17.5 The Common Ion Effect and Precipitation

Cool. So we've just dealt with solubility, or molar solubility, in water. It never said in water. But the only reason I bring up it's in water is it's always in water. But there's nothing else there. And the reason I bring up nothing else there, is let's look at, you know. Part D here. Part D is listed under what? Common ion effect. If you notice, does pure water already have silver in it? Does it already have chloride ions in it? No. But what would happen to this equilibrium if I dumped a bunch of silver ions into the solution. Which way would it shift this equilibrium? To the left. Which means a lot less solid would be dissolved. It would be a lot less soluble then. Okay. That's called the common ion effect. If I added either one of these ions into my solution it would shift it left, meaning the solid is less soluble now. Common ion effect. So instead of looking at solubility in water, part D says "What is the molar solubility of AgCl in 0.1 molar HCl?" Well what do you know about HCl? Strong acid dissociates how much? All the way. Which of these does it actually have in its solution? Cl. How much? 0.1. Which means before the AgCl even dissociates, there's a ton of chloride already present. Even before this gets going. How much is already in my solution then? 0.1. That's your big difference in the problem here. One of your initial concentrations in a common ion problem is not zero. Before our sparingly soluble compound even dissociates, HCl already had dissociated completely. It's a strong electrolyte. And so in this case, at equilibrium, the chloride concentration is not just X, it's 0.1 plus X. And so when I go back to solve my solution here, it's not just x squared anymore. The Ksp hasn't changed. But what actually gets plugged in here? X and 0.1 plus X. Turns out we're going to ignore the plus X because the Ksp is so small we should expect X to be small. And 0.1 plus a really small number is still 0.1. And so it actually makes the solution a little bit easier. Question. Oh I sure didn't. What were you given, the equilibrium constant, what was it called? Because Ksp of AgCl will have anything to do with hydrochloric acid? No. It always refers to exactly that. Notice, if it's the Ksp- Notice a Ka is always an acid dissociating in water. Kb is always base dissociating in water. Ksp is always ionic solid dissociating into water. I didn't actually, actually include the HCL as part of a reaction. If I did, I don't know what that reaction would be called though. But it wouldn't be called Ksp. And I don't have a constant for that. But where I can account for it, is right there. The fact that HCl dissociates completely and puts a ton of chloride in my solution. So don't try and include it in a reaction here. If you're given Ksp, whether there's anything else there, Ksp would always mean this. No matter what else you add. Ksp of AgCl is always that reaction. Cool. And so in this case we're going to ignore the little plus X right there, because that's going to be insignificant. And it makes it actually much easier to solve for X. What do you get when you solve for X here? And if you recall, the molar solubility without the AgCl was 1.3 times 10 to the minus 5. With a little bit of HCl in there, the solubility is now roughly ten thousand times lower. Big decrease in solubility here because we had one of these ions already present in the solution. A common ion present. Again we call it the common ion effect. Cool. Let's do one more of these. Actually let's do another problem. The last type of problem is called precipitation. What happens in a precipitation reaction, back from first semester? You form a what? You form precipitate. What does a precipitate look like? It's a solid. It's actually the exact opposite of solubility. Solubility shows a solid dissolving. Precipitation would show a solid forming. And here's the deal. The Ksp gives you a measure of the greatest concentrations of ions that could be present in the solution and be at equilibrium. If you go past that, you're not at equilibrium. And that equilibrium would have to shift back this way, to lower these concentrations to being back at equilibrium and being saturated. So the idea is this: cations have a positive charge, anions have a negative charge, and they like each other. If you put them in a solution together, a solution can usually handle a certain amount of both. But once you put too much in, they're going to get close enough to where they're just too attracted each other and they'll bond together and form a precipitate. And for different solids there's different measures of attraction. Kind of think of it this way, if I put 10 men and 10 women in a room, and the 10 men were attracted to the 10 women, and the 10 women were attracted to the 10 men, what is the likelihood we're going to get some couples out of this? Pretty good. But what if I put 10 men and 10 women on the jungles of Asia, spread out over the entire continent. What's the likelihood of anybody finding anybody? Not so good. But once those concentrations get high enough, those cations and anions might actually find each other. And the point at which they find each other is anything higher than the Ksp value. Once you exceed the Ksp, you got too many ions and this equilibriums going to have to shift back to the solid form a precipitate to lower them back down. So if you go over the Ksp, you form a precipitate. Cool. So part E. Precipitation. To a 0.001 molar solution of magnesium nitrate- What do you know about nitrate salts? They're completely soluble. They're strong electrolytes. Okay. To a 0.0001 molar solution of Mg(NO3)2, NaOH was added to a final concentration of 0.001 molar. Did a precipitate form? And then you're given a Ksp in parenthesis. What precipitate am I worried about forming? Well precipitates are typically rather insoluble compounds. What rather insoluble compound are they giving me some indication could possibly form here? Yeah. Mg(OH)2. How do I know it's rather in soluble? They gave me a Ksp for it. And the moment you see a Ksp what should you do? First, two things. Write the reaction and write the expression. And we've already done Mg(OH)2. But the parts I'm going to keep here are write the reaction and write the expression. Is this actually the reaction we're worried about doing? As it's written left to right? No. But this is what Ksp means and if I'm given a Ksp this is what I have to write. I'm actually going to see though, if it goes backwards. But if they give you Ksp, you better write the reaction in the correct direction the way it actually works for Ksp and then write the Ksp expression. And so in this case though, the Ksp really applies to a solution at equilibrium that's saturated. But if you're not at equilibrium, we really shouldn't talk about a K value. We should talk about what instead? A Q instead. That reaction quotient if you recall from the last exam. If you're not at equilibrium, this ratio would equal Q not K. And so we're going to talk about a Qsp in this case. Because honestly, we don't know if we're at equilibrium yet. We added magnesium nitrate. Which common ion does magnesium nitrate put in a solution? Magnesium ions. And magnesium nitrate, all nitrate salts are soluble, so it dissociates completely. And so how much magnesium is actually in our solution right now? Well that's the K- that's the Q- that's the Ksp value. But again magnesium nitrate was added to what final concentration? Yeah. 0.0001. And what was the final concentration of hydroxide in the solution? Well yeah, it came from the NaOH. And what do you know about NaOH? It also dissociates completely to give us a hydroxide concentration of 0.001. Now here's the tricky part. A lot of students can't figure out why we don't do X and 2X. Well X + 2 X assumes that these ions are coming from where? From an X molar concentration of solid dissolving. Is that where these ions are coming from in this problem? No. Where are the ions coming from? We're pouring them into the beaker ourselves. They're not coming from solids at all. We don't even start with solid to begin with. We're trying to see if these ions react together to form some solid. And so students don't realize, "Well why don't I double this?" Well these coefficients only apply a 1:2 ratio assuming they're coming from the reactant. But that's not where they're coming from. They're coming from where we put them in. And that's exactly how much we put in, and we're just going to plug those numbers in. What do you get for Qsp? Anybody get a QSP value? So Qsp and Ksp. What was the Ksp value again? So K and Q, if you write them in alphabetical order, if you guys recall that little trick, which one's bigger? Which one's bigger? 10 to the minus 10 is bigger than 10 to the minus 12, right? Tricky, I know. And so in this case which way does this equilibrium need to shift? To the left. So did any solid precipitate form? You bet. When we exceed higher, go higher than the Ksp, and we have gone higher than it right now, that's when a precipitate forms. So on a question like this, we just said "Did the precipitate form?" But you might get a question like this and give you all this data, and instead of saying "Did a precipitate form?" It might say "Which of the following is true?" And it said "A precipitate forms because K is greater than Q." Is that true? Nope. "A precipitate does not form because K is less than Q." Is that true in this case? No. We just figured out that a precipitate really did form, so that wouldn't be true. But why did the precipitate form? Because Q was greater than K. That would be the correct answer they would be looking for. Cool. We've done all the math. Those are the three principal types of problems. You're usually dealing with just plain old solubility, or then you're dealing with solubility in the presence of a common ion, or now you're dealing with precipitation. Those are the three types of solubility questions, or at least calculations. We have one little conceptual thing to talk about and we'll get out of here.