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

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

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Answer 1 · 2018-05-22T17:51:25

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

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....