you add excess sodium sulfate to a solution of a soluble barium compound in order to precipitate all of the barium ion as barium sulfate, BaSO4. Suppose a 458 mg sample of the barium compound is dissolved to create the solution. Then when the sodium is added, 513 mg of barium sulfate precipitates.
a) how many milligrams of barium ion are in the solution, and therefore in the original barium compound sample?
mg Ba in original sample = mg BaSO4 x (atomic mass Ba/molar mass BaSO4)= ? mg Ba.
% Ba in the sample = (? mg Ba from above/458 mg sample)*100 = ??posted by DrBob222