17.4 Solubility and Ksp

So we've dealt with a ton of equilibria. We dealt with a whole chapter of equilibrium on your last exam. We've now dealt with weak acid/weak base and buffers involved in equilibria. And titration are going to get thrown in the mix. But now we're gonna deal with solubility. So solubility deals with when a salt, that's normally rather insoluble, dissociates into its ions. So we can do this for a couple different things. So we typically and most commonly talk about solubility for things that are rather insoluble in this context. So if something's soluble, like if I say NaCl, well we just assume that dissociates all the way well. It's not really true. Once you hit like a 5 molar concentration of NaCl, you pretty much hit your limit and you add more in it's just going to sink to the bottom. But 5 molar is a ton. And so we just assume it's really soluble. Nothing is really 100 percent soluble. And nothing is really 100 percent insoluble. When I say something is soluble, I just mean it's pretty darn soluble. But never completely. When I say that something's insoluble though, we mean that it's just below a certain point in terms of solubility. So in this case, when we talk about solubility equilibria, we'll be talking about compounds that are sparingly soluble. That aren't very soluble. They dissociate only a tiny amount. And so, if you look here, when this dissociates, what ions do we actually get as products? Awesome. Ag plus and Cl minus. When you're doing a solubility calculation, the first thing you should ever do is write out the reaction and write out the equilibrium constant expression. What expression should I get here? And that's it. Why? Because the reactant is a solid. In a solubility equilibrium, the reactant will always be a solid. And solids don't show up. So essentially we're dividing by one, right? And so in this case, since your reactant is always a solid you never have a denominator here, and they call these Ksp. Sp here stands for solubility product. They call it product because you're only multiplying, not dividing. So that's indicative of Ksp. So if you look here we talked about Ka's, and they were for weak acids dissociating into ions in water. Kb was for weak bases dissociating into ions in water. We even talked about Kw, which was water dissociating into ions in water. And now we're talking about Ksp's, which have ionic solids dissociating into ions in water. And so they're just special reactions. So we say Ksp, you should be like "Oh yeah, the reactants are solid, and the product, those are the ions." Okay. So we did this for AgCl. We could do this for a variety of compounds. We could do this for, in fact lets do it for magnesium hydroxide. What ions would magnesium hydroxide dissociate into? Cool. What would the Ksp expression be? Cool. I'm going to rewrite the Ksp for AgCl over here real quick. Just to kind of save a little room. Cool. Let's do it for one more. What about for bismuth iodide? If I was talking about the Ksp reaction, I would know the solid is the reactant, and what ions does it break up or dissociate into? So, bismuth ions and three iodide ions. And what would the KSP expression be? Times? The biggest mistake students make when they're setting up the Ksp expression, they always want to stick a big fat three in the brackets right there. Is that where coefficients go? No. The reason they want to do that though, is once we set up an ice chart here, you'll find out that his concentration is x and his is 3x, and we'd substitute x right there and 3x right there. But that's after we've plugged in from an ice chart. Originally, there is no three in the brackets. But that's why students often make the mistake. A simple question that doesn't involve a calculation is it might just be, "Which of the following is the Ksp expression for bismuth iodine?" Well there it is. If you put a three in the bracket that's wrong. And that'll be one of the answer choices for sure that they're trying to get you to pick. Cool. You never have a number in brackets. For any equilibrium constants like that. You have charges, I guess, but no coefficients. Okay. Whenever you're doing a calculation involving solubility, again the first two things you should ever do, and I only stress this because students get confused at what type of problem they're doing, but if you start off like this it really simplifies it a lot. Write the reaction. Write the expression. Start there. First two things you do every time. Because after you've done that, it'll make your life a lot easier. So let's look at a couple types of problems we might get. A. A says "What is the molar solubility of AgCl?" And then it gives you the Ksp in parenthesis there. It tells you the Ksp is equal to 1.8 times 10 to the minus 10. Okay it's asking for the molar solubility. Well here's the deal, if I told you that 10 AgCl's dissociated. How many Ag's would you get? Ten. How many Cl's would you get? Ten. What if I told you that a hundred AgCl's dissociated. How many silver ions would you get? 100. And how many chlorides? 100. Cool. The thing is I don't know how many of these dissociates, and rather I don't know what concentration of solid dissociates. But the amount that dissociates is called the solubility. Just whatever concentration of solid dissolves, that's the solubility, or more properly, the molar solubility. And since I don't know how much that is, guess what we call it. X. We're going to call it X. Don't know what it is, so we call it X. And if an X molar concentration of AgCl dissociates, how much silver ions will I form? Plus X. And chloride? Plus X. Before this dissociates though, pure water doesn't have any silver ions in it. And it doesn't have any chloride in it. Their initials are zero. And so at equilibrium, what should be the concentration of silver ions? X. And chloride? X. But notice X we defined as the concentration of solid that dissolves. And it's always defined that. No matter what the ratio is in any case. It's always the concentration of solid that dissolves. X is the molar solubility, in this case. The same X that we've already seen in ice charts for these, is the molar solubility. Because of that a lot of teachers will teach this, and a lot of books will teach this, for just these ice charts instead of using X's, they'll use S's. Just so students remember that here it represents the solubility or molar solubility. But I'm just going to keep using X's. I don't want to change anything up from everything we've done. But X from the ice chart is the molar solubility. So question A here says, "What is the molar solubility?" They're really saying, "What is X from your ice chart?" Great. So if we go and plug in to our Ksp expression, what was the Ksp value given again? And concentration of silver at equilibrium is X. Chloride is also equal to X. And we get equals X squared. And so x equals- Awesome. And that's molarity. Notice, it said what is the molar solubility. The question could have said something slightly different. It could have said, what is the concentration of silver ions in that saturated solution? Well it would have been the same number. Or it could have said what is the chloride ion concentration in that saturated solution. Cool. If we look at question B, question B says what is the molar solubility of Mg(OH)2 and it gives you a Ksp for that as well. Now it says what is the molar solubility, what are we actually solving for? We're solving for x again. Awesome. We've already got the equation for magnesium hydroxide up there. We set it up already. We've already got the Ksp expression. We're halfway done. I'll set up the ice chart again. Pure water technically does have a little bit of hydroxide, but it's usually low enough that we don't care. So what is the change in the magnesium ion concentration when an X molar concentration of Mg(OH)2 dissolves? Plus X. And hydroxide? Plus 2x. And so at equilibrium X and 2X, and that's what gets plugged into our expression. So here the Ksp given was 1.6 times 10 to the negative 12, equals- setting everything up exactly as it is- but my Mg concentration is X and my hydroxide concentration is 2x. And this is why students mistakenly try and put a 2 in right there. Because it often gets substituted in from the ice chart, but itself there's no two there. Awesome. You got to square the 2 as well. And you will get 4X cubed. So that's one thing students often forget as well, is they forget this squares the X and the two. And so here you've got to divide by four and then take the cubed root. How do you take the cube root with your calculator? To the 1/3 power. To the 1/3 power. Or use your cube root button. To take a fourth root? 1/4 power. So but divide by four first and then take the cubed root. Can anybody get X for me here? Seven point what? Oh five, my bad. Awesome. What if instead of saying, what is the molar solubility, they said "What is the magnesium ion concentration in the saturated solution?" It's the same number. What if they said "What is the hydroxide ion concentration in your saturated solution?" It'd be twice as much. What if in this one example they said, "What's the pH in a saturated solution?" Wait a minute are we in the acid/base chapter still? No but as long as hydroxide is involved, we can answer it. Because what is the hydroxide concentration again in a saturated solution? Twice that. And then you can take the negative log to get the pOH and then subtract from 14 to get the pH. Notice I didn't have that option back here because it didn't involve OH. But now that it involves OH, I could totally tie this into the last acid/base chapter. Connect two chapters into one question. Cool. Yes. So yes. In every case X from the ice chart is always the molar solubility regardless. Because it's always the amount- concentration of solid that dissolved. Every time. Question. Or they could ask for either one of these concentrations in the saturated solution. So notice if I said, "What's the magnesium ion concentration in the saturated solution?" Well at equilibrium it's equal to X. So it would actually get the same number. But if I said what's the hydroxide concentration in the saturated solution. At equilibrium it's actually equal to twice that number according to the ice chart. And so they could ask you for the molar solubility, or for either ion concentration, which is related to the molar solubility in some way, shape, or form. Solubility or molar solubility that's it. All right, part C or question C, rather: "The molar solubility of bismuth iodide is 1.32 times to ten to the minus five." What did they just give you? No. But what did they just give you? The molar solubility of Bi(I)3 is 1.32 times ten to the- they gave you X. A lot of students just read the numbers and never realize, "Oh they told me X." They're not asking you for X now, they gave it to you. The molar solubility is 1.32 times ten to the minus 5. What are they asking to calculate? Ksp. And so if you notice, if we look at short cutting this process here, bismuth. How many bismuth ions did we form? One. And so it's 1X to the 1 power. How many iodides did we form? And so that's 3X to the 3 power. Kind of a shortcut. Only works in these types of problems where there's no common ions present already. We'll see that example later. But in this case, we see that our Ksp is equal to 27X to the 1/4. But they told us that X was what? And so in this case what is our Ksp actually come out to? That is a small number. So if you notice, which brings up one thing. Typically the lower the KSP, the lower the solubility. However, you got to be careful. Because notice here, if I actually am asking for which compound has the lowest solubility, I'm asking which compound has the lowest value for x. And notice here we had x squared. Here we had 4x cubed. Here we had 27x to the fourth. And so in this case, because they end up with different solutions involving X, you couldn't just say the one with the lowest Ksp would have the lowest X. You might actually have to solve it out. However this one only broke up into two ions and that resulted in x squared. If you were comparing a bunch of compounds that only broke up into two ions exactly, one of each. Then those you could compare directly because they would always end up with x squared equals the Ksp and whichever one had the lowest Ksp, would have the lowest X. But when they start- notice we get one ion here two ions here. One cation. Two anions. You end up with 4x cubed instead. And you have to divide the Ksp by four and then take a cube root, and that's a different way of solving for x mathematically than we did here. And so just by looking at the Ksp's is not enough to compare them. You might have to actually solve for X to actually compare their solubilities. But as long as they break up into the same number of ions, just compare the Ksp's, don't actually solve for x. Question in the back. Cool. Do you remember PEMDAS? Parentheses, exponents, multiplication, division, addition, subtraction. But so, parentheses- well it's already by itself in the parentheses. Exponents. And then multiplication.