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I am relearning some chemistry in order to understand pH for a particular application. My last encounter with the subject was in high school several decades ago. I have the following question:

I read (for example at this libretexts page) that during the auto-ionization of water at room temperature and normal pressure, about 1 out of 500 million water molecules ionize to form hydronium and hydroxide ions. Now, the equilibrium constant for water under those conditions is $10^{-14}$. If we assume that the ionization is an elementary reaction, then the equilibrium constant expression is $K_{w} = \frac{[H_{3}O^{+}][OH^{-}]}{[H_{2}O)]^{2}}$. In that case we would get $\frac{[H_{3}O^{+}]}{[H_{2}O]} = 10^{-7}$ . Since the concentration of $OH^{-}$ and $H_{3}O^{+}$ ions is equal, this implies that the molar ratio of $H_{3}O^{+}$ to water is $10^{-7}$ or 1 in 10 million. Hence the proportion of water molecules that dissociate is also 1 in 10 million and not 1 in 500 million. Which one is correct or what accounts for the discrepancy? Alternatively, is something wrong with my reasoning?

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  • $\begingroup$ Concentration of water is already implicitly contained in the constant, so Kw=[H+(aq)][OH-(aq)]=10^-14. $\endgroup$
    – Poutnik
    Commented Apr 27 at 8:01
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    $\begingroup$ $[\ce{H2O}(l)]=\frac{\rho}{M}=\frac{\pu{1000g/L}}{\pu{18g/mol}}=\pu{55.56M}\;\;\;\;\;;$ $\frac{[\ce{H+}]}{\ce{[H2O]}}=\frac{10^{-7}}{55.56}=1.8\cdot 10^{-9}$ $\endgroup$
    – Sam202
    Commented Apr 27 at 8:14
  • $\begingroup$ OK, I do know that the equilibrium constant is actually defined in terms of the activities of the substances involved and the activity of water is 1. So to get the ratio of molar concentrations, I would have to divide by 55.56 - that would give roughly 2 parts per billion. Thank You. $\endgroup$
    – skm
    Commented Apr 27 at 13:44
  • $\begingroup$ One question is: Does this mean that the water ionization is not an elementary reaction - otherwise the equilibrium constant would be expressible in terms of the concentrations and the stoichiometric coefficients. $\endgroup$
    – skm
    Commented Apr 27 at 13:46
  • $\begingroup$ It is elementary, but as an approximation, only diluted species in aqueous solution or gas phase are included in the equilibrium constant expression. Same reason for why for: $\ce{AgCl(s)<=>Ag+ (aq) +Cl- (aq)}$, the constant is expressed as: $K_{sp}=[Ag^+ (aq)][Cl^- (aq)]$ (i.e. the solid is excluded). $\endgroup$
    – Sam202
    Commented Apr 27 at 18:07

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I believe OP's confusion is related to the difference between the equilibrium constant in terms of concentration $K_c$ and modified equilibrium constant forms that derive from $K_c$ multiplied by a constant concentration of a solid or liquid.

The water auto-ionization reaction is:

$$\ce{H2O(l)<=>H+(aq) +OH-(aq)}$$

$K_c$ for this reaction would be:

$$K_c=\frac{[\ce{H+(aq)}][\ce{OH-(aq)}]}{[\ce{H2O(l)]}}$$

But since $\ce{H2O}$ is a liquid and is in excess compared to either $\ce{H+(aq)}$ or $\ce{OH-(aq)}$, the change that $[\ce{H2O}]$ experiences is negligible compared to the change that both $[\ce{H+(aq)]}$ and $[\ce{OH-(aq)}]$ do:

$$[\ce{H2O}]=55.56-x$$

$$[\ce{H+]}=x$$

$$[\ce{OH-]}=x$$

$$K_c=\frac{x^2}{55.56-x}=1.8\cdot10^{-16}$$

Solving for $x$:

$$x=\pu{10^{-7}M}$$

So the final concentrations of all species would be:

$$[\ce{H2O}]=\pu{55.5599999}\approx \pu{55.56M}$$

$$[\ce{H+]}=\pu{10^{-7}M}$$

$$[\ce{OH-]}=\pu{10^{-7}M}$$

Since the concentration of water practically remains constant after the equilibrium shift, it is considered as such and grouped up with $K_c$ to form another constant $K_w$, and $[\ce{H2O}]$ is excluded from this new constant's expression:

$$K_c[\ce{H2O}(l)]=K_w={[\ce{H+(aq)}][\ce{OH-(aq)}]}=(1.8\cdot10^{-16})(\pu{55.56M})=10^{-14}$$

Now, to answer your specific question, the molar ratio of hydronium to water in pure water is:

$$\frac{[\ce{H+}]}{[\ce{H2O}]}=\frac{\ce{[H+]}\;M_{\ce{H2O}}}{\rho_{\ce{H2O}}}=\require {cancel}\frac{(\pu{10^{-7}\cancel{mol/L})}(\pu{18\cancel{g/mol}})}{\pu{1000\cancel{g/L}}}=1.8\cdot10^{-9}=\pu{1.8ppb}$$

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  • $\begingroup$ Thank you very much for your explanation. Please also see my reply to your second comment above. $\endgroup$
    – skm
    Commented Apr 27 at 18:58

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