I have a pre-lab calculation: not entirely sure how to do it: Assume the mass pe
ID: 514330 • Letter: I
Question
I have a pre-lab calculation: not entirely sure how to do it:
Assume the mass percentage of vitamin C (as ascorbic acid, MW 176.12 g/mol) in a tablet is in 70-85 w % range (the mass of a tablet is around 1.2 gram). Base on the data you got in this lab, design the step 1) of Part II, i.e. what size volumetric flask you will use to make the Vitamin C solution in order to control titration volume in reasonable region (like between 10 mL ~ 30 mL KI3).
So Caclualted molarity of KI3 is: 1.042 M KI3
Procedure:
1) Preparation of KI3 (aq) Solution:
Weigh 5 grams of KI (MW 166.00) and dissolve it in 50 ml of DI in a 400 mL beaker. Weigh 1.6 grams of I2 (MW 253.81) and add it to your beaker. Gently swirl your beaker until the potassium iodide and iodine have dissolved (Keep your operation in Hood. Be prepared that this step may take longer time than you think). After all the iodine is dissolved in KI solution (KI3 formed), transfer the solution to a clean amber reagent bottle.
2) Preparation of sodium thiosulfate solution:
Weigh 2 grams of Na2S2O3 (sodium thiosulfate MW 158.09) accurately by analytical balance and transfer it to another 250 ml volumetric flask. Add about 100 ml of water and dissolve your sodium thiosulfate. Once the sodium thiosulfate is dissolved, fill with DI to 250 ml mark.
3) Standardization of KI3 solution:
Transfer 10 ml of the stock sodium thiosulfate solution via a 10 mL volumetric pipette to a beaker. Add about 10-15 drops of the fresh starch indicator solution. Transfer KI3 solution to a 50 ml burette and record the initial volume. Titrate sodium thiosulfate solution until the color of the solution turns blue. The color should persist for at least a 1 min
Our results:
Data Results
Results of KI3 titrated:
Mass of KI was 5.0908 grams
Mass of I2 was 1.6095 grams
Mass of Na2S2O3 was 2.0030 grams
Flask number
1
2
3
4 (if any)
Initial reading, mL
0.520 mL
6.69 mL
13.18 mL
19.29 mL
Final reading, mL
6.60 mL
13.18 mL
19.29 mL
25.35 mL
Vol. of titrant, mL
6.08 mL
6.490 mL
6.110 mL
6.060 mL
I3 - + 2S2O32- 3I- + S4O62-
Hence 1 mole of I3 is paired to 2 moles 2S2O32-
So one mole of KI3 is paired to 2 moles of Na2S2O3
Moles of KI3= (2.0030 grams Na2S2O3) (1 mol Na2S2O3 / 158.09 g Na2S2O3) (1 mol KI3 / 2 mol Na2S2O3)= 0.006334991 mol KI3
Place this over the liters used in the titrant to get molarity
6.083 mL = 0.006083 L
0.006334991 mol KI3 / 0.006083 L = 1.042 M KI3
So 1.042 M KI3
Flask number
1
2
3
4 (if any)
Initial reading, mL
0.520 mL
6.69 mL
13.18 mL
19.29 mL
Final reading, mL
6.60 mL
13.18 mL
19.29 mL
25.35 mL
Vol. of titrant, mL
6.08 mL
6.490 mL
6.110 mL
6.060 mL
Explanation / Answer
Part 1
KI3 is at equilibrium with iodine and iodate ion.
I2(aq) + I-(aq) <-------> I3-(aq)
From above equation we know that 1 mol triodide was produced from one mole of iodine or from one mole of iodide
Number of moles of iodine = mass of iodine / molar mass of iodine
= 1.6095 g / 253.81 g mol-1 = 0.0063 mol
Number of moles of potassium iodide = mass of KI / molar mass of KI
= 5.0908 g / 166.00 g mol-1 = 0.0307 mol
Thus potassium iodide was used in excess and iodine is the limiting reagent i.e. amount of triodide produced depends on the amount of iodine used.
Number of moles of triodide = 0.0063 mol
KI3 reacts with sodium thiosulfate to give sodium thionate.
I3-(aq) + 2S2O32-(aq) 3I-(aq) + S4O62-(aq)
From the above equation we know that for every mole of triodide, 2 mol of thiosulfate are consumed
Number of moles of thiosulfate = 2 X Number of moles of triodide
= 2 X 0.0063 mol = 0.0126 mol
Alternately,
Number of moles of thiosulfate in the prepared solution = mass of Na2S2O3 / molar mass of Na2S2O3
= 2.0030 g / 158.09 g mol-1 = 0.0126 mol
Part 2
Let us look at the chemical reactions in the actual titration. In this iodometric titration following reactions occur:
The triodide oxidises the ascorbic acid
C6H8O6(aq) + I3-(aq)-------> C6H6O6(aq) + 3I-(aq) + 2 H+(aq)
KI3 reacts with sodium thiosulfate to give sodium thionate.
I3-(aq) + 2S2O32-(aq) 3I-(aq) + S4O62-(aq)
The overall reaction is given as:
2I3-(aq) + C6H8O6(aq) + 2S2O32-(aq) -------> C6H6O6(aq) +6I-(aq) + 2H+(aq) + S4O62-(aq)
Number of moles of triodide = Molarity of triodide X volume of triodide
= 1.042 mol L-1 X 30 X 10-3 L (maximum volume of KI3 in reasonable range)
= 0.03126 mol
From the above equation we know that for 2 mol of triodide, 1 mol of ascorbic acid is consumed
number of moles of ascorbic acid required for maximum volume of KI3 = ½ Number of moles of triodide
= 0.03126 /2 = 0.01563 mol
Concentration of ascorbic acid required for maximum volume of KI3 should be 0.01563 mol in 1 L or 0.01563 M
The sample contains 70-85 wt% of ascorbic acid
mass of ascorbic acid in the sample (range) = wt% of ascorbic acid X weight of sample
= 70 /100 X 1.2 g to 85 /100 X 1.2 g
= 0.84- 1.02 g
number of moles of ascorbic acid (range) = mass of ascorbic acid / molar mass of ascorbic acid
= 0.84 g / 176.12 g mol-1 to 1.02 g / 176.12 g mol-1 = 0.0048 to 0.0058 mol
The dilution volume (range) = number of moles of ascorbic acid / Concentration of ascorbic acid
= 0.0048 mol / 0.01563 mol L-1 to 0.0058 / 0.01563 mol L-1
= 0.307 L to 0.371L
= 307 mL – 371 mL
Hence 400 mL volumetric flask needs to be used to dilute ascorbic acid.
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