Qualification > Sciences
Edexcel CHEMISTRY DOUBTS!!!!
Meticulous:
People sometimes don't see things. No need to go around showing off your smartness.
ksitna:
why do i keep getting 0.6 then???
ksitna:
no actually 0.0006
Saladin:
--- Quote from: The Gladiator on May 10, 2010, 04:58:35 pm ---I need help with this question:
" 25 cm3 of potassium iodate (V) solution was added to excess potassium iodide solution. The iodine liberated required 30 cm3 of 0.5 mol dm-3 sodium thiosulphate. Calculate the concentration of the potassium iodate (V) solution."
Needed ASAP.
Thanks in advance.
--- End quote ---
Can you please give the exact question paper for this, I need to see the equations of the reaction to give you the answer.
Summer :]:
--- Quote from: The Mysterious Dude on May 10, 2010, 09:09:12 am ---You guys are wrong.
You see, the Bromine and Iodine ions are very powerful reducing agennts compared to that of Chlorine, and would result in further reactions, that would not form other products. http://www.chemguide.co.uk/inorganic/group7/halideions.html#top. This site will help u understand the whole thing.
--- End quote ---
with Bromine, it would just form HBR, MHSO4 (m for the metal), Br2, SO2, H20
check the sulphur's oxidation number in MHSO4 its (+6) and in S02 (its +4) - it has been reduced furthermore where there was a decrease in the oxidation number
with Iodine because its a stronger reducing agent it would cause the sulphur to be reduced further more
so H2S is formed..
again check the sulphur's oxidation number MHSO4 (+6) and in S02 (+4) and H2S (-2) - so has been reduced alot due the to the stronger reducing agent which is Iodine.
but you dont need that for the organic part, just the part about redox reactions of halogens..
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