# What is the pKa of water?


In most general chemistry textbooks, the pKa of water at 25 ºC is listed as 14.0. In many organic chemistry textbooks and some biochemistry texts, however, the pKa of water at 25ºC is listed as 15.7. This module describes the derivation of these two values and describes why the value of 15.7 should not be used. Much of the information in this module comes from two review papers with more comprehensive coverage of the topic.1,2

## Two Values for the Same Equilibrium Constant?

The value of 14.0 for the pKw of water at 25ºC has been experimentally measured.3,4 This value can also be obtained by the same process used to calculate the pKa of all other water-soluble compounds that can act as acids in aqueous solution: from the analysis of thermodynamic or electrochemical data for these aqueous solutions.

The origin of the value of 15.7 for the pKa of water can be tracked back at least as far as Brønsted in 1928, and possibly earlier than that. A common source of this value is the incorrect explanation of why the concentration of water, as the solvent, does not appear in the law of mass action for a reaction. This misunderstanding, and the corresponding miscalculation, will be fully described and remedied later in this module.

A second, deliberately developed, source of the value of 15.7 for the pKa of water was a theoretical proposal made by physical organic chemists who were under the false impression that solvents would not appreciably affect the relative strength of acids. These chemists experimentally determined the pKa of methanol dissolved in water at 25°C to be 15.5.5 They knew that previous research had shown that when 2-propanol was used as the solvent, methanol was roughly 3 times more acidic than water.6 They also knew that the commonly accepted pKa of pure water was 14.0. Thus, as it stood, the relative acidity of methanol and water dissolved in 2-propanol (methanol > water) differed from the relative acidity of water and methanol dissolved in water (water > methanol). In an attempt to make their experimental data conform to the incorrect assumption that the solvent identity does not affect relative acid strength, the chemists who measured the pKa of methanol mathematically manipulated the pKa of water so that the relative acidities in the two different solvents would agree. It will be shown below that this mathematical manipulation is incorrect and unnecessary because solvent is known to affect relative acid strength.7,8

## Thermodynamically Correct Derivation of $$K_a$$ and $$pK_a$$

Thermodynamic equilibrium constants ($$K_{eq}$$) are defined based on the activity of each of the substances involved in the chemical reaction.9 The activity of a species is a measure of the “effective concentration” of that species that results from the attractive and repulsive interactions of particles in non-ideal mixtures. Thus, for the reaction:

$bB + cC \rightleftharpoons dD + eE \label{1}$

The law of mass action is written as

$K_{eq} = \dfrac{a_D^d · a_E^e}{a_B^b · a_C^c} \label{1eq}$

where $$a_D^d$$ indicates activity of species $$D$$ raised to the $$d$$ power.

In an ideal dilute solution, the solutes are treated with Henry’s law so that activities can be approximated with concentrations. It is most appropriate to approximate the activity of a solute in a dilute solution with the molality of that solute (m = moles of solute per kilogram of solvent). However, for an ideal dilute aqueous solution at 25 ºC, the molality of a solute is often approximated with the molarity of that solute (M = moles of solute per Liter of solution) because the density of water at 25 ºC is 0.997 kilograms per Liter:

$a_{D} \approx {molality_D} \approx {[D]} \nonumber$

where $${[D]}$$ indicates the concentration of the species $$D$$ in units of molarity.

The solvent in an ideal dilute solution is treated with Raoult’s law so that the solvent can be approximated as the pure liquid. When the solvent is approximated as a pure liquid, its activity is approximated as unity:

$a_{solvent} \approx 1 \nonumber$

The concentration of the solvent does not explicitly appear in the equilibrium expression (Equation \ref{1eq}). The approximation being made is that intermolecular forces are negligible. If the solution is non-ideal, then the effects of the intermolecular forces must be accounted for by employing activity coefficients as correction factors.

The use of activity coefficients is the thermodynamically valid method to describe nonideal solutions, and is unrelated to the method of involving the solvent concentration in the calculations described below.

Thus, for a reaction in which water is a participant and also the solvent, the activity of water is not “left out” of the law of mass action. Rather, it remains in the law of mass action, but it is assumed to have a value of unity (the number 1), and so its value has no effect on the value of $$K_{eq}$$. This process is demonstrated in Example $$\PageIndex{1}$$.

Example $$\PageIndex{1}$$: Dissociation of Hypochlorous Acid

Consider the acid dissociation of hypochlorous acid

$HClO_{(aq)} + H_2O_{(l)} \rightleftharpoons ClO^-_{(aq)} + H_3O^+_{(aq)} \label{2a}$

$K_a = \dfrac{a_{ClO^-} · a_{H_3O^+}}{a_{HClO} · a_{H_2O}} \approx \dfrac{[ClO^-][H_3O^+]}{[HClO]\{1\}} = \dfrac{[ClO^-][H_3O^+]}{[HClO]}$

Using values of $$\Delta G_f^o$$ at 25ºC10 for the species involved, and the equation $$ΔGº = -RT\ln K$$, it can be shown that for Equation $$\ref{2a}$$:

\begin{align*} \Delta G_{reaction}^o &= {\{(\Delta G_f^o ClO_{(aq)}^-) + (\Delta G_f^o H_3O_{(aq)}^+)}\} - {\{(\Delta G_f^o HClO_{(aq)}) + (\Delta G_f^o H_2O_{(l)})}\} \\[4pt] &= {\{(-36.8 kJ) + (-237.1 kJ)}\} - {\{(-79.9 kJ) + (-237.1 kJ)}\} \\[4pt] &= 43.1 kJ \end{align*}

thus

$K_a = 2.8 \times 10^{-8} \nonumber$

or

$pK_a=7.55 \nonumber$

Some chemists prefer to use the short hand notation of $$H^+$$ rather than $$H_3O^+$$ to represent the proton exchanged during the acid/base reaction, resulting in the reaction equation:

$HClO_{(aq)} \rightleftharpoons ClO^-_{(aq)} + H^+_{(aq)} \label{2b}$

for which

$K'_a = \dfrac{a_{ClO^-} · a_{H^+}}{a_{HClO}} \approx \dfrac{[ClO^-][H^+]}{[HClO]} \nonumber$

Using values of $$\Delta G_f^o$$ at 25ºC10 for the species involved, and the equation $$\Delta G^o = -RT\ln K$$, one can show that for Equation $$\ref{2b}$$:

\begin{align*} \Delta G_{reaction}^o &= {\{(\Delta G_f^o ClO_{(aq)}^-) + (\Delta G_f^o H_{(aq)}^+)}\} - {\{(\Delta G_f^o HClO_{(aq)})}\} \\[4pt] &= {\{(-36.8 kJ) + (0 kJ)}\} - {\{(-79.9 kJ)}\} \\[4pt] &= 43.1\, kJ \end{align*}

thus

$K'_a = K_a = 2.8 \times 10^{-8}. \nonumber$

or

$pK_a=7.55 \nonumber$

As expected, the value of $$K_a$$ equals the value of $$K'_a$$ and thus does not depend on the manner in which the exchanged proton is represented.

A similar discussion applies to the autoprotolysis of water, in which one water molecule acts as an acid, and the other water molecule acts as a base:

$H_2O_{(l)} + H_2O_{(l)} \rightleftharpoons OH^-_{(aq)} + H_3O^+_{(aq)} \label{3a}$

or

$H_2O_{(l)} \rightleftharpoons OH^-_{(aq)} + H^+_{(aq)} \label{3b}$

resulting in:

$K_{eq} = \dfrac{a_{OH^-} · a_{H_3O^+}}{a_{H_2O}^2} \approx \dfrac{[OH^-][H_3O^+]}{1^2} = [OH^-][H_3O^+] \nonumber$

or

$K'_{eq} = \dfrac{a_{OH^-} · a_{H^+}}{a_{H_2O}} \approx \dfrac{[OH^-][H^+]}{1} = [OH^-][H^+] \nonumber$

Using values of $$\Delta G_f^o$$ at 25ºC10 for the species involved, and the equation $$\Delta G^o = -RT\ln K$$, one can show that for reaction $$\ref{3a}$$:

\begin{align*} \Delta G_{reaction}^o &= {\{(\Delta G_f^o OH_{(aq)}^-) + (\Delta G_f^o H_3O_{(aq)}^+)}\} - {\{2 \times (\Delta G_f^o H_2O_{(l)})}\} \\[4pt] &= {\{(-157.2\, kJ) + (-237.1\, kJ)}\} - {\{2(-237.1\, kJ)}\} \\[4pt] &= 79.9\, kJ \end{align*}

thus

$K_{eq} = 1 \times 10^{-14} = K_a \nonumber$

or

$pK_a =14.0 \nonumber$

Using values of $$\Delta G_f^o$$ at 25ºC10 for the species involved, and the equation $$\Delta G^o = -RT\ln K$$, one can show that for Equation $$\ref{3b}$$:

\begin{align*} \Delta G_{reaction}^o &= {\{(\Delta G_f^o OH_{(aq)}^-) + (\Delta G_f^o H_{(aq)}^+)}\} - {\{(\Delta G_f^o H_2O_{(l)})}\} \\[4pt] &= {\{(-157.2\,\, kJ) + (0 kJ)}\} - {\{(-237.1\, kJ)}\} \\[4pt] &= 79.9\, kJ \end{align*}

thus

$K'_{eq} = 1 \times 10^{-14} = K_{eq} = K_a \nonumber$

or

$pK_a=14.0 \nonumber$

In addition, it is possible to show that for both reactions $$\ref{3a}$$ and $$\ref{3b}$$,

$K_{eq} = K_a = 1 \times 10^{-14} \nonumber$

using values of Eº at 25ºC and the equation $$nFE^o = RT\ln K$$.1 It is a standard and accepted practice to calculate $$K_a$$ values in both of these ways, and experimentally measured values (or extrapolated estimates)9 confirm these calculated values.

Once again, the value of $$K_{eq}$$ (and thus $$K_a$$) does not depend on the manner in which the exchanged proton is represented. It will be shown below that this equivalency of equations is not possible if the $$pK_a$$ of water is set as 15.7.

## Non-Thermodynamic Derivations

Some general chemistry textbooks list the thermodynamically correct values of the $$K_a$$ for species in aqueous solutions, but use an incorrect explanation as to why water does not appear in the law of mass action equation. These textbooks state that the concentration of water is constant and then claim that the constant value of the concentration has been included in the value of Ka. Other texts explicitly state that the concentration of water is 55.33 Molar at 25ºC (1.000L of water at a density of 0.9970 g/mL, and a molar mass of 18.02 grams/mole), and then explicitly multiply $$K_{eq}$$ by 55.33 to obtain what they claim is a different constant, Ka. As pointed out above, the concentration of water does not have a place in the determination of the value of Ka for water, or for the K of any reaction involving water that occurs in aqueous solution. Thus, although the correct Ka values are shown in these texts, the assumptions that led to the simplification of the law of mass action are lost, especially the assumption that the activity of the water, as solvent, is unity.

The mathematical and chemical rationales presented by the chemists attempting to reconcile the relative acidity of substances in water and in non-aqueous solvents contain similar mistakes. The correct and commonly accepted ways of writing the equations for the autoprotolysis of water are Equations $$\ref{3a}$$ or $$\ref{3b}$$. Instead, Ballinger and Long claim that is possible to differentiate in solution between the water molecules that are acting as solvent molecules and the water molecules that are acting as solute molecules in the role of Brønsted acids. This claim suggests the following equations for the autoprotolysis of water:

$H_2O_{(aq)} + H_2O_{(l)} \rightleftharpoons OH^-_{(aq)} + H_3O^+_{(aq)} \label{4a}$

or

$H_2O_{(aq)} \rightleftharpoons OH^-_{(aq)} + H^+_{(aq)} \label{4b}$

Note the difference in the phases of water between Equations $$\ref{4a}$$ or $$\ref{4b}$$ with Equations $$\ref{3a}$$ or $$\ref{3b}$$. Ballinger and Long make the commonly accepted assumption that the solvent water molecules be treated as the solvent part of an ideal dilute solution, with an activity of 1. They then assume that the solute water molecules be treated as the solute part of an ideal solution whose activity is approximated as the numerical value of concentration of water, 55.33.

With these assumptions, the law of mass action for Equations $$\ref{4a}$$ would be expressed as:

$K'_{a} = \dfrac{a_{OH^-} · a_{H_3O^+}}{a_{H_2O_{(aq)}} · a_{H_2O_{(l)}}} \approx \dfrac{[OH^-][H_3O^+]}{[H_2O_{(aq)}] · 1} = \dfrac{[OH^-][H_3O^+]}{[H_2O_{(aq)}]}$

and the law of mass action for Equation $$\ref{4b}$$ would be expressed as:

$K''_{a} = \dfrac{a_{OH^-} · a_{H_3O^+}}{a_{H_2O_{(aq)}}} \approx \dfrac{[OH^-][H_3O^+]}{[H_2O_{(aq)}]}$

At 25 ºC, the product of $$[OH^-_{(aq)}][H_3O^+_{(aq)}] = 1 \times 10^{-14}$$ and $$[H_2O_{(aq)}] = 55.33$$, thus

$K'_{a}=K''_{a }\approx\dfrac{[OH^-_{(aq)}][H_3O^+_{(aq)}]}{[H_2O_{(aq)}]}= \dfrac{K_{eq}}{55.33} = \dfrac{1 \times 10^{-14}}{55.33} = 2 \times 10^{-16} = K_a^* \label{5}$

Therefore

$pK^*_a=15.7 \nonumber$

Although there are no rules forbidding the assumptions made in this derivation, they require the acceptance of the following conundrums:

1. The idea that the few water molecules acting as the solute have a concentration of 55.33 M and that the water molecules acting as the solvent are pure liquid implies that we have two sets of water molecules (solute ones and solvent ones) both with a concentration of 55.33 M. This is false. The sum of the two water concentrations would have to be 55.33 M: $[H_2O_{(l)}] + [H_2O_{(aq)}]=55.33\;M$ Additionally, setting the activity of the “solute” water molecules as 55.33 and the activity of the “solvent” water molecules as 1 implies that one set of water molecules is 55.33 times more reactive than the other set. This is an absurd assertion. Activity is not merely a convention. It is a description of the potential for a chemical species to react.
2. As stated above, it is common practice to represent the autoprotolysis of water as Equations $$\ref{4a}$$ or $$\ref{4b}$$. The fact that these two reactions are equivalent is absolutely dependent upon the fact that all of the water molecules are identical.
##### Note: Regarding the Erroneous claim that Ka = [H2O]Keq

Many documents make the erroneous claim that the $$K_a$$ for any acid is equal to $$[H_2O]K_{eq}$$. Not only is this statement incorrect because of its improper use of $$[H_2O]$$, but it is also not the equation that Ballinger and Long derived (incorrectly!) in their attempt to make the pKa of water equal 15.7.

Notice that in equation$$\ref{5}$$, $$K_a^* = \dfrac{K_{eq}}{55.33}$$. Rearranging this equation gives $$K_a^*[55.33] = K_{eq}$$, not $$K_a^*=[55.33]K_{eq}$$.

The first instance of the use of $$K_a=[H_2O]K_{eq}$$ has not yet been tracked down, but it must have been made by someone who not only misunderstood thermodynamics, but also misread the Ballinger and Long paper.

## Notation Issues

The following discussion shows that keeping track of the phase notation of $$H_2O_{(l)}$$ versus $$H_2O_{(aq)}$$ allows one to show that the $$H_2O_{(aq)}$$ notation is untenable.

### Using the Correct Notation

The process used to prove the equivalency of Equations $$\ref{3a}$$ and $$\ref{3b}$$ employs a basic rule of equilibrium chemistry; if two reactions are added together to make a new reaction, the equilibrium constant of the new reaction is the product of the equilibrium constants of the two original reactions. Applying that rule to convert from equation $$\ref{3a}$$ to equation $$\ref{3b}$$ requires a third equation:

$H_3O^+_{(aq)} \rightleftharpoons H_2O_{(l)} + H^+_{(aq)}\label{5a}$

Equation $$\ref{5a}$$ is developed from equation $$\ref{3a}$$ as follows. One of the H2O(l) molecules in equation $$\ref{3a}$$ acts as an acid, and its conjugate base is OH-. The other H2O(l) molecule must then act as a base, and its conjugate acid is H3O+. (There is no chemical difference between the two H2O(l) molecules in the equation. The likelihood of a particular water molecule’s acting as an acid or as a base is governed by the statistics of random events.) The value of K for equation $$\ref{5a}$$ can be obtained from the equation Ka · Kb = 1 x 10-14 which holds for conjugate acid/base pairs in water at 25ºC. Because Kb = 1.0 x 10-14 for H2O, the Ka = 1 for H3O+ . It is also true that pKa + pKb = 14.0 for conjugate acid/base pairs in water. Thus, because the pKb of H2O is 14.0, the pKa of H3O+ is 0.0

The conversion of Equation $$\ref{3a}$$ to Equation $$\ref{3b}$$ is then:

 $$H_2O_{(l)} + H_2O_{(l)} \rightleftharpoons OH^-_{(aq)} + H_3O^+_{(aq)}$$ $$+ H_3O^+_{(aq)} \rightleftharpoons H_2O_{(l)} + H^+_{(aq)}$$ $$K_{6} = 1 \times 10^{-14}$$ $$K_{14} = 1$$ pK = 14.0 pK = 0.0 $$H_2O_{(l)} \rightleftharpoons OH^-_{(aq)} + H^+_{(aq)}$$ $$K_{7} = K_{6} · K_{14} = 1 \times 10^{-14}$$ pK = 14.0

Equations $$\ref{3a}$$ and $$\ref{3b}$$ are balanced, result in equivalent products, and because $$K_{6} = K_{7}$$, the two equations have the same value of $$K_{eq}$$. These are equivalent equations.

### Using the Incorrect Notation

Attempting this same conversion with Equations $$\ref{4a}$$ and $$\ref{4b}$$ leads to at least two impossible scenarios caused by the differentiation between $$H_2O_{(l)}$$ and $$H_2O_{(aq)}$$.

To obtain an equation equivalent to Equation $$\ref{5a}$$, conjugate acid/base pairs must be assigned in Equation $$\ref{4a}$$:

$\underbrace{H_2O_{(aq)}}_{\text{acid}} + \underbrace{H_2O_{(l)}}_{\text{base}} \rightleftharpoons \underbrace{OH^-_{(aq)}}_{\text{conjugate base}} + \underbrace{H_3O^+_{(aq)}}_{\text{conjugate acid}} \label{4aEx}$

For example, if $$H_2O_{(aq)}$$ is the acid, then its conjugate base is $$OH^-$$. This means that $$H_2O_{(l)}$$ and $$H_3O^+$$ must be a conjugate pair. There is no conjugate acid related to $$H_2O_{(aq)}$$ in this reaction. Without a conjugate acid for $$H_2O_{(aq)}$$, the conversion from $$\ref{4a}$$ to $$\ref{4b}$$ cannot be carried out.

One possible solution is to obtain a conjugate acid for $$H_2O_{(aq)}$$ by proposing that the very active H2O(aq) molecules are also able to act as Brønsted bases. These H2O(aq) molecules would thus have H3O+ as their conjugate acid. The equation for this reaction would be:

$H_3O^+_{(aq)} \rightleftharpoons H_2O_{(aq)} + H^+_{(aq)}\label{5b}$

Although this proposal is not unreasonable, it is still a part of an impossible scenario because the conversion of Equation $$\ref{4a}$$ to Equation $$\ref{4b}$$ would be:

$$H_2O_{(aq)} + H_2O_{(l)} \rightleftharpoons OH^-_{(aq)} + H_3O^+_{(aq)}$$

$$+ \;H_3O^+_{(aq)} \rightleftharpoons H_2O_{(aq)} + H^+_{(aq)}$$

$$H_2O_{(l)} \rightleftharpoons OH^-_{(aq)} + H^+_{(aq)}$$

The resulting equation is not $$\ref{4b}$$, but $$\ref{3b}$$. This is a glaring error that clearly and simply shows that Ballinger and Long’s proposal is unsupportable.

A second method for obtaining a conjugate acid for H2O(aq) is commonly found in most of the textbooks espousing Ballinger and Long’s value of pKa. In this method, equation $$\ref{5a}$$ is accepted as the equation describing H3O+ as the conjugate acid of H2O(aq), with complete disregard for the phase designation of the H2O. Nevertheless, using the fact that

$pK_a + pK_b = 14.0$

in water at room temperature, and the claim that because the pKa of water is 15.7, the pKb of water must also be 15.7, the pKa of H3O+ is calculated to be -1.7. To clarify, the Equation $$\ref{5a}$$ is (incorrectly!) assigned a pKa = -1.7, with a corresponding value of 50 for the Ka.

The resulting combination of $$\ref{4a}$$ and $$\ref{5a}$$, with their (incorrectly!) assigned K values gives:

 $$\; H_2O_{(aq)} + H_2O_{(l)} \rightleftharpoons OH^-_{(aq)} + H_3O^+_{(aq)}$$ $$+\; H_3O^+_{(aq)} \rightleftharpoons H_2O_{(l)} + H^+_{(aq)}$$ $$K_{8} = 2 \times 10^{-16}$$ $$K_{14} = 50$$ pK = 15.7 pK = -1.7 $$\; H_2O_{(aq)} \rightleftharpoons OH^-_{(aq)} + H^+_{(aq)}$$ $$K_{9} = K_{8} · K_{14} = 1 \times 10^{-14}$$ pK = 14.0

In this case, the two equations are balanced and they result in the equivalent products, but because $$K_{8}$$ does not equal $$K_{9}$$, the two equations do not have the same value of $$K_{eq}$$. These are not equivalent equations. This is a second error that clearly and simply shows that Ballinger and Long’s proposal is unsupportable.

For Further Consideration: Dissociation Constant of Methanol

The great irony behind all of this discussion is that the Ballinger and Long5 paper not only presents data that clearly show that methanol is a weaker acid than water in aqueous solution, but also employs the value of 14.0 for the pKa of water to calculate the pKa of methanol.

The authors determined that for the reaction

$CH_3OH_{(aq)} + OH^-_{(aq)} \rightleftharpoons CH_3O^-_{(aq)} + H_2O_{(l)} \label{41}$

the equilibrium constant, K, has a value of 0.029 at 25.0oC. This value clearly shows that water is the stronger acid than methanol, because the methanol is unlikely to protonate the hydroxide ion (the conjugate base of water). Instead, the opposite is true; water is likely to protonate the methoxide ion (the conjugate base of the methanol.)

To obtain the direct reaction between CH3OH and H2O,

$CH_3OH_{(aq)} + H_2O_{(l)} \rightleftharpoons CH_3O^-_{(aq)} + H_3O^+_{(aq)} \label{42}$

along with the value of the equilibrium constant for this reaction, Ballinger and Long add reactions $$\ref{41}$$ and $$\ref{3a}$$ to obtain $$\ref{42}$$:

 $$\; CH_3OH_{(aq)} + OH^-_{(aq)} \rightleftharpoons CH_3O^-_{(aq)} + H_2O_{(l)}$$ $$+\; H_2O_{(l)} + H_2O_{(l)} \rightleftharpoons OH^-_{(aq)} + H_3O^+_{(aq)}$$ $$K_{18} = 0.029$$ $$K_{6} = 1 \times 10^{-14}$$ pK = 1.54 pK = 14.0 $$\; CH_3OH_{(aq)} + H_2O_{(l)} \rightleftharpoons CH_3O^-_{(aq)} + H_3O^+_{(aq)}$$ $$K_{19} = K_{18} · K_{6} = 2.9 \times 10^{-16}$$ pK = 15.54

Thus, there are two more pieces of evidence in the paper that seems to have started the discussion, that the entire discussion concerning the relative strength of methanol and water as acids in aqueous solution has been completely misunderstood.

All experimental data show that water has a pKw = pKa = 14.0 at 25.0oC, and all experimental data show that methanol is a weaker acid than water in aqueous solution. There is absolutely no experimental evidence to contradict either of these two statements.

## Conclusions

There would be little or no confusion in this matter if it were made clear from the outset that solvents can affect the relative acidity of compounds.7,8 A straightforward method of clarifying the effect of solvent on relative acid strengths would be to present a separate table for each solvent. There would also be less confusion in this matter if it were made clear that the solvent does belong in the law of mass action in its complete form, but that the solvent does not usually appear in the final form of the law of mass action because the solvent in an ideal dilute solution has an activity of unity.

It is incorrect and misleading to present the value of 15.7 for the pKa of water, yet this value has entered the fields of organic chemistry and biochemistry. The proposed value of 1.8 x 10-16 for the Ka of water cannot be justified with thermodynamic data, nor are there any experimental data to support this value. In fact, 1.8 x 10-16 is a hypothetical value that was arrived at using specious arguments in order to justify an incorrect assumption that the relative strengths of acids were not affected by changes in solvent. There is no reason to use this value.

The Ka of water at 25 ºC is $$1 x 10^{-14}$$ and the pKa is 14.0.

## References

1. Meister, E.C.; Willeke, M.; Angst, W.; Togni, A.; Walde, P. Helv. Chim. Acta 2014, 97, 1.
2. Silverstein, T.P.; Heller, S.T. J. Chem. Educ., 2017, 94 (6), pp 690–695.
3. Marshall, W. L.; Franck, E. U. J. Phys. Chem. Ref. Data, 1981, 10, 295.
4. Bandura, A. V.; Lvov, S. N. J. Phys. Chem. Ref. Data, 2006, 35, 15.
5. Ballinger, P.; Long, F.A. J. Am. Chem. Soc. 1960, 82, 795.
6. Hine, J.; Hine, M. J. Am. Chem. Soc. 1952, 74, 5266.
7. Cox, B.G. Acids and Bases: Solvent Effects on Acid–Base Strength, Oxford University Press, 2013. (Note: Even though the author clearly describes solvent effects, he, too, follows the incorrect convention of including concentrations in calculating K values.)
8. Heller, S.T., Silverstein, T.P. pKa values in the undergraduate curriculum: introducing pKa values measured in DMSO to illustrate solvent effects. ChemTexts 6, 15 (2020).
9. Harris, D. C. Quantitative Chemical Analysis 8th ed. New York: W. H. Freeman and Co., 2010.
10. The NBS Tables of Chemical Thermodynamic Properties, J. Phys. Chem. Ref. Data, 1982, 11, Supplement No. 2. (as referenced in Noggle, J. H. Physical Chemistry 3rd ed. New York: HarperCollins, 1996)

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