 Okay, so this, like I said, this one's a follow-up to that last question we did. And I noticed I raised the k, kA expression so we could figure it out again. Okay? So the first thing we need to do for this problem, what is the molar concentration of hypochromous acid if the hydromium ion concentration is 4.01 molar and the anion concentration is 0.125 molar. Well, the first thing we need to do is figure out the kA. And remember, the kA is going to be given to you but you also need to figure out the expression for kA. Okay? So we know that the kA from the table is 2.3 times 10 to the negative 9. And the expression for kA, remember, is the products over the reactants. So we're going to have the concentration of VRO minus the concentration of hydromium ion divided by, so that again equals to kA. In order to get the concentration of hypochromous acid, we're just going to manipulate this equation and isolate this variable. Okay? So we'll just take this, multiply it over there and divide it by kA. Okay? So our new expression will be the concentration of HVRO equals divided by. So now all we need to do is plug those numbers in. VRO minus is 0.125. H3 plus is 0.401. And kA is 2.3 negative 9. So that should give us the concentration of HVRO concentrations for this problem. But if we wanted to do this, it would be a huge concentration for the acid. Unfortunately, this problem isn't going to work out very well, but this is what you get for making up the problem right on the spot. So the number I got was 2.1, 1, 2, 3, 4, 5, 6, 7. 7 mole. Yeah, that's an enormous problem.