I'm not sure how to set this question up...

A solution containing 0.015M Sr^2+ and Ba^2+ ions is treated with Na2SO4 and a percipitate containing one metal sulfate forms. Assition of the sulfate ion continues until the more soluble sulfate salt just begins to percipitate. What % of the less soluble metal ion remains in solution?
ksp(SrSO4)= 2.5x10^-7 and ksp(BaSO4)=1.1x10^-10

BaSO4 ---> Ba2+ + SO42¯
Since BaSO4 is more soluble it will percipitate last?

Here is the way to think through this. Follow closely and put in the numbers where they belong.

BaSO4 ==> Ba^2+ + SO4^2-
Ksp = (Ba^2+)(SO4^2-)
So when you start adding sulfate into the mixture, where will BaSO4 start to ppt? That's when Ksp for BaSO4 is first exceeded and that will be when
(SO4^2-) = Ksp/(0.015) = ?
(NOTE: Since you BaSO4 is less soluble than SrSO4 you don't need the above step--I include it because it helps explain what's going on;)

What happens next. As you continue to add sulfate ion drop by drop, BaSO4 will continue to ppt until what? Until the Ksp for SrSO4 is exceeded then SrSO4 starts coming down with BaSO4. When will that first molecule of SrSO4 appear. When the sulfate ion is
(SO4^2-) = Ksp SrSO4/0.015 = ?

At that exact point, what is the (Ba^2+)? That will be Ksp BaSO4/(SO4^2-) for when SrSO4 starts.
That will tell you (Ba^2+) at that point. The remainder of the BaSO4 has pptd. So what's left in terms of percent?
[(Ba^2+) at this point/0.015)]*100 = % un-precipitated. I did a quicky estimate and it's about 0.05% but you should confirm that and 0.05% is NOT the answer; just close if I didn't make a mistake. I only went through it once.

I answered the S + NO problem back on page 3 or 4.

I really suck at Chemistry = [

Thank you for taking the time to explain it to me and answer my many questions! It is very helpful!

I'm surprise AND I disagree with you. Your posts show more work and better thinking than many of the students that post and in most cases you've worked through most of the problem but just stuck on one point near the end and need to jump that one hump. So don't get discouraged. I think you have been doing great.

To determine the percentage of the less soluble metal ion that remains in solution, you need to compare the solubility product constants (Ksp) of the two metal sulfates, SrSO4 and BaSO4.

First, you need to find the concentration of sulfate ions (SO42-) in the solution when the less soluble metal sulfate just begins to precipitate. This is done by assuming that the less soluble sulfate has reached its maximum solubility and the Ksp expression is satisfied.

For SrSO4:
Ksp = [Sr2+][SO42-] = 2.5x10^-7
Assuming the concentration of sulfate ions (SO42-) is 'x', the concentration of Sr2+ can be expressed as (0.015 - x). Substituting these values into the Ksp expression and solving for 'x' gives:
(0.015 - x)x = 2.5x10^-7
x^2 - 0.015x + 2.5x10^-7 = 0
Using the quadratic formula, you can solve for 'x' and find the concentration of SO42- ions when SrSO4 begins to precipitate.

Repeat the same process for BaSO4 to find the concentration of SO42- ions when BaSO4 begins to precipitate, using the given Ksp value of 1.1x10^-10. Assume the concentration of Ba2+ is (0.015 - x) and solve the quadratic equation.

Once you have the concentrations of SO42- ions for both sulfates, you can calculate the percentage of the less soluble metal ion (Sr2+) that remains in solution. This can be determined by expressing the concentration of Sr2+ as (0.015 - x) and finding the percentage relative to the initial concentration of Sr2+ (0.015 M).

By setting up and solving these equations, you can find the concentrations of SO42- ions for each sulfate precipitate and calculate the percentage of the less soluble metal ion remaining in solution.