Prelab Assignment for Lab 12: Analysis of Vitamin C- Part II Name: Date: Lab sec
ID: 551749 • Letter: P
Question
Prelab Assignment for Lab 12: Analysis of Vitamin C- Part II Name: Date: Lab section: A 25.0 mL volume of 0.010 M KIO, containing an excess of KI, is added to a 0.246 g sample of a Real Lemon™ solution containing vitamin C. The yellow-brown solution, caused by the presence of excess 12, is titrated to a colorless starch endpoint with 1 1 .2 ml. of 0.100 M Na,5203- a) How many moles of K103 were added to the Real Lemon™ solution? b) Calculate the moles of I2 that are generated from the KIO c) How many moles of I2 reacted with the 0.100 M Na:S 0? d) How many moles of l2 reacted with the vitamin C in the Real LemonTM sample? e) Calculate the number of moles and grams of vitamin C in the sample. ) Calculate the percent (by mass) of vitamin C in the Real LemonTM sample.Explanation / Answer
5KI + KIO3 + 6HCitr = 6KCitr + 3I2 + H2O
a)
mol of KIO3 = MV = 0.01*25*10^-3 = 0.00025 mol of KIO3 used
b)
from
5KI + KIO3 + 6HCitr = 6KCitr + 3I2 + H2O
3 mol of I2 requires = 3 mol of KIO3
then
0.00025 mol of KIO3 used implies 3*0.00025 = 0.00075 =mol of I2 produced
c)
mol of I2 left when KIO3 is added
mol of Na2S2O3 = MV = 0.1*1
1.2*10^-3 = 0.00112 mol of Na2S2O3
2 Na2S2O3 + I2 = Na2S4O6 + 2 NaI
ratio is 1:2 so
0.00112 mol of Na2S2O3 = 1/2*0.00112 = 0.00056 mol of I2
d)
mol of I2 acutally reacted in I2 = 0.00075 -0.00056 = 0.00019
e)
mol of vitamin C
from
5KI + KIO3 + 6HCitr = 6KCitr + 3I2 + H2O
0.00019 mol of I2 is used in Citric acid
0.00019/2 = 0.000095 mol of Hcitric acid
mass = mol*MW = 0.000095*192.124 = 0.01825 g of citric acid
f)
% Acid = 0.01825/0.246*100 = 7.42%