potassium iodate

2009-11-17 12:49 am

我剛做完experiment
但我5知個d結果點解
希望大家可以幫下手~~

1)after adding zinc and dil sulphuric acid into KIO3
a brown solution and black ppt are formed
a purple gas evolved when heating

2)after adding potassium iodide and dil sulphuric acid
gives a brown solution and a purple gas evolved when heating

3)after adding silver nitrate,forms white ppt
and no observation after adding 0.1M nitric acid

5該俾理equations等我理解好d~thx

更新1:

If AgIO3 dissolves in nitric acid,why there is still white precipitates in the solution?

更新2:

點解5可以用HCI一定要用H2SO4??

回答 (1)

老爺子

2009-11-17 1:59 am

✔ 最佳答案

1)
In an acidic medium, potassium iodate is an oxidizing agent.
Zinc reduces KIO3 in an acidic medium to give zinc ions and iodine.
5Zn(s) + 2IO3^-(aq) + 12H+(aq) → 5Zn2+(aq) + I2(aq/s) + 3H2O(l)
Iodine dissolves in water to give a brown solution. When iodine is in excess, the undissolved iodine gives a black precipitate. When heated, a purple gas of iodine vapour is formed.

2)
Potassium iodide reduces KIO3 in an acidic medium, and a brown solution of iodine is formed.
5I^-(aq) + IO3^-(aq) + 6H^+(aq) → 3I2(aq) + 3H2O(l)
When heated, a purple gas of iodine vapour is formed.

3)
Potassium iodate solution and silver nitrate solution react to give a white precipitate of silver iodate.
Ag^+(aq) + IO3^-(aq) → AgIO3(s)
or AgNO3(aq) + KClO3(aq) → AgIO3(s) + KNO3(aq)
The silver iodate precipitate redissolves in nitric acid as soluble iodic acid is formed.
AgIO3(s) + HNO3(aq) → AgNO3(aq) + HIO3(aq)

2009-11-16 20:37:38 補充：
3)
The nitric acid used is too dilute (0.1 M). Therefore, AgIO3 ppt still exists.

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