AP Chemistry Solubility Equilibria

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

Today, we are going to continue our discussion of aqueous equilibria, and we are going to discuss solubility equilibria.0004

We are going to discuss the equilibria involved in salts that actually don't dissolve very much, unlike the salts that you are accustomed to, like table salt, for example.0010

When you drop it into water, it completely dissociates into free sodium and free chloride; there are some salts that--they just basically--you drop them in water, and they just sink to the bottom as solids.0020

Now, they do dissolve a little bit, and we are actually able to measure how much they do dissolve; but for all practical purposes, they dissolve so little that, to the naked eye, it looks like they just sink to the bottom like sand.0031

OK, let's jump in and start with some definitions and see what we can do.0043

OK, so some salts (and again, "salt" is just a generic term for an ionic compound--metal-nonmetal) dissolve completely (oh, I really need to learn to spell here); those are the ones we know from our solubility chart way back when; those are soluble.0048

For example, sodium chloride (like we mentioned)--when you drop it into water (I'll just put water on top of the arrow here), you end up with (so this is a solid)--when you drop it into water--you get aqueous sodium ion and chloride ion; the aq means that they are floating around in solution.0083

Or, let's say, potassium nitrate (so KNO₃, solid): drop it in water; you end up with free potassium ion, plus free nitrate ion.0101

I'm not going to put the subscripts; again, I think they're sort of superfluous at this point--I mean, we know what we are talking about; we are talking about soluble salts; we know that it's floating around in solution; any time you have an ion like this, we are talking about aqueous equilibria, so we know that it's in water.0115

OK, now, others dissolve only very little, where mostly the solid stays intact.0130

OK, so notice: these arrows go one way; that means that, when you drop this in water, all of it dissolves; when you drop this in water, all of it dissolves.0158

With certain salts that don't dissolve completely, an equilibrium is established; here there is no equilibrium--I mean, there is, but basically, there is no solid sodium chloride left anywhere (until you get to a saturate solution, which we will talk about in a subsequent lesson); but really, you are talking about complete dissociation.0169

Well, let's take an example like magnesium fluoride: it turns out, magnesium fluoride doesn't dissolve very much.0187

As a solid, when you drop it in water, it is true that some magnesium ion and fluoride ion do break free; but it's actually very, very little that does so.0194

So, there is this equilibrium that is established, and most of the equilibrium is, in fact, over here; which is why, when you drop this in the water, and you stir it around, it just sort of sinks to the bottom; it looks like nothing has dissolved.0210

Well, we know that at the molecular level, some has; and we are able to measure it, but most of it just sort of stays intact; but there is an equilibrium that is established.0222

So now, because of this equilibrium situation--because we have some that is dissolved, but not a lot of it--we can write an equilibrium constant for it.0232

Well, the equilibrium constant is just Mg₂⁺ (the concentration--remember, it's just reactants over the products), times F^-, raised to its coefficient, 2 (because we know that, when this dissociates, it produces 1 Mg ion and 2 fluoride ions; that is what happens here.0243

Well, we call this K_sp; instead of K_eq, when we are dealing with solubility problems where we have a solid that is in equilibrium, in solution with its free ions, we call this the K_sp.0265

It is the called the solubility product constant.0289

The best part is: you are just talking about equilibrium: there is no new math that is going on here--there is no new concept.0298

All of the equilibrium that you have learned before applies here; it's all we have done--just sort of changed the label; instead of K_eq, we call it K_sp.0303

It is called the solubility product constant, or just the solubility product; you will see both of those terms used.0312

Now, I just want to describe what is happening physically, because I want you to actually be able to think about this when you are doing your problems.0323

If I take some magnesium fluoride, solid, and if I drop it in solution, it basically piles up at the bottom, because it's not very soluble.0333

However, some of it does dissolve: you have a little bit of magnesium ion and a little bit of fluoride ion actually floating around in solution.0341

There is this equilibrium between the solid magnesium fluoride and its ions.0351

This is called a saturated solution--you will often hear it referred to as that: a saturated solution is one that is in equilibrium with the solid that is present in solution, at the bottom of the beaker.0359

That is what they mean by "saturated": I can take sodium chloride, and I can drop it into a beaker; I can dissolve it and dissolve it and dissolve it; at some point, no matter how much more salt I add, it's just going to sink to the bottom, because it's completely saturated.0371

There is going to be so much--there are going to be so many sodium ions and chloride ions floating around in solution that the solution just doesn't have enough capacity to carry any more ions.0384

The salt, the sodium chloride, which is normally soluble, will actually sit at the bottom of the beaker as a solid; this is called a saturated solution.0399

So, for these problems, you will often see--it will often start by saying "a saturated solution of sodium chloride," "a saturated solution of sodium phosphate"; that is what this means.0408

It means that the ions are in equilibrium with their solid; that is what this equation says.0419

I just wanted you to know that you will often hear it as the "saturated solution."0427

That means no more ions are going into solution; it's staying as a solid.0432

OK, now, as it turns out (let me rewrite this Mg): magnesium fluoride is in equilibrium with magnesium ion...solid...plus fluoride ion; as it turns out, the K_sp for this is 6.4x10^-9.0436

This is a small number: that means that...so we said that the K_sp is equal to the magnesium ion concentration in moles per liter, times the fluoride ion concentration squared, is equal to 6.4x10^-9; that is a very tiny number.0462

You know what that means--that means the concentration of free magnesium and free fluoride in solution is very, very tiny; that is the whole idea.0481

OK, let's do a barium hydroxide: BaOH₂, solid--it is going to be in equilibrium with a little bit of barium ion, plus 2 moles of hydroxide ion; and the K_sp for this equals 5.0x10^-3.0491

Again, it's a tiny number, which means that there is very little barium and very little hydroxide floating around.0512

Also notice: for every molecule of barium hydroxide that dissociates, 1 ion of barium and 2 ions of hydroxide...the stoichiometry is very, very important; and you will see that in a minute.0518

OK, now we want to talk about the two things that are most important when discussing solubility equilibria.0530

These are the things that we do not want to confuse; and it is the single biggest problem for students when they read their problems; they are in such a hurry, or they are so stressed out, that they actually end up mixing up these two things that I'm going to write down.0537

Solubility: solubility is the amount, in either moles or grams--the amount of solid that dissolves under a given set of conditions.0551

For example, if I said 10.0 grams per 100 milliliters of water, this is an expression of solubility; it is telling me that, if I have 100 milliliters of water, I can dissolve 10 grams of solid; that means 10 grams of a salt will completely dissociate in that solution.0582

Not more than 10 grams: if I put 10.2 grams, 10 grams will dissolve; .2 grams will sit at the bottom as a saturated solution.0607

This is an expression of saturation: it says that I can't squeeze, at a given temperature, more than 10 grams of a solid into solution--into 100 milliliters of solution.0616

It might say it this way: 1.2x10^-3 moles per liter of, let's say...let's just pick something at random: potassium nitrate.0627

Now, obviously, potassium nitrate is going to be more soluble than this; but I'm just using this as an example.0640

This means that, if I have solid potassium nitrate, and if I have 1 liter of water, that means I can only dissolve 1.2x10^-3 moles of this potassium...0646

You know what, I'm not going to use potassium nitrate; I know it's going to end up sticking in somebody's mind, and it's going to end up confusing you kids, so we don't want to do that--just some random salt.0657

That means 1.2x10^-3 moles of this salt will dissolve in 1 liter of water.0667

If I have 2 liters of water, well, it's going to be 2.4x10^-3 moles.0672

But, this is an expression of solubility; it tells me an amount of the solid salt that dissolves under a given set of conditions.0677

This can change, depending on the conditions.0686

Now, solubility product is what we just defined; it is a constant--a constant that expresses the relationship between the concentrations of free ion in solution.0689

It is a constant; it doesn't change; that is the whole idea behind K_sp.0749

A certain amount of something might dissolve in solution; a certain different amount might dissolve under a different set of circumstances; however, what doesn't change...0755

Remember when we talked about an equilibrium position, versus the equilibrium constant?0764

Equilibrium positions can change, but the overall equilibrium constant does not; so, the solubility can change--can be different--but the solubility product is a constant; the K_sp expresses the relationship in a given aqueous solution.0768

MgF₂, Mg²⁺ + 2 F^-; we said that the K_sp was, what, 6.4x10^-9.0786

Different amounts of Mg and F^- might be in solution, depending on how the solution was made; but the product of the magnesium ion concentration in solution, times the fluoride ion concentration, squared, always equals the same constant.0802

That is the difference: do not mistake solubility for solubility product, or solubility product constant; that is the single biggest mistake that kids make.0821

Other than that, the problems to deal with this are actually quite simple.0830

We have been doing a ton of these equilibrium problems; we should be well-versed by now.0833

We do Before, Change, After; we do Initial, Change, Equilibrium; nothing is different, except now we just have a different constant to work with, K_sp instead of K_eq or K_a or K_b.0838

OK, let's just jump into some examples, and I think it will start to bring everything to light.0849

So, Example 1: Copper (1) bromide has a solubility of 1.9x10^-4 moles per liter at 25 degrees Celsius.0856

So, if I have one liter of water at 25 degrees Celsius, I can only dissolve 1.9x10^-4 moles of copper bromide in that solution; no more will go in there.0886

Anything more than that, and it will have an equilibrium--it will be saturated.0899

OK, what is its K_sp?0903

Well, let's write out the reaction first; let's write out the equilibrium: Copper bromide (it's copper (1) bromide, so it's CuBr)--when it dissociates, it dissociates into copper +1, plus a bromide -.0911

Let's write its K_sp expression: chemistry--start with a reaction--write the equilibrium expression; it will get you started, and it will keep you on track.0929

Copper +, times Br^-; this is solid; these are aqueous; we know that--OK.0943

The solubility of a salt means how much has dissolved in a given volume.0951

Now, how much of a given salt has dissolved in a given volume?0958

This is telling me that 1.9x10^-4 moles per liter of copper (1) bromide dissociates; well, for every 1 mole of copper bromide that dissociates, 1 mole of copper ion is produced; 1 mole of bromide ion is produced.0963

What does that mean?--that means that 1.9x10^-4 moles per liter, at 25 degrees Celsius--that means, when it dissolves, it produces 1.9x10^-4 Molar copper ion and 1.9x10^-4 bromide ion.0984

So, the solubility of a salt means how much has dissolved; we know that--in this case, the copper and a bromide--they are the solubility, because for every amount that dissolves, it produces (because the ratio is 1:1, 1:1) that much of the free ion.1004

Again, if 1.9x10^-4 moles dissolves in 1 liter, well, that produces 1.9x10^-4 moles of free copper ion in that liter.1055

It produces 1.9x10^-4 moles of free bromide ion in that liter.1065

Those are our solubilities, moles per liter; we have our concentrations, so we're done.1070

Let's just see what this looks like in an ICE chart.1076

CuBr in equilibrium with Cu⁺ + Br^-; OK, our Initial, our Change, our Equilibrium...1079

So, this is a solid, so it doesn't matter; it doesn't actually show up in the K_sp expression; and because it doesn't show up in the K_sp expression, we don't care about it.1092

Before anything happens, we drop this into water; before it dissociates, there is none of this, there is none of this, and there is a whole bunch of this.1103

How much dissociates?--well (let's do this in red), that much dissociates, so -1.9x10^-4.1110

Now, we said it doesn't matter, but it is good to put it here to remind us what is dissolving; and the negative sign means it's dissolving, it's breaking up; that much of this copper bromide is disappearing.1122

+1.9x10^-4, +1.9x10^-4; this does not matter; we have 1.9x10^-4 moles per liter of copper ion floating around; we have 1.9x10^-4 moles per liter bromide ion floating around.1135

So now, our K_sp is equal to our copper ion times our bromide ion; it equals 1.9x10^-4, times 1.9x10^-4, and we get a K_sp equal to 3.61x10^-8.1162

We have calculated a K_sp from a solubility; the solubility is the amount of solid that dissolves; well, for every amount of solid that dissolves, the reaction creates a certain amount of one ion and a certain amount of the other ion, depending upon the stoichiometric coefficients.1184

Let's do another example.1202

Notice: we omitted the units; this is moles per liter, moles per liter; we don't do moles squared per liter squared--we just, for the K_sp, ignore the units; we don't write it down.1210

OK, Example 2: all right, the solubility of bismuth (3) iodide is 7.78x10^-3 grams per liter.1220

So notice, this time they gave it to us in grams per liter.1252

Calculate its K_sp.1255

Well, let's start with our reaction: bismuth (3) iodide (that is a solid; that is going to be...you know what, let me go ahead and do this in another color here--let me go back to black)...so, we have bismuth iodide, solid; it is in equilibrium with a bismuth 3+ ion; but notice, for every "molecule"--for every unit--of bismuth iodide that dissolved, it produces one mole of bismuth ion, and it produces 3 moles of iodide.1263

Now, our stoichiometric coefficients are going to be different.1306

That is our reaction; let's write our K_sp expression.1310

Our K_sp is going to be bismuth 3+, times iodide -, cubed; yes, that's exactly right; OK.1313

Now, one of the things that you have probably noticed with these equilibrium problems, acid-base problems, is the sheer notational busy-ness--not complexity, but you have charges; you have exponents; you have coefficients; just write slowly and keep track of them all.1326

Otherwise, it will drive you crazy, and it will throw you off.1347

If you are going to make a mistake, you want to make a conceptual mistake; you don't want to make a notational or arithmetic mistake--that is what is important.1351

OK, so now, this is our K_sp expression; we have our reaction; 1 mole of this produces 1 mole of bismuth ion; it produces 3 moles of that ion; but notice, this is in grams per liter.1359

These have to be in moles per liter, so we need to change this solubility to molarity--not a problem; we'll just divide it by the molar mass of the bismuth iodide.1374

Let's go ahead and do that first; so let's do 7.78x10^-3 (because again, we are looking for K_sp, so we need to find this, and we need to find that; but it's in moles per liter, but the solubility is in grams).1388

This is in grams per liter, times...well, let's see: we want: 1 mole of the bismuth iodide happens to be 589.7 grams; and when we do this, we get 1.32x10^-5 moles per liter; so this is the solubility.1403

Once again, that means 1.32x10^-5 moles of bismuth iodide will dissolve in 1 liter of water.1439

OK, well, let's do our chart: BiI₃ is in equilibrium with Bi³⁺ + 3 I^-; our Initial; our Change; and our Equilibrium (this Equilibrium one is the one that we are going to put back into our K_sp expression).1446

OK, we don't care about this (no, I'm not going to have any stray lines here, because these problems are confusing enough without these stray lines), so nothing there.1468

Before anything happens, 00: this is: we just drop it into water, before it comes to equilibrium.1479

Well, a certain amount of the bismuth iodide is going to dissolve; that is the solubility; that is this number, right here; so it is -1.32x10^-5.1485

Well, for every 1 mole of this, 1 mole of that is created; so it's going to be +1.32x10^-5.1500

But, for every mole of this that dissolves, 3 moles of iodide are produced; so, if 1.32x10^-5 moles per liter dissolve, this is going to be 3 times 1.32x10^-5; what we end up with (this doesn't matter)--we end up with 1.32x10^-5 (that's 0 plus that); 0 plus this equals 3.96x10^-5--that is the concentration of our bismuth; this is the concentration of our iodide.1510

And now, we go ahead and we put these numbers into this expression, and we get (let me write it again on this page) K_sp is equal to bismuth 3+ ion concentration, times the iodide concentration cubed, equal to 1.32x10^-5, times 3.96x10^-5, cubed.1553

We end up with 8.20x10^-19; that is our K_sp.1587

Now, notice what we did: this 3 here shows up in two places--it shows up in the K_sp as the exponent; it shows up for this value, based on the stoichiometry.1594

For every mole of bismuth iodide that dissolves, 3 moles of iodide go into solution.1610

Therefore, 1.32x10^-5 moles per liter of bismuth iodide that dissolve release 3.96x10^-5 moles per liter of iodine.1618

When you put this value into this, this "cubed" shows up, now, differently.1629

The stoichiometry gives you this number, but you still have to use this number, because it's part of the K_sp expression; don't think that it is one or the other--it is both.1635

Be very, very careful when you are doing these solubility products.1644

OK, let's do Example #3: OK, so in this one, we're going to go in reverse--we gave you the solubility and you found the K_sp; now we're going to give you the K_sp; let's see what the solubility is.1647

The K_sp of magnesium hydroxide is 8.9x10^-12 at 25 degrees Celsius; calculate its solubility.1671

OK, so "calculate its solubility"--they are saying, at 25 degrees Celsius, given a certain moles per liter, given 1 liter of water, how much magnesium hydroxide will actually dissolve?1696

That is what they are asking; OK.1710

Simple enough: well, let's write the equation.1712

Magnesium hydroxide (this is chemistry; you have to have an equation) is in equilibrium, when it dissociates, with one magnesium ion, plus two hydroxide ions.1716

OK, well, we have the K_sp; we want the solubility.1729

Well, the solubility is the amount of this that actually dissolves; OK, so let's do our ICE chart--here is what an ICE chart looks like now.1734

It doesn't matter how much we start with: 0, 0--this is before anything happens.1742

As it comes to equilibrium, as some of this dissolves, a certain amount is going to dissolve; that is -x.1750

Well, for every x that dissolves, x of this shows up; so this is going to be +x.1756

For every x of this that dissolves, it produces 2 hydroxides; so this is 2x.1762

Well, this doesn't matter; this is x; this is 2x; these are the two things that show up in the K_sp expression.1771

So, let's go ahead and write: K_sp is equal to...well, we said it was 8.9x10^-12; that is equal to the magnesium ion concentration, times the hydroxide ion concentration squared; it's equal to x, times 2x squared (be very careful how you do this), equals x times 4x squared, equals 4x cubed.1780

We have 4x³=8.9x10^-12; when we divide by 4 and take the cube root, we end up with 1.3x10^-4.1818

This is our solubility, right here.1834

You remember: it's the -x; 1.3x10^-4: solubility--we include the unit, so this is moles per liter--which means that, in 1 liter of water, 1.3x10^-4 moles of magnesium hydroxide will dissolve.1839

There you go; OK, now let's talk about something called relative solubilities.1863

This is going to be more of a qualitative thing, so we're not really going to be doing any math here.1872

Relative solubilities: if you have a group of salts, you can use K_sp values, of course, to decide the order of solubilities--only for salts that produce the same number of ions when they dissociate.1878

OK, so if you have a group of salts, you can use the K_sp values; you can compare them to decide the order of solubilities, only for salts that produce the same number of ions upon dissociation.1941

OK, so let's just do a quick example here.1953

If I have silver iodide, well, the K_sp of silver iodide is 1.5x10^-16; if I have barium sulfate, the K_sp is equal to 1.5x10^-9.1956

And, if I have nickel (2) carbonate (or nickel carbonate--I guess nickel is always +2 for the most part), the K_sp is equal to 1.4x10^-7.1973

Well, each of these produces...when Ag dissociates, it's 1 mole of Ag ion and 1 mole of iodide ion; here, barium sulfate--1 mole of barium ion and 1 mole of sulfate ion (sulfate ion--the S, the O₄--they don't dissociate; SO₄ is an ion itself); nickel carbonate--when it dissociates...1 mole of nickel; 1 mole of carbonate.1986

So, in each case, each is producing 2 moles of ion total.2011

Because we can do that, we can compare these K_sps.2016

Well, the higher the K_sp...that means the equilibrium is farther to the right; if it's farther to the right, that means the concentration of the individual ions is higher.2019

Well, a higher concentration of free ion means that more of a salt has dissolved.2029

Therefore, that has a bigger solubility than this; this has a bigger solubility than that.2035

The order of solubilities, in terms of greatest to least, is: nickel carbonate is more soluble than barium sulfate, is more soluble than silver iodide.2041

You can do this, because they produce the same number of ions.2053

Because they produce the same number of ions, you can just compare K_sps; the one with the biggest K_sp is the most soluble--it dissolves the most.2056

Now, let's try...what about copper (2) sulfide (which has a K_sp equal to 8.5x10^-45); how about barium phosphate (it has a K_sp of 6x10^-39--very, very little dissolution); and iron (3) hydroxide (notice, it's FeOH₃, not FeOH₂; OK, K_sp is equal to 4.0x10^-38).2066

When this dissolves, it produces 1 mole of copper, 1 mole of sulfide; this produces 3 moles of barium, 2 moles of phosphate; this produces 1 mole of iron, 3 moles of hydroxide.2110

You can't do it--you cannot compare; you cannot say that this -38 is bigger than this -39, is bigger than this 10^-45.2125

Numerically, it's bigger; but, because these produce different numbers of ions upon dissolution, you cannot do it.2135

If you have a group of ions that produce the same number of ions upon dissolution, then you can use your K_sp values to qualitatively decide which is the most soluble.2143

You will see questions like this, absolutely, on the AP exam.2155

OK, so this was our general introduction to solubility equilibria; we are going to spend, actually, 2 or 3 more lessons on solubility equilibria, because they do tend to get a little bit more involved.2160

Again, it's a lot like buffers and titrations; I definitely want you to understand this, because, if you can understand these equilibria, then most of the rest of chemistry is actually an absolute breeze.2171

So, we'll see you next time; thank you for joining us here at Educator.com; goodbye.2182