4.3.1. Reaction and Stoichiometry

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 discuss the reaction between a weak acid and a strong base. The equation HA + BOH → A⁻ + B⁺ + H₂O shows how a weak acid (HA) reacts with a strong base (BOH). The important aspect of this reaction is the stoichiometry, which indicates a 1:1 ratio: for every one mole of weak acid, one mole of strong base reacts. Consequently, at the equivalence point, all the weak acid has been neutralized, resulting in the formation of its conjugate base and the cation from the strong base.

Think of this reaction like a cooking recipe. If you have a recipe that requires one ingredient to balance with another, you need exactly one of each for it to work. If you are baking a cake (weak acid), you need a certain amount of sugar (strong base) to neutralize the acidity and create the final product (conjugate base). Just like in cooking, if you have too much of one ingredient, it won't turn out the same way!

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

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:

3. Moles HA remaining = initial moles HA – moles BOH added.

4. 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.

This chunk explores the initial conditions and equilibrium of the weak acid before any strong base has been added. The initial pH of a weak acid like acetic acid can be calculated using its dissociation constant (Ka). When the strong base is added, the solution acts as a buffer; this means it can resist changes in pH. By using the Henderson-Hasselbalch equation, students can calculate the pH based on the amounts of undissociated weak acid (HA) and its conjugate base (A⁻), helping them understand how buffer solutions work.

Imagine you're riding a bike down a hill. If the hill gets too steep, you might lose control and fall (equivalent to the pH changing drastically). But if you have brakes that allow you to gradually slow down and maintain control (the buffer), then you can navigate the descent much more safely. In this scenario, the weak acid and its conjugate base are like the brakes, helping maintain stability in pH as more of the strong base is added.

3. At 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:
4. Determine concentration of A minus at equivalence: initial moles HA ÷ total volume (Va + Ve).
5. Set up Kb expression: Kb = ([HA] × [OH minus]) ÷ [A minus] (equilibrium concentrations).
6. Solve for [OH minus], calculate pOH, then pH.

In this section, we focus on the equivalence point during the titration, where all the weak acid has reacted with the strong base. At this point, only the conjugate base (A⁻) and the cation from the strong base are in the solution. Importantly, the conjugate base can hydrolyze, which produces OH⁻ ions and raises the pH, making the solution basic. To find the pH, you need to know the concentration of A⁻ created and use the Kb expression derived from Ka.

Consider this situation like finishing a puzzle. When you add the last piece (strong base), everything clicks together (the equivalence point). After this, if the completed puzzle stays laid out on the table (the solution), any breeze (hydrolysis) that comes through can slightly shift the pieces apart (increasing pH). The stronger the breeze, the more it can impact your puzzle's appearance—just like the contribution of A⁻ to the pH.

4. After Equivalence (Vb > Ve): ○ Excess strong base dominates; pH determined by the leftover OH minus concentration as in the strong acid-base case.

This chunk covers the phase after reaching the equivalence point in the titration. Once you've added more strong base than is required to completely react with the weak acid, the pH calculation relies on the concentration of the excess hydroxide ions (OH⁻) present. The pH will increase significantly in this phase, reflecting the dominating presence of the strong base.

Think of this scenario like adding too much sugar to your coffee. At first, it's just the right sweetness (equivalence point), but once you keep adding sugar (excess strong base), your coffee becomes overly sweet—hard to drink because the balance is off. In the same way, after equivalence, the basicity of the solution can overshadow the previous flavor of the weak acid.

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