 Okay, so this one says the solution consists of 0.2 molar magnesium chloride and 0.1 molar copper 2 chloride. Calculate the hydroxide ion concentration that would separate the metal ions as their hydroxide. And that gives the Ksp of magnesium hydroxide as this, Ksp of calcium hydroxide as this. Okay, so with the calcium hydroxide's Ksp being lower, you would expect that to selectively precipitate first. Okay, so even though we're going to be adding the hydroxide's hydroxide ion, the magnesium will not precipitate until all the copper has done precipitate. Is everybody okay with that? Okay, wonderful. So the first thing we want to do is write out the two equations. Okay, so maximum amount of hydroxide ion we can add. Let's think about that first. So in order to think about that, we're going to have to use the Ksp of magnesium hydroxide. Okay, so the Ksp of magnesium hydroxide is going to equal concentration of an Mg2 plus OH minus square like that. So we're looking for the concentration of OH minus. Is everybody okay with what we're doing here? Okay, so let's rearrange this equation to solve for OH minus. So OH minus equals Ksp by the bottom. I think that's squared there. It contains the square root of both sides. And that should give us the hydroxide ion concentration. Is everybody okay with that? So let's just plug each other. So Ksp is 6.3 times 10 to the negative 10 divided by 0.20, based to the 1. 5.6 times 10 to the negative 5th hole. Okay, so that is the maximum amount of hydroxide ion you can add before magnesium will start precip- the magnesium hydroxide will start precipitating. So let's calculate, well, what's the concentration of copper ions left in solution if that's the case? If we have added that much hydroxide, is everybody okay with what we're about to do? Let's calculate the amount of copper ions still left in solution. So I'm going to just replace that question mark with 5.6 times 10 to the negative 5th hole. Can I raise this bottom? Yeah. Okay, so we wanted to figure out, well, copper? Well, we know Ksp equals concentration of Cu2 plus times the concentration of OH minus squared. So we know OH minus concentration, it's right there. We know the Ksp of copper hydroxide, so we should be able to figure out how much copper is left in solution, Cu2 plus copper 2. So if we do this, it's going to be Ksp divided by the concentration of OH minus squared. So Ksp of the copper 2 hydroxide, 2.2 times 10 to the negative 20, divided by 5.6 times 10 to the negative 5 squared, 7.0 is 10 to the negative 12th hole. So a very, very tiny amount of copper still left. So hopefully that convinces you that we did selectively precipitate the copper to hydroxide. So just a tiny, tiny, tiny bit amount of what? Okay, so it's effectively nothing. Is everybody okay? No? Why did we have the copper? If what? So what do we do, concentration of copper in the beginning? Well, I mean, you could have made an ice table and done the same thing if you wanted to. Yeah, this is the unique way of doing it though, okay? Yeah, so it's selective precipitation, so you can do it this way, okay? So the copper just came out, all came out of solution. So if you want to think about the ratio of this relative to this, right? How much copper is left from what we started with? Tiny, tiny, tiny amount, you know? So hopefully that, does that answer your question? Okay, good enough, right?