Autoprotolysis Constant of Water 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
$$\ce{HCl + H2O (l) &lt;=&gt; H3O+ + OH- + Cl- }$$
Of course this isn't balanced, but why doesn't the hydrogen ion on the left affect the equilibrium constant?
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.