Water can transfer a proton between two water molecules:

$$\mathrm{2H_2O(l)\rightleftharpoons H_3O^+(aq)+OH^-(aq)}.$$

This process is called the autoionization of water. One water molecule acts as a Brønsted-Lowry acid and the other as a base.

Because liquid water is present in enormous excess compared with the tiny amounts ionized, its effective concentration is treated as constant. The equilibrium relation is therefore written using the ion-product constant of water,

$$K_w=[\mathrm{H_3O^+}][\mathrm{OH^-}].$$

At $25,^\circ\mathrm C$,

$$K_w\approx1.0\times10^{-14}.$$

In pure water, each autoionization event produces one hydronium ion and one hydroxide ion, so

$$[\mathrm{H_3O^+}]=[\mathrm{OH^-}].$$

Combining this equality with $K_w$ gives

$$[\mathrm{H_3O^+}]=[\mathrm{OH^-}]=1.0\times10^{-7},\mathrm M$$

at $25,^\circ\mathrm C$.

The same product relation holds in aqueous solutions. If acid increases $[\mathrm{H_3O^+}]$, equilibrium requires $[\mathrm{OH^-}]$ to decrease so their product remains $K_w$ at the same temperature. Likewise, increasing hydroxide lowers hydronium.

A solution is acidic when

$$[\mathrm{H_3O^+}]>[\mathrm{OH^-}],$$

basic when

$$[\mathrm{H_3O^+}]<[\mathrm{OH^-}],$$

and neutral when the two concentrations are equal.

Neutrality does not fundamentally mean $[\mathrm{H_3O^+}]=10^{-7},\mathrm M$. The value of $K_w$ changes with temperature, so the neutral concentrations change too. Neutrality means equal hydronium and hydroxide concentrations.