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

1 Answer
Apr 10, 2017

Answer:

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#