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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?

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  • $\begingroup$ 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 at 10:12
  • $\begingroup$ @tschoppi yup i read the paper but it just confuses me a lot more $\endgroup$ – Advil Sell Jan 25 at 10:20
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    $\begingroup$ 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 at 12:29
  • $\begingroup$ Similar to the paper you provide, but perhaps more condensed: chem.libretexts.org/Bookshelves/Organic_Chemistry/… $\endgroup$ – Night Writer Jan 26 at 13:23
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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 $$\mathrm pK_\mathrm{a}=15.74$$

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  • $\begingroup$ 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 at 6:44
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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.

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  • $\begingroup$ 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 at 10:53
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    $\begingroup$ 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 at 16:11

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