Calculating pH During a Titration of a Weak Base with a Strong Acid

In this guide, we will explore how to calculate pH during the titration of a weak base with a strong acid. We'll provide step-by-step examples to illustrate the process, making it easy to understand the underlying concepts of acid-base chemistry.
By Jamie

Understanding pH Calculation in Titration

Titration is a laboratory method used to determine the concentration of an unknown solution by reacting it with a solution of known concentration. When titrating a weak base with a strong acid, the pH changes gradually until it reaches the equivalence point, where the amount of acid equals the amount of base.

In this article, we will cover:

  • Key concepts in titration
  • How to calculate the pH at different stages of the titration

Key Concepts

  • Weak Base: A base that does not completely ionize in solution (e.g., ammonia, NH₃).
  • Strong Acid: An acid that completely ionizes in solution (e.g., hydrochloric acid, HCl).
  • Equivalence Point: The point in a titration where the number of moles of acid equals the number of moles of base.

Example 1: Titration of Ammonia with Hydrochloric Acid

Given:

  • Concentration of NH₃: 0.50 M
  • Volume of NH₃ solution: 50.0 mL
  • Concentration of HCl: 0.50 M

Step 1: Calculate moles of NH₃

Moles of NH₃ = Concentration × Volume

$$
ext{Moles of NH₃} = 0.50 ext{ M} imes 0.050 ext{ L} = 0.025 ext{ moles}
$$

Step 2: Determine the volume of HCl needed for neutralization

At the equivalence point, moles of HCl = moles of NH₃.

$$
ext{Moles of HCl} = 0.025 ext{ moles}
$$

Volume of HCl = Moles of HCl / Concentration of HCl

$$
ext{Volume of HCl} = 0.025 ext{ moles} / 0.50 ext{ M} = 0.050 ext{ L} = 50.0 ext{ mL}
$$

Step 3: Calculate pH before adding HCl

For a weak base like NH₃, we can use the Kb to find the pH:

  • Kb for NH₃: 1.8 × 10⁻⁵
  • Use the formula:

$$
ext{pOH} = - ext{log}[ ext{OH}⁻]
$$

At the start,

$$
[ ext{OH}⁻] = ext{sqrt}(K_b imes C) = ext{sqrt}(1.8 imes 10^{-5} imes 0.50)
$$

Calculating gives:

$$
[ ext{OH}⁻] = 0.0057 ext{ M}
$$

Now, find pOH:

$$
ext{pOH} = - ext{log}(0.0057) = 2.24
$$

Then calculate pH:

$$
ext{pH} = 14 - ext{pOH} = 14 - 2.24 = 11.76
$$

Step 4: Calculate pH at the equivalence point

At the equivalence point, the solution contains only the conjugate acid (NH₄⁺):

  • Concentration of NH₄⁺ = 0.025 moles / (50.0 mL + 50.0 mL) = 0.25 M

Using Ka for NH₄⁺ (Kw/Kb), we find:

$$
K_a =
rac{1.0 imes 10^{-14}}{1.8 imes 10^{-5}} = 5.56 imes 10^{-10}
$$

Now, calculate pH:

$$
ext{pH} =
rac{1}{2} imes (pK_a - ext{log}[ ext{NH}_4^+])
$$

$$ ext{pH} =
rac{1}{2} imes (9.25 - 0.60) = 4.32
$$

Summary

In this example, we have covered how to calculate the pH at various stages of titration, specifically for a weak base (ammonia) being titrated with a strong acid (hydrochloric acid). Understanding these calculations is essential for mastering acid-base chemistry and titration techniques.