15.0 mL of an iodine solution of unknown molarity was titrated against a .001 M solution of S2o32-. The blue black iodine-starch complex just disappeared when 20.2 mL of thiosulfate was added. What is the exact molarity of the Ia solution
I2 + 2S2O3^2- ==> S4O6^2- + 2I^-
mols S2O3^2- = M x L = ?
Convert mols S2O3^- to mols I2 using the coefficients in the balanced equation.
Then M S2O3^2- = mols S2O3^2-/L S2O3^2-
To determine the exact molarity of the iodine (I2) solution, we can use the concept of stoichiometry. In this titration, we know that thiosulfate (S2O32-) reacts with iodine in a 1:2 ratio. That means one mole of thiosulfate reacts with two moles of iodine.
Given the molarity (M) of the thiosulfate solution, which is 0.001 M, and the volume of thiosulfate used, which is 20.2 mL (0.0202 L), we can calculate the number of moles of thiosulfate used:
Moles of thiosulfate = Molarity × Volume
= 0.001 M × 0.0202 L
= 0.0000202 moles of thiosulfate
Since the reaction between thiosulfate and iodine occurs in a 1:2 ratio, the number of moles of iodine is double the moles of thiosulfate used:
Moles of iodine = 2 × Moles of thiosulfate
= 2 × 0.0000202 moles
= 0.0000404 moles of iodine
Now, we can determine the molarity of the iodine solution by dividing the moles of iodine by the volume of the iodine solution used:
Molarity of iodine = Moles of iodine / Volume of iodine solution
= 0.0000404 moles / 0.0150 L
= 0.002693 M
Therefore, the exact molarity of the iodine solution is 0.002693 M.