posted by Jematormal91 on .
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?
What Ksp values are you using? Did they come from your text/notes?
From my textbook,
I'm actually not sure how to solve it, now I think the [Ca+]= 0.0500M
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?
I think I did it wrong...
1.20x10^-5 = [SO42-][0.0350]^2
Then I substituted into:
4.93x10^-5 = [9.80x10^-3][Ca2+]
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.
Thank you for your help!
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