posted by Anon on .
Sodium sulfate is slowly added to a solution containing 0.0500 M Ca^2+ (aq) and 0.0390 M Ag^+ (aq). What will be the concentration of Ca^2+ (aq) when Ag2SO4(s) begins to precipitate?
What percentage of the Ca^2+ (aq) can be separated from the Ag^+ (aq) by selective precipitation?
ksp for CaSO4 is 4.93*10^-5
ksp for Ag2SO4 is 1.20*10^-5
The concentration of Ca^2+ is 9.86*10^-4. Is this correct?
How do you find the percentage?
No although you got part way home with that answer. I do the following.
Ksp CaSO4 = (Ca^2+)(SO4^2-) = 4.93E-5
What must SO4 be to ppt CaSO4?
(SO4^2-) = (4.93E-5/0.05) = 9.86E-4M.
Ksp Ag2SO4 = (Ag^+)^2(SO4^2-) = 1.2E-5
What must SO4 be to ppt Ag2SO4. That's
(SO4^2-) = [1.2E-5/(0.039)^2] = 7.89E-3 M.
Now this is where the chemical reasoning comes in.
If you have a solution as described in the problem and you add tiny crystals of Na2SO4 (or liquid) a bit at a time, the first salt that exceeds Ksp will ppt first. The above calculation shows CaSO4 requires LESS SO4 than Ag2SO4; therefore, CaSO4 will start pptng first and it will continue pptng as more Na2SO4 is added UNTIL Ksp for Ag2SO4 is exceeded. When will that occur? When the SO4 gets up to where? 7.89E-3 of course. (That's the point at which Ag^2)^2(SO4^2-) > 1.2E-3. So we take 7.89E-3, plug that into CaSO4 Ksp for SO4 and solve for (Ca^2+). Check my work but I obtained about 0.006 and that's approximate at that. That's the answer to part a.
b. Therefore you can obtain pure CaSO4 until Ksp for Ag2SO4 is exceeded but not after that. At that point, you start getting the two mixed sulfates. At that point, Ca^2+ = 0.006M, you had 0.05 to start so you must have pptd 0.05-0.006 = about (remember this is approximate) 0.044. Percent recovery = about (0.044/0.05)*100 = ? about a little less than 90%.