In pure water, [H$^+$] = 10$^{-7}$ M and [OH$^-$] = 10$^{-7}$ M. So, we conclude pH of water is 7 and so is the pOH. We define the ionic product of water as Kw = [H$^+$][OH$^-$] = 10$^{-14}$ M$^{2}$. Till this, everything is fine.

Now, without any premise, it is concluded that the product [H$^+$][OH$^-$] of all concentrations in solutions of all sorts is the same, i.e. 10$^{-14}$.

This leads to the conclusion that for all acidic, basic and neutral solutions, pH + pOH = 14.

So, why is it concluded that [H$^+$][OH$^-$] = 10$^{-14}$ M$^2$? What is the nature of Kw, the ionic product of water, that mysteriously is applicable to all solutions?

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    $\begingroup$ This is just an equilibrium constant, like any other. $\endgroup$ – Ivan Neretin May 29 '17 at 12:52
  • $\begingroup$ We derive it for water. Why is it true for all solutions? $\endgroup$ – ThemysteryOflife May 29 '17 at 13:01
  • $\begingroup$ You don't have to derive it using water and anybody who says that Kw can only be determined by measuring the pH of water is wrong. You seem to have been taught something that makes no logical sense. $\endgroup$ – orthocresol May 29 '17 at 13:17
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    $\begingroup$ @ThemysteryOflife: All (aqueous) solutions contain water. At equilibrium, water coexists with $\ce{H+}$ and $\ce{OH-}$ in a ratio determined by the equilibrium constant. This is true regardless of the presence of other solutes. $\endgroup$ – a-cyclohexane-molecule May 29 '17 at 17:21
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    $\begingroup$ Very roughly speaking, given that pure water has an unusually high concentration of 55.5 mol of water molecules per litre, to a first approximation you can consider the vast majority of common aqueous solutions (up to about 0.1 mol of solute per litre) to be "pure water plus some rounding error". Thus the ionic product of water is essentially unchanged. For more concentrated solutions, though, the solute does interfere with the ionization of water. See the wiki page. $\endgroup$ – Nicolau Saker Neto May 29 '17 at 22:49

The solution product for the reaction $\ce{H2O -> H+ + OH-}$ is a constant, $10^{-14}$, which we call $K_w = [\ce{H+}][\ce{OH-}]$. To make any aqueous solution, we put a solute into water. Because the dissociation constant (which is basically an equilibrium constant) of a substance only changes with temperature, $K_w$ remains the same. Thus, for any aqueous solution at a given temperature, $K_w$ is constant.

You need to treat $K_w$ as an equilibrium constant. For example, if I have a reaction $\ce{A -> 2B + C}$, $K_{eq} = \frac{[\ce{B}]^2[\ce{C}]}{[\ce{A}]}$; let's say I measured at equilibrium $[\ce{B}] = 1$ M, $[\ce{C}] = 2$ M, and $[\ce{A}] = 0.1$ M . This means $K_{eq} = 20$. Now, if I increased the concentration of $[\ce{A}]$ to $5$ M, $K_{eq}$ wouldn't change: the concentrations would have to adapt to my change. This is the same for water: once I determine $K_w$, it is the same for all concentrations of $\ce{H+}$ and $\ce{OH-}$.

  • $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. $\endgroup$ – pentavalentcarbon May 29 '17 at 17:50
  • $\begingroup$ @QingQuan Xia This is not what I meant $\endgroup$ – ThemysteryOflife May 30 '17 at 1:24
  • $\begingroup$ Kw remains same, I understand. It is applicable for all solutions, I don't understand $\endgroup$ – ThemysteryOflife May 30 '17 at 1:26

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