What is the $\text{p}K_\text{a}$ of water? A simple google search yields the value $15.74$, but this site and this paper say it's $14.0$.

According to my understanding, the correct answer should be $14.0$:

$$\text{p}K_{\text{a}}= -\log([\ce{H+}][\ce{OH-}])$$

For $25~\text{°C}$: $$[\ce{H+}][\ce{OH-}] = 10^{-14} = K_{\text{w}}$$

Thus follows $\text{p}K_{\text{a}}=14$.

Can you tell me which value is correct and why?

  • 1
    $\begingroup$ Did you read the paper you link to? I glanced through it and it seems to try to explain where this discrepancy comes from. $\endgroup$
    – tschoppi
    Jan 25, 2019 at 10:12
  • $\begingroup$ @tschoppi yup i read the paper but it just confuses me a lot more $\endgroup$
    – Advil Sell
    Jan 25, 2019 at 10:20
  • 4
    $\begingroup$ Use 14 as this is what all the (thermodynamic) tables of values are based on. The difference between values is in the 'standard states' used. The normal (thermodynamic) one is to use the pure solvent (water) as the standard state, effectively this means, via the activity, replacing water concentration with 1 in the equilibrium constant equation. From page 20 in the paper you quote is a summary of how we do this. The 15.74 arises by including the water concentration as 55.3 molar. $\endgroup$
    – porphyrin
    Jan 25, 2019 at 12:29
  • $\begingroup$ Similar to the paper you provide, but perhaps more condensed: chem.libretexts.org/Bookshelves/Organic_Chemistry/… $\endgroup$
    – Buck Thorn
    Jan 26, 2019 at 13:23
  • $\begingroup$ See also this question $\endgroup$
    – Jan
    Oct 15, 2019 at 4:54

4 Answers 4


The ion product of water is usually expressed as $$K_\mathrm{w}=[\ce{H+}][\ce{OH-}]$$ At a temperature of $25\ ^\circ\mathrm C$, its value is approximately $K_\mathrm{w}=10^{-14}$, or $\mathrm pK_\mathrm{w}=14$.

However, the ion product of water is not to be confused with the acid dissociation constant of water.

Generally, the dissociation constant for the simplified reaction $$\ce{HA <=> A- + H+}$$ is defined as $$K_\mathrm{a}=\frac{[\ce{A-}][\ce{H+}]}{[\ce{HA}]}$$ Thus in case of water $$K_\mathrm{a}=\frac{[\ce{OH-}][\ce{H+}]}{[\ce{H2O}]}$$ The concentration of pure water at a temperature of $25\ ^\circ\mathrm C$ is $c=55.345\ \mathrm{mol\ l^{-1}}$. Therefore, $$\begin{align}K_\mathrm{a}=\frac{[\ce{OH-}][\ce{H+}]}{[\ce{H2O}]}=\frac{10^{-14}}{55.345}=1.807\times10^{-16}=10^{-15.74}\end{align}$$ or $$\mathrm pK_\mathrm{a}=15.74$$

  • $\begingroup$ hello Sir , I know this calculation but i was asking that which value of the 2 will be used generally for comparision of acidic strength $\endgroup$
    – Advil Sell
    Jan 27, 2019 at 6:44

If you read the full paper, you should see the discussion whether the $\mathrm{p}K_\mathrm{a}$ is $14$ or $15.7$.

From a thermodynamic point your assumption is correct, and $14$ is the "correct" value. In the paper, they mention that

... listed values often are based on conventions which are different from the thermodynamic ones.

  • $\begingroup$ So which value is correct for everyday use , beacuse it creates a lot of diffrence one value make it less acidic than methanol and other don't. $\endgroup$
    – Advil Sell
    Jan 25, 2019 at 10:53
  • 2
    $\begingroup$ As long as you work with your water as a solvent you should use 14. There might be minor cases where water is a reactant where it has a higher pKa but usally it is considered 14. $\endgroup$
    – Inselino
    Jan 25, 2019 at 16:11

The situation is the choice of units. When molarity is the chosen unit water must be expressed as a molarity and the pKa = 15.74. The approximation is then made that the concentration of water in dilute solutions is [almost] constant and the equation is modified by multiplying both sides by the concentration of pure water resulting in subtracting 1.74 from 15.74 to give pKw [Not pKa] =14.00 at some T near 298K. This is usually glossed over by saying the activity of pure water is defined as 1..

An activity of 1. is best defined as a mole fraction of 1. now the situation is that the solvent is expressed as a mole fraction when the solutes are expressed as molarities. In comparisons one must examine the units involved. Usually alcohols are expressed as molarities while water is mixed mole fraction [activity] and molarity. Correct units are necessary for evaluations and the correct mole fractions or molarities must be used. This applies to ideal solutions; nonideality brings activity coefficients into consideration.


It is common practice to ignore the change in pH caused by the addition of NaCl to aqueous solutions. This practice is valid for Cl- ions (pKb ≈ 20) because it is a much weaker base than water is. But is this practice valid for the Na+ ion, which has a pKa = 13.9?

If we were to take the pKa of water to be 15.7, the Na+ ion would be over 60 times as strong an acid as water itself is. If that were the case, a 0.1 M solution of NaCl would have a decrease in pH of about 0.5 pH units below the pH of pure water. Ignoring the change in pH caused by the addition of Na+ ions would not be valid in this scenario.

However, if we take the pKa of water to be 14.0, then the Na+ ion is only 1.3 times as strong an acid as water itself is, and the decrease in pH caused by the added Na+ can be ignored in most situations. This is the common experience of everyone who has measured the pH of a solution of NaCl and the expectation of anyone who has made a solution of NaCl.

The pKa of water is 14.0. The molar concentration of water should never be used in equilibrium calculations of aqueous solutions


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.