2.2 Ionization Constants: Ka, Kb, and Kw

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Because acid/base solutions are systems at equilibrium, we can write equilibrium constant expressions for these systems.

K_a and K_b

Here are several examples, including some that show water as a reactant. Pay attention to the physical states and whether or not a particular substance is included in the equilibrium expression. Also notice that we now identify the equilibrium constant as K_a for acids and K_b for bases.

\(\ce{HCl(aq) -> H^{+}(aq) + Cl^{-}(aq)}\)

\(Ka = \frac{[\ce{H+}][\ce{Cl-}]}{[\ce{HCl}]} = 1.3 \times 10^6\)

\(\ce{HCl(aq) + H2O(l) -> H3O^{+}(aq) + Cl^{-}(aq)}\)

\(Ka = \frac{[\ce{H3O+}][\ce{Cl-}]}{[\ce{HCl}]} = 1.3 \times 10^6\)

\(\ce{HCHO2(aq) -> H^{+}(aq) + CHO2^{-}(aq)}\)

\(Ka = \frac{[\ce{H+}][\ce{CHO2-}]}{[\ce{HCHO2}]} = 1.8 \times 10^{-4}\)

\(\ce{Mg(OH)2(aq) -> Mg^{2+}(aq) + 2OH^{-}(aq)}\)

\(Kb = \frac{[\ce{Mg2+}][\ce{OH-}]^{2}}{[\ce{Mg(OH)2}]}\)

\(\ce{NH3(aq) + H2O(l) -> NH4^{+}(aq) + OH^{-}(aq)}\)

\(Kb = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} = 1.8 \times 10^{-5}\)

Recall that the value of the equilibrium constant indicates which side of a reaction is favored. When K_eq is greater than 1 the product side is favored; when K_eq is less than 1 the reactants are favored.

So we now have a numerical way to indicate acid and base strength with K_a and K_b

A large value of K_a
means there are many
\(\ce{H+}\) ions in solution -
in other words, a strong acid

A large K_b indicates
many \(\ce{OH-}\) ions -
a strong base

A table of acid and base strengths gives K_a and K_b values for a number of acids and bases. K_a values for very strong acids are typically not given as there is little use in expressing them numerically.

K_w

We usually do not think of water as producing ions, but that isn't the case. Water does ionize, although not very well.

We can write the ionization equation for water in two ways (both mean essentially the same thing). The equilibrium constant, K_w, can also be written for both equations as shown.

Notice that \(\ce{H2O}\) does not appear in the K_w expression because it is a liquid, and you'll remember that liquids and solids are not included in equilibrium constant expressions.

\(\ce{H2O(l) -> {H+(aq)} + {OH-(aq)}}\)

\(Kw = {[\ce{H+}][\ce{OH-}]} = 1.0 \times 10^{-14}\)

\(\ce{2H2O(l) -> {H3O+(aq)} + {OH-(aq)}}\)

\(Kw = {[\ce{H+}][\ce{OH-}]} = 1.0 \times 10^{-14}\)

Values for K_w are given for 25°C.

Two more key items to note:

In pure water, the balanced equation tells us that the concentrations of \(\ce{H+}\) and \(\ce{OH-}\) will be equal to one another (this is what makes water neutral). Solving the equations shown in the table above, we find that [\(\ce{H+}\)] = [\(\ce{OH-}\)] = 1.0×10^-7
As long as temperature remains constant K_w is a constant - it's value will not change.

The value of K_w is very small, meaning that very few ions are present. Most water remains "intact" as \(\ce{H2O}\), and few ions form. Why, then, do we even mention it? For a very important reason that we will examine in the next section.