4.2. Strong Acid vs. Strong Base Titration

Generic reaction for titrating a strong acid (HA) with a strong base (BOH):
HA + BOH → AB + H₂O

• Stoichiometry is 1:1 since one mole of HA reacts with one mole of BOH.

The reaction between a strong acid and a strong base can be represented as HA (acid) reacting with BOH (base) to form a neutral salt (AB) and water (H₂O). In this reaction, the mole ratio is 1:1; this means that for every mole of acid, one mole of base is required to reach neutralization at the equivalence point.

Think of this titration like cooking where you need equal parts of salt (the acid) and water (the base) to make a perfect brine solution. If you over or under add either component, the result won’t be what you need, just like how the reaction has to be balanced to create a neutral solution.

1. Initial Region (Before Any Base Added): pH equals that of the strong acid. For example, if initial concentration of HCl is 0.100 M, pH = 1.00.
2. Buffer-Like Region (Small Additions of Base): There is no significant buffer region; pH gradually rises as small amounts of base are added.
3. Equivalence Point (Half-Volume Point): At this point, solution contains only the salt (A⁻ from acid and B⁺ from base) dissolved in water. Therefore, pH = 7.00 (at 25 °C).
4. Beyond Equivalence (Excess Base): Once past equivalence, added OH⁻ dominates; pH rises gradually toward that of the pure strong base.

The titration curve of a strong acid against a strong base displays distinct regions: The initial region where only the strong acid is present gives a very low pH. As you begin to add the strong base, there’s a slow and gradual increase in pH until the equivalence point is reached, where the acid has been completely neutralized and the solution is neutral with a pH of 7. After the equivalence point, the pH jumps sharply as additional base is added, indicating that the solution is becoming increasingly basic.

Imagine this process like adding lemonade concentrate to water: initially, when you have mostly concentrate, the mixture is very sour (low pH). As you slowly add water (the base), the sourness (acidity) decreases gradually. Once you’ve added the perfect amount of water, the taste becomes just right (neutral), and adding more water makes it taste more diluted (basic).

1. Before Equivalence (Volume of base added = Vb, less than Ve):
   • Moles of H⁺ remaining = initial moles HA - moles base added = (Ca × Va) - (Cb × Vb).
   • Volume of solution = Va + Vb.
   • [H⁺] = remaining moles H⁺ ÷ total volume.
   • pH = – log₁₀([H⁺]).
2. At Equivalence (Vb = Ve):
   • Moles HA initially = Ca × Va.
   • Equivalence volume of base Ve = (Ca × Va) ÷ Cb.
   • pH = 7.00 (neutral).
3. After Equivalence (Vb > Ve):
   • Moles of OH⁻ in excess = (Cb × Vb) - (Ca × Va).
   • Volume of solution = Va + Vb.
   • [OH⁻] = (excess moles OH⁻) ÷ total volume.
   • pOH = – log₁₀([OH⁻]).
   • pH = 14.00 – pOH.

During the titration, you calculate the moles of acid and base to determine how the pH changes. Before the equivalence point, you will need to subtract the moles of base added from the moles of acid to find how much acid remains, and then you can use this to calculate the pH. At the equivalence point, there are no excess H⁺ or OH⁻ ions, resulting in a neutral pH of 7. After the equivalence point, excess OH⁻ is present, and you calculate the pH based on the concentration of excess hydroxide ions.

Consider titration like balancing a scale: before equal weights are added (the equivalence point), if you add extra weight on one side (base), you calculate how much more weight is needed to balance the scale (neutral point). Once balanced, any further weights you add correlate to knowing how much heavier one side has become compared to the other (excess base), which helps you gauge how unbalanced your system (pH) has become.

Example: Titration of 25.0 mL of 0.100 M HCl with 0.100 M NaOH.

• Initial moles HCl = 0.0250 L × 0.100 M = 2.50 × 10⁻³ mol.
• Equivalence volume Ve = (2.50 × 10⁻³ mol) ÷ (0.100 M) = 0.0250 L = 25.0 mL.
• Case A (10.0 mL NaOH added):
  • Moles OH⁻ added = 0.0100 L × 0.100 M = 1.00 × 10⁻³ mol.
  • Moles H⁺ remaining = 2.50 × 10⁻³ – 1.00 × 10⁻³ = 1.50 × 10⁻³ mol.
  • Total volume = 25.0 mL + 10.0 mL = 35.0 mL = 0.0350 L.
  • [H⁺] = (1.50 × 10⁻³) ÷ 0.0350 = 4.29 × 10⁻² M.
  • pH = – log₁₀ (4.29 × 10⁻²) = 1.37.
• Case B (At Equivalence, 25.0 mL NaOH added):
  • Moles OH⁻ = 2.50 × 10⁻³.
  • pH = 7.00 (neutral).
• Case C (40.0 mL NaOH added):
  • Moles OH⁻ added = 0.0400 L × 0.100 M = 4.00 × 10⁻³ mol.
  • Moles OH⁻ in excess = 4.00 × 10⁻³ – 2.50 × 10⁻³ = 1.50 × 10⁻³ mol.
  • Total volume = 25.0 mL + 40.0 mL = 65.0 mL = 0.0650 L.
  • [OH⁻] = (1.50 × 10⁻³) ÷ 0.0650 = 2.31 × 10⁻² M.
  • pOH = – log₁₀ (2.31 × 10⁻²) = 1.64 → pH = 14.00 – 1.64 = 12.36.

In this example, the titration of HCl with NaOH is examined to demonstrate how calculations of moles and pH change at different steps. You first determine the initial moles of HCl, calculate the equivalence volume, and evaluate the pH before, at, and after the equivalence point using formulas. This thorough calculation shows exactly how many H⁺ ions remain in the solution and allows you to determine the pH throughout the various stages of the reaction.

Think of this example as budgeting money: you start with a set amount (the initial HCl), and as you 'spend' (add NaOH), you calculate how much is left (remaining H⁺). At some point, you've spent exactly what you had (equivalence), and after that, you can only have leftovers (excess base). Each step in the budgeting process is like calculating the pH at that moment.

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