How many ml of water should be added to 5000 ml of sulfuric acid solution with a titer of 0.009001 so that 1ml of the resulting solution corresponds to 0.01 g of barium sulfate?
BaSO4->H2SO4, so 0.01 g of baso4 corresponds to 98*(0.01/233) = 4.291845*10^-5 g of H2SO4 - that's the resulting solution titer (i. e. 4.291845*10^-5 g/ml),
One must dilute original solution as much as 0.009001/4.291845*10^-5 = 209.723 times higher,
The resulting solution volume is 5000*209.723 = 1048615 ml,
BaSO4->H2SO4, so 0.01 g of baso4 corresponds to 98*(0.01/233) = 4.291845*10^-5 g of H2SO4 - that's the resulting solution titer (i. e. 4.291845*10^-5 g/ml),
One must dilute original solution as much as 0.009001/4.291845*10^-5 = 209.723 times higher,
The resulting solution volume is 5000*209.723 = 1048615 ml,
Water to be added (ml) = 1048615-5000 = 1043615.