#### Learning Objective

- Calculate the pH of a buffer made only from a weak acid.

#### Key Points

- The strength of a weak acid (buffer) is usually represented as an equilibrium constant.
- The acid-dissociation equilibrium constant, which measures the propensity of an acid to dissociate, is described using the equation: [latex]{ K}_{a }=\frac { { [H }^{ + }][{ A }^{ - }] }{ [HA] }[/latex] .
- Using K
_{a}and the equilibrium equation, you can solve for the concentration of [H^{+}]. - The concentration of [H
^{+}] can then be used to calculate the pH of a solution, as part of the equation: pH = -log([H^{+}]).

#### Term

- equilibriumThe state of a reaction in which the rates of the forward (reactant to product) and reverse (product to reactant) reactions are the same.

## What Does pH Mean in a Buffer?

In chemistry, pH is a measure of the hydrogen ion (H^{+}) concentration in a solution. The pH of a buffer can be calculated from the concentrations of the various components of the reaction. The balanced equation for a buffer is:

[latex]HA \rightleftharpoons H^+ + A^-[/latex]

The strength of a weak acid is usually represented as an equilibrium constant. The acid-dissociation equilibrium constant (K_{a}), which measures the propensity of an acid to dissociate, for the reaction is:

[latex]K_{a} = \frac{\left [H^{+} \right ]\left [A^{-} \right ]}{\left [HA \right ]}[/latex]

The greater [H^{+}] x [A^{–}] is than [HA], the greater the value of K_{a}, the more the formation of H^{+} is favored, and the lower the pH of the solution.

## ICE Tables: A Useful Tool For Solving Equilibrium Problems

ICE (Initial, Change, Equilibrium) tables are very helpful tools for understanding equilibrium and for calculating the pH of a buffer solution. They consist of using the initial concentrations of reactants and products, the change they undergo during the reaction, and their equilibrium concentrations. Consider, for example, the following problem:

Calculate the pH of a buffer solution that initially consists of 0.0500 M NH_{3} and 0.0350 M NH_{4}^{+}. (Note: K_{a} for NH_{4}^{+} is 5.6 x 10^{-10}). The equation for the reaction is as follows:

[latex]NH_4^+ \rightleftharpoons H^+ + NH_3[/latex]

We know that initially there is 0.0350 M NH_{4}^{+} and 0.0500 M NH_{3}. Before the reaction occurs, no H^{+} is present so it starts at 0.

During the reaction, the NH_{4}^{+} will dissociate into H^{+} and NH_{3}. Because the reaction has a 1:1 stoichiometry, the amount that NH_{4}^{+} loses is equal to the amounts that H^{+} and NH_{3} will gain. This change is represented by the letter x in the following table.

Therefore the equilibrium concentrations will look like this:

Apply the equilibrium values to the expression for K_{a}.

[latex]{ 5.6 \times 10^{-10}} = \frac { { [H }^{ + }][{ NH_3 }] }{ [NH_4^+] } = \frac { x (0.0500+x)}{ 0.0350-x }[/latex]

Assuming x is negligible compared to 0.0500 and 0.0350 the equation is reduced to:

[latex]{ 5.6 \times 10^{-10}} = \frac { { [H }^{ + }][{ NH_3 }] }{ [NH_4^+] } = \frac { x (0.0500)}{ 0.0350 }[/latex]

Solving for x (H^{+}):

x = [H^{+}] = 3.92 x 10^{-10}

pH = -log(3.92 x 10^{-10})

pH = 9.41