# Dissociation of weak bases

## Basicity constant $K_b$

The law of mass action (Guldberg and Waage) applies to the equilibrium of dissociation of weak bases:
$B$ $+$ $H_2O$ $\rightleftarrows$ $ OH^-$ $+$ $HB$
$K$ $=$ $\frac{[OH^-][HB]}{[B][H_2O]}$
In dilute solutions of bases, the molarity of $H_2O$ differs very little from what we have seen for pure water $[H_2O]$ $=$ $55,5 \frac{mol}{L}$,
therefore:
$K\cdot[H_2O]$ $ =$ $\frac{[OH^-][HB]}{[B]}$
Under these conditions, $K\cdot[H_2O]$ is a new constant which is like $K$, depending only on the temperature. It's called $K_b $: basicity constant of the couple $(HB, B)$

Basicity constant of the couple $(HB, B)$:
$K_b$ $=$ $\frac{[OH^-][HB]}{[B]}$

## Relative strength of weak bases

Let's take two weak bases 1 and 2 with $K_{b1}\gt K_{b2} $:
The numerator of the expression $\frac{[OH^-][HB]}{[B]}$ is greater for 1 than 2 (or the denominator smaller ), that means that in the aqueous solution of 1 there will be more $OH^-$ and $HB$ than in that of 2 and less $B$ .
1 is more dissociated (more "strong") that 2

$K_{b1}$ $\gt$ $ K_{b2} \;$ $\Rightarrow$
$B_1$ "stronger" than $B_2$

Example:
The sulfide ion $S^{2-} $ with $K_b= 10$ is a (weak!) base a lot stronger than the hydrosulfide ion $HS^-$ with $K_b = 10^{-7}$

## The $pK_b$

By analogy with the pOH it is defined:

$pK_b$ $=$ $-logK_b$
$K_b$ $=$ $10^{-pK_b}$

As the function $-log$ is decreasing, we get for the two weak bases 1 and 2:

$pK_{b1}$ $\lt$ $ pK_{b2} \;$ $\Rightarrow$
$B_1$ "stronger" than $B_2$

## Relationship between acidity and basicity constants of the same acid-base couple

$K_a \cdot K_b$ =
$\frac{[H_3O^+][B]}{[HB]}\frac{[OH^-][HB]}{[B]}$ =
$[H_3O^+][OH^-]$ =
$10^{-14}$
therefore:
$pK_a + pK_b$ =
$-logK_a -logK_b$ =
$-(logK_a + logK_b)$ =
$-log(K_aK_b)$ =
$-log(10^{-14})$ =
$14$

$K_a\cdot K_b$ $=$ $10^{-14}$
$pK_a$ $+$ $pK_b$ $=$ $14$

Example:
Given an acid with $pK_a=3$:
Then
$K_a$ $=$ $10^{-3}$
and for its corresponding base:
$pK_b$ $=$ $11$
and
$K_b$ $=$ $10^{-11}$