How to Calculate the pH of a Weak Acid

pH of a Weak Acid Worked Chemistry Problem

scientist pouring liquid into beaker
Determine the pH of a weak aid using the acid dissociation constant and quadratic equation. Glow Images, Inc / Getty Images

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 Ka= 6.5 x 10-5


Benzoic acid dissociates in water as

C6H5COOH → H+ + C6H5COO-

The formula for Ka is

Ka = [H+][B-]/[HB]

[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 Ka equation

Ka = x · x / (C -x)
Ka = x²/(C - x)
(C - x)Ka = x²
x² = CKa - xKa
x² + Kax - CKa = 0

Solve for x using the quadratic equation

x = [-b ± (b² - 4ac)½]/2a

x = [-Ka + (Ka² + 4CKa)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 Ka and C

Ka = 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


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 Ka equation,

Ka = x²/(C - x)
Ka = x²/C

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

x² = Ka·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.