Students performed Part I of this experiment (Standardizing the DCP solution) as
ID: 554625 • Letter: S
Question
Students performed Part I of this experiment (Standardizing the DCP solution) as described in the procedure section. The weight of the reagent grade AA that was 40.21 mg. The AA was diluted to 50.00mL in a volumetric flask. Three 5.00 mL portions of the AA solution were titrated with 22.64, 24.47, and 23.18 mL of DCP respectively. Calculate the average molarity of the DCP solution and the standard deviation associated with these measurements Check your two answers. (MM Ascorbic Acid-176.124 g/mol) Select Answer(s): 1.01x10-3 M 01.01x10-6 M 2.28x104 M 2.28x10'7 M 9.33x10-4 M 9.33x10 M 9.75x10-4 M 9.75x10-7 M 9.85x10-4NM 9.85x10 M 3.15x10-5 M 3.15x10-8 M 3.86x10-5 M 3.86x10-8 M 4.57x103 M 04.57x10-6 MExplanation / Answer
AA (ascorbic acid) is oxidized by DCP as below.
AAred + DCPox ----------> AAox + DCPred
As per the stoichiometry of the reaction,
1 mole AA = 1 mole DCP.
Mass of AA taken for the experiment = 40.21 mg = (40.21 mg)*(1 g/1000 mg) = 0.04021 g.
Molar mass of AA = 176.124 g/mol; mole(s) of AA taken for the experiment = (0.04021 g)/(176.124 g/mol) = 2.2830*10-4 mol.
Volume of AA prepared = 50.00 mL = (50.00 mL)*(1 L/1000 mL) = 0.05 L; molar concentration of AA prepared = (2.2830*10-4 mole)/(0.05 L) = 4.566*10-3 mol/L = 4.566*10-3 M.
Since AA and DCP react on a 1:1 molar ratio, we can use the dilution equation to calculate the molarity of DCP. Take the first entry as an example.
The dilution equation is
M1*V1 = M2*V2 where M1 = 0.8042 M, V1 = 5.00 mL and V2 = 22.64 mL.
Plug in values and get
(4.566*10-3 M)*(5.00 mL) = M2*(22.64 mL)
====> M2 = 1.0084*10-3 M
Fill in the rest of the table as below.
Molarity of AA (M)
Volume of AA (mL)
Volume of DCP (mL)
Molarity of DCP (M)
Average Molarity of DCP (M)
0.8042
5.00
22.64
1.0084*10-3
1/3*(1.0084*10-3 + 9.3297*10-4 + 9.8490*10-4) = 9.7542*10-4
0.8042
5.00
24.47
9.3297*10-4
0.8042
5.00
23.18
9.8490*10-4
The closest match is 9.75*10-4 M (ans).
The standard deviation is given as
SD = [(1.0084*10-3 – 9.7542*10-4)2 + (9.3297*10-4 – 9.7542*10-4)2 + (9.8490*10-4 – 9.7542*10-4)2]/(3 – 1) M = 3.8598*10-5 M.
The closest match is 3.86*10-5 M.
Therefore, the average molarity of DCP with standard deviation is 9.75*10-4 3.86*10-5 M (ans).
Molarity of AA (M)
Volume of AA (mL)
Volume of DCP (mL)
Molarity of DCP (M)
Average Molarity of DCP (M)
0.8042
5.00
22.64
1.0084*10-3
1/3*(1.0084*10-3 + 9.3297*10-4 + 9.8490*10-4) = 9.7542*10-4
0.8042
5.00
24.47
9.3297*10-4
0.8042
5.00
23.18
9.8490*10-4