Ionization of Water, Weak Acids, and Weak Bases: -The Ionization of Water Is Expressed by an Equilibrium Constant
The degree of ionization of water at equilibrium (Eqn 2–1) is small; at 25 °C only about two of every 109 molecules in pure water are ionized at any instant. The equilibrium constant for the reversible ionization of water (Eqn 2–1) is
In pure water at 25 C, the concentration of water is 55.5 M (grams of H2O in 1 L divided by its gram molecular weight: (1,000 g/L)/ (18.015 g/mol)) and is essentially constant in relation to the very low concentrations of H and OH, namely, 1x10-7 M. Accordingly, we can substitute 55.5 M in the equilibrium constant ex pression (Eqn 2–3) to yield
which, on rearranging, becomes
(55.5 M) (Keq)= [H+] [OH-]=Kw
where Kw designates the product (55.5 M) (Keq), the ion product of water at 25 °C. The value for Keq, determined by electrical-conductivity measurements of pure water, is 1.8x10-16 M at 25 C. Substituting this value for Keq in Equation 2–4 gives the value of the ion product of water:
Kw=[H+] [OH-]= (55.5 M) (1.8x10-16 M)= 1.0x10-14M2
Thus the product [H+] [OH-] in aqueous solutions at 25 C always equals 1x10-14 M2. When there are exactly equal concentrations of H+ and OH-, as in pure water, the solution is said to be at neutral pH. At this pH, the concentration of H+ and OH- can be calculated from the ion product of water as follows:
Kw=[H+] [OH-]=[H+]2
Solving for [H+] gives
As the ion product of water is constant, whenever [H+] is greater than 1 than 1x10-7 M, [OH-] must become less 10 7M, and vice versa. When [H|+] is very high, as in a solution of hydrochloric acid, [OH-] must be very low. From the ion product of water we can calculate [H+] if we know [OH-], and vice versa (Box 2–2).
