Iodine, I2, reacts with aqueous thiosulfate ion in neutral solution according to the balanced equation I2(aq)+2S2O2−3(aq)→S4O2−6(aq)+2I−(aq).
How many grams of I2 are present in a solution if 33.00 mL of 0.165 M Na2S2O3 solution is needed to titrate the I2solution?
I would look at that equation if I were you. Isn't that S2O3^2- and S4O6^2-?
To determine the number of grams of I2 present in the solution, we need to first calculate the number of moles of Na2S2O3 used in the titration. Then we can use the stoichiometry of the reaction to determine the number of moles of I2. Finally, we can convert the moles of I2 to grams using its molar mass.
Here's how you can calculate it step by step:
Step 1: Calculate the number of moles of Na2S2O3 used
Given:
Volume of Na2S2O3 solution (V) = 33.00 mL = 0.03300 L
Concentration of Na2S2O3 solution (C) = 0.165 M
Using the formula: moles = concentration × volume
moles of Na2S2O3 = 0.165 M × 0.03300 L
Step 2: Use stoichiometry to find moles of I2
The balanced equation tells us that 1 mole of I2 reacts with 2 moles of Na2S2O3. Therefore, the number of moles of I2 is half the number of moles of Na2S2O3 used in the titration.
moles of I2 = (moles of Na2S2O3) / 2
Step 3: Convert moles of I2 to grams
To convert from moles to grams, we need to multiply by the molar mass of I2.
The molar mass of I2 is approximately 253.8 g/mol.
grams of I2 = (moles of I2) × (molar mass of I2)
Now, you can plug in the values into the equations above to find the answer.