Determining Vitamin C in Foods

Some chefs recommend boiling green beans, broccoli, and other vegetables with a little sodium bicarbonate to help keep them bright green. Some claim, however,that this treatment can destroy much of their vitamin C.

Can chemistry support or refute these claims? Careful measurements based on titrations may provide the required information.

Vitamin C Levels

Broccoli and Green beans are reasonably good sources of vitamin C (80 and 16 mg per 100g respectively), but like all foods, they can lose it quickly during cooking. They lose vitamin C somewhat less quickly on exposure to air than other sources like orange juice, because their skin provides a barrier.

The following vitamic C levels are a sampling of those found elsewhere: The US Recommended daily allowance (RDA) is 75 milligrams per day (United States' National Academy of Sciences)^[1]. Higher values have been suggested^[2], but there is no evidence that it cures colds or is otherwise beneficial at levels several times the RDA.

Table \(\PageIndex{1}\) Plant Sources of Vitamin C

Plant source Amount (mg / 100g) Rose hip 2000 Red pepper 190 Parsley 130 Broccoli 90 Brussels sprouts 80 Elderberry 60 Strawberry 60 Orange 50 cantaloupe 40 grapefruit 30

Plant source Amount (mg / 100g) Spinach 30 Potato 20 Green Beans 16 Tomato 10 Watermelon 10 Banana 9 Carrot 9 Apple 6 Lettuce 4 Raisin 2

But vitamin C is notoriously easy to destroy. Orange juice may lose half of its vitamin C in a week in the refrigerator, but cut fruit lost much less after 6 days (Mango, strawberry, and watermelon: less than 5%, Pineapple: 10%,Kiwi: 12%, and Cantaloupe: 25%, all after 6 days). ^[3] </p> The US Department of Agriculture (USDA) site reports the following vitamin C levels:

Table \(\PageIndex{2}\) Amount of Vitamin C in mg/100g

Vegetable	Raw	Frozen	Frozen then Boiled
Green Beans	12.2	9.7	4.1
Broccoli	89.2	64.9	40.1

But if sodium bicarbonate is added to preserve the green color of the vegetable, the vitamin C may be almost completely destroyed^[4]. High temperatures, or the enzyme ascorbic acid oxidase promote the oxidation of ascorbic acid to dehydro ascorbic acid (DHAA), which is still a useful vitamin. But in a basic solution like water + sodium bicarbonate, the DHAA is rapidly converted to an inactive form, diketogulonic acid (DKGA):

Figure \(\PageIndex{1}\) Ascorbic Acid

Figure \(\PageIndex{2}\) Vitamin C Dehydro Ascorbic Acid

Figure \(\PageIndex{3}\) (DHAA) 1,2-diketo-L-gulonic acid (DKGA)

Preserving the Green Color of Vegetables

Green beans change color when they're cooked in acidic water, with a higher concentration of hydrogen ions (H⁺), which replace the magnesium ion (Mg²⁺) at the center of the chlorophyll molecule (see Figure). This destroys the color.

Figure \(\PageIndex{4}\) Cut Green Beans, showing some losing their color

Figure \(\PageIndex{5}\) chlorophyll

Sodium bicarbonate (or sodium hydrogencarbonate, NaHCO₃), removes hydrogen ion and saves the chlorophyll:

\(\ce{H+ + NaHCO3 -> Na+ + H2CO3 -> Na+ + CO2 + H2O}\)

So preserving the green color of broccoli with sodium bicarbonate promotes the destruction of vitamin C.

How was this proven? With a titration to measure the vitamin C and sodium bicarbonate.

Titrations

A titration is a volumetric technique in which a solution of one reactant (the titrant) is added to a solution of a second reactant (the "analyte") until the equivalence point is reached. The equivalence point is the point at which titrant has been added in exactly the right quantity to react stoichiometrically with the analyte. If either the titrant or analyte is colored, the equivalence point is evident from the disappearance of color as the reactants are consumed. Otherwise, an indicator may be added which has an "endpoint" (changes color) at the equivalence point, or the equivalence point may be determined from a titration curve. The amount of added titrant is determined from its concentration and volume:

\(\text{n(mol)} = \text{C(mol/L)} \times \text{V(L)}\)

and the amount of titrant can be used in the usual stoichiometric calculation to determine the amount of analyte.

The titration process can be observed in a Chemistry Comes Alive Titration Videos, or if these videos are not available from your workstation, on YouTube Videos. Using "titration" as the keyword in YouTube finds many videos, including of an acid titrated with base and phenolphtalein indicator.

First we'll see how the vitamin C content can be determined by titration with base if there are no other acids present (in a vitamin C tablet, for example). Later we'll see that in foods a more specific titration is required.

A vitamin C tablet can be dissolved in water to give a colorless ascorbic acid solution.

A measured volume of the solution to be titrated, in this case, colorless aqueous acetic acid, HC₆H₇O₆(aq) is placed in a beaker. The colorless sodium hydroxide NaOH(aq), which is the titrant, is added carefully by means of a buret. The volume of titrant added can then be determined by reading the level of liquid in the buret before and after titration. This reading can usually be estimated to the nearest hundredth of a milliliter, so precise additions of titrant can be made rapidly.

Figure \(\PageIndex{6}\) The initial reading of the buret. Placing a white card behind the buret can help the precision of the reading.

As the first few milliliters of titrant flow into the flask, some indicator briefly changes to pink, but returns to colorless rapidly. This is due to a large excess of acetic acid. The limiting reagent NaOH is entirely consumed.

The added indicator changes to pink when the titration is complete, indicating that all of the aqueous acetic acid has been consumed by NaOH(aq). The reaction which occurs is

\(\ce{HC6H7O6 (aq) + NaOH (aq) -> Na+ (aq) + C6H7O6- (aq) + H2O (l)}\)

Only one of the hydrogen atoms in ascorbic acid acts as an acid, and this one is written before the rest of the formula. Eventually, all the ascorbic acid is consumed. Addition of even a fraction of a drop of titrant produces a lasting pink color due to unreacted NaOH in the flask. The color change that occurs at the endpoint of the indicator signals that all the ascorbic acid has been consumed, so we have reached the equivalence point of the titration. If slightly more NaOH solution were added, there would be an excess and the color of the solution in the flask would get much darker. The endpoint appears suddenly, and care must be taken not to overshoot the endpoint.

After the titration has reached the endpoint, a final volume is read from the buret. Using the initial and final reading, the volume added can be determined quite precisely:

Figure \(\PageIndex{6}\) The final reading of the buret. If the liquid is colorless, placing a white card behind the buret can aid in precise readings.

The object of a titration is always to add just the amount of titrant needed to consume exactly the amount of substance being titrated. In the NaOH—HC₆H₇O₆(aq)reaction [Eq. (1)], the equivalence point occurs when an equal molar amount of NaOH has been added from the graduated cylinder for every mole of HC₆H₇O₆(aq) originally in the titration flask. That is, at the equivalence point the ratio of the amount of NaOH, added to the amount of HC₆H₇O₆(aq) consumed must equal the stoichiometric ratio

\(\dfrac{n_{NaOH} \text{(added from graduated cylinder)}}{n_{CH_3COOH} \text{(initiallly in flask)}} = \text{S} \dfrac{\text{NaOH}}{\text{CH}_3\text{COOH}} = \dfrac{\text{1 mol NaOH}}{\text{1 mol CH}_3\text{COOH}}\)

 

Example \(\PageIndex{1}\): Vitamin C

Vitamin C tablets contain ascorbic acid (C₆H₈O₆) and a starch “filler” which holds them together. To determine how much vitamin C is present, a tablet can be dissolved in water andwith sodium hydroxide solution, NaOH(aq).

\[ \text{C}_{6} \text{H}_{8} \text{O}_{6} (aq) + \text{NaOH} (aq) \rightarrow \text{ Na C}_{6} \text{H}_{7} \text{O}_{6} (aq) + \text{H}_{2} \text{O} (l) \]

If titration of a dissolved vitamin C tablet requires 16.85 cm³ of 0.1038 M NaOH, how accurate is the claim on the label of the bottle that each tablet contains 300 mg of vitamin C?

Solution

The known volume and concentration allow us to calculate the amount of NaOH(aq) which reacted with all the vitamin C. Using the stoichiometric ratio

\[\text{S}\left( \dfrac{\text{C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}{\text{NaOH}} \right)=\dfrac{\text{1 mmol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}{\text{1 mmol NaOH}}\]

we can obtain the amount of C₆H₈O₆. The molar mass converts that amount to a mass which can be compared with the label. Schematically

\[ \begin{align} & V_{\text{NaOH}}\rightarrow{c_{\text{NaOH}}}n_{\text{NaOH}}\rightarrow{\text{S(C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}\text{/NaOH)}}n_{\text{C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}\rightarrow{M_{\text{C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}}\text{m}_{\text{C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}} \\ & \text{m}_{\text{C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}=\text{16}\text{.85 cm}^{\text{3}}\times \dfrac{\text{0}\text{.1038 mmol NaOH}}{\text{1 cm}^{\text{3}}}\times \dfrac{\text{1 mmol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}{\text{1 mmol NaOH}}\times \dfrac{\text{176}\text{.1 mg }}{\text{mmol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}} \\ & = 308.0 \text{ mg} \end{align}\]

Note that the molar mass of C₆H₈O₆

\[\dfrac{\text{176}\text{.1 g }}{\text{1 mol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}=\dfrac{\text{176}\text{.1 g }}{\text{1 mol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}\times \dfrac{\text{10}^{\text{-3}}}{\text{10}^{\text{-3}}}\]

\[=\dfrac{\text{176}\text{.1 g}\times \text{10}^{\text{-3}}\text{ }}{\text{10}^{\text{-3}}\text{ mol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}=\dfrac{\text{176}\text{.1 mg }}{\text{1 mmol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}\]

can be expressed in milligrams per millimole as well as in grams per mole.

The 308.0 mg obtained in this example is in reasonably close agreement with the manufacturer’s claim of 300 mg. The tablets are stamped out by machines, not weighed individually, and so some variation is expected.

Titrating Vitamin C in Foods

In foods where other acids are present, vitamin C is determined its reaction with iodine. This reaction depends on the same "reducing" power of Vitamin C that makes it react with oxygen.

A solution with an exactly known concentration of iodine (I₂) is prepared by adding a weighed aliquot of iodine to a solution containing excess potassium iodide (KI). The excess iodide dissolves the iodine by forming I₃^- (I₂ •I^-, which reacts just like I₂). This solution is added to the food, which reacts with most of the iodine:

\(\ce{C6H8O6 (aq) + 2H2O + I- (aq) -> C6H6O6 (aq) + 3I- (aq) + 2H3O+(aq)}\)

Here two of the I^- (aq) ions come from reduction of I₂ in the I₃^- ion, and the third is the extra I^- ion.

The excess iodine (as I₃^- ) is "back-titrated"^[5] with sodium thiosulfate (Na₂S₂O₃) solution that was already standardized.

\(\ce{I3- + 2S2O2 (2-) -> 3I- + S4O6(2-)}\)

Example \(\PageIndex{2}\): Vitamin C in Food

As 25.00 mL of a solution which is 0.0200 M in I₃^- is added to a food with stirring, it initially is decolorized by the vitamin C, but in the end remains brown indicating the presence of excess I₃^-. This is titrated, requiring 8.54 mL of 0.00500 M S₂O₃^2-. Calculate the amount of vitamin C in the food.

The total amount of iodine added to the food is:

\(n_{I2 tot} = \text{C}_M \times \text{V} = \text{0.020 M} \times \text{ 0.0250 L} = \text{0.000500 mol}\)

The amount of iodine left after the vitamin C reacts is calculated from the titration with thiosulfate. The amount of thiosulfate added is:

\(n_{S_2O_2(2-)} = \text{C} \times \text{V} = \text{0.00500 M} \times \text{0.00854 L} = \text{ 0.0000427}\)

The amount of iodine is calculated using the stoichiometric ratio:

\(n_{iodine} = n_{thiosulfate} \times \dfrac{n_{iodine}}{n_{thiosulfate}} = \text{0.0000427 mol thiosulfate} \times \dfrac{\text{1 mol iodine}}{\text{2 mol thiosulfate}} = \text{0.0000214 mol}\)

The amount of iodine consumed by the vitamin C was then:

\(\text{0.000500 mol} - \text{0.0000214 mol} = \text{0.000479 mol}\)

Since the vitamin C reacts with 1:1 stoichiometry with the iodine, there is also 0.000479 mol of ascorbic acid, and the mass is:

\(m_{\text{ascorbic acid}} = n (g) \times \text{M} (g/mol) = \text{0.000479 mol} \times \text{176.12} \dfrac{g}{mol} = \text{0.0843}\)

Example \(\PageIndex{3}\): Concentration of Solution

A few drops of phenolphthalein were added to 200 mL sample of the hot cooking water containing sodium bicarbonate used by a chef to cook broccoli. The solution turned pink, proving that it was basic. Then 50.00 mL of 0.0100 M HCl was added, and the solution turned colorless. Finally, the solution was titrated with 0.0100 M NaOH until it again turned pink, requiring 7.92 mL. What was the concentration of bicarbonate in the original solution?

Solution The HCl react with NaHCO₃ to form H₂O and CO₂:

\(\ce{HCl (aq) + NaHCO3 (aq) -> CO2 (aq) + H2O}\)

but some HCl was left over. It was titrated with NaOH:

\(\ce{HCl (aq) + NaOH (aq) -> Na+ (aq) + H2O}\)

The total amount of HCl added was

\(n_{HCl} = \text{C} \times \text{V} = \text{0.0100} \dfrac{mol}{L} \times \text{0.0500 L} = \text{0.000500 mol}\)

The amount of HCl that remained after reacting with the sodium bicarbonate was determined by the titration with NaOH:

\(n_{NaOH} = \text{C} \times \text{V} = \text{0.0100} \dfrac{mol}{L} \times \text{0.0092 L} = \text{0.0000792 mol}\)

So the amount of HCl that reacted with the sodium bicarbonate was:

\(\text{0.000500 mol} - \text{0.0000792 mol} = \text{0.000421 mol}\)

And since the reaction of HCl with sodium bicarbonate is 1:1, this is also the amount of sodium bicarbonate. The concentration of sodium bicarbonate in the cooking water was

\(\text{C}(M) = \dfrac{\text{n} (mol)}{\text{V} (L)} = \dfrac{\text{0.000421 mol}}{\text{0.200 L}} = \text{0.0021 M}\)

Standardization of Solutions

Titration is often used to determine the concentration of a solution. In many cases it is not a simple matter to obtain a pure substance, weigh it accurately, and dissolve it in a volumetric flask as was done in Example 1 of Solution Concentrations. NaOH, for example, combines rapidly with H₂O and CO₂ from the air, and so even a freshly prepared sample of solid NaOH will not be pure. Its weight would change continuously as CO₂(g) and H₂O(g) were absorbed. Hydrogen chloride (HCl) is a gas at ordinary temperatures and pressures, making it very difficult to handle or weigh. Aqueous solutions of both of these substances must be standardized; that is, their concentrations must be determined by titration.

Example \(\PageIndex{4}\): Titration Reaction

A sample of pure potassium hydrogen phthalate (KHC₈H₄O₄) weighing 0.3421 g is dissolved in distilled water. Titration of the sample requires 27.03 ml NaOH(aq). The titration reaction is

\[ \text{NaOH} (aq) + \text{KHC}_{8} \text{H}_{4} \text{O}_{4} (aq) \rightarrow \text{NaKC}_{8} \text{H}_{4} \text{O}_{4} (aq) + \text{H}_{2} \text{O} \]

What is the concentration of NaOH(aq) ?

Solution To calculate concentration, we need to know the amount of NaOH and the volume of solution in which it is dissolved. The former quantity could be obtained via a stoichiometric ratio from the amount of KHC₈H₄O₄, and that amount can be obtained from the mass

\[m_{\text{KHC}_{\text{8}}\text{H}_{\text{4}}\text{O}_{\text{4}}}\text{ }\xrightarrow{M_{\text{KHC}_{\text{8}}\text{H}_{\text{4}}\text{O}_{\text{4}}}}\text{ }n_{\text{KHC}_{\text{8}}\text{H}_{\text{4}}\text{O}_{\text{4}}}\text{ }\xrightarrow{S\text{(NaOH/KHC}_{\text{8}}\text{H}_{\text{4}}\text{O}_{\text{4}}\text{)}}\text{ }n_{\text{NaOH}}\]

\[n_{\text{NaOH}}=\text{3}\text{.180 g}\times \dfrac{\text{1 mol KHC}_{\text{8}}\text{H}_{\text{4}}\text{O}_{\text{4}}}{\text{204}\text{.22 g}}\times \dfrac{\text{1 mol NaOH}}{\text{1 mol KHC}_{\text{8}}\text{H}_{\text{4}}\text{O}_{\text{4}}}\]

\[=\text{1}\text{.674 }\times 10^{\text{-3}}\text{ mol NaOH}=\text{1}\text{.675 mmol NaOH}\]

The concentration is

\[c_{\text{NaOH}}=\dfrac{n_{\text{NaOH}}}{V}=\dfrac{\text{1}\text{.675 mmol NaOH}}{\text{27}\text{.03 cm}^{\text{3}}}=\text{0}\text{.06197 mmol cm}^{\text{-3}}\]

or 0.06197 M.

By far the most common use of titrations is in determining unknowns, that is, in determining the concentration or amount of substance in a sample about which we initially knew nothing. The next example involves an unknown that many persons encounter every day.

From ChemPRIME: 3.12: Titrations

References

1. ↑ http://onlinepill.in/order-vitamin_c-online-en.html?q=vitamin_c
2. ↑ http://onlinepill.in/order-vitamin_c-online-en.html?q=vitamin_c
3. ↑ http://onlinepill.in/order-vitamin_c-online-en.html?q=vitamin_c
4. ↑ Coultate, T. Food: The Chemistry of its Components, RSC Publishin, 2009, p. 337
5. ↑ http://onlinepill.in/order-vitamin_c-online-en.html?q=vitamin_c

Contributors and Attributions

• Ed Vitz (Kutztown University), John W. Moore (UW-Madison), Justin Shorb (Hope College), Xavier Prat-Resina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn.