4.3. Weak Acid vs. Strong Base Titration

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

HA + BOH → A minus + B plus + H₂O

• Stoichiometry is 1:1 for HA and BOH.
• At equivalence, all HA has been converted to its conjugate base A minus plus B plus (from the strong base cation) present in water.

In this chunk, we focus on the fundamental reaction that occurs during the titration of a weak acid with a strong base. The weak acid (represented as HA) reacts with the strong base (represented as BOH) in a neutralization reaction. The result is a salt (A minus) and water. The '1:1' stoichiometry denotes that one mole of the weak acid reacts with exactly one mole of the strong base. This indicates a balanced chemical reaction, which is key to determining how much strong base is required to fully react with a given amount of weak acid. At the equivalence point in this titration, all the weak acid has been converted into its conjugate base.

Think of this titration like making a smoothie. The weak acid (HA) is like a fruit puree, and the strong base (BOH) is like a sweetener. When you mix them, just like sweetening the smoothie to your desired taste (the equivalence point), you want just the right amount of sweetener to balance out the sourness of the fruit puree, achieving the perfect smoothie blend without overpowering either flavor.

1. Initial pH (Before Base Added): pH depends on the weak acid equilibrium. Calculate using Ka.

   • Example: 0.100 M acetic acid (CH₃COOH, Ka = 1.8 × 10⁻⁵). Initial pH ≈ 2.87 (calculated earlier).

2. Before Equivalence (0 < Vb < Ve):

   • The solution is a buffer containing a mixture of undissociated HA and its conjugate base A minus (formed in part by reacting HA with BOH).
   • At any addition point:
     1. Moles HA remaining = initial moles HA – moles BOH added.
     2. Moles A minus formed = moles BOH added.
   • Use the Henderson-Hasselbalch equation: pH = pKa + log₁₀([A minus] ÷ [HA])
   • where pKa = – log₁₀(Ka). Concentrations of HA and A minus are moles divided by total solution volume. Often the volume factors cancel if the ratio of moles is used directly.

3. At Half-Equivalence Point (Vb = ½ Ve):

   • Moles BOH added = ½ initial moles HA.
   • Moles HA remaining = ½ initial moles HA.
   • Moles A minus formed = ½ initial moles HA.
   • Therefore [A minus] = [HA].
   • So pH = pKa (since log₁₀(1) = 0).

4. The half-equivalence point is useful to determine the pKa of a weak acid experimentally by measuring pH at that volume.

This chunk explains the features of the titration curve when a weak acid is titrated with a strong base. The initial pH is determined by the weak acid's equilibrium (where we can calculate it using the acid dissociation constant, Ka). Before the equivalence point, as the strong base is added, a buffer solution is created, which can be quantified using the Henderson-Hasselbalch equation. This equation relates the pH of a buffer solution to the concentration of the acid and its conjugate base. At the half-equivalence point, the concentrations of the acid and its conjugate base are equal, leading to a pH that equals pKa. This is important as it provides a way to experimentally determine the pKa of the weak acid being studied.

Imagine making a salad dressing with vinegar (the weak acid) and oil (the strong base). At first, the vinegar has a strong flavor, but as you add oil, the dressing becomes a balanced mix of flavors; the equilibrium shifts. The half-equivalence point in this case represents a perfect blend where the flavors are equally balanced. You can figure out how much oil to add for the best taste (pKa), which is similar to how chemists find the pKa of an acid through observations in their titration curves.

3. Equivalence Point (Vb = Ve):
   • All original HA is converted to A minus. The solution contains the conjugate base A minus (and B plus from the strong base).
   • Conjugate base A minus hydrolyzes in water: A minus + H₂O ⇌ HA + OH minus
   • The solution is basic at the equivalence point because A minus generates OH minus.
   • The pH at equivalence depends on Kb for A minus, i.e., Kb = Kw ÷ Ka. To calculate pH:
     1. Determine concentration of A minus at equivalence: initial moles HA ÷ total volume (Va + Ve).
     2. Set up Kb expression: Kb = ([HA] × [OH minus]) ÷ [A minus] (equilibrium concentrations).
     3. Solve for [OH minus], calculate pOH, then pH.

At the equivalence point in the titration of a weak acid with a strong base, all of the weak acid has been converted into its conjugate base (A minus). The strong base (B plus) that was added reacts completely with the weak acid, leading to a solution where A minus can react with water to generate hydroxide ions (OH minus), indicating a basic solution. To find the pH at this point, the base dissociation constant (Kb) of the conjugate base is used. This involves calculating the concentration of A minus at the equivalence point and setting up the Kb expression to solve for hydroxide ion concentration and ultimately the pH.

Think of the equivalence point like reaching a tipping point in a relationship. At this point, both partners have contributed equally, and the dynamics shift towards a new balance. In our titration, once all weak acid (HA) has been neutralized and converted to its conjugate base (A minus), the solution transforms into a new entity, possessing basic properties. This parallels how relationships evolve when both sides fully engage, leading to different outcomes than when they started.

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