4.4. Weak Base vs. Strong Acid Titration

Generic reaction of a weak base (B) with a strong acid (HA):

B + HA → BH plus + A minus

• 1:1 stoichiometry for B and HA.
• At equivalence, all base has been converted to its conjugate acid BH plus. The resulting solution is acidic because BH plus donates H plus in water.

In this chunk, we understand the basic chemical reaction during the titration of a weak base with a strong acid. The weak base, represented as 'B', reacts with a strong acid 'HA'. In this reaction, one mole of weak base reacts with one mole of strong acid, which we describe as 1:1 stoichiometry. At the equivalence point of the titration, all of the weak base has reacted to form its conjugate acid, which is represented as 'BH plus'. This denotes that the resulting solution becomes acidic because this conjugate acid can donate protons (H plus) in water, leading to a decrease in pH.

Think of this reaction as a game of tag where 'B' is someone who is 'it' and 'HA' is the person they are trying to tag. When 'B' tags 'HA', it becomes a new person 'BH plus'. Now, 'BH plus' is like a new player in the game who can also tag others (donate H plus) but is now on the other team (the acid side).

1. Initial pH: pH determined by the weak base equilibrium. B + H₂O ⇌ BH plus + OH minus

• Use Kb to calculate initial pH.

2. Before Equivalence: Buffer region where B and BH plus are in equilibrium. Use Henderson-Hasselbalch form for bases: pOH = pKb + log₁₀ ([BH plus] ÷ [B])

• or equivalently for pH: pH = pKw – pOH

3. Equivalence Point: All base has become BH plus. BH plus hydrolyzes in water: BH plus + H₂O ⇌ B + H₃O plus

• The solution is acidic. Calculate [H plus] from the acid dissociation constant Ka for BH plus, where Ka = Kw ÷ Kb.

4. After Equivalence: Excess strong acid controls pH. Use straightforward calculation for strong acid in remaining volume.

The titration curve shows how the pH of the solution changes as we add strong acid to a weak base. Initially, we calculate the pH based on the weak base's equilibrium before adding any strong acid. The pH will rise gradually until we reach the equivalence point. Using the Henderson-Hasselbalch equation allows us to account for the buffer region, where both the weak base and its conjugate acid coexist. At the equivalence point, all weak base turns into its conjugate acid. This conjugate acid will interact with water, producing H plus ions, which makes the solution acidic. Beyond this point, any added acid will determine the final pH since it will dominate the solution's acidity.

Imagine filling a glass with water (representing the weak base) before adding lemonade (the strong acid). Initially, the water has a certain taste (pH), and as you begin to add lemonade, the taste changes slightly, and the glass is still a mix until you pour in just the right amount of lemonade (equivalence point). After that point, the lemonade taste becomes overwhelming (excess strong acid), dominating the flavor of the drink.

Initial volume of NH₃ solution = 25.0 mL at 0.100 M → moles NH₃ = 0.00250 mol.

Initial pH:

• NH₃ (Kb = 1.8 × 10⁻⁵).
• x ≈ sqrt(Kb × C₀) = sqrt[(1.8 × 10⁻⁵) × 0.100] = 1.34 × 10⁻³ M = [OH minus].
• pOH = – log₁₀ (1.34 × 10⁻³) = 2.87 → pH = 14.00 – 2.87 = 11.13.

Half-Equivalence (12.5 mL HCl added):

• Moles HCl = 0.0125 L × 0.100 M = 0.00125 mol.
• Moles NH₃ remaining = 0.00250 – 0.00125 = 0.00125 mol.
• Moles NH₄ plus formed = 0.00125 mol.
• Ratio [BH plus] ÷ [B] = [NH₄ plus] ÷ [NH₃] = 1.00.
• pOH = pKb = – log₁₀ (1.8 × 10⁻⁵) = 4.74 (Kb = 1.8 × 10⁻⁵).
• pH = 14.00 – 4.74 = 9.26 at half-equivalence.

Equivalence Point (25.0 mL HCl added):

• Volume total = 25.0 + 25.0 = 50.0 mL.
• Moles NH₄ plus = 0.00250 mol. [NH₄ plus] = 0.00250 ÷ 0.0500 = 0.0500 M.
• NH₄ plus is a weak acid with Ka = Kw ÷ Kb = (1.0 × 10⁻¹⁴) ÷ (1.8 × 10⁻⁵) = 5.56 × 10⁻¹⁰.
• Let x = [H plus] from hydrolysis: NH₄ plus ⇌ NH₃ + H plus
• Then Ka = x² ÷ (0.0500 – x) ≈ x² ÷ 0.0500 → x = sqrt(Ka × 0.0500) = sqrt[(5.56 × 10⁻¹⁰) × 0.0500] = 5.27 × 10⁻⁶ M.
• pH = – log₁₀ (5.27 × 10⁻⁶) = 5.28. So the equivalence point is acidic (around pH 5.3).

In this chunk, we walk through a specific titration example where ammonia (NH₃) is titrated with hydrochloric acid (HCl). Initially, we calculate the pH based on the concentration of NH₃. As we add HCl, we reach the half-equivalence point where half of the ammonia has converted to ammonium ion (NH₄ plus), resulting in a specific pH. Finally, at the equivalence point, all ammonia has reacted to form NH₄ plus. We need to determine the pH based on the weak acid behavior of NH₄ plus in water, leading to the solution being acidic.

Consider a seesaw balancing act. Here, ammonia is one side of the seesaw trying to counterbalance the weight of the hydrochloric acid. When you add just enough acid to fully convert the ammonia to ammonium ion, the seesaw tips to the acidic side. If you slightly increase the acid amount beyond that equilibrium, the seesaw remains tilted, accentuating that acidic balance.

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