This problem is going to have several parts as we examine a beaker or a flask that contains some silver ions and some lead ions in the solution. So, let's get a kind of a picture in our mind of what's going on. We've got ourselves a flask, and I'm going to draw a beaker because I'm just a little bit better at drawing beakers than I am at drawing flasks. And in this beaker, there are some silver ions in solution and there's some lead ions in solution. And it tells me the concentration of those two. Now you can have these cations without anions in there as well. So, there's some anions floating around in this solution as well. But we're really not interested in what they are. Then we're going to come along, and we're going to add some sodium chloride to our flask or to our beaker. Now as soon as we add the sodium chloride, that's going to give us some sodium ions in there, and some chloride ions in there. Okay, soon as we add those, the sodium and the silver, and the sodium and the lead won't do anything, they're cations, they're not going to form a compound. But the chloride on the other hand is going to be able to bump into the silver ions or the chloride ions and precipitate out, because these two things, right here, are mentioned as insoluble salts, if you're given a Ksp for them. So, the question is, with their different Ksp values here, which one will precipitate first? Well, if we look at these values, we see that the Ksp of lead chloride is bigger because of this part, is bigger than the Ksp of this. 10 to the -5 is much bigger than 10 to the -10. As a matter of fact, it's a 100,000 times bigger. So, it is much more soluble than the Ksp. So, generally speaking, as Ksp increases, the molar solubility increases. So, when you're comparing salts, their Ksp is not changing, one salt to another. The one with the larger Ksp would have the better, higher molar solubility. Now this could only absolutely be said, if they have the same formula ratio of cation to anion, which we don't have here. This has twice as many chlorides as it has cations, twice as many anions as it has cations. And the silver chloride is a one to one ratio, and so when you've got that difference, If the Ksp's were kind of close, you'd have a hard time saying which one which one was more and which one was least soluble. But for these values, they are so drastically different that I feel very comfortable at saying, that the silver chloride is least soluble. Therefore, this is most soluble. It's least soluble and the silver chloride will therefore precipitate first. It will require less chloride to start off precipitating. So, now we're going to ask, well if that's the case, what's going to be the minimum concentration of sodium chloride that is going to get this to precipitate? Well, for that, we're going to take the Ksp of the silver chloride, and we're going to write what Ksp should be equal to, it would be the silver concentration times the chloride concentration, and we're going to be able to solve for the chloride concentration. That's what we're going to try to determine. That would be the amount of sodium chloride it would take to precipitate. So, let's really ask it that way, what is the minimum concentration of chloride ions that will initiate the precipitation? because it's not the sodium that initiates it, it's the chloride. The Ksp is 1.77 x 10 to the -10. The silver was 0.02 molar and we can therefore solve for the chloride concentration. And that gives me a CO minus concentration of 8.9 x 10 to the -9. That's how much it takes chloride to precipitate, and this will be the first one to precipitate because we established it was the least soluble. But let's verify that it really is the least soluble with certainty. Let's figure out how much chloride it takes to precipitate the lead chloride. If it takes more, then the lead chloride is more soluble. So, we'll write the expression for lead chloride, Ksp, is the lead concentration times the chloride concentration squared. We will put our values in. Ksp for lead chloride was 1.17 x 10 to the -5, lead is 0.02 and we can solve for the Cl-. We divide both sides by 0.02 and that's going to leave me with the chloride concentration. And then, if I squared and then if I take the square root, I will know the chloride concentration. So, Cl- would be equal to 2.4 x 10 to the -2. So, this takes a whole lot more chloride to precipitate the lead chloride than it took chloride to precipitate the silver chloride. So, that verifies what we did in this part where we said the silver chloride precipitate first, it certainly will precipitate first, if it take less of the chloride to get it to precipitate. Now, as we move on with this problem, we're now asked, what is going to be the remaining concentration of that first cation, once the second cation starts to precipitate? Well, what was the first cation to precipitate? We could change this question, what is the remaining concentration of silver ions, once the lead ions start to precipitate, okay? So, that would be the second ion, this would be the first ion. So, that would be the way it could change, based on what we've learned already. What is the remaining concentration of silver ions once the lead ions start to precipitate? So, we're going to take the, since we only know about silver, we're going to take the Ksp expression for silver chloride. And we're going to put in our Ksp value of 1.77 x 10 to the -10. And that is going to, we're trying to figure out the silver concentration. And so we had to put in a number for the chloride concentration. Well, once the lead starts to precipitate, what was the chloride concentration? Let's look back here. Once the led starts to precipitate, Here is the chloride concentration. So, let's put that number in, 2.4 x 10 to the -2. So we divide both sides by 2.4 x 10 to the -2. We will have the silver concentration, and the silver concentration is equal to 7.3 x 10 to the -9. So, once again, kind of get a picture in our mind of what's going on. Okay, so we had this beaker. Okay, and we had some silver ions and lead ions in there. We're going to start adding that sodium chloride to the flask and the first thing to precipitate is the silver chloride. So, it starts settling on the bottom, it's a solid, starts to settle at the bottom. And a little bit more, a little bit more, starts to settle on the bottom. Eventually, as we keep on adding the sodium chloride to here, eventually, we'll start getting our first hint of lead chloride to start to precipitate. As soon as that first hint of lead chloride starts to precipitate, we'll stop and say, are there any more silver ions still in solution? There certainly are lead. Because it just started to precipitate, are there anymore? Well, let's figure it out, so as soon as this first starts to precipitate, we knew it took this much chloride to get it to precipitate, so we put that in, we solve for the silver. So, the silver is going to go from 0.20 all the way down to 7.3 x 10 to the -9, without any lead chloride precipitating. And then finally, some lead chloride will start to precipitate. So, we've separated, we've pulled out most of the silver, not all, but that's what selective precipitation is all about. You're getting one to precipitate first before the next one starts, you get most, not all, you never get all, get most of the first one out before the second one starts to precipitate.
