# Vitamin C Dcpip

Topics: Ascorbic acid, Acid, Titration Pages: 2 (413 words) Published: February 25, 2013
Vitamin c in fruit juice using dcpip titration

Preparing the standard ascorbic acid solution
1. Weigh out accurately about 0.2 g ascorbic acid and make up to 1 L with distilled water. 2. Calculate the concentration of the ascorbic acid solution: C = n/V = m/M/V.

Sample calculation: mass of ascorbic acid = 0.205 g
C (ascorbic acid) = 0.205/176.12/1.00 = 0.00116 M

Preparing the DCPIP solution
Weigh out accurately approximately 0.24 g DCPIP (Mr = 268.1g/mol) and make up to 1 L with distilled water. This is very hard to dissolve. Leave overnight and check for undissolved powder before continuing. Filter if necessary. C = n/V = m/Mr/V

Sample calculation: mass DCPIP = 0.24 g
Approximate C (DCPIP) = 0.24/268.1/1 = 0.000895 M.

Standardizing the DCPIP
1. Pipette a 10 mL aliquot of the ascorbic acid solution into a flask, and titrate against the DCPIP solution (in the burette) to a persistent pink end point (that lasts for 30 seconds). 2. Calculate the concentration of the DCPIP solution:

CV (ascorbic)= CV (DCPIP)
Sample calculation: Titre = 21.25 mL (DCPIP)
0.00116 x 10.00 = C x 21.25
C (DCPIP) = 0.000545 M

Titration of Vitamin C in solution with DCPIP
1. Pipette 10.00 mL fruit juice into a conical flask and add about 10 mL distilled water. 2. Titrate the juice against the DCPIP in the burette to a pink end point that lasts for 30 seconds. 3. Calculate the mass of ascorbic acid in 100 mL juice. Reaction is 1:1 n (ascorbic acid) = CV (DCPIP)

m/M= CV
m= M(ascorbic) x C(DCPIP) x V(DCPIP)
Sample calculations: titre = 11.90 mL of DCPIP solution m (ascorbic acid) = 268.1 x 0.000895 x 11.90/1000 = 0.002855 g (in 10 mL) m (ascorbic acid)= 0.002855 x 10 = 0.02855 g/100mL (or 28.55 mg/100mL)

Richard Walding

Comment from Daniel Bischa, Pioneer SHS, Mackay, about ensuring any ascorbate salt is in its acidic form: The pKa’s of ascorbic acid are...