How to Calculate the pH of a Weak Acid

Calculating the pH of a weak acid is a bit more complicated than determining the pH of a strong acid because weak acids don't completely dissociate in water. Fortunately, the formula for calculating pH is simple. Here's what you do.

pH of a Weak Acid Problem

What is the pH of a 0.01 M benzoic acid solution?

Given: benzoic acid K_a= 6.5 x 10^-5

Solution

Benzoic acid dissociates in water as

C₆H₅COOH → H⁺ + C₆H₅COO^-

The formula for K_a is

K_a = [H⁺][B^-]/[HB]

where
[H⁺] = concentration of H⁺ ions
[B^-] = concentration of conjugate base ions
[HB] = concentration of undissociated acid molecules
for a reaction HB → H⁺ + B^-

Benzoic acid dissociates one H⁺ ion for every C₆H₅COO^- ion, so [H⁺] = [C₆H₅COO^-].

Let x represent the concentration of H⁺ that dissociates from HB, then [HB] = C - x where C is the initial concentration.

Enter these values into the K_a equation

K_a = x · x / (C -x)
K_a = x²/(C - x)
(C - x)K_a = x²
x² = CK_a - xK_a
x² + K_ax - CK_a = 0

Solve for x using the quadratic equation

x = [-b ± (b² - 4ac)^½]/2a

x = [-K_a + (K_a² + 4CK_a)^½]/2

**Note** Technically, there are two solutions for x. Since x represents a concentration of ions in solution, the value for x cannot be negative.

Enter values for K_a and C

K_a = 6.5 x 10^-5
C = 0.01 M

x = {-6.5 x 10^-5 + [(6.5 x 10^-5)² + 4(0.01)(6.5 x 10^-5)]^½}/2
x = (-6.5 x 10^-5 + 1.6 x 10^-3)/2
x = (1.5 x 10^-3)/2
x = 7.7 x 10^-4

Find pH

pH = -log[H⁺]

pH = -log(x)
pH = -log(7.7 x 10^-4)
pH = -(-3.11)
pH = 3.11

Answer

The pH of a 0.01 M benzoic acid solution is 3.11.

Solution: Quick and Dirty Method to Find Weak Acid pH

Most weak acids barely dissociate in solution. In this solution we found the acid only dissociated by 7.7 x 10^-4 M. The original concentration was 1 x 10^-2 or 770 times stronger than the dissociated ion concentration.

Values for C - x then, would be very close to C to seem unchanged. If we substitute C for (C - x) in the K_a equation,

K_a = x²/(C - x)
K_a = x²/C

With this, there is no need to use the quadratic equation to solve for x

x² = K_a·C

x² = (6.5 x 10^-5)(0.01)
x² = 6.5 x 10^-7
x = 8.06 x 10^-4

Find pH

pH = -log[H⁺]

pH = -log(x)
pH = -log(8.06 x 10^-4)
pH = -(-3.09)
pH = 3.09

Note the two answers are nearly identical with only 0.02 difference. Also notice the difference between the first method's x and the second method's x is only 0.000036 M. For most laboratory situations, the second method is 'good enough' and much simpler.