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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 CaCl_{2} was titrated with 0.015 M sodium sulfate. For what volumes of added titrant solution is precipitate free?
CaSO_{4} pK_{sp} = 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} + SO_{4}^{2} > CaSO4
Ksp = 10^{4.6}
K = Q to initiate precipitation.
Q = Ksp = [Ca^{+2}][SO_{4}^{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 SO_{4}^{2} in 0.015 H_{2}SO_{4 } is about 0.006 M. Assumption 0.015 M H_{2}SO_{4 } is fully dissociated is not correct.

Concentration of SO_{4}^{2} in 0.015 H_{2}SO_{4 } is about 0.006 M. Assumption 0.015 M H_{2}SO_{4 } 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 [Ca^{2+}][SO_{4}^{2}] is again below K_{sp}. That's the meaning of the second root that you ignored.