What is the solubility of calcium hydroxide in 0.10 molar calcium nitrate? The KSP of calcium hydroxide is given to us at 6.5 times 10 to the negative 6. So get into your mind that calcium and the lower group 2s are insoluble. So how do we do this particular problem? We're going to have to first write a reaction equation just how we do every other problem, equal to re-improving. So calcium, so we're not going to use the calcium nitrate because the nitrate is a spectator ion. All with spectator ion. So it's the calcium hydroxide portion that we're interested in. Okay, let's write the KSP expression. Like that. So in order to do this one, it says, well, it gives us the concentration of CA and of 32 is 0.10 molar. So if you recall, like we said, the HN3 is the spectator ion. So that's going to be the same as the initial concentration of CA2 plus. So everybody okay with that? We're not okay with that. Let's do another reaction, which just to prove to you. Okay, so this thing is soluble. So what is it going to go to? So for every one of these, we have one of these. So for every point one molar of this, we have point one molar of that. Is that true? Okay, so let's do a nice table. Like that. So here we're going to put 0.10 molar. Here, nothing. Here. So nothing here, but plus S, remember, for solubility. And plus S, for solubility. How do you choose? Thank you. So here we're going to have 0.10 plus S. And here we're going to have two S. Okay, so we can simplify this one by removing that. Okay, just because of the 5%. If you do 5% at the end, you can check it out. You'll see it's lower than 5%. So we're going to remove the S there. And so now we have that set. Okay, so let's plug into our equilibrium constant expression and see what S is. So the KSB is going to be the same. So we're going to have S equals 0.10. And that drug side is 2S. Okay, so we're going to have S equals 0. Okay, so. Okay, so. Okay, and the solubility I get is 4.0 times 10. Okay, so. So as you might imagine, when you added to a solution with a common ion, a solubility will be lower than what it was without that common. Because of the Lich-Apliance principle, it's going to be pushing the reaction the other way. So everybody okay with thinking like that? Any questions about this one? Okay, wonderful.