## Calculating the pH of a Buffer Solution

#### 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: ${ K}_{a }=\frac { { [H }^{ + }][{ A }^{ - }] }{ [HA] }$ .
• Using Ka 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:

$HA \rightleftharpoons H^+ + A^-$

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

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

The greater [H+] x [A] is than [HA], the greater the value of Ka, 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 NH3 and 0.0350 M NH4+. (Note: Ka for NH4+ is 5.6 x 10-10). The equation for the reaction is as follows:

$NH_4^+ \rightleftharpoons H^+ + NH_3$

We know that initially there is 0.0350 M NH4+ and 0.0500 M NH3. Before the reaction occurs, no H+ is present so it starts at 0.

During the reaction, the NH4+ will dissociate into H+ and NH3. Because the reaction has a 1:1 stoichiometry, the amount that NH4+ loses is equal to the amounts that H+ and NH3 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 Ka.

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

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

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

Solving for x (H+):

x = [H+] = 3.92 x 10-10

pH = -log(3.92 x 10-10)

pH = 9.41