a solid sample of impure Ba(OH02 is added to .4000L of .5000M of HBr. the remaining solution was acidic, it was then titrated to equivalence point with .1200L of .5000M NaOH. what mass of pure Ba(OH)2 was present in the sample?

Answer 1

The idea here is that an excess of HBr was added to Ba(OH)2. Some reacted with Ba(OH)2 and some was left over. How much was left over? That amount is what was titrated with the NaOH.
Ba(OH)2 + 2HBr ==> 2H2O + BaBr2 +(xs HBr)
mmoles HBr initially = 400.0 x 0.5000M = 200.0 mmoles.
Amount back titrated = 120.0 mL x 0.5000M NaOH = 60 mmoles.
Amount HBr used up with the Ba(OH)2 reaction is 200.0-60.0 = 140.0 mmoles.
Looking at the reaction, mmoles Ba(OH)2 = 1/2 mmoles HBr = 1/2*140.0 = 70 mmols Ba(OH)2 that reacted.
grams Ba(OH)2 = moles x molar mass = ??

Answer 2

the grams of Ba(OH)2 that reacted was 11.99g. but isn't this still of the impure Ba(OH)2 sample? how do i find out the mass of pure Ba(OH)2?

thanks for you're help so far

Answer 3

No. That's the mass of Ba(OH)2 in the sample, unless of course, the HBr titrated something else in addition to the Ba(OH)2. If we knew the mass of the impure sample, and we don't, the percent Ba(OH)2 could be determined.
%Ba(OH)2 in the sample = (11.99/mass sample)*100 = ??