[iamthewalrus]

Question: A 75.0 mL sample of 1.48×10⁻² M Na₂SO₄(aq) is added to 75.0 mL of 1.24×10⁻² M Ca(NO₃)₂ (aq)
What percentage of the Ca²⁺ remains unprecipitated?
(Ksp = 9.1x10^-6)
I tried solving for [Ca2+] using Ksp and then divided moles of Ca2+ in the solution by # of original moles then multiplied by 100. Can't seem to get it though. Help would be much appreciated, thanks!

[Borek]

Calculate initial concentrations of both Ca²⁺ and SO₄^2-. When CaSO₄ precipitates both concentrations go down by the same amount (because of the stoichiometry). They go down till their product hits Ksp. Try to express it as an equation - with one unknown (amount of precipitate). Then it should be a breeze.

[iamthewalrus]

Calculate initial concentrations of both Ca²⁺ and SO₄^2-. When CaSO₄ precipitates both concentrations go down by the same amount (because of the stoichiometry). They go down till their product hits Ksp. Try to express it as an equation - with one unknown (amount of precipitate). Then it should be a breeze.

I'm still having lots of trouble with this question. I got [SO4] = 0.0074 M and [Ca2+] = 0.0062
Then I did ([SO4][Ca2+]) - x = 9.1 x 10^-6
so x = 3.68 x 10^-5 M = amount both concentrations go down by correct?
So I substracted this from the initial concentration, then divided by total, then x 100... my answers make no sense

[Borek]

You have not expressed what is happening as an equation. During precipitation concentration of both Ca²⁺ and SO₄^2- goes down by x. Can you calculate this x?

[iamthewalrus]

(0.00754M-x)(0.0062M-x) = 9.1 x 10^-6 correct!?
using quadratic, solved for x which turned out to be 9.8757 x 10^-3M
So if Ca2+ goes down by this much, I'd subtract it from 0.0062M but this is a negative value?!? :S

[XGen]

Quadratics have two answers. Have you checked the second one?

[iamthewalrus]

Tried that too the second answer turned out negative

[Borek]

Check your math. 9.87x10^-3 is OK, but I got positive second root.

[iamthewalrus]

FINALLY.
Thank you so much for the help everyone.