e ,@ A 1. For each titration of the fresh orange juice, record the following dat
ID: 1019395 • Letter: E
Question
e ,@ A 1. For each titration of the fresh orange juice, record the following data in the table below. If you had to perform the fine titration three times, discard the volume that is different. Record volumes using all the digits provided by the lab even if they are zero ra initial volume in the burette50ml 50ml final volume in the burette42.74m!43.34ml volume dispensed in the titration 6.05ml 6.66ml 40ml 6.66ml volume of fresh orange juice40.0ml volume of indicator 4.19 2. For each titration the table below. s different If you had to perfo ecord volumes using all the digits provided vy the lab even it they are zero Coarse Titration Fine Titration #1 Fine Titration #2 50ml 47.85ml 2.16ml 40ml itial v final volume in the burette 45.66ml in the titeati volume of fresh orange juice 40ml div»ol li> table > tbody » tr » td Words: 751Explanation / Answer
The reaction of titration is
ascorbic acid + I2 2 I + dehydroascorbic acid
so each mole of ascorbic acid will react with one mole of Iodine molecule
at the end point
Moles of ascorbic acid = Moles of Iodine molecule
Given:
Fine titration 1: the readings given are
a) the volume of orange juice used should = 40mL
volume of iodine soluiotn used = 6.66
Molarity of acid = 6.66 X 0.015 / 40 = 0.0025 Molar
b) using old orange juice, as per the text given in the last picture
the volume of orange juice used should = 40mL
volume of iodine soluiotn used = 4.19mL
Molarity of acid = 4.19 X 0.015 / 40 = 0.0016 Molar
Molecular weight = 176.12 g / mole
So concentration in mg/mL = 0.0016 X 176.12 = 0.282 mg / mL
Other questions:
1) the concentration of orange juice = 0.25mg / mL
requirement = 75mg
For 1mg volume needed = 1/0.25 = 4mL
Volume needed for 74 mg = 4 X 75 mL = 300mL
the volume in cup =240mL
So the cup of juice required = 300/240 = 1.25
2) The ascorbic acid gets oxidized in presence of air which will decrease its concentration in orange juice
3) the vitamin C per cup = 93.90mg (for grape fruit juice)
the ascorbic acid in half cup = 93.9 / 2 = 46.95 mg
Moles of ascrobic acid = mass of ascorbic acid / molecular weight = 46.95 / 176.12 = 0.267 millimoles
molarity of iodine solution = 0.0150 molar
moles of iodine solution used = moles of ascorbic acid present
milliMoles of iodine = molarity X volume = 0.0150 X volume in mL
0.267 millmoles = 0.0150 X volume in mL
Volume in mL = 17.8 mL of iodine will be required
4) The balanced equation will be:
C6H8O6 + I2 = C6H6O6 + 2H+ +2I-
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