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 kw for the above reaction is known as the ion product of water. The value of kw at room temperature is 1.007 x 10-14.
kw = [H3O+] * [OH-] / [H2O]2= [H3O+] * [OH-] = 1.007 x 10-14 (2)
In pure water, at 25 ℃, since [H3O+] = [OH-], equation (2) becomes:
kw = [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] = kH * PCO2 = 0.032 * PCO2
The equilibrium that takes place is given in Fig. I.1:
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+ |
Hiç yorum yok:
Yorum Gönder