[Borek]

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 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.

[Sophia7X]

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?

[AWK]

Concentration of SO₄^2- in 0.015 H₂SO₄ is about 0.006 M. Assumption 0.015 M H₂SO₄ is fully dissociated is not correct.

[Borek]

Concentration of SO₄^2- 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.

[Borek]

< 0.37 L H2SO4 soln?

This is NOT a correct answer.

[DrCMS]

 Up to 27ml and then after 371ml of the sodium sulphate solution?

[Borek]

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⁺][SO₄^2-] is again below K_sp. That's the meaning of the second root that you ignored.