Okay, hopefully you saw how part A and part B was accomplished already, and now you're ready to look for how a solution buffered at a pH of three, which is acidic, would affect the solubility of this salt. So let's begin by writing the reaction, copper hydroxide which is a solid in equilibrium with the copper ions and the hydroxide ions and we solved that before. Do an I,C,E, table, now some of the solid will be place into the solution and there won't be any copper ions, but there will be some hydroxide ions present. We know for, it is buffered at a pH of 3, and we want to know the hydroxide concentration. We can get the pOH, which we know would be 14 minus the 3 or 11, and that would give me a hydroxide of 10 to the minus 11. So we can put that in here, 1 x 10 to the -11. And we would know it, because there's two places after the decimal point, we could know this to two significant figures. So now some of this will dissolve, and I will have some copper ions, and I will have. Don't know why I'm being carried away with that, but it's 2 s, so we still would have some solid in the bottom. We have s for the amount of copper ion, and we're going to put this value here because it is buffered. Now what does that mean? If it's buffered to a pH of three, then even if we're changing the hydroxide there is a side reaction with the buffer going on that is consuming any added and maintaining this pH of three in the system. So I'm not going to change that number. Now I'am ready to write the KSP, KSP which we know is the copper concentration times the hydroxide concentration squared. We can plugin what we know, KSP is 2.2 times 10 to the minus 20, and we can put S in for copper and we can put 1 times to the minus 11 and for the hydroxide and squared. So the molar solubility would be 2.2 times 10 to the -20 divided by 1.0 times 10 to the -22, which is what this squared would be, 1 times 10 to the -22. This would give me a molar solubility of 2.2 times 10, well negative 20 minus 22 would be to the second power. Where that would be 220 molar, okay, so let's think about that number. If we compare it to the others, this is really high, and what we learned is that adding an acid to an insoluble salt will increase the solubility. Well, in pure water, we look up here at this number, this was the molar solubility in pure water. We can really affect the solubility of an insoluble salt by changing the pH. Raising the pH to pH of 3 increased the solubility of the salt very much so.