Homework Help

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

xkat123's profile pic

Posted via web

dislike 1 like

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

1 Answer | Add Yours

llltkl's profile pic

Posted (Answer #1)

dislike 1 like

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.

Sources:

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes