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

Question

Olivia Anton · Accepted 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.

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#

Caleb Adams · Answer

undefined To find the undissolved mass of salt, we need to determine how much of the salt dissolves in the water. The solubility of strontium chromate in water at room temperature is approximately 0.68 grams per 100 milliliters of water. So, for 1/2 L (500 mL) of water, the maximum amount of strontium chromate that can dissolve is 3.4 grams. Since 3.06 grams of strontium chromate is added, the amount dissolved will be 3.06 grams. Therefore, the undissolved mass of salt is 3.06 grams.