We are ready to see how the silver chromate dissolves in the solution that contains silver nitrate. It's a very similar problem as part b, just a few subtle differences. We'll write the same reaction that we wrote for part a and part b, and that is that the silver chromate. Solid is on the left side of the equation and the ions it breaks apart into are on the right side of the equation. So nothing different here. In the I C E table, we start out by putting some of the solid into the solution. The solution already contains silver ions. We look up here, we want to figure out how much silver ions are present. Well there's one silver in this silver nitrate, so this is the concentration of the silver ions that goes here. Common mistake for students is for them to see the 2 and want to double this number. This is how much silver's presence before the Silver Chromate has ever had a chance to dissolve. So it has nothing to do with this coefficient. And there's no chromate in part c. So, once again, molar solubility. This is how much the Silver Chromate dissolves. We will produce two S here, and + S here. If you dissolved a little bit, you will still have some left as a solid, you have 0.15 + 2s here, and you have s here. Once again we will do our Ksp expression. Ksp is a silver concentration squared times a chromate concentration. We will plug in the values that we have Ksp is 1.12 x 10 to the -12. Silver's concentration is 0.15 + 2s and that is squared. And chromate is just s. Now here we're going to assume that just a very small amount dissolves. So we can ignore this term, or at least we think we can. We'll check that before we're finished, and we are going to square the 0.15. And then we'll have times s. So we divide both sides by 0.225. And this will give me s. And s = 4.98 times 10 to the -11. So the silver makes a big impact. On the Ksp, on the solubility, sorry, it doesn't change the Ksp at all, has a big impact on the molar solubility, much more so than in part b. Part b the molar solubility just dropped down ever so slightly. It went from what it was in pure water, which was 6.5 times 10 to the -5, and this became 1.0 times 10 to the -5, so it dropped a little bit. Because of this square term, having some silver concentration already present in solution really effected the molar solubility, and this is way lower than what it is in pure water. Now let's check our assumption, was it a valid assumption? Let's look right here. If we double s, okay, so that's going to be about, let's say, 1 times 10 to the-10. That's when you double this number. We double it, it's going to be way too small to even think about when you add it there to change this number of 0.15, it's still definitely 0.15. So that was a very good assumption for that part of the problem.