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Self-ionization of water - Autoionization of water - The ion product of water (kw)

A water molecule can act as an acid and as a base according to:

In the above reaction – autoionization of water - one water molecule acts as an acid donating a proton to the other water molecule that acts as a base – proton acceptor. From the equation we see that for every hydronium ion H₃O⁺ that is formed a hydroxyl ion OH^-  is formed also.

Therefore, at equilibrium: [H₃O⁺] = [OH^-]

The equilibrium constant k_w for the above reaction is known as the ion product of water. The value of k_w at room temperature is 1.007 x 10^-14.

k_w = [H₃O⁺] * [OH^-] / [H₂O]² = [H₃O⁺] * [OH^-] = 1.007 x 10^-14     (2)

In pure water, at 25 ℃, since [H₃O⁺] = [OH^-], equation (2) becomes:

k_w = [H₃O⁺]² = 1.007 x 10^-14  and therefore,   [H₃O⁺] = [OH^-] ≈ 10^-7 M    (25 ℃)

The pH of pure water at 25 ℃:  pH = -log [H⁺] = -log (10^-7) =  -(-7) = 7  

However, the pH of pure water would slightly decrease over time - to 6.6 – 6.8 -due to the dissolved CO₂ from the air.

Carbon dioxide CO₂ is slightly soluble in pure water and the solubility decreases with temperature. At pressures up to about 4 atm, the solubility follows Henry’s law:

[CO₂] = k_H * P_CO2 = 0.032 * P_CO2

  The equilibrium that takes place is given in Fig. I.1:

 

Fig. I.1: CO₂ equilibrium between the gaseous and liquid phase

Once it has dissolved, a small portion of the CO₂ reacts with water to form carbonic acid H₂CO₃ that dissociates in water to produce H₃O⁺ decreasing the pH of the solution (Fig. I.2):

Fig. I.2: CO₂ reacts with water to form carbonic acid H₂CO₃ that dissociates in water to produce H₃O⁺