1. The iodide in a sample that also contained chloride was converted to iodate by treatment with an excess of bromine: 3H,O + 3Br, + I¯ → 6Br + I0,¯ + 6H+ The unused bromine was removed by boiling; an excess of barium ion was then added to precipitate the iodate: Ba?+ + 2IO, Ba(IO3), In the analysis of a 2.0375 g sample, 0.0678 this analysis as percent potassium iodate. g barium iodate was recovered. Express the results of

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1. The iodide in a sample that also contained chloride was converted to iodate by treatment with an excess
of bromine:
3H,O + 3Br2 + I¯ → 6Br + I0,¯ + 6H†
The unused bromine was removed by boiling; an excess of barium ion was then added to precipitate the
iodate:
Ba?+ + 2IO, → Ba(IO3)2
In the analysis of a 2.0375 g sample, 0.0678 g of barium iodate was recovered. Express the results of
this analysis as percent potassium iodate.
Transcribed Image Text:1. The iodide in a sample that also contained chloride was converted to iodate by treatment with an excess of bromine: 3H,O + 3Br2 + I¯ → 6Br + I0,¯ + 6H† The unused bromine was removed by boiling; an excess of barium ion was then added to precipitate the iodate: Ba?+ + 2IO, → Ba(IO3)2 In the analysis of a 2.0375 g sample, 0.0678 g of barium iodate was recovered. Express the results of this analysis as percent potassium iodate.
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