Question

A solution contains 0.0500 M Ca2 (aq) and 0.0350 M Ag (aq). What will be the concentration of Ca2 (aq) when Ag2SO4(s) begins to precipitate?

What percentage of the Ca2 (aq) can be separated from the Ag (aq) by selective precipitation?

-Solving the 1st one I got [Ca2+]= 5.03x10^-3 M

But, now how do I get the percentage by selective precipitation?

Answer 1

What Ksp values are you using? Did they come from your text/notes?

Answer 2

From my textbook,

Ksp(Ag2SO4)= 1.20x10^-5

Ksp(CaSO4)= 4.93x10^-5

Answer 3

I'm actually not sure how to solve it, now I think the [Ca+]= 0.0500M

Answer 4

Thanks. The values in my old text are different. I can't get your answer to the first part of 5.03E-5. How did you do that?

Answer 5

I think I did it wrong...

Ksp[Ag2SO4]= [SO42-][Ag+]^2

1.20x10^-5 = [SO42-][0.0350]^2

[SO42-}=9.80x10^-3M

Then I substituted into:

Ksp[CaSO4]= [SO42-][Ca2+]

4.93x10^-5 = [9.80x10^-3][Ca2+]

[Ca2+]= 5.03x10^-3M

Answer 6

You did it correctly. I forgot and used my Ksp values and not yours.

The way you do part b is this.
You have Ca^2+ when Ag2SO4 first starts to ppt (that's the 0.00503M). If we had a liter of that stuff, it would contain 0.00503 mols Ca^2+. So you have pptd at that point all but 0.00503; therefore, 0.05-0.00503 is the amount pptd at that point so the percent pptd is [(0.05-0.00503)/0.05]*100 = ? If I punched in the right numbers its approximately 90% (not a good separation technique if our goal was 100% selectivity). But then we should have realized from the close proximity of the Ksp values that we probably could not separate them completely. And we can't.

Answer 7

Thank you for your help!

Answer 8

What is the silver ion concentration in a solution prepared by mixing 375 mL of 0.363 M silver nitrate with 381 mL of 0.430 M sodium chromate? The Ksp of silver chromate is 1.2 × 10-12