Question

Describe the procedure you would follow to determine the concentration of Vitamin C in a sample if the volume of DCPIP required to reach the endpoint was more than the 25-mL flask could hold. Note the titration must still be carried out in the flask. Be sure to include any calculations you would need to perform.

Answer 1

DCPIP is an indicator for acid - base reaction. In basic solution it is blue in color and is represented as DCPIP^-1 and in acidic solution it is pink in color and represented by DCPIPH.

Vitamin C is ascorbic acid, The reaction between ascorbic acid and DCPIPH is a redox reaction in which DCPIPH acts as an electron acceptor and ascorbic acid as an electron donor. So, reduction of DCPIPH and oxidation of asbcorbic acid takes place.

DCPIPH + Ascorbic acid ----------> DCPIPH2 + Ascorbate ion

(pink color)                                 (colorless)

From this equation, they react in 1:1 ratio.

Let's say we have a 15 g sample that contains ascobic acid is dissolved in 30 ml of water. let's say we take 1 ml of this solution and titrate it with 0.1% DCPIP solution. Say, 1.5 ml of DCPIP are used to reach the end point.

0.1% DCPIP means 0.1 g of it in 100 ml solution. We could calculate the moles of DCPIP as...

1.5 ml x (0.1g/100ml) x (1mol/290.08g) = 5.17 x 10^-6 mol

Since, they react in 1:1 mol ratio. So,

5.17 x 10^-6 mol DCPIP x (1mol ascorbic acid/1mol DCPIP) = 5.17 x 10^-6 mol ascorbic acid

Since, we used 1 ml that is 0.001 L of ascorbic acid and it has 5.17 x 10^-6 mol. So, the concentration of ascorbic acid is, 5.17 x 10^-6 mol/0.001L = 5.17 x 10^-3 mol/L that is 5.17 x 10^-3 M