Question

If `3.06*g` `"strontium chromate"` is added to `1/2*L` of water, what is the UNDISSOLVED mass of salt?

Answer 1

We interrogate the equilibrium:

`SrCrO_4(s) rightleftharpoonsSr^(2+) + CrO_4^(2-)`

I get approx. `2.5*g` strontium chromate as the undissolved mass.

Explanation:

Now we have the volume of the solvent, and we can calculate the molar quantity of strontium chromate. We need another parameter, `K_(sp)` for strontium chromate.......

This site gives `K_(sp)` `SrCrO_4=2.2xx10^-5` (which is higher than I would have expected, I hope the source is kosher).

Given `K_(sp)`, `[Sr^(2+)][CrO_4^(2-)]=2.2xx10^-5`

If `S="solubility of strontium chromate"`, then `S^2=K_(sp)`;

and `S=sqrtK_(sp)=sqrt([Sr^(2+)][CrO_4^(2-)])=4.69xx10^(-3)*mol*L^-1`.

Which leads to a dissolved mass of............

`=1/2*Lxx203.61*g*mol^-1xx4.69xx10^-3*mol*L^-1`

`=0.478*g` with respect to `"strontium chromate"`.

And thus, for a saturated solution, the mass of strontium chromate remaining is `(3.06-0.478)*g=2.58*g`