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The water dissociation constant was first defined with respect to dissociation of water into hydroxyl and hydronium ions. How then can one say that it is valid for an acidic solution, that is, even in an acidic solution the product of the concentrations of the hydronium and hydroxyl ions is $10^{-14}\ \mathrm{M}$?

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    $\begingroup$ That's the nature of chemical equilibrium. There is a reason why constant is defined as $\ce{[H+]\cdot[OH-]}$ and not some other way. $\endgroup$ – Ivan Neretin May 31 '16 at 4:18
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If Kw is written in terms of concentration (as opposed to activity), as [H+][OH-], it is only $10^{-14} M^2$ in the limit of zero ionic strength, but it is approximately correct in general.

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Acidity and alkalinity in aqueous solutions is based upon the autoionization of water:

$$\ce{HOH + HOH -> H3O+ + OH-}$$

For pure water, the following equation is valid: $$[\ce{H3O+}] = [\ce{OH-}] = 1 \times 10^{-7}~\mathrm{M}$$

Thus, to calculate $K_\mathrm{w}$: $$K_\mathrm{w} = [\ce{H3O+}][\ce{OH-}] = (10^{-7})(10^{-7}) = 10^{-14}$$

If a solution is acidic, then $[\ce{H3O+}]>[\ce{OH-}]$ and if it is alkaline $[\ce{H3O+}]<[\ce{OH-}]$. However, even if the concentrations are not equal the product of molar concentrations $[\ce{H3O+}][\ce{OH-}]$ will always be $10^{-14}$ at $25~\mathrm{^\circ C}$.

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