[captaincharismaccc]

I am not sure if I have done this correctly.  Any hints would be helpful.

Question:  A 100.0mL solution of 1.00mol/L aqueous sodium sulfate, is added to a 200.0mL solution of 1.00mol/L aqueous barium chlordie.  Barium sulfate precipitates from the mixture .Ksp of barium sulfate is 1.1x10^-10.  Calculate the mass of the precipitate of barium sulfate.


I have calculated the Q/trial ion product and used that to find the mass.

Thanks for any help/hints.


First I set up initial conc of barium ions to be 1.0M and sulfate is 1.0M
Then the final concentrations are barium = (1.0)(200/300) =0.67M
                                               sulfate = (1.0)(100/300) =0.33M

trial ion product = [Ba][SO4]                            n = cv = (0.2211M)(0.3L) = 0.06633 mol
                      =0.67 x 0.33                           MM = 233.40 for barium sulfate
                      = 0.2211 M
                 
m = n (MM) = 15.5 g
                 

[Borek]

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