Polyprotic acids and bases in Foods - ChemPRIME

# Polyprotic acids and bases in Foods

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What do polyprotic acids have to do with cola beverages?

Cola drinks contain phosphoric acid

Phosphoric acid and its salts account for 25% of all the acid used in the food industries.[1] The main use of phosphoric acid is in the soft drink industry, particularly cola and root beer beverages. Phosphoric acid functions as an acidulant and provides the specific tart note to the flavor of these products.

Regular intake of cola beverages has been associated to low bone mineral density (BMD) in women. This is an issue of public health due to the popularity of cola drinks and the fact that BMD is strongly linked with fracture risks. Although the specific role of phosphoric acid present in cola drinks on low BMD is not clear, phosphoric acid has been shown to interfere with calcium absorption and to contribute to imbalances that lead to additional loss of calcium.[2]

The following is an old recipe for Coca Cola published in "Beverages and their adulteration" in 1919 by Harvey Wiley, considered by many the father of the FDA.[3]

Ingredient / Compound (units) Amount
Caffeine (grains* per fluid ounce) 0.92-1.30
Phosphoric acid, H3PO4 (percent) 0.26-0.30
Sugar, total (percent) 48.86-58.00
Alcohol (percent by volume) 0.90-1.27
 Caramel, glycerin, lime juice, essential oils, and plant extractive
Present
Water (percent) 34.00-41.00
*1 grain=65 mg

According to the same reference,[4] the beverage coca cola is made by mixing one ounce of this syrup per glass.

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For polyprotic acids (i.e. phosphoric and malic), we can write down an equilibrium constant for each proton lost. These constants are subscripted 1, 2, etc., to distinguish them. The dissociation equilibria of phosphoric acid, a triprotic acid, are:

$\text{H}_{3}\text{PO}_{4} + \text{H}_{2}\text{O} \rightleftharpoons \text{H}_{3}\text{O}^{+} + \text{H}_{2}\text{PO}_{4}^{-}$

$K_{a\text{1}}=\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }[\text{ H}_{\text{2}}\text{PO}_{4}^{-}\text{ }]\text{ }}{\text{ }[\text{ H}_{\text{3}}\text{PO}_{\text{4}}\text{ }]\text{ }}=\text{6}\text{.9 }\times \text{ 10}^{-\text{3}}\text{ mol dm}^{-\text{3}},$

$\text{H}_{2}\text{PO}_{4}^{-} + \text{H}_{2}\text{O} \rightleftharpoons \text{H}_{3}\text{O}^{+} + \text{H}\text{PO}_{4}^{2-}$

$K_{a\text{2}}=\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }[\text{ H}\text{PO}_{4}^{2-}\text{ }]\text{ }}{\text{ }[\text{ H}_{\text{2}}\text{PO}_{\text{4}}^{-}\text{ }]\text{ }}=\text{6}\text{.1 }\times \text{ 10}^{-\text{8}}\text{ mol dm}^{-\text{3}}$

and

$\text{H}\text{PO}_{4}^{2-} + \text{H}_{2}\text{O} \rightleftharpoons \text{H}_{3}\text{O}^{+} + \text{PO}_{4}^{3-}$

$K_{a\text{3}}=\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }[\text{PO}_{4}^{3-}\text{ }]\text{ }}{\text{ }[\text{ H}\text{PO}_{\text{4}}^{2-}\text{ }]\text{ }}=\text{4}\text{.8 }\times \text{ 10}^{-\text{13}}\text{ mol dm}^{-\text{3}}$

A general treatment of the pH of solutions of polyprotic species is beyond our intended scope, but it is worth noting that in many cases we can treat polyprotic species as monoprotic. In the case of H3PO4, for example, Ka1 is very much larger than Ka2 and this much larger than Ka3 indicating that H3PO4 is a very much stronger acid than H2PO4 and this is much stronger than HPO42–. This difference means that when H3PO4 is dissolved in water, we can treat it as a monoprotic acid and ignore the possible loss of a second proton.

EXAMPLE 1 Cola beverages contain between 50 and 70 mg of phosphoric acid per 100 mL (references 1 and 5). What is the expected pH value for a cola beverage containing 0.07% phosphoric acid?

Solution 0.07% phosphoric acid equal 70 mg of phosphoric acid per 100 mL of beverage. Calculating the corresponding number of moles, we have

\begin{align} \text{n}_{\text{H}_{3}\text{PO}_{4}}&=\frac{\text{0.07 g}\text{ H}_{3}\text{PO}_{4}}{\text{98 g}\text{ mol}^{-1}}\\ \text{ }&=\text{7.14}\times\text{10}^{-4}\text{mol}\text{ H}_{3}\text{PO}_{4}\\ \end{align}

and its concentration is then

\begin{align} \text{ }[\text{H}_{3}\text{PO}_{4}]\text{ }&=\frac{\text{n}_{\text{H}_{3}\text{PO}_{4}}}{V_{\text{solution}}}=\frac{\text{7.14}\times \text{10}^{-4}\text{ mol }}{\text{1.0}\times \text{10}^{-1}\text{ dm}^{3}}\\ \text{ }&=\text{7.14}\times 10^{-3}\text{mol dm}^{-3}\\ \end{align}

Using the equation

\begin{align} {[\text{H}_{3}\text{O}^{ +}]} &\approx \sqrt{K_{a}c_{a}}\\ \end{align}

discussed in the pH of Solutions of Weak Acids and the Ka1= 6.9 X 10-3mol dm-3 for the dissociation of phosphoric acid we can calculate de concentration of hydronium-ions.

\begin{align} {[\text{H}_{3}\text{O}^{ +}]} &\approx \sqrt{\text{6.9} \times \text{10}^{-3} \text{mol dm}^{-3}\left(7.14\times\text{10}^{-3}\text{mol dm}^{-3}\right)}\\ \text{ }&\approx \sqrt{\text{4.93} \times \text{10}^{-5} \text{mol}^{ 2} \text{dm}^{-6}}\\ \text{ }&\approx \text{7.02}\times\text{10}^{-3}\text{mol dm}^{-3}\\ \end{align}

Checking the accuracy of the approximation we find

$\frac{[\text{H}_{\text{3}}\text{O}^{\text{+}}]\text{ }}{c_{a}}=\frac{\text{0.00702}}{\text{0.00714}}=\text{0.98}, \,that \,is, \, \text{98 percent}$

Given that the ratio [H3O+]/ca is much more than 5%, the above approximation is not valid. In fact, since the ratio is almost 100% even multiple approximations using the equation

$[\text{H}_{3}\text{O}^{+}] = \sqrt{K_{\text{a}}\text{(}c_{\text{a}}-\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ )}}$

number (7) in The pH of Solutions of Weak Acids, will not provide an acceptable value for the concentration of hydromiun-ions in this cola beverage.The reason is that the concentration and the Ka1 of phosphoric acid in this example have the same order of magnitude (10-3) and the use of a quadratic equation to find the concentration of hydronium-ions is actually necessary.

Considering the equilibrium

$\text{H}_{3}\text{PO}_{4} + \text{H}_{2}\text{O} \rightleftharpoons \text{H}_{3}\text{O}^{+} + \text{H}_{2}\text{PO}_{4}^{-}$

and the expression of its constant

$K_{a\text{1}}=\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }[\text{ H}_{\text{2}}\text{PO}_{4}^{-}\text{ }]\text{ }}{\text{ }[\text{ H}_{\text{3}}\text{PO}_{\text{4}}\text{ }]\text{ }}=\text{6}\text{.9 }\times \text{ 10}^{-\text{3}}\text{ mol dm}^{-\text{3}}$

With an initial concentration of phosphoric acid of 7.14 x 10-3mol dm-3, we can build the following table expressing the equilibrium concentrations of all species in terms of [H3O+]:

 Species Initial concentrationa Change in concentrationa Equilibrium concentrationa H3O+ 10-7 b [H3O+] [H3O+] H2PO4- 0 [H3O+] [H3O+] H3PO4 7.14 x 10-3 -[H3O+] 7.14 x 10-3-[H3O+]
 amol dm-3 bCan be ignored because the concentration of hydronium-ions produced by phosphoric acid will be much larger.

Rewriting the equilibrium constant for the first dissociation of phosphoric acid in terms of [H3O+] and the initial concentration of the acid we have:

\begin{align} K_{a\text{1}}&=\frac{[\text{ H}_{3}\text{O}^{+}][\text{ H}_{3}\text{O}^{+}]}{\text{ 7.14}\times \text{10}^{-3} - [\text{ H}_{3}\text{O}^{+}]}=\text{6}\text{.9 }\times \text{ 10}^{-\text{3}}\text{ mol dm}^{-\text{3}}\\ \text{ }\\ \text{ }&=\frac{[\text{ H}_{3}\text{O}^{+}]\text{ }^{2}}{\text{ 7.14}\times \text{10}^{-3} - [\text{ H}_{3}\text{O}^{+}]}=\text{6}\text{.9 }\times \text{ 10}^{-\text{3}}\text{ mol dm}^{-\text{3}} \end{align}

Performing the calculations and rearranging the above expression we have

\begin{align} -[\text{ H}_{3}\text{O}^{+}]\text{ }^{2}-\text{6.9}\times \text{10}^{-3}[\text{ H}_{3}\text{O}^{+}]+ \text{4.93}\times\text{10}^{-5}=\text{0} \end{align}

A quadratic equation for which the root values can be found using the formula:

\begin{align} \text{x}&=\frac{-\text{b}\pm\sqrt{\text{b}^{2}-\text{4ac}}}{\text{2a}} \end{align}

Where,

\begin{align} \text{x}&=[\text{ H}_{3}\text{O}^{+}]\\ \text{a}&=-\text{1}\\ \text{b}&=-\text{6.9}\times \text{10}^{-3}\\ \text{c}&=\text{4.93}\times\text{10}^{-5}\\ \end{align}

From which, the possible values of [H3O+] are:

\begin{align} \left[\text{H}_{3}\text{O}^{+}\right]_{1} &=-\text{1.13}\times\text{10}^{-2}\text{mol dm}^{-3}\\ \text{ }\\ \left[\text{H}_{3}\text{O}^{+}\right]_{2} &=\text{4.37}\times\text{10}^{-3}\text{mol dm}^{-3}\\ \end{align}

The first root has a negative value and we discard it. So, the concentration of hydronium-ions in the cola beverage is 4.37 x 10-3 mol dm-3 and its pH is

$\text{pH}=-\text{log}\left(\text{4.37}\times\text{10}^{-3}\right)=\text{2.36}$

This is a rather low pH. However, when we taste cola beverages, we do not perceive them as acidic. The reason is that they tend to be very sweet (10-11% sugar) and the perception of the acid taste is diminished. We also need to keep in mind that in carbonated beverages exists CO2 which will form carbonic acid modifying in a less significant way the pH of the solution.

Phosphoric acid is also used in fruit jellies, processed cheese, buttermilk, and fermentation processes where it is employed to adjust or maintain specific pH values (buffering agent). The salts derived from phosphoric acid have numerous applications in food processing, the table below shows some of them.[5]

Application of phosphates in food processing

 Phosphate Food Function Ca(H2PO4)2:H2   Na3Al2H15(PO4)8 Baked goods Leavening agent Na5P3O10 Seafood Retain moisture, inhibit color and  lipid oxidation, protect native       protein, reduce drip Na2HPO4 and   Na3Al2H15(PO4)8 Processed cheese and   cheese sauces "Melting salts": Emulsion   development and stability (NaPO3)n Salad dressings Thickening aid, sequestrant,   stabilizer K2HPO4 Imitation coffee creamers Adjust pH and prevent   "feathering" Na2HPO4 Evaporated, condensed, or   dried milk Inhibit protein coagulation Ca(H2PO4)2 Canned fruit and vegetables Firming of texture Na2H2P2O7 French fries Sequestrant of iron to inhibit   blackening after frying (NH4)2HPO4 Beer and wine Yeast nutrient (NaPO3)n Beer and wine Prevent clouding KH2PO4 (or Na)   (NaPO3)n   Na5P3O10 Processed eggs Preserve color, protect protein   from coagulation, improve   whipping ability (whites)

In the cases of polyprotic bases, we can establish the equation for the equilibrium constant for each proton gained. The carbonate ion, CO32–, is an example of a diprotic base for which the appropriate base constants are

$\text{CO}_{3}^{2-} + \text{H}_2\text{O} \rightleftharpoons \text{HCO}_{3}^{-} + \text{OH}^{-}$

$K_{b\text{1}}=\frac{\text{ }[\text{ HCO}_{\text{3}}^{-}\text{ }]\text{ }[\text{ OH}^{-}\text{ }]\text{ }}{\text{ }[\text{ CO}_{\text{3}}^{\text{2}-}\text{ }]\text{ }}=\text{2}\text{.1 }\times \text{ 10}^{-\text{4}}\text{ mol dm}^{-\text{3}}$

and

$\text{HCO}_{3}^{-} + \text{H}_2\text{O} \rightleftharpoons \text{H}_{2}\text{CO}_{3} + \text{OH}^{-}$

$K_{b\text{2}}=\frac{\text{ }[\text{ H}_{\text{2}}\text{CO}_{\text{3}}\text{ }]\text{ }[\text{ OH}^{-}\text{ }]\text{ }}{\text{ }[\text{ CO}_{\text{3}}^{-}\text{ }]\text{ }}=\text{2}\text{.4 }\times \text{ 10}^{-\text{8}}\text{ mol dm}^{-\text{3}}$

Solutions of salts containing the carbonate ion, such as Na2CO3 or K2CO3 can be treated similarly.

EXAMPLE 2 Find the pH of a 0.100-M solution of sodium carbonate, Na2CO3. Use the base constant Kb1 = 2.10 × 10–4 mol dm–3.

Solution We ignore the acceptance of a second proton and treat the carbonate ion as a monoprotic base. We then have

\begin{align} \text{ }[\text{ OH}^{-}\text{ }]\text{ }=\sqrt{K_{b}c_{b}}&=\sqrt{\text{2}\text{.10 }\times \text{ 10}^{-\text{4}}\text{ mol dm}^{-\text{3}}\text{ }\times \text{ 0}\text{.100 mol dm}^{-\text{3}}} \\ \text{ }&=\text{4}\text{.58 }\times \text{ 10}^{-\text{3}}\text{ mol dm}^{-\text{3}} \\ \end{align}

Checking, we find that

$\frac{\text{ }[\text{ OH}^{-}\text{ }]\text{ }}{c_{b}}=\frac{\text{4}\text{.58 }\times\text{ 10}^{-\text{3}}}{\text{0}\text{.1}}\approx \text{4}\text{.6 percent}$

so that our approximation is only just valid.

We now find

$\text{pOH}=-\text{log}\left(\text{4.58}\times\text{10}^{-3}\right)=\text{2.34}$

while          $\text{pH}=\text{14}-\text{2.34}=\text{11.36}\,$

Since the carbonate ion is a somewhat stronger base than NH3, we expect a 0.1-M solution to be somewhat more basic, as actually found.

A glance at the Ka and and Kb tables reveals that most acid and base constants involve numbers having negative powers of 10. As in the case of [H3O+] and [OH], then, it is convenient to define

$\text{p}K_{a}=-\text{log}\frac{K_{a}}{\text{mol dm}^{-\text{3}}}$...and...$\text{p}K_{b}=-\text{log}\frac{K_{b}}{\text{mol dm}^{-\text{3}}}$

Using these definitions, the larger Ka or Kb is (i.e., the stronger an acid or base, respectively), the smaller pKa or pKb will be. For a strong acid like HNO3, Ka = 20 mol dm–3 and

$\text{p}{K}_{a}=-\text{log 20}=-\left(\text{1.30}\right)=-\text{1.30}$

Thus for very strong acids or bases pK values can even be negative.

## References

1. Food Chemistry, 3rd Ed. 2004, Belitz, et al.
2. Tucker, K.L., Morita, K., Ning Qiao, Hannan, M.T., Cupples, L.A., Kiel, D.P. 2006. Colas, but not other carbonated beverages, are associated with low bone mineral density in older women: The Framingham Osteoporosis Study. Am. J. Clin. Nutr. 84:4:936-942.