitate first; this gives us a fantastic, fantastic analytical method for actually taking a solution with a whole bunch of different cations in it, and using different reagents (like, in this case, iodide) to actually separate out these ions--because that is what we do in chemistry: most chemistry is just a mixture of things.1711

A huge part of chemistry and biology involves the notion of separation; I need to purify things; I need to separate them into their individual components.1736

In nature, things don't come just individually; they come in one big mess that is all mixed up.1746

Half of chemical science is based on the notion of separation; so there you go.1752

Iodide--if there is a solution that is a mixture of lead and copper, iodide is a fantastic reagent (in the form of sodium iodide, potassium iodide...any iodide source will do) for separating out these two ions, if I happen to want to separate out these two ions.1757

This is what we mean by selective precipitation: I'm going to precipitate this before I precipitate that; that is the idea.1773

OK, so now, let's go ahead and talk about the last part, which is not going to involve any mathematics; it is going to be qualitative discussion, based on this notion of selective precipitation.1781

And, in fact, it is called classical qualitative analysis.1794

It is definitely one of the best labs that you can ever do, and it is also one of the most tedious labs that you can ever do; but, if you are good at this, you will be a highly-respected and highly-valued chemist and/or biologist.1798

Qualitative analysis: qualitative just means identity; quantitative means how much.1814

So, when I say I'm doing a qualitative analysis to find out if there is sulfide in something, that means I am trying to identify the sulfide.1821

If I'm doing a quantitative analysis, that means I want to find out how much sulfide.1827

Qualitative analysis means just the separation part, not the measurement of the amount that we have actually separated.1832

OK, so let me write out what it actually is: it involves the separation of a mixture of common cations by selective precipitation, by what we just did.1840

I guarantee you, there is a wonderful discussion of this in all of your textbooks (if you happen to be using textbooks); those of you that are, I would definitely suggest looking at it, because it gives you some really, really nice...I'm going to give you a nice pictorial representation, but it goes into a really beautiful discussion of this--qualitative analysis--very beautiful area of chemistry.1872

It is actually still one of the few things that impresses me.1890

OK, well, so we have a mixture; we have this collection of ions in one solution; so I'm going to write out all of these ions, and we are going to talk about the procedures (sort of big-picture, broad-strokes procedures) that we use to separate these ions into groups.1895

I have Hg₂²⁺; this is the iron (1) ion (remember, iron (1) is one of those ions that is a dimer; it actually sticks together in 2s), so that is why it's Hg₂²⁺--each Hg carries a 1+ charge.1912

I have silver ion; I have lead 2+ ion; I have Hg²⁺; these are different, OK--Hg₂²⁺ and Hg²⁺...those are different ions.1929

Cd²⁺, bismuth...so again, this is just one big beaker with all of these cations floating around; I am going to use selective precipitation to separate them out.1941

Bismuth 3+; copper 2+; tin 4+; cobalt 3+ (is that cobalt 3+? I think so, yes); zinc 2+; you might want to confirm this for me; I think this is cobalt 3+, but it might be cobalt 2+; I'll see if I can recall in just a moment.1953

Nickel 2+; iron 2+; chromium 3+; no, it's not cobalt 3+, it's cobalt 2+; and this is aluminum 3+; calcium; barium; magnesium; ammonium--wow; sodium; and potassium.1981

So, if I have a solution that has all of these ions in it, I want to use the method of selective precipitation, adding different reagents, so I can pull out this one, this one, this one, this one, this one...how can I do that?2012

Well, here is how you do it: we are going to break this up into five different groups.2027

The first thing you want to do, when you have a mixture like this...the first thing you want to do is: you want to add dilute hydrochloric acid.2032

When you do that, you will precipitate out these three things: Hg₂Cl₂, AgCl, and PbCl₂.2043

This is group 1.2057

So, that goes away; that goes away; that goes away; they are pulled out of this solution, leaving behind the remaining cations (in other words, groups 2 to 5).2062

I am actually going to list the groups after I draw this picture.2086

I have a mixture of ions in a particular solution; I need to find a way to selectively precipitate the ones that I want; I'm going to do this in groups.2090

The first thing I do is: I add dilute hydrochloric acid; I pull out the mercury (1) chloride, the silver chloride, and the lead (2) chloride.2098

Those immediately precipitate out, leaving the other cations behind.2105

OK, so now, let me go to the next page; let me go this way; the next thing that I actually do is: I add (oops, I don't want these stray lines coming again, especially when we are trying to draw a picture)...the next step, when I have these remaining cations that are left over: I add hydrogen sulfide to the acidic mixture, right?2110

I have added dilute hydrochloric acid to this; I have precipitated this out, but I have added hydrochloric acid; so now, the solution is acidic.2145

To an acidic solution...when I add hydrogen sulfide to this, here is what is going to precipitate out.2152

Let me go to the next page; and it's going to be like this.2160

I'll write again: Add H₂S: what you precipitate out is the following: HgS; CdS; Bi₂S₃; CuS; and SnS₂.2165

This is group 2; OK.2191

And again, I have what is remaining, which is groups 3 to 5.2198

OK, to the remaining cations, the next thing I do to the solution is: I add hydroxide.2208

OK, in other words, I make the solution basic.2218

When I do that (I should write precipitates; I should write precipitates), what falls out are the following: cobalt sulfide; zinc sulfide; manganese sulfide; nickel sulfide; iron sulfide; chromium (3) hydroxide; aluminum hydroxide (always a 3 charge on aluminum).2228

This forms group 3.2267

What you have left over are groups 4 and 5, the remaining ions.2272

To this, you add the carbonate ion.2282

When you do that, you precipitate out copper (2) carbonate (that is CuCO₃; yes...let's see; we have CuCO₃); we have BaCO₃, barium carbonate; and we have magnesium carbonate; this forms group 4.2287

What you are left with is group 5; so now, we are going to discuss what these groups are.2320

Groups: 1--you have the insoluble chlorides (mercury (1) chloride, silver chloride, lead (2) chloride--those are the insoluble chlorides); those are the first ones to precipitate out by the addition of hydrochloric acid--the insoluble chlorides.2336

Now, the next thing we did was: we added H₂S (hydrosulfuric acid, or hydrogen sulfide) to the acidic solution.2356

When you add hydrogen sulfide to an acidic solution, group 2--all of those things--precipitate out as a solid.2364

They are the sulfides, insoluble in acidic solution.2374

Once we have pulled those ions out by binding them with sulfide, well, next what we do is...remember what we did?--we added OH^- to the solution.2387

We take this acidic solution that is left over; we make it basic--we raise the pH above 7 (pretty significantly above 7).2397

And now, what falls out are the sulfides that are insoluble in basic solution (plus a couple of the hydroxides, like the chromium and the aluminum hydroxide).2404

Well, what is left over--the next thing we did was: we added carbonate.2421

What ends up falling out are the insoluble carbonates.2426

And then, what you are left with is the final group, which is going to be made up of the alkali metals; now the only thing in solution, at this point, is alkali metals and ammonium ion.2434

So, we started off with this mixture of a bunch of cations, and we had this procedure for actually separating out groups of ions.2452

Now, once we separate this out into this group, this group, this group, this group, this group--within each group, we can use more subtle separation techniques to actually separate out the individual things themselves.2461

What we will probably do is: in the laboratory, more often than not, what happens is: let's say you take the group 3--you end up actually acidifying the solution, making it soluble again, and then you use other reagents (more subtle reagents) to actually separate out within the group.2473

But this is sort of a broad classical qualitative analysis to break things up--to separate out cations that fall into groups.2491

So, you start by adding HCl; you separate out the insoluble chlorides.2499

Then, you add H₂S to separate out the sulfides that are insoluble in acidic solution.2505

And then, you go ahead and you turn that solution basic; therefore, you precipitate out the sulfides that are insoluble.2512

Those fall to the bottom; you filter those out.2518

And then, you add carbonate to the solution, and then the insoluble carbonates will fall to the bottom; you filter that out.2521

Now, what you are left with is a bunch of alkali metals and ammonium.2527

Now, we are not saying that you are always going to find all of these cations in solution; we are saying that this is a broad-strokes method for a nice classical qualitative analysis--how to identify which species, based on separation into groups.2532

I would definitely recommend you taking a look, again, at this picture and the previous couple of slides; it's a nice little flowchart for what you add when and what precipitates--what stays in solution, what precipitates and what stays in solution...2547

And also, take a look at that in your book.2560

Just sort of stare at it for a while, and hopefully, it will make sense.2562

OK, this closes out our discussion of solubility equilibria.2567

When we see each other next time, we are going to start discussing complex ion equilibria.2571

As you can imagine, it is just a bunch of equilibrium problems with...just things that look a little different.2575

But, the techniques that we have been using are the same techniques we are going to continue to use.2581

Thank you for joining us here at Educator.com.2585

We'll see you next time; goodbye.2588