Question

When `3.06* g` of `SrCrO_4` is added to `500 *mL` of water, how many grams of `SrCrO_4` precipitate?

I know the answer,but I can't figure out the solution.

Answer 1

Well, what is the solubility product that governs the solubility of strontium chromate...?

Explanation:

This site here says `K_"sp"=3.60xx10^-5`...these DATA SHOULD have been included in the question...

And so we address the solubility equilibrium...

`Sr^(2+) + CrO_4^(2-) rightleftharpoonsSrCrO_4(s)`

And if `S_(SrCrO_4)="solubility of the salt"`

`K_"sp"=[Sr^(2+)][CrO_4^(2-)]=S^2=K_"sp"=3.60xx10^-5`

And so `S=sqrt(3.60xx10^-5)=6.00xx10^-3*mol*L^-1`..

And this represents a gram solubility of....

`6.00xx10^-3*mol*L^-1xx203.61*g*mol^-1=1.22*g*L^-1` or `0.611*g*"500 mL"`...

And according to this, `(3.06-0.611)*g=2.45*g`

..And this mass will remain UNDISSOLVED....