How many grams of I2 are present in a solution if 38.55ml of 0.160M Na2S2O3 solution is needed to titrate I2 the solution?
I2 + 2Na2S2O3 ==> Na2S4O6 + 2NaI
mols Na2S2O3 = M x L = ?
mols I2 = 1/2 mol Na2S2O3 (from the coefficients in the balanced equation).
g I2 = mols I2 x molar mass I2 = ?
To determine the number of grams of I2 present in the given solution, we need to use the concept of stoichiometry. We can set up a balanced equation to represent the reaction between I2 and Na2S2O3:
I2 + 2Na2S2O3 -> 2NaI + Na2S4O6
From the balanced equation, we can see that it takes 1 mole of I2 to react with 2 moles of Na2S2O3. Therefore, the molar ratio between I2 and Na2S2O3 is 1:2.
First, we need to find the number of moles of Na2S2O3 used in the titration. The volume of Na2S2O3 solution used is given as 38.55 mL, and the concentration of Na2S2O3 is 0.160 M (Molarity = moles/volume in liters).
Converting the volume to liters:
38.55 mL = 38.55 mL * (1 L / 1000 mL) = 0.03855 L
Calculating the number of moles of Na2S2O3:
moles of Na2S2O3 = concentration * volume
moles of Na2S2O3 = 0.160 M * 0.03855 L = 0.00617 moles
Since the molar ratio between I2 and Na2S2O3 is 1:2, we can conclude that half the number of moles of Na2S2O3 used will be the moles of I2 present in the solution:
moles of I2 = 0.00617 moles / 2 = 0.00309 moles
Finally, we need to convert the moles of I2 to grams. The molar mass of I2 is approximately 253.8 g/mol:
grams of I2 = moles of I2 * molar mass
grams of I2 = 0.00309 moles * 253.8 g/mol ≈ 0.784 grams
Therefore, approximately 0.784 grams of I2 are present in the solution.