Standardization of iodine solution was prepared using 10.0 mL of a standard, 1.00 mg/mL, ascorbic acid solution. Titration with the iodine solution required 35.80 mL to reach the end point. What is the molarity of the iodine solution?
ascorbic acid + I2 ==> 2I^- + dehydroascorbic acid
mg ascorbic acid = 10 mL x 1 mg/mL = 10 mg.
mols ascorbic acid = 0.01 g/molar mass ascorbic acid
Using the coefficients in the blanced equation, convert mols ascorbic acid to mols I2.
Then M I2 = mols I2/L I2
To find the molarity of the iodine solution, we need to use the concept of stoichiometry, which involves the balanced equation for the reaction.
The balanced equation for the reaction between iodine and ascorbic acid can be represented as follows:
C6H8O6 (ascorbic acid) + I2 (iodine) → C6H6O6 (dehydroascorbic acid) + 2HI (iodide ion)
From the balanced equation, we can see that the ratio between iodine and ascorbic acid is 1:1.
Given that 10.0 mL of a 1.00 mg/mL ascorbic acid solution was used, we can calculate the number of moles of ascorbic acid:
Number of moles of ascorbic acid = volume (in L) × concentration (in mol/L)
= 0.010 L × 1.00 × 10^-3 mol/L
= 1.00 × 10^-5 mol
Since the ratio between iodine and ascorbic acid is 1:1, the number of moles of iodine used can also be considered as 1.00 × 10^-5 mol.
Next, we need to calculate the molarity (concentration) of the iodine solution using the moles of iodine and the volume of the iodine solution used (35.80 mL = 0.03580 L):
Molarity (concentration) of iodine solution = moles of iodine / volume of iodine solution (in L)
= (1.00 × 10^-5 mol) / 0.03580 L
≈ 2.79 × 10^-4 mol/L, or 2.79 × 10^-4 M
Therefore, the molarity of the iodine solution is approximately 2.79 × 10^-4 M.