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 $(HB, B)$: $K_b$ $=$ $\frac{[OH^-][HB]}{[B]}$
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 $K_b = 10^{-7}$
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 $B_2$
$K_a \cdot K_b$ = $\frac{[H_3O^+][B]}{[HB]}\frac{[OH^-][HB]}{[B]}$ = $[H_3O^+][OH^-]$ = $10^{-14}$ 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 $K_b$ $=$ $10^{-11}$