Asked by Jim
how wud u prepare 1L of a 1.30M aqueous solution of barium sulfide
explain
explain
Answers
Answered by
DrBob222
Theoretically I would do it this way.
You want how many mols? That's M x L = mols.
Then mols = grams/molar mass. You have molar mass and mols, solve for grams.
Place that many grams in a 1L volumetric flask, add some water, swirl until dissolved, then make to the mark with distilled water and mix thoroughly.
NOTE: Having said all of that, this won't work because BaS is so insoluble. You simply will not dissolve that much BaS in 1 L of water. You could add some HCl to it to make it dissolve and that will do it; however, that will evolve most if not all of the sulfide ion as H2S and you will have 1.30M BaCl2 and not BaS. My best guess is that this is a made up problem to see how it's done and the author of the problem didn't know enough chemistry to realize that BaS is not soluble.
You want how many mols? That's M x L = mols.
Then mols = grams/molar mass. You have molar mass and mols, solve for grams.
Place that many grams in a 1L volumetric flask, add some water, swirl until dissolved, then make to the mark with distilled water and mix thoroughly.
NOTE: Having said all of that, this won't work because BaS is so insoluble. You simply will not dissolve that much BaS in 1 L of water. You could add some HCl to it to make it dissolve and that will do it; however, that will evolve most if not all of the sulfide ion as H2S and you will have 1.30M BaCl2 and not BaS. My best guess is that this is a made up problem to see how it's done and the author of the problem didn't know enough chemistry to realize that BaS is not soluble.
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