ata Table Molarity of NaOH Initial Volume on burette 010 M Final Volume of Buret
ID: 1036019 • Letter: A
Question
ata Table Molarity of NaOH Initial Volume on burette 010 M Final Volume of Burette Volume of NaOH used Volume of Ascorbic Acid solution used aort Calculations Calculate the number of moles of NaOH used in the titration. 010 molus Use the balanced chemical equation to calculate the number of moles of ascorbic acid in the 20.0 mL sample. Use the chemical formula to calculate the molar mass of ascorbic acid Calculate the mass of ascorbic acid in the 20.0 ml sample in mg. Use a simple proportion to calculate the mass of ascorbic acid in a tablet if one tablet was used to make 200:0-mL of solution. Calculate the percent error of your measured value compared against the manufacturer's reported value of 200 mg Ascorbic Acid per tablet Page 4 of 5Explanation / Answer
Ans. #1. 1 mol NaOH yields 1 mol OH-.
Moles of OH- consumed = Molarity of NaOH x Volume of solution in liters
= 0.010 M x 0.04808 L
= 0.0004808 mol
#2. Balanced reaction: C6H7O6 (aq) + NaOH (aq) ----> H2O (l) + NaC6H7O6 (aq)
Stoichiometry: 1 mol Vitamin C (ascorbic acid, C6H7O6) is neutralized by 1 mol NaOH.
So, at equivalence point, total number of moles of NaOH consumed is equal to total number of moles of ascorbic acid present in solution.
# formula I: Moles of ascorbic acid = Moles of NaOH = 0.0004808 mol
# Formula II:
M1V1 (NaOH) = M2V2 (ascorbic acid)
Where, M and V represent respective molarities and volumes.
Or, M2 = (M1V1) / V2
Or, M2 = (0.010 M x 48.08 mL) / 20.0 mL = 0.02404 M
Now,
Moles of ascorbic acid = Molarity x Volume in liters
= 0.02404 M x 0.020 L
= 0.0004808 mol
#3. Ascorbic acid: Chemical formula = C6H7O6
Molar mass = 175.12 g/ mol
#4. Mass of ascorbic acid = Moles x Molar mass
= 0.0004808 mol x (175.15 g/ mol)
= 0.084197696 g
= 84.197696 mg
#5. Total mass of ascorbic acid in 250 mL (= 1 tablet) = Mass per 20 mL x 250 mL
= (84.197696 mg / 20 mL) x 250 mL
= 1052.4712 mg
#6. Error = Calculated value – Actual value
= 1052.4712 mg – 1000.0 mg
= 52.4712 mg
Now,
% error = (error / Actual value) x 100
= (52.4712 mg / 1000 mg) x 100
= 5.25 %
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