In the previous learning objective, we worked through part a where we determine the molar solubility of copper hydroxide in just water. In this problem, we're going to determine the molar solubility in a solution that's buffered at pH of 9. Now this is a type of common ion effect problem. But it's seems a little strange by then talking about us having a solution that's buffered at a pH of 9. But let's beginning by writing the reaction of the copper hydroxide and water. Okay, what does it do? It dissolves so this is the dissolution into its ions. So this is how we did part A previously. Where it changes is in the I line, because we're not dissolving it in water, we're dissolving it in a buffer with a pH of 9. So let's see what the hydroxide concentration would be if the pH were 9. Well we know that if the PH is 9, then the POH is 14-9. And that would be 5. If we wanted to know the Concentration, that would be 10 to the -5 or, 1.0 x 10 to the -5. So, when we dump some of the solid into the solution it's not into water, but there is no copper present, but there is already some hydroxide present. So we have some of these ion present. Now we will have -s, which would be once we find s that's the molar solubility. This is how much we dissolve, we have s on the side and we'd have +2s on this side. So there's still some of the solid sitting at the bottom, there'd be s here for the copper 2+, and I'm going to put 1.0 x 10 to the -5 and that's all, and not add +s. Why is that? Well, because it tells me it's buffered to that pH. So it is resisting change to pH, it is maintaining this hydroxide concentration. Now we're ready to calculate the s value so we take our 2.2 x 10 to the -20, and that would be the concentration of copper times the concentration of hydroxide squared. We plug in values that we know. Which we have s for this and we have (1.0 x 10 to -5) squared for this, and that will give me s = 2.2 x 10 to -20 divided by 1.0 x 10 to -10. Because -5 squared is to -10. This will give me a solubility of 2.2 x 10 to the -10. That will be the concentration of the copper hydroxide that would dissolve. Now, what did we learn in this unit? We learned that if you have a solution that contains a common ion, it will always decrease the solubility of that salt in the solution. What I've written up here, what we obtained for the value of copper hydroxide solubility in pure water. 1.9 x 10 to the -7. This is one 2.2 x 10 to the -10. That is lower. It has decreased the solubility by having a common ion present in solution.
