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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

Title: Problem of the week - 09/04/2012
Post 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.
Title: Re: Problem of the week - 09/04/2012
Post by: Sophia7X on April 10, 2012, 06:21:31 PM
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?
Title: Re: Problem of the week - 09/04/2012
Post by: AWK on April 11, 2012, 08:29:39 AM
Concentration of SO42- in 0.015 H2SO4 is about 0.006 M. Assumption 0.015 M H2SO4 is fully dissociated is not correct.
Title: Re: Problem of the week - 09/04/2012
Post by: Borek on April 11, 2012, 09:15:53 AM
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.
Title: Re: Problem of the week - 09/04/2012
Post by: Borek on April 13, 2012, 11:11:09 AM
< 0.37 L H2SO4 soln?

This is NOT a correct answer.
Title: Re: Problem of the week - 09/04/2012
Post by: DrCMS on April 13, 2012, 01:44:51 PM
 Up to 27ml and then after 371ml of the sodium sulphate solution?
Title: Re: Problem of the week - 09/04/2012
Post by: Borek on April 13, 2012, 01:52:07 PM
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.