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# A iodometric titration calculation?

1 g of KIO3{potassium iodate} is dissolved in water to gives a 250.00 cm^3 solution.
25.00cm^3 of potassium iodate is pippeted into a conical flask and add about 1.00cm^3 of 0.10gcm^-3 potassium iodine solution. and also add 0.1M sulphuric acid to acidified the mixture. {THUS, Iodine solution is liberatd}

KIO3 + 5KI+3H2SO4====> 3K2SO4 + 3I2 + 3H2O

then the mixture solution is titrated with 25.00cm^3 of
sodium thiosulphate solution to reach the end point.
THe 1 cm^3 of 0.1 Gcm^-3 of starch is also added to the mixture when it is near the end point

HOW CAN I FIND THE the following a,b coefficient in the

equation?
aNa2S2O3 + bI2 ===> products where a and b are the

stoichiometric coefficients

we do not have the molarity of sodium thiosulphate, how can

I find the number of mole of sodium thiosulphate?
also why starch solution should be also freshly prepared?

potassium iodide turns yellow on standing in air.
account for this phenomenon with the u
by Hahaha
Member since:
August 24, 2007
Total points:
42,889 (Level 7)

I tend to believe what you mean by "potassium iodine" is actually "potassium iodide". So the two reactions are:
KIO3 + 5KI + 3H2SO4 → 3K2SO4 + 3I2 + 3H2O
2Na2S2O3(aq) + I2(aq) → Na2S4O6(aq) + 2 NaI(aq)
As you stated in the question title, the second reaction is called iodometry. In the reaction, the thiosulfate anion reacts stoichiometrically with iodine, reducing it to iodide as it is oxidized to tetrathionate.
You are right, we can not find the number of mole of sodium thiosulphate without knowing its concentration.
Starch solution should be freshly prepared, since an old solution may be deteriorated by bacteria, thus may not work as required.
Potassium iodide slowly turns yellow on standing in air, due to the oxidation of the negative iodide ions by dissolved oxygen to iodine.

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