Question

A selection of commercial fruit juices and fruit extracts were analysed for ascorbic acid content.

A student took kiwifruit juice to analyse. This juice is reputed to have ascorbic acid concentrations about 3 times greater than that of orange juice. The student diluted the juice by taking a 25.00 mL sample to a final volume of 100.0 mL. The student analysed 25.00 mL aliquots of the diluted juice solution for ascorbic acid by titration using a standardised iodine solution.

Data:
Volume juice for dilution: 25.00 mL
Final diluted volume: 100.0 mL
Volume of diluted juice analysed: 25.00 mL
Standard Iodine solution concentration: 0.00496 mol L-1
Average titration volume 18.44 mL

Calculate the concentration of ascorbic acid in the original juice in units of mg/100 mL. Enter your answer as a whole number of mg.

Answer 1

Ans. Balanced reaction: Ascorbic acid (Vitamin C, C₆H₈O₆) titration with Iodine

IO₃^-(aq) + 5I^-(aq) + 6H⁺(aq) ------> 3 I₂(aq) + 3H₂O(l)               - Reaction 1

C₆H₈O₆(aq) + I₂(aq) --------> 2I^-(aq) + 2H⁺(aq) + C₆H₆O₆(aq)   - Reaction 2

3 C₆H₈O₆(aq) + IO₃^-(aq) ------> I^- (aq) + 3 H₂O + 3 C₆H₆O₆(aq) – Net Reaction

# Step 1: Note the two reactions – the 1^st reaction shown formation of iodine from iodate ions, whereas the second reaction shows titration of ascorbic with iodine formed in the first step. Since the question mentions “standard iodine solution” but no “Iodate”, it’s assumed that the [I₂] is known in the solution. So, we proceed with [I₂] as the standard iodine solution.

# Following the stoichiometry of balanced reaction, 1 mol ascorbic acid is reacts with 1 mol iodine.

Now,

            Moles of Iodine (I₂) consumed = Molarity x Volume in liters

                                                            = 0.00496 mol L^-1 x 0.01844 L

                                                            = 0.0000914624 mol

# Moles of Ascorbic acid in 25 mL aliquot = Moles of Iodine consumed

                                                            = 0.0000914624 mol

# Mass of Ascorbic acid in 25 mL aliquot = Moles x MW

                                                            = 0.0000914624 mol x 176.12592 g mol^-1

                                                            = 0.016108899345408 g

# So, [Ascorbic acid] in aliquot = Mass of ascorbic acid / Volume of aliquot titrated

                                                            = 0.016108899345408 g / 25 mL

                                                            = 0.00064435597381632 g / mL

                                                            = 0.64435597381632 mg/ mL

# Step 2: The ascorbic acid is aliquot is prepared by diluting 25.0 mL of the original fruiting juice to a volume of 100.0 mL. [Note that the concentration of ascorbic acid in the diluted aliquot is 0.64435 mg mL^-1 irrespective of the volume of aliquot]

Now, using    C1V1 (total aliquot) = C2V2 (Total original Juice sample)

            Or, (0.64435597381632 mg mL^-1 x 100 mL) = C2 x 25 mL

            Or, C2 = (0.64435597381632 mg mL^-1 x 100 mL) / 25 mL

            Hence, C2 = 2.5774 mg mL^-1 = 257.74 mg / 100 mL

# Therefore, [Ascorbic acid] in original fruit juice = 258 mg / 100 mL