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How do you prepare 50mL of 0.20 M potassium iodate in 0.080 M sulfuric acid?

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xkat123 | Student, Undergraduate | Honors

Posted September 10, 2013 at 1:05 AM via web

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How do you prepare 50mL of 0.20 M potassium iodate in 0.080 M sulfuric acid?

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llltkl | College Teacher | Valedictorian

Posted September 10, 2013 at 7:45 AM (Answer #1)

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Potassium iodate, `KIO_3` in dilute acid solution is a fairly strong oxidizing agent used in analytical chemistry.

Potassium iodate has a relative molar mass of 214.00.

50 mL 0.20 M solution requires (50*0.2*214)/1000, i.e. 2.14 g salt to be dissolved in the given volume.

Again, molarity of concentrated  sulphuric acid is about 18.4, so 50 mL 0.080 M acid solution requires (50*0.08)/18.4, i.e. 0.22 mL of concentrated sulphuric acid to be dissolved in the given volume.

To prepare 50 mL of 0.20 M potassium iodate in 0.080 M sulfuric acid, take about 30 mL of cold, boiled out distilled water in a 50 mL volumetric flask, dissolve in it 2.14 g of pure and dried `KIO_3` , then add to it 0.22 mL concentrated sulphuric acid. Mix thoroughly, cool and make upto the 50 mL mark with cold, boiled out distilled water, mixing the solution uniformly throughout.

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