Question

1. Let’s assume that during last lab you prepared a solution of NaOH. After standardization you determined the concentration to be 0.113 M. This NaOH was then used to determine the concentration of ascorbic acid in a vitamin C tablet. To do this, a Vitamin C tablet was crushed and 0.200 g were placed in a flask with 50 mL of water. The vitamin C solution was then titrated with the standard NaOH. This was repeated twice, for a total of 3 trials. For each trial, the following volumes of NaOH were used.

Trial 1: 3.55 mL

Trial 2: 3.57 mL

Trial 3: 3.60 mL

Ascorbic acid and NaOH react according to the following balanced equation:

HC₆H₇O₆(aq) + NaOH(aq) à NaC₆H₇O₆ + H₂O(l)

Use this information to complete the table below. Show all work for calculations below

Average Volume of NaOH used:
Moles of NaOH used:
Moles of ascorbic acid:
Moles of Ascorbic acid per gram: (divide the moles from above by the sample mass)
Moles of ascorbic acid in the tablet (If a tablet weighs 1.210 g, use the acid/g from above to convert these grams to moles)
Grams of ascorbic acid in the tablet. (convert the moles in the tablet to grams, hint…the chemical formula for ascorbic acid is above

Answer 1

Balanced equation:
HC6H7O6(aq) + NaOH(aq) ====> NaC6H7O6(aq) + H2O(l)

Trial 1

Voluem of NaOH = 3.55 ml

Concentration of NaOH = 0.113 M

Moles of NaOH = 3.55 x 0.113 /1000 =  0.00040115‬ Moles

Moles of ascorbic acid =    0.00040115‬ Moles

Mass of ascorbic acid presents in 0.2 g =     0.00040115‬ Mol x 176.12 g/mol =  0.07065 g

Moles of Ascorbic acid per gram =  0.07065 g x 1 /0.2= 0.35325 g

Moles of ascorbic acid in the tablet =   0.00040115‬ Mol x 1.210 /0.2 = 0.0024269 Moles

Mass of ascorbic acid in the tablet = 0.0024269 Mol x 176.12 g/mol = 0.4274 g

Trial 2

Voluem of NaOH = 3.57 ml

Concentration of NaOH = 0.113 M

Moles of NaOH = 3.57 x 0.113 /1000 =  0.00040341 Moles

Moles of ascorbic acid =    0.00040341‬ Moles

Mass of ascorbic acid presents in 0.2 g =     0.00040341‬ Mol x 176.12 g/mol =  0.0710485 g

Moles of Ascorbic acid per gram =  0.0710485 g x 1 /0.2= 0.3552g

Moles of ascorbic acid in the tablet =   0.00040341Mol x 1.210 /0.2 = 0.00244057 Moles

Mass of ascorbic acid in the tablet = 0.00244057 Mol x 176.12 g/mol = 0.4298 g

Trial 3

Voluem of NaOH = 3.60 ml

Concentration of NaOH = 0.113 M

Moles of NaOH = 3.60 x 0.113 /1000 =  0.0004068 Moles

Moles of ascorbic acid =    0.0004068‬ Moles

Mass of ascorbic acid presents in 0.2 g =     0.0004068‬ Mol x 176.12 g/mol =  0.0716456 g

Moles of Ascorbic acid per gram =  0.0716456 g x 1 /0.2= 0.35822 g

Moles of ascorbic acid in the tablet =   0.0004068 Mol x 1.210 /0.2 = 0.00246114 Moles

Mass of ascorbic acid in the tablet = 0.00246114 Mol x 176.12 g/mol = 0.4334 g