#### Learning Objective

- Calculate the Kw (water dissociation constant) using the following equation: Kw = [H+] x [OH−] and manpulate the formula to determine [OH−] = Kw/[H+] or [H+]=Kw/[OH-]

#### Key Points

- The base dissociation constant K
_{bE}measures a base’s basicity, or strength. - K
_{b}is related to the acid dissociation constant, K_{a}, by the simple relationship pK_{a}+ pK_{b}= 14, where pK_{b}and pK_{a}are the negative logarithms of K_{b}and K_{a}, respectively. - K
_{b}and K_{a }are also related through the ion constant for water, K_{w}, by the relationship [latex]K_W=K_a\times K_b[/latex] .

#### Term

- conjugate acidthe species created when a base accepts a proton

In chemistry, a base is a substance that can accept hydrogen ions (protons) or, more generally, donate a pair of valence electrons. The base dissociation constant, K_{b}, is a measure of basicity—the base’s general strength. It is related to the acid dissociation constant, K_{a}, by the simple relationship pK_{a} + pK_{b} = 14, where pK_{b} and pK_{a} are the negative logarithms of K_{b} and K_{a}, respectively. The base dissociation constant can be expressed as follows:

[latex]K_b = \dfrac{[\text{BH}^+][\text{OH}^-]}{\text{B}}[/latex]

where [latex]\text{B}[/latex] is the base, [latex]\text{BH}^+[/latex] is its conjugate acid, and [latex]\text{OH}^-[/latex] is hydroxide ions.

## The Base Dissociation Constant

Historically, the equilibrium constant K_{b} for a base has been defined as the association constant for protonation of the base, B, to form the conjugate acid, HB^{+}.

[latex]B(aq) + H_2O(l) \leftrightharpoons HB^+(aq) + OH^-(aq)[/latex]

As with any equilibrium constant for a reversible reaction, the expression for K_{b} takes the following form:

[latex]K_{b} = \frac{[OH^{-}][HB^{+}]}{[B]}[/latex]

K_{b} is related to K_{a} for the conjugate acid. Recall that in water, the concentration of the hydroxide ion, [OH^{−}], is related to the concentration of the hydrogen ion by the autoionization constant of water:

[latex]K_W=[H^+][OH^-][/latex]

Rearranging, we have:

[latex][OH^{-}] = \frac{K_{w}}{[H^{+}]}[/latex]

Substituting this expression for [OH^{−}] into the expression for K_{b }yields:

[latex]K_{b} = \frac{K_{w}[HB^{+}]}{[B][H^{+}]} = \frac{K_{w}}{K_{a}}[/latex]

Therefore, for any base/conjugate acid pair, the following relationship always holds true:

[latex]K_W=K_aK_b[/latex]

Taking the negative log of both sides yields the following useful equation:

[latex]pK_a+pK_b=14[/latex]

In actuality, there is no need to define pK_{b} separately from pK_{a}, but it is done here because pK_{b} values are found in some of the older chemistry literature.

## Calculating the pH of a Weak Base in Aqueous Solution

The pH of a weak base in aqueous solution depends on the strength of the base (given by K_{b}) and the concentration of the base (the molarity, or moles of the base per liter of solution). A convenient way to find the pH for a weak base in solution is to use an ICE table: ICE stands for “Initial,” “Change,” and”Equilibrium.”

Before the reaction starts, the base, B, is present in its initial concentration [B]_{0}, and the concentration of the products is zero. As the reaction reaches equilibrium, the base concentration decreases by *x* amount; given the reaction’s stoichiometry, the two products increase by *x *amount. At equilibrium, the base’s concentration is [B]_{0} – *x, *and the two products’ concentration is *x.*

The K_{b} for the reaction is:

[latex]K_{b} = \frac{[BH^+][OH^-]}{[B]}[/latex]

Filling in the values from the equilibrium line gives:

[latex]K_{b} = \frac{x^2}{[B]_{0}-x}[/latex]

This quadratic equation can be solved for *x*. However, if the base is weak, then we can assume that *x* will be insignificant compared to [B]_{0}, and the approximation [B]_{0}– *x* ≈ [B]_{0} can be used. The equation simplifies to:

[latex]K_{b} = \frac{x^2}{[B]_{0}}[/latex]

Since *x* = [OH]^{–}, we can calculate pOH using the equation pOH = –log[OH]^{–}; we can find the pH using the equation 14 – pOH = pH.