evelync12
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A student added 6 cm3 of 0.25 mol dm–3 barium chloride solution to 8 cm3 of 0.15 mol dm–3 sodium sulfate solution.
The student filtered off the precipitate and collected the filtrate.
Give an ionic equation for the formation of the precipitate.
Show by calculation which reagent is in excess.
Calculate the total volume of the other reagent which should be used by the student so that the filtrate contains only one solute.

i don't understand this part (i have all the answers up until this part):
"Calculate the total volume of the other reagent which should be used by the student so that the filtrate contains only one solute."


ANSWERS:
Ba2+ + SO42–  BaSO4
n BaCl2 (6/1000 x 0.25) = 1.5 x 10–3 and n Na2SO4 = (8/1000 x 0.15) = 1.2 x 10–3
and BaCl2 /barium chloride in excess
10 cm3 (of 0.15 mol dm-3 sodium sulfate)
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Pigster
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(Original post by evelync12)
A student added 6 cm3 of 0.25 mol dm–3 barium chloride solution to 8 cm3 of 0.15 mol dm–3 sodium sulfate solution.
The student filtered off the precipitate and collected the filtrate.
Give an ionic equation for the formation of the precipitate.
Show by calculation which reagent is in excess.
Calculate the total volume of the other reagent which should be used by the student so that the filtrate contains only one solute.

i don't understand this part (i have all the answers up until this part):
"Calculate the total volume of the other reagent which should be used by the student so that the filtrate contains only one solute."


ANSWERS:
Ba2+ + SO42–  BaSO4
n BaCl2 (6/1000 x 0.25) = 1.5 x 10–3 and n Na2SO4 = (8/1000 x 0.15) = 1.2 x 10–3
and BaCl2 /barium chloride in excess
10 cm3 (of 0.15 mol dm-3 sodium sulfate)
How many mol are there in 10 cm3 of the Na2SO4?

Since the amount of Na2SO4 is now = the amount of BaCl2, all the Ba2+ is converted to BaSO4(s) and the solution now only contains NaCl(aq).
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evelync12
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(Original post by Pigster)
How many mol are there in 10 cm3 of the Na2SO4?

Since the amount of Na2SO4 is now = the amount of BaCl2, all the Ba2+ is converted to BaSO4(s) and the solution now only contains NaCl(aq).
why in 10cm3?
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Pigster
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(Original post by evelync12)
why in 10cm3?
You have worked out that BaCl2 is in excess.
The Q states, "of the other reagent", i.e. what volume of 0.15 mol dm-3 Na2SO4 is needed to react with all of the BaCl2 (you worked out 0.0015 mol). Go calculator... 0.010 dm3.
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