Titration of Citric Acid in 7-UP and Vitamin C
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Titration is a convenient method for determining the concentration of some critical substance in a solution. For example, we'll show below how to determine the concentration of Vitamin C in fruit, or the concentration of citric acid in 7-Up.
In chemistry, "titration" has a related definition: A common technique of volumetric analysis in which a standard solution of one reagent is added little by little from a burette to a second reagent whose amount is to be determined. The equivalence point, at which an exactly stoichiometric amount of reagent has been added, may be determined by using an indicator or by measurements of color, resistance, current flow, or potential, whose variation with added reagent changes abruptly when the equivalence point is reached [1]. The end point of an indicator is the point in the titration where the color changes. It's important to design a titration so that the endpoint of the indicator occurs at the equivalence point of the titration, as we'll see later.
For example, the concentration of a sample of vinegar can be determined by titration as follows: A measured volume of the solution to be titrated is placed in a beaker along with a few drops of indicator.
In this case, the beaker contains colorless aqueous vinegar or acetic acid, CH3COOH(aq), and the indicator might be phenolphthalein, which changes to pink in a basic solution.
The colorless sodium hydroxide NaOH(aq), which is the titrant, or the solution to be added, is added carefully from a buret. The added indicator changes to pink when the acid has all reacted with the NaOH(aq), and a slight excess of base remains. The reaction which occurs is
\[ \text{C} \text{H}_{3} \text{COOH} (aq) + \text{ NaOH} (aq) \rightarrow \text{ Na}^{+} (aq) + \text{CH}_{3} \text{COO}^{-} (aq) + \text{H}_{2} \text{O} (l) \label{1} \]
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 the buret can enhance a scientist's ability to be precise about volumes:
As the first few cubic centimeters 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.
Using "titration" as the keyword in YouTube finds many videos, including this one of an acid titrated with base and phenolphtalein indicator.
Eventually, though, all the acetic acid is consumed. Addition of just a fraction of a drop of titrant produces a lasting pink color due to unreacted NaOH in the flask. The change in color is called the endpoint, and it indicates that all the acetic acid has been consumed and that the equivalence point of the titration has been reached. If more NaOH solution were added, there would be an excess and the color of the solution in the flask would get much darker. The darker color would show that we had overtitrated, or overshot the equivalence point, by adding more than enough NaOH to react with all the acetic acid. Because of overtitrating, a experimenter must be particularly careful at the end of a titration.
After the titration has reached the equivalence point, the buret can again be used to determine a final volume reading. Using the initial and final reading, the volume added can be determined quite precisely:
In this case, the final volume is read as 37.28 mL, and the initial volume (above) was 1.09 mL, so the volume added was 37.28 - 1.09 = 36.19 mL or 36.19cm3.
In some titrations, intense colors of either the titrant or solution to be titrated can indicate the endpoint. Usually, however, it is necessary to add an indicator—a substance which combines with excess titrant to produce a visible color or to form an insoluble substance which would precipitate from solution at the equivalence point, as seen in the previous example. No matter what type of reaction occurs or how the equivalence point is detected, however, 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—CH3COOH 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 CH3COOH originally in the titration flask. That is, at the equivalence point the ratio of the amount of NaOH, added to the amount of CH3COOH consumed must equal the stoichiometric ratio
\[\dfrac{n_{\text{NaOH}}\text{(added from graduated cylinder)}}{n_{\text{CH}_{\text{3}}{\text{COOH}}}\text{(initially in flask)}}=\text{S}( \dfrac{\text{NaOH}}{\text{CH}_{\text{3}}\text{COOH}} )\]
\[=\dfrac{\text{1 mol NaOH}}{\text{1 mol CH}_{\text{3}}\text{COOH}} \]
When the equivalence point has been reached in any titration, the ratio of the amounts of substance of the two reactants is equal to the stoichiometric ratio obtained from the appropriate balanced chemical equation. Therefore we can use the stoichiometric ratio to convert from the amount of one substance to the amount of another.
Example \(\PageIndex{1}\): Titration Calculation
Citric acid is added to sodas and other foods to make them sour, or as a preservative. What volume of 0.5215 M NaOH would be needed to reach the equivalence point when titrating 25.00 ml of 7-Up, which is 0.0233 M in citric acid (H3C6H5O7), given S (NaOH/H3C6H5O7)) = 3/1?
Citric Acid
Citrus fruits are high in citric acid
\[\ce{H3C6H5O7 (aq) + 3NaOH (aq) -> 3H2O (l) + Na3H3C6H5O7 (aq)}\nonumber\]
Solution At the equivalence point, the stoichiometric ratio will apply, and we can use it to calculate the amount of NaOH which must be added:
\(n_{NaOH}(added) = n_{H_3C_6H_5O_7} (in flask) \times \text{S} \dfrac{\text{NaOH}}{\text{H}_3\text{C}_6\text{H}_5\text{O}_7}\)
The amount of H3C6H5O7 is obtained from the volume and concentration:
\(n_{H_3C_6H_5O_7} (in flask) = \text{25.00 cm}^3 \times \text{0.0633} \dfrac{mmol}{cm^3} = \text{1.583} \text{H}_3 \text{C}_6 \text{H}_5 \text{O}_7 \)
Then
\(n_{NaOH} (added) = \text{1.583 mmol} \text{H}_3 \text{C}_6 \text{H}_5 \text{O}_7 \times \dfrac{\text{3 mol NaOH}}{\text{1 mol} \text{H}_3 \text{C}_6 \text{H}_5 \text{O}_7 } \times \dfrac{10^{-3}}{10^{-3}} = \text{1.583 mmol} \text{H}_2 \text{O}_2 \times \dfrac{\text{3 mmol NaOH}}{\text{1 mol} \text{H}_3 \text{C}_6 \text{H}_5 \text{O}_7 } = \text{4.748 mmol NaOH}\)
To obtain VNaOH(aq) we use the concentration as a conversion factor:
\(\text{V}_{NaOH} = \text{4.748 mmol NaOH} \times \dfrac{\text{1 cm}^3}{\text{5.2154}\times \text{10}^{-1}\text{mmol NaOH}} = \text{9.10 cm}^3\)
Note that overtitrating [adding more than 9.10 cm3 of NaOH(aq) would involve an excess (more than 1.272 mmol) of NaOH.
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 \(\PageIndex{2}\): Concentration of Solution
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.
Example \(\PageIndex{3}\): Vitamin C Tablets
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). The equation is
\[ \text{C}_{6} \text{H}_{8} \text{O}_{6} (aq) + \text{NaOH} (aq) \rightarrow \text{ Na C}_{6} \text{H}_{7} \text{O}_{6} (aq) + \text{H}_{2} \text{O} (l) \]
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.
From ChemPRIME: 3.12: Titrations