General Chemistry/Chemical Equilibria/Acid-Base Equilibrium

Consider an acid, HA, in water. According to the Brønsted-Lowry acid/base theory, the acid should protonate the water to form hydronium and the conjugate base, A^-. There will be an equilibrium between the acid and water, and hydronium and the conjugate base.

${\hbox{HA}}+{\hbox{H}}_{2}{\hbox{O}}\leftrightarrow {\hbox{H}}_{3}{\hbox{O}}^{+}+{\hbox{A}}^{-}$

This equilibrium can be used to calculate the concentrations of species in the solution.

Acid Dissociation Constant

Helpful Hint!
Strong acids/bases dissociate completely, so their equilibrium constants are overwhelmingly large.

Like all equilibria, an acid/base dissociation will have a particular equilibrium constant which will determine the extent of the reaction (whether it lies to the left or right of the equation). As the equilibrium constant approaches zero, the reaction tends to form 100% reactants. As the equilibrium constant approaches infinity, the reaction tends to form 100% products. The equilibrium constant K=1 states that there will be 50% products and 50% reactants

Because the equilibrium is used for calculating the concentrations of weak acids, very little water actually reacts. The concentration of water during the reaction is, therefore, a constant, and can be excluded from the expression for K. This gives rise to a special equilibrium constant, K_a, known as the acid dissociation constant. It is simply K multiplied by the concentration of water.

$K_{a}={\hbox{K}}[{\hbox{H}}_{2}{\hbox{O}}]={\frac {[{\hbox{H}}_{3}{\hbox{O}}^{+}][{\hbox{A}}^{-}]}{[{\hbox{HA}}]}}$

The K_a of a weak acid determines how acidic it is, i.e., how far its equilibrium lies to the right. The K_a values of weak acids can and have been determined experimentally.

Base Dissociation Constant

A similar equilibrium exists when a weak base is dissolved in water. The base will accept a proton from water and form a conjugate acid, BH⁺.

${\hbox{B}}+{\hbox{H}}_{2}{\hbox{O}}\leftrightarrow {\hbox{OH}}^{-}+{\hbox{BH}}^{+}$

This equilibrium has its own special constant, K_b, known as the base dissociation constant. Like the acid dissociation constant, it is defined as the equilibrium constant multiplied by the concentration of water.

$K_{b}=K[{\hbox{H}}_{2}{\hbox{O}}]={\frac {[{\hbox{BH}}^{+}][{\hbox{OH}}^{-}]}{[{\hbox{B}}]}}$

Ion Product Constant

A special equilibrium exists between ordinary water molecules. Occasionally, one water molecule will act as an acid, and donate a proton to another water molecule (which acts as a base). This is the auto-ionization of water.

${\hbox{H}}_{2}{\hbox{O}}+{\hbox{H}}_{2}{\hbox{O}}\leftrightarrow {\hbox{H}}_{3}{\hbox{O}}^{+}+{\hbox{OH}}^{-}$

By the Le Chatelier Principle, we can see that if the hydronium ion concentration is increased (by adding an acid), the equilibrium will move to the left and there will be a lower concentration of hydroxide. Thus, hydronium and hydroxide concentration are inversely related—an increase in one will result in a decrease in the other, and vice versa.

The equilibrium expression for this reaction is given a special name, K_w. Because it simplifies to the product of hydronium and hydroxide concentration, it is sometimes called the ion product constant. The value of this constant is 1.0 × 10^-14 at 25°C.

$K_{w}=[{\hbox{H}}_{3}{\hbox{O}}^{+}][{\hbox{OH}}^{-}]=1.0\times 10^{-14}$

This expression can be used to find the pH of pure water. Recall that pH is the negative logarithm of hydronium ion concentration. If we set the hydronium ion concentration in the above expression to be $x$, we can derive the pH. The hydroxide ion concentration must also be $x$ (since each molecule of hydroxide is a result of a molecule of hydronium forming). We have:

${\begin{matrix}x^{2}&=&1.0\times 10^{-14}\\x&=&{\sqrt {1.0\times 10^{-14}}}\\x&=&1.0\times 10^{-7}\\{\hbox{pH}}&=&-\log {1.0\times 10^{-7}}\\{\hbox{pH}}&=&7.0\end{matrix}}$

This technique can be used to determine the pH of any solution if either one of the ion concentrations are known.

Conjugate Base Expressions

The conjugate bases of weak acids have a relationship with their parent acids. Consider the equilibrium expression of the conjugate base, A^-, of the weak acid HA.

$K_{b}={\frac {[{\hbox{HA}}][{\hbox{OH}}^{-}]}{{\hbox{A}}^{-}}}$

If we multiply the expression for an acid by the expression for its conjugate base, the concentrations of the acid and conjugate base cancel and we obtain the ion product constant for water! This allows us to calculate the K_b of a base if the K_a of its conjugate acid is known (and vice versa).

${\begin{matrix}{\frac {[{\hbox{H}}_{3}{\hbox{O}}^{+}][{\hbox{A}}^{-}]}{[{\hbox{HA}}]}}&\times &{\frac {[{\hbox{HA}}][{\hbox{OH}}^{-}]}{[{\hbox{A}}^{-}]}}&=&[{\hbox{H}}_{3}{\hbox{O}}^{+}][{\hbox{OH}}^{-}]\\K_{a}&\times &K_{b}&=&K_{w}\\\end{matrix}}$

Summary

The definitions of the acid and base dissociation constants are very important. They are presented here for reference.

$K_{a}={\frac {[{\hbox{H}}_{3}{O}^{+}][{\hbox{A}}^{-}]}{[{\hbox{HA}}]}}$	The acid dissociation constant measures the strength of an acid dissolved in water.
$K_{b}={\frac {[{\hbox{BH}}^{+}][{\hbox{OH}}^{-}]}{[{\hbox{B}}]}}$	The base dissociation constant measures the strength of a base dissolved in water.
$K_{w}=K_{a}\times K_{b}$	The ion product constant relates the concentrations of acids/bases and their conjugates in a solution.