By John Brennan
Updated Aug 30, 2022

A titration is a classic laboratory technique used to determine the concentration of a substance in solution. Acid–base titrations, where an acid and a base neutralize each other, are the most frequent. The point at which all of the acid or base in the analyte has been neutralized is called the equivalence point. For diprotic acids or bases, a second equivalence point also occurs, and calculating the pH at this point is straightforward with the right data.

Step 1

Identify the analyte’s nature (acid or base), the specific type (e.g., oxalic acid), and its concentration and volume. In homework problems this information is supplied; in a lab you record it as you titrate.

Step 2

Diprotic species—those that can donate or accept more than one proton—exhibit a second equivalence point. Retrieve the second dissociation constant (K_a2 for acids or K_b2 for bases) from a reliable source such as a standard reference text or reputable online database.

Step 3

Calculate the total moles of the analyte. Multiply its molarity by its volume (in liters). For example, 40 mL of 1 M oxalic acid contains:

(0.040 L × 1 mol L⁻¹) = 0.04 mol.

Step 4

Determine the final volume by adding the titrant volume to the original analyte volume. In the example, 80 mL of 1 M NaOH added to 40 mL of 1 M oxalic acid yields 120 mL total.

Step 5

Compute the concentration of the conjugate species at the second equivalence point:

0.04 mol ÷ 0.120 L = 0.333 M.

Step 6

Determine the K_b of the fully deprotonated conjugate base (or K_a of the conjugate acid). For a diprotic acid, K_b = 1×10⁻¹⁴ ÷ K_a2. Using oxalic acid (K_a2 = 5.4×10⁻⁵):

1×10⁻¹⁴ ÷ 5.4×10⁻⁵ = 1.852×10⁻¹⁰.

Step 7

Set up the base equilibrium expression:

K_b = ([OH⁻][conjugate acid])/[conjugate base].

Step 8

Assuming the change in concentration is small, let x represent [OH⁻] and [conjugate acid]. The equation becomes:

1.852×10⁻¹⁰ = x² ÷ 0.333.

Solving: x² = 6.167×10⁻¹¹ → x = 7.85×10⁻⁶ M.

Step 9

Convert the hydroxide concentration to pH. First, find pOH = –log₁₀(7.85×10⁻⁶) = 5.105. Then, pH = 14 – pOH = 8.90.

Step 10

Thus, the pH at the second equivalence point is 8.90.

Things Needed

• Pencil
• Paper
• Calculator

TL;DR (Too Long; Didn't Read)

This calculation omits water autoionization, which can affect very dilute solutions of weak bases or acids. Nonetheless, it provides a reliable estimate for most practical purposes.