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.

Check your work before reporting a value. The pH of a weak acid should be less than 7 (not neutral) and it's usually less than the value for a strong acid. Note there are exceptions. For example, the pH of hydrochloric acid is 3.01 for a 1 mM solution, while the pH of hydrofluoric acid is also low, with a value of 3.27 for a 1 mM solution.

Sources

• Bates, Roger G. (1973). Determination of pH: theory and practice. Wiley.
• Covington, A. K.; Bates, R. G.; Durst, R. A. (1985). "Definitions of pH scales, standard reference values, measurement of pH, and related terminology". Pure Appl. Chem. 57 (3): 531–542. doi:10.1351/pac198557030531
• Housecroft, C. E.; Sharpe, A. G. (2004). Inorganic Chemistry (2nd ed.). Prentice Hall. ISBN 978-0130399137.
• Myers, Rollie J. (2010). "One-Hundred Years of pH". Journal of Chemical Education. 87 (1): 30–32. doi:10.1021/ed800002c
• Miessler G. L.; Tarr D .A. (1998). Inorganic Chemistry (2nd ed.). Prentice-Hall. ISBN 0-13-841891-8.