<p>For the Molarity of Pb2+ problem, I don't see why you have to add the product, because PbSO4 is going to form a precipitate, not any ions. So wouldn't it be only the amount left, not reacted = 0.025M</p>
<p>Yeah its .025M. I actually did the problem this time and noticed that you don't need to know the Ksp per se. You only know that .005 moles of the PbCl2 reacted and formed precipitate. Therefore .005 moles of PbCl2 are left unreacted, which dissolves into Pb+2 ions... which is in .2L of solution (both 100ml solutions were mixed together)!!</p>
<p>Yeah, that's true^^^. Reading my post above now, it seems kinda confusing. The ksp is not needed because Pb2+ as a result of some PbSO4 dissolving is negligible.</p>
<p>I think you would need to know the ksp, mainly because PbCl2 is insoluble as well (Iodide, bromide, and chloride are insoluble with Ag, Hg, Pb). In essence, would PbCl2 not dissociate seeing as it is a precipitate already? </p>
<p>Therefore, you would need to know the exact equilibrium constant to figure this out.</p>
<p>Oh crap, I completely forgot my solubility rules! I was thinking of PbCl2 like some normal soluble compound. This is definitely a trick question if it asked for the lead ion concentration and didn't give any Ksps. UNLESS, it was a MC question and one of the answer choices was "0" or "negligible" or something.</p>
<p>Ugh. I keep thinking of more things when I read my post. Actually, since it said 100ml of a .10M solution of PbCl2.. it could be right before the saturation point, before the precipitate forms. As in Q<Ksp at that moment, would definitely work if the Ksp is really large.</p>
<p>Yes, I guess if you consider that the PbCl2 was found in ions, that would make sense but those free ions would then form PbSO4, which would then form another precipitate, leaving no Pb2+ ions left....</p>
<p>No dude. Look at post #62. .025M of Pb+2 ions would be in the solution if the PbCl2 was right at the saturation point and was found in ions. Like I said only .050 moles of it could react with the .050 moles of H2SO4 to form the PbSO4 precipitate. Gosh we are making this problem much harder than it is. I believe that is how you do it unless, like I said both Pb compounds are above their ksp and then the answer is NONE.</p>
<p>hey do you guys want to do a chat today??</p>
<p>Sure .</p>
<p>pm me abt the chat, thanks :D</p>
<p>a chat would be epic.</p>
<p>Can anyone help me answer these questions?</p>
<p>-a solution of potassium iodide is electrolyzed</p>
<p>-an excess of nitric acid soln is added to a solution of tetraamminecopper(II)sulfate</p>
<p>Thank you.</p>
<p>Electrolyzation (if I'm not mistaken) refers to dissociation by electrical current.
As it doesn't give the metals used for the poles of the apparatus, I'm going to assume just KI goes to K + I.
I have no idea on the second part, and I could very easily be wrong on the first.</p>
<p>KI would yield K+ and I-.</p>
<p>HNO3 + ((NH3)4)(CuSO4) would yield ??????? maybe, HCuSO4 + NH3(NO3)</p>
<p>I'm not sure about the answer the 2nd question. I did not balance the equations.</p>
<p>Okay. the answers are apparently
I- + H2O --> K+ + OH-
and
H+ + Cu(NH3)4 --> Cu +NH4</p>
<p>I had no idea why.</p>
<p>The one about the copper tetraammine is a complex ion formation, which probably won't be on there anymore since they only do three chem reactions now. (Basics, probably one redox, one spectator ion reaction, and one varied one)</p>
<p>i think there will probably be a ppt reaction. they love to see if you know your ions and your solubility rules. :)</p>
<p>any body have rleased exams or audit?</p>
<p>^^Isn't it a bit too late for that?</p>