The Ksp values of BaSO4 and SrSO4 are 1.1x10^-10 and 3.8x10^-7 respectively. Calculate the concentrations of Ba2+, Sr2+, and SO4 2- in a solution that is saturated with BaSO4 and SrSO4.

Answer 1

I would do this. Since the SrSO4 is relatively more soluble than BaSO4 (Ksp about 3,500 times) I would calculate the solubility of SrSO4 in a saturated solution ignoring the solubility of BaSO4 and figure the small solubility of BaSO4 makes very little difference.

So Ksp = (Sr^2+)(SO4^2-) = 3.8E-7 and solve for (Sr^2+) = (SO4^2-) = approx (but you need to do it more accurately) 6E-4. Then that SO4^2- acts as a common ion to make the more insoluble BaSO4 even more insoluble; i.e.,
..........BaO4 ==> Ba^2+ + SO4^2-
I.........solid....0.......6E-4
C.........solid....x........x
E.........solid....x.......6E-4+x

Then plug the E line into Ksp for BaSO4 and solve for (Ba^2+) and use the quadratic formula to solve it; i.e. don't call 6E-4+x = 6E-4 because x will make a small difference. When you finish you will have x = (Ba^2+), you have SO4^2- = 6E-4 or whatever it really is +x, and Sr you have from the first calculation.

Answer 2

Hello,Sorry, May I know what if the Ksp of both substances no different much??

Answer 3

I think that simplifies the problem somewhat but the process is the same.

Answer 4

After thinking about this for 5-6 hours I don't think it simplifies the problem and I don't think the process is the same.