For more information, please see full course syllabus of AP Chemistry

For more information, please see full course syllabus of AP Chemistry

## Discussion

## Study Guides

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books & Services

### Related Articles:

### Solubility Equilibria

- Insoluble salts do dissolve, but very, very little. Their Equilibrium constants are called Ksp.
- ICE charts are still used, as these are Equilibrium problems.

### Solubility Equilibria

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Solubility Equilibria 0:48
- Solubility Equilibria Overview
- Solubility Product Constant
- Definition of Solubility
- Definition of Solubility Product
- Example 1
- Example 2
- Example 3
- Relative Solubilities

### AP Chemistry Online Prep Course

### Transcription: 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 KNO _{3}, 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 Mg _{2}^{+} (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: BaOH _{2}, 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

*MgF _{2}, Mg^{2+} + 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: BiI _{3} is in equilibrium with Bi^{3+} + 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 4x ^{3}=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 O _{4}--they don't dissociate; SO_{4} 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 _{sp}s.*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 _{sp}s; 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 FeOH_{3}, not FeOH_{2}; 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

1 answer

Last reply by: Professor Hovasapian

Wed Dec 20, 2017 7:28 AM

Post by Vishal Raman on December 20, 2017

Just for a clarification, you can still compare the solubilities of the ions with not 1:1 ratios, right? You just need to calculate the solubility of each one individually.

1 answer

Last reply by: Professor Hovasapian

Fri Mar 4, 2016 9:46 PM

Post by Nadan Cha on March 4, 2016

Hello Professor,

Firstly, thank you for your lecture -- it is fantastic as always =).

For relative solubilities you said we can decide the order of solubilities only for salts that produce the same number of ions when they dissociate, right?

Do you mean the TOTAL number of ions they produce? So, if a salt produces 1 mol of ion A and 3 mol of ion B, can I compare solubility with a salt that produces 2 mol of ion C and 2 mol of ion D (because they both produce 4 mol of ions in total)?

Thank you so much!

1 answer

Last reply by: Professor Hovasapian

Tue Dec 30, 2014 11:48 PM

Post by Rafael Mojica on December 30, 2014

Please do and Organic Chemistry course, you will save thousands of premed lives!

0 answers

Post by Mohamed Yassin on November 30, 2013

Hi there, why is the connection to the site (Educator.com) is so slow even though i have good connection with other site. is your site under maintenance? why is continuously buffing and reversing as you watch to to the start of the video it is so annoying! please advise what to do.

Thank you

1 answer

Last reply by: Professor Hovasapian

Wed May 15, 2013 1:37 AM

Post by Kendrick Miyano on May 14, 2013

What do you mean by "The higher the Ksp, the farther the equilibrium is to the right,"?