You're right. (CrO4^2-) = 0.00114M
Your c1v1 = c2v2 is ok but you should be solving for c2 and not v. You know volume.
c1v1 = c2v2
0.01M*3.00mL = c2*6.00 mL.
c2 = 0.005 = [Cu(NO3)2] initially
Do the same for K2CrO4. You should get
1/2 * 0.006 = 0.003 = K2CrO4.
...Cu(NO3)2 + K2CrO4 ==> CuCrO4 + 2KNO3
I..0.005......0.003........0
C...-x..........-x
E..0.005-x...0.00114
You know Eq is 0.00114 for K2CrO4 you can calculate x and that will tell you how much Cu(NO3)2 was used and that will give you Eq for Cu(NO3)2.
Then plug in and calculate Ksp.
"When 3.00 ml of a 0.00600 M K2CrO4 solution and 3.000 ml of a 0.0100 M Cu(NO3)2 solution are mixed, some solid forms. The mixture is centrifuged and the supernatant is removed. The absorbance of the supernatant, measured at 400 nm in a 1.00 cm cuvette, is 0.286. The extinction coefficient of Cr04^2- at 400 nm is 250.0 M^-1 cm^-1. Calculate Ksp for CuCrO4."
So, I'm not sure how to go about this. My teachers suggested strategy is to find the concentration of CrO4^2-, then the equilibrium amount, initial amount and amount of CrO4^2- used to form CuCrO4 and then move on to Cu^2+.
I rearranged the absorbance equation to get the concentration of CrO4^2- = 1.14x10^-3 M.
Then I attempted using C1V1 = C2V2 to figure out the equilibrium amount of CrO4^2-. I ended up with a V2=15 mls.
That's where I stopped because it didn't sound correct. Any advice would be appreciated.
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