AP Chemistry Solubility Equilibria, Part II

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

Today, we are going to continue our discussion of solubility; it's going to be the second part of solubility, and there is going to be one more part after this.0004

This is a very, very important application of the equilibrium concept--of the aqueous equilibrium concept.0011

So, we want to make sure (the problems are sufficiently different among themselves that we want to make sure) that we have a really solid grasp of this, because again, if this acid-base equilibria, solubility equilibria, and subsequent lessons (complex ion equilibria)--if we get a good grasp of what is going on, and if we are able to handle the chemistry, then virtually everything in chemistry and biology will be an absolute walk in the park, I promise you.0017

It really just does get easier when you understand this.0042

OK, let's get started.0044

Today, we are going to talk about something called the common ion effect for solubility.0047

Now, we discussed the common ion effect when we talked about acid-base equilibria earlier; this is pretty much the same thing--it's just an application of Le Chatelier's Principle.0052

I'll go ahead and define it and discuss it really briefly, and we'll just launch into some examples.0062

Common ion effect: the solubility of an ionic solid (salt, in other words) decreases if the solution into which you drop the solid already contains an ion (or ions, actually--I should say "already contains ions," because you might have one or both) common to the solid.0067

OK, so again, the solubility of an ionic solid decreases if the solution into which you drop the solid already contains ions common to the solid.0140

For example, let's say, if you have a beaker, and in there you decide to throw in some sodium chloride (salt), which is completely soluble; so you mix it up, and now you have some free chloride ion and some free sodium ion floating around.0148

Well, now if you end up dropping, let's say, lead chloride or silver chloride (which are not soluble salts--there is a little bit of solubility)--if you drop them in there, because the chloride ion is already in there, it's going to actually suppress the dissociation of the silver chloride.0160

Let's just go ahead and do a problem, and I think it will make more sense, as opposed to doing sort of a run-down explanation, and then doing the problem.0180

As we do the problem, I'll talk about what it is that is actually going on, and I'll draw you a picture of the solution.0188

Example 1: Calculate the solubility (in other words, how much dissolves--an amount per volume) of calcium fluoride (which is CaF₂) in a 0.030 Molar sodium fluoride solution.0194

OK, now: up here in the definition, I said that you have this solution that has a common ion, and then you drop some of the salt in it; it could be the other way around--you could drop the silver chloride in there, and then add, let's say, the sodium chloride to that.0224

It doesn't matter the order in which you do it, because again, once the system comes to equilibrium, that K_sp expression--it is a measure of equilibrium concentration.0237

It really doesn't matter which order you do it; just for the purposes of definition, I decided to choose the fact that the solution already has the chloride ion, and into it I'm dropping the silver chloride.0248

It doesn't matter the order in which you do that.0262

Calculate the solubility of a calcium fluoride in a .030 Molar sodium fluoride solution.0265

OK, so the K_sp of calcium fluoride, which we do need, is equal to 4.0x10^-11; so just by looking at the K_sp, we know that the calcium fluoride is very, very insoluble--very little of it actually dissociates.0272

Well, when we are doing solubility product, as we do any kind of chemistry, we write a reaction, just to get a sense of it, so we know what is going on, so we can follow it.0292

We don't want to do things in our head.0301

Oftentimes, just by writing things down, pieces come together.0303

CaF₂ is going to be in equilibrium with one calcium ion (which is a 2+ charge), and it's going to be in equilibrium with two...I'm sorry, it dissociates into one calcium and two fluoride ion.0306

Now, the K_sp expression for this is equal to Ca²⁺, times F^- squared (right?--law of mass action, equilibrium constant).0322

Here is what is happening: we have this solution, which is .030 Molar sodium fluoride; so I have some sodium ion floating around and fluoride ion floating around, because sodium fluoride is fully soluble salt, like sodium chloride.0334

OK, so there is my .03 Molar; now, I'm going to drop some calcium fluoride into here: that is what I'm doing--I'm taking this calcium fluoride, and I'm dropping it into this solution, and I want to see--I want to calculate how much of it actually dissolves in this solution.0350

Well, we did a solubility problem earlier, in the last lesson, but that was in pure water; now, there is this common ion, fluoride.0373

You see, that is what we mean by common ion: the ion that is common to the salt is already in solution.0383

So, what is going to happen, by Le Chatelier's Principle: since there is fluoride ion already in there, it is as if you take this equilibrium, and it's as if you have increased the concentration of this fluoride ion.0390

What it is actually going to do is: it's going to push the equilibrium that way.0402

What that means is that it's going to suppress the dissociation; another way of thinking about it is: because of the presence of the fluoride ion already in solution, this calcium fluoride, solid--when you drop it in solution, it's not going to dissociate very much--even less that what it normally does (based on this), simply because there is fluoride ion already in solution.0406

If there is fluoride ion already in solution, it's not going to allow it to dissolve.0428

That is the whole idea behind the common ion effect: it tends to push a reaction back, or it tends to pull it forward, depending on what is happening, chemically.0432

Let's go ahead and just run this problem; this is a simple ICE chart, except now, the ICE chart looks slightly different.0442

Again, just let the chemistry--the situation--decide what the ICE chart looks like.0449

Let me write my dissociation again: CaF₂ dissociates into calcium 2+, plus 2 F^-; I have an Initial; I have a Change; I have an Equilibrium.0454

Well, this--it doesn't matter what it is; it's a solid, so we don't really care what happens to it; there is no calcium to begin with in the solution; but the fluoride ion--before (again, Initial--this is before) anything happens, before...when you drop it in, before the system comes to equilibrium, before it dissociates, the fluoride ion concentration is the concentration of fluoride ion already in there.0470

Well, we know that that is 0.030 Molar.0497

Well, now the Change: it starts to dissociate; a certain amount is going to dissolve; a certain amount is going to show up, 1:1; and this is 1:2--2x is going to show up in terms of the fluoride ion (right?--this is 1:1; this is 1:2).0503

For every mole of this that dissociates, for every unit of this, it creates 2 units of fluoride ion; that is what this x and 2x are.0521

Our equilibrium concentrations are: this one doesn't matter; this one is x; and this one is 0.030+2x.0529

These are our equilibrium concentrations of calcium and fluoride respectively; so now, we go ahead, and we put them into the K_sp expression.0539

I don't know if I should do this on the same page or not; let me see...I don't know; that is OK--I'll go ahead and do it on the next page--it's not a problem.0549

OK, so now the K_sp (let me write the expression again)--it equals the calcium 2+ concentration (that is calcium, not copper), and the F^- concentration squared.0563

Well, the calcium concentration is x; the fluoride ion concentration is 0.030+2x (let me make that 2x a little more clear), squared; OK.0576

Now, we are going to have a "squared" here; we are going to end up multiplying it, x; we are going to have a cubic equation; again, for those of you that are dealing with graphing calculators, a cubic equation is not a problem.0593

Or, if you are dealing with mathematical software, a cubic equation is definitely not a problem--you will get your answer back in a microsecond.0603

You know what, when you are sort of in a fast test, and you don't...you sort of want some approximation techniques.0610

We do the same approximation that we did before: we know that calcium fluoride is already not very soluble (we have a K_sp which is 10^-11); we know that the common ion, fluoride, is going to suppress that dissociation even more.0616

So, this x--this x right here--compared to the .030, it's probably going to be really, really small; so for all practical purposes, we can just ignore it.0630

We are going to rewrite this as...put it approximately, and you will see it is a fantastic, fantastic approximation...x times 0.030 squared.0640

And again, it is this x that you drop, not this x.0652

Well, when we do this calculation, we get x equal to 4.4x10^-8; that is the solubility--remember, x--that was the amount that dissolved; that was the amount of calcium that showed up; twice that, is the fluoride that showed up.0656

So, this is in moles per liter--molarity--so 4.4x10^-8 moles of calcium fluoride dissolve if you happen to drop it in a solution which is already .030 Molar sodium fluoride.0680

Again, the presence of the fluoride suppresses the dissociation--doesn't allow the formation--because when it dissociates, it forms fluoride ion.0696

There is already fluoride ion in solution, so it's going to keep it from dissociating.0704

Now, let's go ahead and check the validity of this approximation.0708

We go by the 5% rule; so we'll take 4.4x10^-8, divided by the 0.030 (that is what we compared it to), times 100.0712

Well, get a look at this: this is 0.000148%.0726

This is a percentage; there are actually 2 more decimal places, if you want to just use the decimal.0733

So, .000148% is a lot less than 5%; it's not even measurable, for all practical purposes; so we are absolutely, absolutely justified, in almost every single solubility problem, for reducing this--eliminating it--and just taking the approximation the way that we have done before.0738

We are not doing anything new here.0758

There you go: common ion effect--the only thing that is different is the ICE chart--in the line that says Initial, instead of having just one thing and then 00 for the ions, one of those ions is going to have a number, because there is going to be a certain amount of it in solution when you drop the salt in there.0759

OK, now let's move on to the second aspect of the common ion effect: pH and solubility.0779

As it turns out, the pH of a solution that you drop something in...depending on what you are dropping in, it has an effect on the extent to which it is actually soluble.0791

For example, something called milk of magnesia--maybe your parents might remember it--it is a suspension of magnesium hydroxide.0803

Magnesium hydroxide is not a very soluble salt; however, in acid solution, it becomes almost completely soluble.0811

The idea is: you drink it as this sort of suspension of solid particles in a liquid--you can taste it; it's kind of chalky and stuff--but when it comes into the acid, the hydrochloric acid is acidic solution; it actually dissolves it according to what is necessary.0819

Now, I'm going to show you what the chemistry actually looks like; but let me go ahead and define what is going on first.0835

The pH of a solution can (that doesn't mean "will"; it can) affect the solubility of a salt, because the H⁺ or OH^- ions in solution (right?--because a less-than-7 pH--that means there is a high concentration of hydrogen ion; a pH of greater than 7--there is a high concentration of hydroxide ion--these are free ions floating around) are going to react with the free ions of the slightly dissociated salt, and it's going to change the chemistry.0842

So, because the hydrogen ion or the hydroxide ions in solution will suppress or induce--so it could push it either way--dissociation (dissociation is just dissolving--dissociation is the fancy term for it)...0900

OK, so here is the example: we are going to use calcium hydroxide, simply because magnesium hydroxide is always used; it's the same chemistry.0925

Calcium hydroxide is Ca(OH)₂; it has a K_sp...I'll actually write it in just a minute; but it dissociates into Ca²⁺ + 2 OH^-.0935

It has a K_sp equal to 1.3x10^-6--not very soluble.0948

Now, watch what happens: if we take this and just sort of drop it into normal solution, we can change the pH a couple of ways: we can add hydroxide solution to it and raise the pH, making it more basic; or we can drop acid into the solution, raising the hydrogen ion concentration and dropping the pH below 7.0956

Watch what happens: now, if we add OH (so adding OH^---in other words, raising the pH), we'll suppress dissociation (I'll call it...yes, I'll call it dissociation) by the common ion effect that we just discussed.0976

See: one of the things that is produced upon dissociation is sodium hydroxide; well, if I add hydroxide to the solution, now I'm increasing this hydroxide ion concentration; I'm raising it.1006

By Le Chatelier's Principle, the system is going to adjust in order to lower this hydroxide back down.1018

The only way to lower the hydroxide is to push the equilibrium this way by forming calcium hydroxide, solid--taking hydroxides out of free solution and sequestering them as the solid.1024

It's going to push the equilibrium this way, so it's going to suppress the dissociation.1036

The dissociation is to the right; it will suppress the dissociation by the common ion effect, right?1041

One of the ions produced is the hydroxide; if you add hydroxide, which is a common, or if there happens to be hydroxide ion in solution (like the previous problem, which was already in solution when we dropped it in there), it's going to suppress the dissociation.1054

However, adding H⁺ will increase the solubility, and here is why (you can almost imagine why).1067

It will increase the solubility by reacting with the OH^- formed from dissociation to form water.1077

Well, if you drop this in solution in a certain amount of hydroxide, and now it's floating around in solution a little bit; well, now, if you add hydrogen ion to it (an acid, lowering the pH), that acid that you drop is going to react with the OH.1116

Remember, H⁺ and OH, whenever they are in proximity--they form water.1131

Well, when they form water, this hydroxide is going to be pulled--is going to be eaten up; as this is eaten up, it's going to leave an emptiness here.1134

In order for this K_sp to be viable, this is going to dissociate more and create more hydroxide.1142

As you add more hydrogen ion, it's going to react with the hydroxide; it's going to form water, leaving a dearth of hydroxide; therefore, some more is going to dissolve.1151

Some more is going to dissolve; so the more acid I actually add...some more is going to dissolve.1161

The acid is not actually dissolving the solid; the acid is reacting with the hydroxide, which creates a depletion of hydroxide ion; therefore, by Le Chatelier's Principle, this reaction will move in a direction to produce more hydroxide ion that has been depleted.1166

There is a very big difference: the acid is not eating up the solid when you drop acid into it: it's actually an application of Le Chatelier's Principle that is taking effect.1184

OK, to form water...let me finish this...removing OH (it might be nice if I actually wrote the word so that you could read it)...removing OH^-...pulling the reaction forward.1195

OK, let's see what this actually looks like.1222

OK, so here is what is actually happening: let's see, let's just use one unit of calcium hydroxide, OH^-, OH^-; OK.1228

There is, of course, a bunch of calcium hydroxide, solid, at the bottom; so this is calcium hydroxide, solid.1240

They are in equilibrium with each other; this is a saturated solution--remember, the saturated solution means that the solid is in equilibrium with the free ions according to the dissociation reaction, and there is a solubility product associated with that.1249

The concentration of calcium, times the concentration of free hydroxide squared (because two units are produced upon dissociation)--those two concentrations, multiplied, equals the K_sp for this particular salt at that particular temperature.1261

That is what the K_sp measures.1275

OK, well, here is what happens: when I take hydrogen ion (so in this case, let's add hydrogen ion)--when I add hydrogen ion--when I make the solution acidic, or when it's already acidic and I drop the salt in there, here is what happens.1277

I'll do 2 H⁺s; one of these H⁺s reacts with one of these OH^-: H⁺ + OH^- forms neutral water.1293

That went away; the second H⁺, plus the second OH^-, forms another molecule of water; that went away.1305

It is reacting, and it's eating up the hydroxide; now, what you end up with is a solution that just has calcium in it, and some--again, this is the calcium hydroxide, solid; that is that; and now, basically, because you have this calcium left over, now some more of this is going to dissolve according to: calcium hydroxide goes to calcium 2+, plus OH^-, 2.1315

Because now there is no hydroxide floating around, more of this will dissolve in order to release OH^-, OH^-; and it will release another calcium--the calcium itself doesn't really matter all that much.1357

But, because there is no OH^- in here, it wants to move in this direction; in order to move in this direction to produce hydroxide ion, this has to dissolve--that is the whole idea.1371

OK, let's see: let's do another example here, Example 2, for this particular thing; let's do barium phosphate.1384

Barium phosphate is Bo₃(PO₄)₂; it is in equilibrium with 3 barium ion, plus 2 phosphate ion.1398

So again, the pH is going to affect the solubility of a salt if either the OH^- that you add or the H⁺ that you add (or if those OH^- or H⁺ are already in solution), if they react with one of these ions.1410

That is the whole idea: when they react with one of these ions, they are going to...well, they are either going to suppress it and push the equilibrium that way, or they are going to react with it and take it out of solution by reacting with it, pulling the reaction that way--in other words, making this more soluble or less soluble.1428

Well, if we have barium phosphate in acidic solution, or upon addition (again, the order doesn't matter) of H⁺, the following reaction takes place.1447

You remember phosphate: hydrogen phosphate is a weak acid, which means the phosphate is a reasonably strong conjugate base.1478

If it comes in contact with H⁺, it's going to react with the H⁺ according to: PO₄^3- + H⁺ is going to almost completely form hydrogen phosphate 2-.1487

Again, because this is this is the conjugate...one of the products of the dissociation of the salt happens to be something that will react with hydrogen ion to produce hydrogen phosphate.1504

When it reacts with hydrogen ion, it pulls this out of solution--it sequesters it; it lowers the concentration of free phosphate ion.1514

When you lower the concentration of free phosphate ion, this equilibrium is going to shift in the direction which increases the phosphate ion.1523

The only way to do that is by creating more phosphate ion; the only way to do that is by dissociating, by dissolving.1531

It is going to pull the reaction that way; that is what is going on with pH.1538

pH is about hydrogen ion-hydroxide ion concentration; if one of the ions from the dissolving salt reacts with the hydrogen ion or the hydroxide, the equilibrium is going to shift one way or the other.1546

It is going to suppress dissociation by common ion effect; it is going to pull the reaction forward to your right--it's going to pull the reaction forward by reacting with one of the species that dissolves.1559

OK, now let's go ahead and talk about something called precipitation, itself.1570

We have talked about solubility; we have talked about K_sp; now what happens when we actually mix solutions that have certain salts that, when they come together, they precipitate?1579

You remember way back, in the early part of the course, we talked about precipitation reactions, using the solubility chart to decide which salts will precipitate and which will not?1590

Well, now we are to discuss it quantitatively; we are going to use the molarities and volumes to decide whether a precipitate will form or will not form.1598

Let's talk about precipitation, and we'll do a problem, and we'll finish up this lesson.1609

Precipitation (oh, I'm going to need to spell it right; let's try this one more time): OK, we're going to define something called the ion product.1615

Define the ion product: well, it's very, very easy--it's the same as the K_sp expression, but it represents initial concentrations--initial concentrations, not equilibrium concentrations; initial--very, very important--initial concentrations, not equilibrium concentrations.1634

Do you remember back when we did equilibrium--we were just discussing equilibrium in general?1681

We had this thing called Q; we said it is the concentration of species in that particular flask (or whatever) at any given time (not necessarily at equilibrium).1685

This is analogous; and, in fact, we use Q again.1697

So, let's do...for example, we have been working with calcium fluoride, so let's go ahead and stick with calcium fluoride.1700

2 F^-: let's write the K_sp expression; this is at equilibrium--equals calcium 2+ concentration, times F^- squared (that is the law of mass action).1709

Q (or Q_sp--you can write Q; I tend to write Q_sp, simply just to sort of remind myself that I'm dealing with solubilities--either one of those is fine): it is the initial calcium ion concentration (so I'll put a little subscript 0, which means "initial")--times the initial fluoride ion concentration, squared.1721

Initial means...when you first drop certain amounts of these ions into solution, when you mix solutions, before any reaction takes place, before anything comes together to form solid calcium fluoride (if it does)...that is the initial concentration.1751

So, here is the mathematics of it--well, not the mathematics--here is how you decide.1771

If Q is greater than the K_sp (because you know that the K_sp is defined for every single salt at a given...), then a precipitate will form, and continue to form until the ion concentrations satisfy the K_sp.1779

So again...well, actually, let me write the other one, and then we'll talk about what this really means.1824

If Q is less than K_sp, then no precipitate forms.1833

OK, now recall: K_sp is a measure of the ion concentration in equilibrium with the solid.1849

I want to go ahead and write that down: recall, the K_sp is a measure of free ion concentrations in equilibrium with the solid from which they come.1859

OK, so, for example, let's go ahead and use...well, you know what, let's just go ahead and actually do a problem; again, I think if we do the problem--and we will discuss what is actually going on as we do the problem--I think it will make more sense.1901

OK, so...but remember, K_sp is a measure of the free ion concentrations in equilibrium with the solid from which it comes.1921

So, in the case of calcium fluoride (which is a solid), calcium 2+, plus 2 F^-, this is what we call a saturated solution.1928

A saturated solution is when you have free ions that make up the particular solid, and you have solid that is visible in the solution.1940

Like we said, if you keep dropping a bunch of salt into a certain fixed amount of water, at some point the salt is not going to dissolve anymore; you are just going to stir it around, and the salt is going to settle at the bottom.1951

Well, that means there is some chloride ion; there is some sodium ion; and there is a bunch of solid sodium chloride; there is an equilibrium that is established there.1961

There is a K_sp associated with that.1970

When you...well, let's just actually do the problem; it will make more sense.1972

OK, so let me do this in blue.1977

Example: 750 milliliters of a 3.0x10^-3 molarity silver nitrate solution is mixed with 300.0 milliliters of a 2.5x10^-2 Molar potassium chloride solution.1984

Will a precipitate form?2034

OK, so notice: earlier, we were talking about just taking a certain solution and dropping some solid into it and seeing how much dissolves.2039

Now, what we are doing is: we are taking some solutions of free ion; we are mixing them together; and, if a couple of those ions form in soluble salts, we want to find out, based on how much of each free ion there is in there, if an actual solid will drop to the bottom--if a precipitate will suddenly appear.2052

Let's draw this out: we have a solution; we have 750 milliliters of a silver nitrate.2074

This is a silver nitrate solution, AgNO₃; well, we know that things that have nitrate in them are completely soluble, so what you have is free silver ion and free nitrate ion floating around; it just looks like some water, because it's fully soluble.2081

Now, we also have another solution: we have 300 milliliters of this potassium chloride solution.2097

Well, potassium chloride is also a completely soluble salt, so we write KCl as a solution; but really, what you have is free potassium and free chloride--it's fully dissolved.2103

And again, these are not saturated solutions; there is no solid anywhere--these are just solutions of potassium chloride and silver nitrate.2114

Now, what we do is: we mix them together.2121

OK, when we mix them together, now what we have is, well, 750 milliliters, plus 300; now we have 1050 milliliters of the KCl and the AgNO₃ solution, mixed.2124

Before any reaction takes place, what you have is free silver ion, free nitrate ion, free potassium ion, and free chloride ion.2141

These are the major species in solution; and again, major species--let's go take a look at these major species, and let's see if a reaction is going to take place.2152

Well, silver nitrate--if they slam back into each other, they are just going to dissolve again--they are fully soluble.2161

If potassium and chloride ion slam into each other, they are just going to dissolve again; that doesn't matter.2168

Positive and positive: they are not going to...when they slam into each other, they are just going to bounce right off.2173

So, the only two possibilities are potassium and nitrate coming together, and silver and chloride coming together.2178

Well, potassium nitrate is also soluble; so, if they slam together and they stick, it's just going to dissolve again; but, silver and chloride, when they slam into each other--the K_sp...let me see...the K_sp happens to be (let me actually write it here): the K_sp of AgCl equals 1.6x10^-10.2187

Well, we know from our solubility chart that it is insoluble, so when these slam together, chances are they will precipitate.2213

However, we need to find out if they will precipitate, based on these amounts.2220

These concentrations of free silver and free chloride--they have to equal (the product has to equal) the K_sp before something will precipitate.2225

Remember, that is the whole idea: if it's equal to the K_sp or bigger, it will precipitate; if it's less than that, it hasn't reached the K_sp value; therefore, it is not a saturated solution.2234

The ions are free; there is no precipitate.2245

In this case, if we didn't know from our solubility chart that this is insoluble, we can just take a look at the K_sp and know that this is one of the possibilities, that AgCl may precipitate here, because we are looking at a small K_sp.2247

OK, so let's actually do the math.2262

So now, let's go ahead and write out the possible reaction: AgCl is going to be in equilibrium with Ag⁺ + Cl^-, 1:1.2265

Our K_sp expression equals Ag⁺, Cl^-; our Q_sp is equal to the initial concentration of Ag, times the initial concentration (it's the same as that, except initial concentrations)...now, let's calculate this.2280

We'll calculate this; we'll multiply them together; we'll see what we get.2299

Let's rewrite the K_sp on this page: K_sp equals 1.6x10^-10, for our reference.2303

OK, go ahead and move on to blue.2311

The silver ion concentration is going to equal...well, concentration is moles per liter; the total number of moles of silver, and the total volume.2316

Well, the total volume has changed; I added 750 milliliters to 300 milliliters; so now, that silver ion is not just floating around in 750 milliliters--it's floating around in 1050 milliliters.2325

I need to know how many moles that was.2336

Well, we had 750.0 milliliters of a 2.0 (I'm sorry, this says 3.0, not 2.0)...volume times molarity: 3.0x10^-3 millimoles per milliliter; that is going to give us moles (millimoles), over the total number of liters/total number of milliliters/total volume.2338

Plus 300 milliliters (right?): the moles of silver, divided by the total volume that it is floating around in--you end up with 2.14x10^-3 molarity.2372

Now, let's calculate the initial chloride ion concentration.2388

Well, it's equal to the total chloride ion and total volume; well, the initial chloride solution--we had 300 milliliters of it, and the molarity was 2.5x10^-2 millimoles per milliliter.2392

When you multiply those two, that will give us the moles of chloride ion that we started off with in the second beaker.2412

And now we have mixed them, and again, our total volume is 750 + 300 milliliters, and we get 7.14x10^-3 Molar.2418

Well, Q_sp is equal to that times that (right?--this expression), and there are no exponents, so we're OK.2433

It is going to be 2.14x10^-3, times 7.14x10^-3; that equals 1.52x10^-5, which is definitely bigger than the K_sp, which is 1.6x10^-10; so yes, AgCl (a little arrow pointing down is an old symbol for a precipitate) will precipitate.2447

That is it: so this is a nice, quantitative method of letting us know whether something actually will precipitate, based on the product of the ion concentrations.2482

If it is bigger than the K_sp, then yes, a precipitate will show up.2495

If it is less than the K_sp...well, K_sp measures equilibrium; something that is in equilibrium with the solid from which those ions come--if these numbers, multiplied together, are less than the K_sp, well, that means there is no solid for there to be any equilibrium reached.2499

OK, thank you for joining us here at Educator.com to discuss solubility equilibria, and we have one more lesson to discuss that same subject.2517

See you next time; goodbye.2525