The autoionization or self-ionization of water is an endothermic ionization reaction where a molecule of pure water gets dissociated into hydrogen ions (H+) and hydroxyl ions (OH-). As water is a very weak electrolyte, it gets ionized minimally.

Autoionization Chemical Equation

When two water molecules collide, a hydrogen ion from one molecule gets transferred to the other, producing a positively charged hydronium ion (H₃O⁺) and a negatively charged hydroxyl ion.

H₂O (l) + H₂O (l) ⇌ H₃O⁺ (aq.) + OH^– (aq.)

We commonly use the simplified form of the reaction:

H₂O (l) ⇌ H⁺ (aq.) + OH^– (aq.)

Acid-Base Properties of Water

As it is amphoteric, water can act as both an acid and a base in solution. When mixed with a base, water donates hydrogen ions and acts as an acid. However, when water is mixed with an acid, it accepts hydrogen ions and acts as a base.

• Acting as a base:

H₂O (l) + CH₃COOH (aq.) ⇌ H₃O⁺ (aq.) + CH₃COO^– (aq.)

• Acting as an acid:

NH₃ (aq.) + H₂O (l) ⇌ NH₄⁺ (aq.) + OH^– (aq.)

During autoionization, water can be both an acid and a base to itself.

H₂O (l) + H₂O (l) ⇌ H₃O⁺ (aq.) + OH^– (aq.)

Autoionization Constant

As mentioned before, H₂O splits into H⁺ and OH^–.

K_c = [H⁺] [OH^–]/[H₂O]

K_c is the dissociation constant of water at equilibrium.

As pure water is a poor electrolyte, it gets slightly ionized. So, the concentration of H₂O is much greater than the concentrations of H⁺ and OH^–. Therefore, [H₂O] can be regarded as constant.

K_c [H₂O] = [H⁺][OH^–]

As K_c and [H₂O] are constants, they are replaced by another constant, ‘K_w’.

K_w = [H⁺][OH^–]

K_w is called the autoionization constant of water.

From the above explanation, ‘K_w’ can be regarded as the ionic product of water, i.e., the product of the molar concentrations of H⁺ and OH^– ions. As H⁺ ions exist as H₃O⁺ ions in water, the ionic product can alternatively be stated as the product of molar concentrations of H₃O⁺ and OH^– ions.

K_w = [H₃O⁺][OH^–]

Relationship Among pH, pOH, and pK_w

The autoionization constant is temperature-dependent and increases with an increase in temperature. It has a fixed value of 1×10^-14 at 25 ͦC.

The concentrations of dissociated H⁺ and OH^– ions are equal. So, let the [H⁺] and [OH^–] be ‘x’.

By substituting the values in the equilibrium expression, we get,

[H⁺] [OH^–] = 1 x 10^-14

=> x . x = 1 x 10^-14

=> x² = 1 x 10^-14

=> x = 1 x 10^-7

Therefore, [H⁺] = [OH^–] = 1 x 10^-7

Taking negative logarithm on both sides of [H⁺] equation:

-log [H⁺] = -log (10^-7)

=> pH = – (-7)

=> pH = 7 [ Since, pH = -log [H⁺]]

Similarly, pOH will also be 7. This result is expected since the pH of pure water is 7.

Now, taking the negative logarithm of K_w

-log (K_w) = -log (1×10^-14)

=> pK_w = – (-14) = 14

Therefore, at 25 ͦC,

pK_w = pH + pOH