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

C_{6}H_{5}COOH → H^{+} + C_{6}H_{5}COO^{-}

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 C_{6}H_{5}COO^{-} ion, so [H^{+}] = [C_{6}H_{5}COO^{-}].

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

_{a}x - CK

_{a}= 0

Solve for x using the quadratic equation

**x = [-b ± (b² - 4ac) ^{½}]/2a**

x = [-K_{a} + (K_{a}² + 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.