My chemistry textbook states that $[\ce{H3O+}][\ce{OH-}] = K_w$ in all aqueous solutions, which we know of as $K_w$. However, I was wondering why this was true. The textbook states that it is true because $K_w$ is an equilibrium constant, but why are we allowed to assume that the other ions take no part in this reaction?

To clarify further, consider the following reaction $$\ce{HCl + H2O (l) <=> H3O+ + OH- + Cl- }$$ Of course this isn't balanced, but why doesn't the hydrogen ion on the left affect the equilibrium constant?


2 Answers 2


Your textbook is correct, the $K_\mathrm{w}$ expression is as follows: $$K_\mathrm{w} = [\ce{H3O+}][\ce{OH-}]=1.0\times10^{-14}$$

Since for every $\ce{OH-}$ there must be one $\ce{H+}$ ($\ce{H+}$ in water is synonomous with $\ce{H3O+}$) the mol ratio is 1:1, if moles are equal the molarities are equal. I'm going to substitute $x$ in for both $$K_\mathrm{w}=[x][x]=[x]^2=1.0\times10^{-14}; x= 1.0\times10^{-7}.$$

So the molarity of $[\ce{OH-}]$ and $[\ce{H3O+}]$ are both $1.0\times10^{-7}$ That's a really really low amount, small enough to ignore.

However, a lot of reactions (especially in Organic Chemistry) you'll see a negative ion, for example, $\ce{Cl-}$ pulling the $\ce{H+}$ off of a water. So when $[\ce{OH-}]$ or $[\ce{H3O+}]$ are needed, water can be torn apart to make it.

  • $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details. $\endgroup$ Commented Mar 15, 2016 at 3:45
  • $\begingroup$ Thanks Martin. I'll be sure to learn Latex before posting again. Sorry you had to go through and correct my post. $\endgroup$
    – Jess L
    Commented Mar 15, 2016 at 5:58
  • $\begingroup$ You are welcome. I am happy when my edits are appreciated. For me it is no problem, and I am just trying to help. Don't worry too much about it, learn from the edits others make on your posts and you'll do just fine. If you want to play a little, you can use the sandbox on meta. $\endgroup$ Commented Mar 15, 2016 at 6:09

!!Bing!! !!Bing!! !!Bing!! Congratulations - You're thinking like chemist!!

Your textbook has stated an oversimplification which isn't really true. Let's define $\ce{K_{w}}$ based on concentrations as: $$\ce{K_{w} = [H^+][OH^{-}]}$$

(1) The value of $\ce{K_{w}}$ changes with temperature.

(2) $\ce{K_{w}}$ changes with pressure.

(3) $\ce{K_{w}}$ depends on the total ionic strength of the solution, which includes spectator ions.

See Wikipedia article "Self-ionization of water" section "Dependence on temperature, pressure and ionic strength."

The "real" relationship at any temperature and pressure is given in the goldbook as noted by Martin: $$\mathrm{K^{*}_{w} = 1.00 \times 10^{-14} = a(H^+) a(OH^-)}$$

where $\mathrm{a(H^+)}$ and $\mathrm{a(OH^-)}$ are the activities of $\ce{H^+}$ and $\ce{OH^-}$. In dilute solution the activity of an ion can be approximated by its concentration. A "dilute solution" is dilute enough if a linear relationship holds between concentration and activity. In general this would be 0.1 N or less.

  • $\begingroup$ Wait sorry, I made a false conclusion off of what the chemistry textbook said. It says $[H_3O^+][OH^-] = K_w$ always, but thank you for taking note of this error. $\endgroup$ Commented Mar 15, 2016 at 2:56
  • $\begingroup$ The self-ionization constant of the reaction $\ce{HA + B <=> A- + HB+}$ is certainly defined as $K_\mathrm{w} = a(\ce{A-})\cdot a(\ce{HB+})$ at any given temperature or pressure. See goldbook). OP makes no statement about the book contradicting that. It just simplifies that instead of activities the concentrations are used, and that it is true for any aqueous solution (which is a stretch). $\endgroup$ Commented Mar 15, 2016 at 3:41
  • $\begingroup$ Trying to be a smartass I tripped over my own feet. I think I have it fixed correctly now. $\endgroup$
    – MaxW
    Commented Mar 15, 2016 at 4:35
  • $\begingroup$ Hm, I'm sorry but now you contradict yourself. Your points 1-3 are absolutely correct. The value of $K_\mathrm{w}$ is only $1.0\times10^{-14}$ for standard conditions. The definition, i.e. product of the activities is generally valid for any pure phase of water. (And maybe you'd try to ping me, so that I can remove my vote) $\endgroup$ Commented Mar 15, 2016 at 6:13
  • $\begingroup$ @Martin - マーチン - Don't know if this will ping you are not. Did you miss that I defined $\mathrm{K_w}$ as depending on absolute concentrations, and $\mathrm{K^{*}_{w}}$ as depending on activities? $\endgroup$
    – MaxW
    Commented Mar 15, 2016 at 6:29

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