Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Problem of the Week Archive => Topic started by: Borek on April 09, 2012, 08:15:51 AM
-
100 mL solution of 0.01 M CaCl2 was titrated with 0.015 M sodium sulfate. For what volumes of added titrant solution is precipitate free?
CaSO4 pKsp = 4.6
Edit: changed sulfuric acid to sodium sulfate, it wasn't my intent to make it complicated because of the sulfuric acid dissociation.
-
Net ionic equation: Ca+2 + SO4-2 --> CaSO4
Ksp = 10-4.6
K = Q to initiate precipitation.
Q = Ksp = [Ca+2][SO4-2]
[0.1 L*0.01M/(0.1 L + V)][0.015 M*V/(0.1 L + V)] = 10-4.6
V = 0.37 L
< 0.37 L H2SO4 soln?
-
Concentration of SO42- in 0.015 H2SO4 is about 0.006 M. Assumption 0.015 M H2SO4 is fully dissociated is not correct.
-
Concentration of SO42- in 0.015 H2SO4 is about 0.006 M. Assumption 0.015 M H2SO4 is fully dissociated is not correct.
Good point. I am slightly modifying the question, otherwise it is getting more complicated than I wanted it to be.
It was obvious sooner or late I am going to miss something.
-
< 0.37 L H2SO4 soln?
This is NOT a correct answer.
-
Up to 27ml and then after 371ml of the sodium sulphate solution?
-
Yes.
Sophia7X - you were quite close. Trick is, when you add more titrant calcium gets diluted - and at some point it is so diluted [Ca2+][SO42-] is again below Ksp. That's the meaning of the second root that you ignored.