Pure water at 25°C ionizes as what?

see explanation

Explanation:

Pure water (H₂O) at 25°C follows an autoionization mechanism. That is, water reacts with itself to yield Hydronium Ions (H₃O⁺) and Hydroxide Ions (OH⁻) according to the following proton transfer mechanism.

2H₂O(l) ⇄ H₃O⁺(aq) + OH⁻(aq) => H₂O(I) ⇄ H⁺(aq) + OH⁻(aq) as a shorthand form of the process primarily for convenience. The H⁺ ion is a proton and would not be found in solution as an independent free form particle, but the shorthand format is generally accepted in the chemical community because of it's ease of application.

Other related facts of water ionization include...

At 25°C/1atm => the molar concentrations of  [H⁺(aq)]  = 1 x 10⁻⁷M and [OH⁻(aq)] = 1 x 10⁻⁷M

Applying to the equilibrium ionization expression for water ...

                H₂O(I) ⇄    H⁺(aq)      +      OH⁻(aq)

C(eq):          ∞*         1 x 10⁻⁷M          1 x 10⁻⁷M

*The molar concentration of water is considered to be infinite [∞] as compared to concentration values of H⁺(aq) and OH⁻(aq) and is for practical purposes assumed to be constant. This leads to the classic Ionization Product constant expression for water (Kw = [H⁺(aq)][OH⁻(aq)].

The value for Kw at 25°C/1atm is then ...

Kw =  [H⁺(aq)][OH⁻(aq)] = (1 x 10⁻⁷M)(1 x 10⁻⁷M) = 1 x 10⁻¹⁴M² (units are typically dropped and Kw listed simply as 1 x 10⁻¹⁴.

With this, the terms acidity, neutral and alkaline acid base systems are defined and related to pH. That is, if ...

[H⁺(aq)] > [OH⁻(aq)] => Acidic System => pH < 7

[H⁺(aq)] = [OH⁻(aq)] => Neutral System => pH = 7

[H⁺(aq)] < [OH⁻(aq)] => Alkaline System => pH < 7

pH = -log[H⁺]  and pOH = -log[OH⁻] and pH + pOH = 14.