Vitamin C DCPIP titration

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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 (M r

= 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/M r

/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 blue 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 m

= CV

= 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)

= 0.002855 x 10 = 0.02855 g/100mL (or 28.55 mg/100mL) m (ascorbic acid)

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 about 4.2 and 11.6. If the method requires dissolution into an extraction solution that is moderately acidic then this extraction solution will protonate the ascorbic acid and you will not have the sodium salt but the free acid in solution. As a general rule of thumb, pH of 2 units below the pKa will give approximately 99% protonation.

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