Question

A 50.0 mL sample of 1.50×10−2 M Na2SO4(aq) is added to 50.0 mL of 1.23×10−2 M Ca(NO3)2(aq).

What percentage of the Ca2+ remains unprecipitated?

What I did

n=CV
Na2SO4 = (1.50×10−2 M)(0.05L)
= 0.00075

Ca(NO3)2 = (1.23×10−2 M)(0.05L)
= 0.000615

0.00075 - 0.000615 = 0.000135

0.000135 / 0.000615 x100% = 21.95%

The answer is incorrect. Thanks

Answer 1

On the surface you pptd ALL of the Ca^2+ didn't you? You had 0.000615 mols to begin and you used 0.000615 so you used all of it and left an excess of 0.000135 mols [SO4]^2-. So the answer is you pptd 100% and 0% remains in solution UNLESS you want to consider the Ksp of CaSO4 in which case you have sulfate as a common ion which decreases the solubility of CaSO4 somewhat.. I don't know what your prof wants but if I gave a problem like this I would expect Ksp to play a role.

Answer 2

There isn't Ksp given in the question. What to do?

Answer 3

Personally, I would look up Ksp for CaSO4, use the excess sulfate as a common ion, calculate the amount of Ca^2+ in solution, then mols Ca^2+/0.000615)100 = % Ca remaining. Remember Ksp uses concentrations so you will need to convert from M to mols before that final calculations is done. Of course, I'm assuming you have had some discussions about Ksp. If not then I would go the simple route and report 0%

Answer 4

Okay I'll try. Thanks for your time