A solution of I2 was standardized with ascorbic acid. Using a .1000 g sample of pure ascorbic acid, 25.32 mL of I2 were required to reach the starch end point. A. What is the molarity of the iodine solution? B. What is the titer of the iodine solution?
See above.
To determine the molarity of the iodine solution and the titer, we need to use the given information and some fundamental concepts in analytical chemistry.
A. To calculate the molarity of the iodine solution, we can use the principle of equivalence between the ascorbic acid and iodine. The balanced equation for the reaction is:
C6H8O6 + I2 -> C6H6O6 + 2HI
From the balanced equation, we can see that one mole of ascorbic acid reacts with one mole of iodine. Therefore, the number of moles of iodine (n) used can be calculated using the equation:
n = weight of ascorbic acid (g) / molar mass of ascorbic acid (g/mol)
Given that the weight of the ascorbic acid sample is 0.1000 g and the molar mass of ascorbic acid is 176.12 g/mol, we can substitute these values into the equation:
n = 0.1000 g / 176.12 g/mol
After calculating, we find that n = 0.000567 mol.
Since 25.32 mL of iodine solution is used, we need to convert it to liters (L):
Volume of iodine solution (L) = 25.32 mL / 1000 mL/L = 0.02532 L
The molarity (M) of the iodine solution can be calculated using the equation:
Molarity (M) = n / Volume of iodine solution (L)
Substituting in the values we have:
Molarity (M) = 0.000567 mol / 0.02532 L
Calculating the result gives Molarity (M) ≈ 0.0224 M.
Therefore, the molarity of the iodine solution is approximately 0.0224 M.
B. The titer of the iodine solution is the number of moles of iodine (I2) per unit volume of the solution. In this case, it can be calculated using the equation:
Titer (M) = Molarity of the iodine solution (M)
Using the molarity we calculated earlier, the titer of the iodine solution is approximately 0.0224 M.