Topic 8.5 Notes – Acid-Base Titrations

1. What an Acid-Base Titration Is

A titration is a controlled neutralization used to determine the concentration of an unknown solution (the analyte) by slowly adding a solution of known concentration (the titrant).

In acid-base titrations, the reaction is a proton transfer:

• Strong acid + strong base $$\ce{H+ + OH- -> H2O}$$
• Weak acid + strong base $$\ce{HA + OH- -> A- + H2O}$$
• Weak base + strong acid $$\ce{B + H+ -> BH+}$$

As titrant is added, we measure pH and graph it vs. volume added. That graph is the titration curve.

Below are typical titration curves for a weak acid with a strong base (left) and a weak base with a strong acid (right). Focus on the overall shape and where the equivalence point falls relative to pH 7.

Titration curves for weak acid-strong base and weak base-strong acid

Key regions to notice on these curves:

• Initial pH → determined only by the analyte.
• Equivalence point → moles titrant added = moles analyte originally present.
• Half-equivalence point (weak systems) → halfway to equivalence.
• After equivalence → excess titrant controls pH.

2. Equivalence Point and Stoichiometry

At equivalence, the reaction has reached stoichiometric equality.

For a monoprotic acid and base:

$$M_a V_a = M_b V_b$$

This works because moles acid = moles base at equivalence.

If coefficients aren’t 1:1, include them from the balanced equation.

Example (quick setup)

Suppose 30.0 mL of an unknown $\ce{HNO3}$ solution requires 18.0 mL of 0.200 M $\ce{NaOH}$ to reach equivalence.

Moles base added: $$(0.200)(0.0180) = 0.00360 \text{ mol}$$

That equals moles acid. So: $$M_a = \frac{0.00360}{0.0300} = 0.120 \text{ M}$$

That’s the classic FRQ-style move.

What Determines pH at Equivalence?

It depends on the major species left in solution.

Equivalence does not automatically mean neutral. It means equal moles reacted.

That mistake shows up constantly on MCQs.

3. Weak Acid or Weak Base Titrations

Before equivalence in a weak system, both the weak species and its conjugate are present. That’s a buffer.

Example: $$\ce{HF + OH- -> F- + H2O}$$

As base is added, you have both $\ce{HF}$ and $\ce{F-}$.

Henderson-Hasselbalch

$$\text{pH} = \text{p}K_a + \log \frac{[A^{-}]}{[HA]}$$

You use this anywhere in the buffer region.

Half-Equivalence Point

At halfway to equivalence:

• $[HA] = [A^{-}]$
• $\log(1) = 0$
• pH = pKa

This is huge.

From a curve:

1. Find equivalence volume.
2. Cut it in half.
3. Read the pH there.
4. That value = $\text{p}K_a$.

For weak base titrations:

• At half-equivalence, $[B] = [BH^{+}]$, so $\text{pOH} = \text{p}K_b$. Equivalently, the pH read from the curve at half-equivalence equals the $\text{p}K_a$ of the conjugate acid $\ce{BH+}$. You can then find $\text{p}K_b = \text{p}K_w - \text{p}K_a$.

The AP loves giving a curve and asking for $K_a$. You read pH at half-equivalence and convert.

4. How to Read a Titration Curve

Think in terms of what’s in excess.

Before equivalence

• Strong systems → leftover strong acid/base controls pH.
• Weak systems → buffer calculations.

At equivalence

• Identify the species remaining.
• Ask whether it reacts with water.
  • Conjugate base → produces $\ce{OH-}$.
  • Conjugate acid → produces $\ce{H3O+}$.

After equivalence

• Excess titrant dominates.
• Ignore the original analyte.

What the shape tells you

• Initial pH very low → strong acid analyte.
• Equivalence pH above 7 → weak acid titrated with strong base.
• Multiple steep jumps → polyprotic acid.

You should be able to look at a curve and identify the system without numbers.

5. Polyprotic Acid Titrations

Polyprotic acids donate more than one proton:

$$\ce{H2A -> HA- -> A^{2}-}$$

Each step has its own $K_a$, and typically: $$K_{a1} > K_{a2} > K_{a3}$$

When a diprotic acid is titrated with a strong base, the titration curve shows separate steps for each proton removed.

Diprotic weak acid titrated with a strong base

What you learn from the curve:

• Number of equivalence points = number of acidic protons.
• Each half-equivalence point gives a different $\text{p}K_a$.
• Between equivalence points → buffer regions.

You are expected to:

• Identify major species in each region.
• Determine how many protons are acidic.
• Read $\text{p}K_a$ values from half-equivalence points.

You are not expected to calculate full equilibrium tables for every species in polyprotic systems. Focus on dominant species reasoning.