pK_{b} is the negative base-10 logarithm of the base dissociation constant (K_{b}) of a solution. It is used to determine the strength of a base or alkaline solution.

pKb = -log_{10}K_{b}

The lower the pK_{b} value, the stronger the base. As with the acid dissociation constant, pK_{a}, the base dissociation constant calculation is an approximation that is only accurate in dilute solutions. Kb can be found using the following formula:

K_{b} = [B^{+}][OH^{-}] / [BOH]

which is obtained from the chemical equation:

BH^{+} + OH^{−} ⇌ B + H_{2}O

### Finding pKb from pKa or Ka

The base dissociation constant is related to the acid dissociation constant, so if you know one, you can find the other value. For an aqueous solution, the hydroxide ion concentration [OH^{-} follows the relation of the hydrogen ion concentration [H^{+}]" K_{w} = [H^{+}][OH^{-}

Putting this relation into the K_{b} equation gives: K_{b} = [HB^{+}K_{w} / ([B][H]) = K_{w} / K_{a}

At the same ionic strength and temperatures:

pK_{b} = pK_{w} - pK_{a}.

For aqueous solutions at 25° C, pK_{w} = 13.9965 (or about 14), so:

pK_{b} = 14 - pK_{a}

### Sample pK_{b} Calculation

Find the value of the base dissociation constant K_{b} and pK_{b} for a 0.50 dm^{-3} aqueous solution of a weak base that has a pH of 9.5.

First calculate the hydrogen and hydroxide ion concentrations in the solution to get values to plug into the formula.

[H^{+}] = 10^{-pH} = 10^{-9.5} = 3.16 x 10^{–10} mol dm^{–3}

K_{w} = [H^{+}_{(aq)}] [OH^{–}_{(aq)}] = 1 x 10^{–14} mol^{2} dm^{–6}

[OH^{–}_{(aq)}] = K_{w}_{/}[H^{+}_{(aq)}] = 1 x 10^{–14} / 3.16 x 10^{–10 }**= **3.16 x 10^{–5} mol dm^{–3}

Now, you have the necessary information to solve for the base dissociation constant:

K_{b} = [OH^{–}_{(aq)}]^{2}_{/}[B_{(aq)}] = (3.16 x 10^{–5})** ^{2}** / 0.50

**=**2.00 x 10

^{–9}mol dm

^{–3}

pK_{b} = –log(2.00 x 10^{–9}) **= **8.70