1) It takes 62.25 ml of .04 M Na2S2O3 solution to completely react with the iodine...
1) It takes 62.25 ml of .04 M Na2S2O3 solution to completely react with the iodine present in a 175 ml idoine solution. How many grams of iodine (I2) were initially present in the sample and what is the final concentration (in mol/L) of I- ions in the solution
Homework Answers
Equation of the reaction: 2
Na2S2O3 + I2
Na2S4O6 + 2 NaI
Moles of Na2S2O3 = molarity * Volume = 0.04 * 62.25 * 10-3 = 2.49 * 10-3 moles
from the equation, moles of I2 = moles of Na2S2O3 / 2 = 2.49 / 2 * 10-3 moles = 1.245 * 10-3 mol
Mass of I2 = moles * molar mass = 1.245 * 10-3 * 254 = 0.316 g
From the equation, [Na2S2O3] = [NaI] = [I-] = 0.04 M
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