Weak acids in foods - pH and beyond - ChemPRIME

# Weak acids in foods - pH and beyond

What do weak acids have to do with the following foods?

The acids we find in foods and other biological systems tend to have small dissociation constants. In other words, they are weak acids. Besides the acids naturally found in food, many of them are added during processing to improve the flavor, appearance, shelf life, and other characteristics of food (food additives [1]).

The table below shows the most common weak acids employed in food processing and their pKa's. You can observe that their ability to lower the pH of the food enables them with a variety of additional functions. A low pH can prevent or slow down the growth of spoilage and pathogen microorganisms. In the case of Clostridium botulinum, a pH below 4.5 will stop the production of the toxin responsible for botulism (known as botox in the cosmetic industry). Some weak acids, however, possess antimicrobial activity due to their ability to interact with cell membranes and disrupt them, or by interfering with microbial metabolic pathways (i.e. benzoic acid). Weak acids are effective antimicrobials in their protonated form. For such reason, they are mostly used in the preservation of acid foods.

The degree of ionization and overall charge of proteins (i. e. casein) and some carbohydrates (i.e. pectins, chitosan) depends on pH, affecting their ability to form aggregates and complexes. This is the reason for the central role that pH plays in the manufacture and textural characteristics of dairy products (cottage and cheddar cheeses, yogurt) and the formation and stability of gels (jellies and marmalades). Weak acids also modify the flavor of foods; they can act as buffering agents, and some of them sequestrate metals helping to prevent oxidation reactions (citric acid).

Weak acids employed as food additives[2]

    Acid pK1; pK2; pK3 Applications Acetic 4.74 Preservative and seasoning agent. More effective against yeast and bacteria than against mold. Used in baked products, pickled vegetables, and mayonnaise Propionic 4.87 Antimicrobial agent against mold and less effective against bacteria. It is used in baked products to inhibit mold and avoid ropiness Sorbic 4.8 Antifungal agent in baked products, cheeses, beverages (fruit, juices, wines), dried fruits, margarine, jellies, and marmalades Benzoic 4.19 Antimicrobial agent mostly against yeast and mold in sour foods (pH 4-4.5 or lower), carbonated beverages, marmalades, jellies, fish preserves, margarine, and pickled vegetables Succinic 4.19; 5.63 Plasticizer in dough making, its esters are used as emulsifiers in baking Adipic 3.00; 4.52 Powdered fruit juice drinks, gelling of marmalades and jellies, improving cheese texture Fumaric 4.19; 5.63 Increasing shelf life of dehydrated foods and promoting gel setting Lactic 3.86 Improving the foaming properties of egg white, flavor of beverages and pickled vegetables. It is used in the form of calcium lactate in milk powders Malic 3.40; 5.05 Manufacturing of marmalades, jellies, beverages, and canning of fruits and vegetables Tartaric 2.98; 4.34 Acidification of wine, it is also used in fruit juice drinks, sour candies, ice cream, and as a synergist for antioxidants Citric 3.09; 4.74 Manufacture of confections, fruit juice, ice cream, marmalade, and jelly. Also used in vegetable canning and dairy products such as butter and processed cheese Phosphoric 2.15; 7.1; ~12.4 Widely used in the soft drink industry (cola). Employed in processed cheese, baking powder, as an active buffering agent and pH adjusting ingredient in diverse processes

 This option will not work correctly. Unfortunately, your browser does not support inline frames. This option will not work correctly. Unfortunately, your browser does not support inline frames. This option will not work correctly. Unfortunately, your browser does not support inline frames.Figure 1. Ball-and-stick model for some weak acids: (a) carbonic acid, H2CO3; (b) acetic acid, CH3COOH; (c) phosphoric acid, H3PO4.

Calculating the pH of solutions of weak acids

When any weak acid, which we will denote by the general formula HA, is dissolved in water, the reaction

$\text{H}{\text{A}}\,\text{ + H}_{\text{2}}\text{O }\rightleftharpoons\,\text{A}^{-}\,\text{ + }\,{\text{H}_{\text{3}}\text{O}^{\text{+}}}$       (1)

proceeds to only a limited extent, and we must allow for this in calculating the hydronium-ion concentration and hence the pH of such a solution. In general, the pH of a solution of a weak acid depends on only two factors, the concentration of the acid, ca, and the magnitude of an equilibrium constant Ka, called the acid constant, which measures the strength of the acid. The acid constant is defined by the relationship:

$K_{c}\text{ }\times \text{ 55}\text{.5 mol dm}^{-\text{3}}=K_{a}=\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }[\text{ A}^{-}\text{ }]\text{ }}{\text{ }[\text{ HA }]\text{ }}$      (2)

(The acid constant is Kc multiplied by the constant concentration of water, as already defined in the section on the law of chemical equilibrium. It is also called the ionization constant or the dissociation constant of the acid.)

Measurements of the conductivity of acetic acid solutions have been used to find the acid constant for acetic acid, Ka(CH3COOH). The Ka for acetic acid is only approximately a constant, varying from a value of 1.81 × 10–5 mol dm–3 in very dilute solutions to a value of 1.41 × 10–7 mol dm–3 in a 1M solution. A similar variation is found for other weak acids, so that most of the calculations we do using Ka are only approximate. Only two or possibly three significant figures should be retained. The table below gives the Ka values for a few selected acids arranged in order of their strength. This table is part of our larger collection of acid-base resources. It is at once apparent from this table that the larger the Ka value, the stronger the acid. The strongest acids, like HCl and H2SO4 have Ka values which are too large to measure, while another strong acid, HNO3, has Ka value close to 20 mol dm–3. Typical weak acids such as HF and CH3COOH have acid constants with a value of 10–4 or 10–5 mol dm–3. Acids like the ammonium ion, NH4+, and hydrogen cyanide, HCN, for which Ka is less than 10–9 mol dm–3 are very weakly acidic.

The Acid Constants for Some Acids at 25°C can be found here.

The hydronium-ion concentration and the pH of a solution of a weak acid depend on the concentration of the acid. At this point, we need to clarify what we mean by the phrase concentration of acetic acid. Is it the initial concentration of the acid? or the remaining concentration after dissociation? In order to resolve this difficulty, we will use the term stoichiometric concentration of acid and the symbol ca to indicate the total amount of acetic acid originally added per unit volume of solution. On the other hand we will use the term equilibrium concentration and the symbol [HA] to indicate the final concentration of this species in the equilibrium mixture.

Let us now consider the general problem of finding [H3O+] in a solution of a weak acid HA whose acid constant is Ka and whose stoichiometric concentration is ca. According to the equation for the equilibrium,

$\text{H}{\text{A}}\,\text{ + H}_{\text{2}}\text{O }\rightleftharpoons\,\text{A}^{-}\,\text{ + }\,{\text{H}_{\text{3}}\text{O}^{\text{+}}}$       (1)

for every mole of H3O+ produced, there must also be a mole of A produced. At the same time, a mole of HA and a mole of H2O must be consumed. Since the volume which all these species occupy is the same, any increase in [H3O+] must be accompanied by an equal increase in [A] and an equal decrease in [HA]. Consequently we can draw up the following table (in which equilibrium concentrations of all species have been expressed in terms of [H3O+]:

 Species Initial concentrationa Change in concentrationa, b Equilibrium concentrationa H3O+ 10-7 [H3O+] [H3O+] A- 0 [H3O+] [H3O+] HA ca -[H3O+] ca -[H3O+]

amol dm-3

bThe hydronium-ion concentration actually increases from 107 mol dm-3 to the equilibrium concentration, and so the change in each of the concentrations is ± ([H3O+ ] –10-7 mol dm-3). However, the concentration of hydronium ions produced by the weak acid is usually so much larger than –10-7 mol dm-3 that the latter quantity can be ignored.

We can now substitute the equilibrium concentrations into the expression

$K_{a}=\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }[\text{ A}^{-}\text{ }]\text{ }}{\text{ }[\!\!\text{ HA }]\text{ }}=\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }^{\text{2}}}{c_{a}-\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}\text{ }$      (3)

This could be solved for [H3O+] by means of the quadratic formula, but in most cases a quicker approximate method is available. Since the acid is weak, only a small fraction of the HA molecules will have donated protons to form H3O+ ions. Therefore [H3O+] is only a small fraction of ca and can be ignored when we calculate ca – [H3O+]. That is,

\begin{align} \text{c}_{a}-[\text{H}_{3}\text{O}^{+}]&\approx\text{c}_{a} \end{align}      (4)

Equation (3) then becomes

$K_{a}=\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }^{\text{2}}}{c_{a}}$

which rearranges to

$[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }^{\text{2}}\approx {K_{a}c_{a}}$

Taking the square root of both sides gives an important approximate formula:

\begin{align} {[\text{H}_{3}\text{O}^{ +}]} &\approx \sqrt{K_{a}c_{a}}\\ \end{align}      (5)

 size=150

EXAMPLE 1 Use Eq. (5) to calculate the pH of a 0.0116 M solution of benzoic acid, C5H6COOH.

Solution

From the table Weak acids employed as food additives above, the pKa of benzoic acid is 4.19 and its Ka = 10–pKa = 6.46 × 10–5 mol dm–3.

Since ca = 1.16 × 10–2 mol dm–3, we have

\begin{align}\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }&=\sqrt{K_{a}c_{a}} \\ \text{ }&=\sqrt{\text{(6}\text{.46 }\times \text{ 10}^{-\text{5}}\text{ mol dm}^{-\text{3}}\text{)(1}\text{.16 }\times \text{ 10}^{-\text{2}}\text{ mol dm}^{-\text{3}}\text{)}} \\ \text{ }&=\sqrt{\text{7}\text{.5 }\times \text{ 10}^{-\text{7}}\text{ mol}^{\text{2}}\text{ dm}^{-6}}=\sqrt{\text{75 }\times \text{ 10}^{-8}\text{ mol}^{2} \text{dm}^{-6}} \\ \text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }&=\text{8}\text{.66 }\times \text{ 10}^{-\text{4}}\text{ mol dm}^{-\text{3}} \\ \text{ pH}&=-\text{log(8}\text{.66 }\times \text{ 10}^{-\text{4}}\text{)}=-\text{(0}\text{.94}-\text{4)}=\text{3}\text{.06} \\ \end{align}

In food analysis, it is common to express the concentration of hydronium-ions in terms of the concentration of the most abundant acid in that food. For example, lactic acid is used for dairy products, citric acid for citrus fruits, tartaric acid for grapes, and malic acid for apples.

Cranberries[3] (Vaccinium vitis-idaea.

EXAMPLE 2 Cranberries are rich in benzoic acid. Assuming that the only source for hydronium-ions in cranberries is benzoic acid, find its concentration in canned cranberry juice, which has an average pH of 2.41.

Solution With a pH of 2.41, the concentration of hydromiun-ions is

[H3O+] = 10–pH mol dm–3 = 10–2.41 mol dm–3 = 3.89 × 10–3 mol dm–3

From Example 1, we know that Ka of benzoic acid is 6.46 × 10–5 mol dm–3

Rearranging equation (5)

$c_{a}=\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }^{\text{2}}}{K_{a}}=\frac{\text{ }[{ 3}\text{.89}\times \text{ 10}^{-3}\text{ mol dm}^{-\text{3}}\text{ }]\text{ }^{\text{2}}}{{6}\text{.46}\times \text{ 10}^{-5}\text{ mol dm}^{-\text{3}}\text{ }}=\text{0}\text{.23}\text{ mol dm}^{-\text{3}}$

Thus, the concentration of benzoic acid in canned cranberry juice is 0.23 mol dm–3.

In a few cases, if the acid is quite strong or the solution very dilute, it turns out that Eq. (5) is too gross an approximation. A convenient rule of thumb for determining when this is the case is to take the ratio [H3O+]/ ca. If this is larger than 5 percent or so, we need to make a second approximation, and then the rules for successive approximations can be applied. Equation (3) can be converted to a convenient form for successive approximations by multiplying both sides by ca – [H3O+]:

\begin{align} {[\text{H}_{3}\text{O}^{ +}]}^{2} &\approx K_{a}\left(c_{a}- [\text{H}_{3}\text{O}^{ +}]\right) \end{align}

or      $[\text{H}_{3}\text{O}^{+}] = \sqrt{K_{\text{a}}\text{(}c_{\text{a}}-\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ )}}$...(6)

To get a second approximation for [H3O+] we can feed the first approximation into the right side of this equation. The exact procedure is detailed in the following example.

Yogurt is a fermented milk product originated in the Middle East, it has been produced for thousands of years. Yoghurt cultures consist of thermophilic lactic acid bacteria, Streptococcus thermophillus and Lactobacillus delbrueckii ssp. bulgaricus, which live together symbiotically. These bacteria ferment lactose to produce lactic acid lowering the pH. The manufacturing process of yogurt involves incubation of milk with 1.5-3% of the culture at 42-45 oC for about 3h. The final product contains between 0.7 to 1.1% of lactic acid.

Casik: Turkish yogurt sauce[4].

EXAMPLE 3 Calculate the pH of a batch of yogurt containing 0.9% (w/w%) of lactic acid. In the event of contamination with spores of Clostridium botulinum, can we ensure that this is a safe product to consume?

Solution First, we need to calculate the concentration of lactic acid in the yogurt. The formula of lactic acid is CH3CH(OH)COOH and its molecular weight, 90.08 g/mol. Assuming a base of 100 g of sample, we have 0.9 g of lactic acid per 100 g of yogurt. This means

\begin{align} \text{n}_{\text{CH}_{\text{3}}\text{CH(OH)COOH}}&=\frac{\text{0.9 g lactic acid}}{\text{90.08 g}\text{ mol}^{-\text{1}}}\\ \text{ }&=\text{9}\text{.99}\times\text{10}^{-\text{3}}{\text{mol lactic acid}}\\ \end{align}

In order to calculate the concentration of lactic acid in mol dm–3, we need to know the volume of the sample. If the density of yogurt is 1.056 g cm-3 and

$\text{Density = }\frac{\text{mass}}{\text{volume}}\text{ or }\rho \text{ = }\frac{\text{m}}{\text{V}}$

the actual volume of this yogurt sample is

$\text{V} \text{ = }\frac{\text{m}}{\rho}=\frac{\text{100 g}}{\text{1}\text{.056 g cm}^{-3}}={\text{94}\text{.7 cm}^{3}}={\text{0}\text{.0947 dm}^{3}}$

Thus, the concentration of lactic acid in this yogurt is

\begin{align}{\text{ }[\text{CH}_{\text{3}}\text{CH(OH)COOH}\text{ }]\text{ }}&=\frac{n_{\text{CH}_{\text{3}}\text{CH(OH)COOH}}}{V_{\text{solution}}}=\frac{\text{9}\text{.99}\times \text{10}^{\text{-3}}\text{ mol }}{\text{94}\text{.7}\times \text{10}^{\text{-3}}\text{ dm}^{\text{3}}}\\ \text{ }&=\text{1}\text{.055}\times 10^{^{\text{-1}}}\text{mol dm}^{\text{-3}}\\ \end{align}

From the table Weak acids employed as food additives above, we have that the pKa of lactic acid is 3.86 and its Ka is therefore, Ka = 10–pKa = 1.38 × 10–4 mol dm–3.

Using equation (5), the concentration of hydronium-ions becomes

\begin{align} \text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }&=\sqrt{K_{a}c_{a}} \\ \text{ }&=\sqrt{\text{1}\text{.38 }\times \text{ 10}^{-\text{4}}\text{ mol dm}^{-\text{3}}\times \text{ 0}\text{.1055 mol dm}^{-\text{3}}} \\ \text{ }&=\text{3}\text{.815 }\times \text{ 10}^{-3}\text{ mol dm}^{-\text{3}} \\ \end{align}

Checking we find

$\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{c_{a}}=\frac{\text{0}\text{.003815}}{\text{0}\text{.1055}}=\text{0}\text{.036, that is, 3.6 percent}$

Since the ratio [H3O+]/ ca is less than 5%, we do not need to use equation (6) for further approximations. The pH of this yogurt is then

$\text{pH}=-\text{log }\left( \text{3.815 } \times \text{10}^{-3} \right) =\text{2.41}$

The calculated pH for yogurt, based on the concentration of lactic acid alone, is very low compared to the measured pH with a reported value of 4.0-4.2. The reason for such difference is that yogurt is a highly complex system, it contains milk proteins (amino acids) and colloidal salts (calcium phosphates, citrates, etc.) which possess acid-base properties and act as buffering agents. The measured pH is then higher than the calculated based on the concentration of lactic acid.

Crystal structure of Botulinum Neurotoxin Serotype A[5].
Going back to the question regarding the safety of this yogurt in the event of contamination with spores of Clostridium botulinum. At pH below 4.5, this microorganism would not grow nor produce the botulism toxin. Many processed foods use what is called multiple hurdle strategy to prevent the growth of pathogenic and spoilage microorganisms. Examples of hurdles include pH, temperature, and salt concentration. In the case of yogurt, appropriate refrigeration would also prevent C. botulinum from growing and producing the botulism toxin.

Example 4 Calculate the pH of the yogurt from Example 4 if it was analyzed in the beginning of the fermentation process when the concentration of lactic acid was only 0.1%.

Taking 100 g as base, if the concentration of lactic acid was only 0.1 g in 100 g of yogurt, the number of moles would be

\begin{align} \text{n}_{\text{CH}_{\text{3}}\text{CH(OH)COOH}}&=\frac{\text{0.1 g lactic acid}}{\text{90.08 g}\text{ mol}^{-\text{1}}} \\ \text{ }&=\text{1}\text{.10}\times\text{10}^{-\text{3}}{\text{mol lactic acid}} \\ \end{align}

Considering that the volume of the hundred grams of sample is 0.0947 dm3 (calculated above), the concentration of lactic acid is

\begin{align}{\text{ }[\text{CH}_{\text{3}}\text{CH(OH)COOH}\text{ }]\text{ }}&=\frac{n_{\text{CH}_{\text{3}}\text{CH(OH)COOH}}}{V_{\text{solution}}}=\frac{\text{1}\text{.10}\times \text{10}^{\text{-3}}\text{ mol }}{\text{94}\text{.7}\times \text{10}^{\text{-3}}\text{ dm}^{\text{3}}}\\ \text{ }&=\text{1}\text{.16}\times 10^{^{\text{-2}}}\text{mol dm}^{\text{-3}}\\ \end{align}

Using equation (5) and a Ka = 1.38 × 10–4 mol dm–3 for lactic acid, the concentration of hydronium-ions becomes

\begin{align} \text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }&=\sqrt{K_{a}c_{a}} \\ \text{ }&=\sqrt{\text{1}\text{.38 }\times \text{ 10}^{-\text{4}}\text{ mol dm}^{-\text{3}}\times \text{ 0}\text{.0116 mol dm}^{-\text{3}}} \\ \text{ }&=\text{1}\text{.27 }\times \text{ 10}^{-3}\text{ mol dm}^{-\text{3}} \\ \end{align}

Checking we find

$\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{c_{a}}=\frac{\text{0}\text{.00127}}{\text{0}\text{.0116}}=\text{0}\text{.109, that is, 10.9 percent}$

Since the ratio [H3O+]/ ca is more than 5%, a second approximation is thus necessary. We feed our first approximation of [H3O+]= 1.27 × 10–3 mol dm–3 into Eq. (6):

\begin{align} \text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }&=\sqrt{K_{a}c_{a}} \\ \text{ }&=\sqrt{\text{1}\text{.38 }\times \text{ 10}^{-\text{4}}\text{ mol dm}^{-\text{3}}\text{(0}\text{.0116}-\text{0}\text{.00127) mol dm}^{-\text{3}}} \\ \text{ }&=\text{1}\text{.19}\times \text{ 10}^{-3}\text{ mol dm}^{-\text{3}} \\ \end{align}

Taking a third approximation

\begin{align} \text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }&=\sqrt{\text{1}\text{.38 }\times \text{ 10}^{-\text{4}}\text{ mol dm}^{-\text{3}}\text{(0}\text{.0116}-\text{0}\text{.00194) mol dm}^{-\text{3}}} \\ \text{ }&=\text{1}\text{.15 }\times \text{ 10}^{-3}\text{ mol dm}^{-\text{3}} \\ \end{align}

Since this differs from the second approximation by less than 5 percent, we take it as the final result. The pH is

$\text{pH}=-\text{log }\left( \text{1.15 } \times \text{10}^{-3} \right) =\text{2.93}$

Cross check: Since lactic acid is a stronger acid than benzoic acid, we expect this solution to have a lower pH than 0.0116 M C6H5COOH and indeed it does.