A iodometric titration calculation?

Question

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

Answer 1

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.

Answer 2

Iodometric Titration Calculations

Answer 3

1 L or 1000 cm3 of 0.100 M hydrochloric acid solution contains 0.1 moles of hydrochloric acid. Therefore, 1 cm3 of 0.100 M hydrochloric acid solution contains 0.1/1000 moles of hydrochloric acid and 13.0 cm3 of 0.100 M hydrochloric acid solution contains 13.0 x (0.1/1000) moles of hydrochloric acid = 0.0013 moles of HCl. HCl reacts with Sr(OH)2 according to the following equation: Sr(OH)2 + 2HCl ---> SrCl2 + 2H2O i.e., each mole of HCl reacts with half the number of moles of Sr(OH)2. So 0.0013 moles of HCl react with 0.00065 moles of Sr(OH)2. The 0.00065 moles of Sr(OH)2 are contained in 10.0 cm3. Therefore 1 cm3 of this same concentration solution will contain 0.00065/10 moles of Sr(OH)2 and 1000 cm3 or 1 L of this same concentration solution will contain 1000 x (0.00065/10) moles of Sr(OH)2 = 0.065 moles of Sr(OH)2. So the concentration of Sr(OH)2 solution is 0.065 M or 0.065 mol/L. (Additionally, if the 10.0 cm3 contained 0.00065 moles of Sr(OH)2, and the molar mass of Sr(OH)2 = 121.63 g/mol, then mass of Sr(OH)2 in 10.0 cm3 = 0.00065 mol x (121.63 g/mol) = 0.079 g of Sr(OH)2.)