Answer to Question #63055 in General Chemistry for manar

Question

0.2037g of NMe4I3 was collected in a 125mL flask and 95% ethanol (24mL) was added in the same flask. this sample then was titrated with Na2S2O3 ( Sodium Thiosulphate 0.09015M). the initial volume of Na2S2O3 was 0.9mL and the final one was 6.7

I need to find the moles of I2 in the sample, NMe4I3 determined weight , impurity weight, %purity

Answer 1

Lets find the volume of thiosulphate solution - 6.7 ml - 0.9 ml = 5.8 ml.
Now, we can find the amount of pure thiosulphate (in mols) - 0.0058 l * 0.09015 mol/l = 0.00052 mol.
According to the following equation I2 + 2Na2S2O3 = Na2S4O6 + 2NaI 1 mol of I2 reacts with 2 mols of Na2S2O3. Thus, 0.00052 mols of Na2S2O3 will react with 0.00026 mols of I2.
[I3]- anion gives 1 mol of I2 and 1 mol of I- anion. Thus there was 0.00026 mols of [I3]- anion in the sample or 0.00026 mols of [Me4N]+[I3]- salt (0.1183 g of it).
The weight of impurities is 0.2037g - 0.1183g = 0.0854g.
The % of purity is 0.1183g/0.2037g * 100% = 58.08 wt.%