Considering the volumes and the concentrations used in mixture 1, what percentag
ID: 935955 • Letter: C
Question
Considering the volumes and the concentrations used in mixture 1, what percentage of the moles of H2O2 present have been consumed during the timed reaction? Calculate the number of moles of H2O2 present initially and determine how many have reacted with iodide when the thiosulfate runs out. How does the I- concentration change during this same time?
Mixture 1: 75mL water, 30mL 0.05M HC2H3O2 - NaC2H3O2 (buffer, 0.05M of each), 25mL 0.05m KI, 5mL 0.1% starch, 5mL 0.05M NaS2O3, 10mL 1.02M H2O2. (standardized in class and this was the average of the class).
Explanation / Answer
Total volume of the mixture = 75+30+25+5+5+10 = 150 mL
Moles of each important species in the mixture is:
KI = 0.05*0.025 = 0.00125 mol
Na2S2O3 = 0.05*0.005 = 0.00025 mol
H2O2 = 1.02*0.01 = 0.0102 mol
I2 and S2O32- reacts according to the following equation:
I2 + S2O32- ----> 2I- + S4O62-
Iodide and H2O2 reacts according to the following equation:
2I- + 2H+ + H2O2 -----> I2 + 2H2O
Overall, one mole of H2O2 consumes one mole of S2O32- . Therefore, when S2O32- ions run out, 0.00025*2 = 0.0005 mol of I- reacted with 0.00025 mole of H2O2.
Initial mole of H2O2 present in the mixture = 0.0102 moles
Amount of H2O2 reacted with I- when the S2O32- ions run out is 0.00025 mol
In the presence of S2O32- ion, the amount of I- ions are oxidised by H2O2 and the Iodine, I2 produced are immediately consumed by S2O32- ions and are converted to I- ions. The amount of I- ions consumed by H2O2 is equal to that produced by S2O32- .