NAME mass of potassium iodate = a sa 3 g Preliminary calculation of molarity of
ID: 552155 • Letter: N
Question
NAME mass of potassium iodate = a sa 3 g Preliminary calculation of molarity of potassium iodate solution DATE 6.5034. m M:: o.anume, O.SL mass of ascorbic acid-2, 251- g Calculation of mg/mL of standard ascorbic acid solution O. a51 a51 360 Ascorbic Acid Titration Calculations Volume of standard ascorbic acid 50 mL Trial 1 Beginning titrant volumea S M,So)-0.oc2ym (ai so) ( S ry· C n CC/M mL M Ending titrant volume 21.5mL M, O.COI mL Total volume 2150 Trial 2 Beginning titrant volume 1.30 mL 90Explanation / Answer
For the redox titration of ascorbic acid, firstly we calculate the molarity of pottasium iodate which is correct and then the concentration of ascorbic acid in standard solution.
Concentration of ascorbic acid in standard = mass/ molar mass/ volume = 251 / 176.12 / 250 = 1.43/250 = .0057 mg/ml
After that we started the titration for the titration calculation we have to calculate the average volume which can be calculated as
Average volume = Total volume / No of trials
Average volume of titrant = 21.50 + 17.90 + 19.50+ 19 / 4
Average volume of titrant = 77.9 / 4 = 19.48 ml
Now average concentration of titrant = 0.0057 X 19.48 = 0.00029 mg
Titration of Ascorbic acid unknown
Average volume of titrant = 11.40 +11.50+11.40 /3
Average volume of titrant used = 34.3 / 3 = 11.43 ml
Now we have to calculate average concentration of ascorbic acid in unknown, so
Volume of titrant in standard = 19.48 ml
Volume of titrant in unknown = 11.43 ml
Volume of titrant in standard X Vitamin C in standard = Volume of titrant in unknown X Vitamin C in unknown
19.48 X 251 = 11.43 X y
19.48 X 251 / 11.43 = y
4889.48 / 11.43 = y
y = 427.78 mg or 0.427 g
Average concentration of ascorbic acid in unknown sample is 0.427 g in 10 ml.
Average concentration of ascorbic acid mg/ml = 427.78 / 10 = 42.78 mg/ml.
Titration of ascorbic acid in fruit juice
Average volume of titrant = 9.20 + 9.35 + 9.45 /3
Average volume of titrant used = 28 / 3 = 9.33 ml
Now we have to calculate average concentration of ascorbic acid in fruit juice, so
Volume of titrant in standard = 19.48 ml
Vitamin C in standard = 251 mg
Volume of titrant in unknown = 9.33 ml
Volume of titrant in standard X Vitamin C in standard = Volume of titrant in unknown X Vitamin C in unknown
19.48 X 251 = 9.33 X y
19.48 X 251 / 9.33 = y
4889.48 / 9.33 = y
y = 524.06 mg or 0.524 g
Average concentration of ascorbic acid in fruit juice is 524.06 mg.
Average concentration of ascorbic acid mg/ml = 524.06 / 30 = 17.47 mg/ml.
Average amount of ascorbic acid in per serving = 17.47 mg
Post lab question
Orange juice or any other juice can not be titrated with acid base titration because redox reaction is better than acid base titration because there are also other acids preasent in fruit juices which interfere in the titration. Triiodide oxidises Vitamin C to dehydroascorbic acid, which react with with starch and form blue-black complex.
In reaction between vitamin C and iodine , iodine is first converted to triiodide by reaction with iodide. Then this triiodide oxidises Vitamin C to dehydroascorbic acid and triiodide get reduced to iodine.
Number of grams of bleach can be calculated as
10.58 X 0.0160 = m/74.44 X 1
0.17 X 74.44 = m
Mass of sodium hypochlorite = 12.65 g/l
In 1000 ml, amount of sodium hypochlorite = 12.65 g
In 1 ml , amount of sodium hypochlorite = 12.65 / 1000 = 0.0126 g
Density of bleach = 1.07 g/ml
Volume of commercial bleach titrated= Volume of diluted bleach X dilution
Volume of commercial bleach titrated = 10 X 1/ 100 = 0.1 ml
Mass of commercial bleach = Density X Volume of commercial bleach titrated
Mass of commercial bleach = 1.07 X 0.1 = 0.107 g
wt% sodium hypochlorite in bleach = Mass of sodium hypochlorite / mass of commercial bleach
wt% sodium hypochlorite in bleach = 0.0126 / 0.107 X 100% = 0.112 X 100%
wt% sodium hypochlorite in bleach = 11.77%
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