In this lesson, we will learn:

- To recall the equilibrium ionization expressions for
*weak*acids and bases. - How to relate K
_{a}and K_{b}for conjugate acid/base pairs. - How to calculate concentration of aqueous ions in weak acid/base solutions.

__Notes:__- In Autoionization of water, we looked at the equilibrium:
2 H _{2}O_{(l)}$\, \rightleftharpoons \,$ H_{3}O^{+}_{(aq)}+ OH^{-}_{(aq)}

The equilibrium constant was expressed as:K _{w}= [H_{3}O^{+}] [OH^{-}] = 1 * 10^{-14}at 25^{o}C

We saw the effect of adding strong acids/bases to equilibrium concentrations of water and dissolved ions. That was straightforward because__strong acids and bases experience 100% dissociation in water__.

We also saw in - In strong and weak acids and bases that
__weak acids and bases do not experience 100% dissociation__. This makes expressions for their dissociation in water more complicated. - Every weak acid has an
and a weak base a*acid dissociation constant, K*_{a}__base dissociation constant,__. These are equilibrium constants showing how much the acid/base dissociates when dissolved in water (aqueous solution).*K*_{b}

For a weak acid HX dissolved in water:*K*= $\frac{[H_3O^{+}] [X^{-}]}{[HX]}$_{a}

For a weak base B dissolved in water:*K*= $\frac{[HB^{+}] [OH^{-}]}{[B]}$_{b}

- Just like in the ionization of water and other equilibria; this is an equilibrium constant expression. This means that
__the higher the K__(because the concentrations of the dissociated ions, in the numerator, are larger values) and therefore_{a}/K_{b}value, the greater the degree of dissociation__the stronger the acid or base__. - Remember that
__for strong acids and bases K__. This is because in the K_{a}/K_{b}values are not normally used_{a}expression, their [HX] or [B] is equal to or almost zero due to complete dissociation, so the values are incredibly large. __For strong acids, pK__and is more appropriate to use for strong acids, instead of the extremely large K_{a}is used instead of K_{a}. pK_{a}is the negative logarithm of the K_{a}value_{a}values they have.

- Just like in the ionization of water and other equilibria; this is an equilibrium constant expression. This means that
- In Conjugate acids and bases, we learned in a conjugate pair that a stronger conjugate acid will have a weaker the conjugate base. This relationship affects the concentration of aqueous ions and therefore affects the K
_{a}and K_{b}values for a conjugate acid/base pair!

Consider the equations for the conjugate pair acid-base pair HX and X^{-}:Conjugate acid: $\qquad$ HX + H _{2}O $\, \rightleftharpoons \,$ X^{-}+ H_{3}O^{+}$\qquad$*K*= $\frac{[H_3O^{+}] [X^{-}]}{[HX]}$_{a}Conjugate base: $\qquad$ X ^{-}+ H_{2}O $\, \rightleftharpoons \,$ HX + OH^{-}$\qquad$*K*= $\frac{[HX] [OH^{-}]}{[X^{-}]}$_{b}

- Both equations depend on [X
^{-}] and [HX] so K_{a}and K_{b}themselves can be related, and terms cancelled out:

As you can see, the result is the product of [H_{3}O^{+}] and [OH^{-}] which is the expression for K_{w}. Therefore for a conjugate pair:K _{a}* K_{b}= K_{w}

- Both equations depend on [X