A neutralisation curve (also called a titration curve) is a graph of pH versus the volume of titrant added during an acid–base titration. The curve compresses stoichiometry, equilibrium, and dilution into a single picture: the pH changes slowly in some regions and extremely rapidly near equivalence when the limiting reagent switches.
Core chemical meaning
Acid–base neutralisation converts an acid and a base into a salt and water, with the dominant proton-transfer reaction written in net ionic form. For a strong acid titrated by a strong base, the reaction is essentially complete: \[ \text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O}(\ell) \]
The neutralisation curve reflects which species remains in excess after reaction and mixing. Away from equivalence, pH is controlled by the concentration of the excess strong acid or strong base. Near equivalence, small additions of titrant cause large swings in pH because the excess becomes very small compared with the total volume.
Key points on a neutralisation curve
“Equivalence point” indicates stoichiometric equality of acid and base (moles of acidic equivalents equal moles of basic equivalents). “End point” indicates the observed indicator change and ideally lies close to equivalence.
| Curve region | Strong acid + strong base | Weak acid + strong base |
|---|---|---|
| Initial solution | pH set by the strong acid concentration (complete dissociation). | pH set by weak-acid equilibrium, \(K_a\), and initial concentration. |
| Mid-titration | pH controlled by excess \(\text{H}^+\) (before equivalence) or excess \(\text{OH}^-\) (after equivalence). | Buffer region present; \(\text{HA}/\text{A}^-\) mixture controls pH and resists change. |
| Equivalence point | pH \(\approx 7\) at \(25^\circ\text{C}\), because the salt is typically neutral and water autoionization dominates. | pH \(> 7\) at \(25^\circ\text{C}\), because \(\text{A}^-\) hydrolyzes water and generates \(\text{OH}^-\). |
| Beyond equivalence | pH controlled by excess strong base concentration after dilution. | pH controlled by excess strong base concentration after dilution. |
Quantitative relationships along the curve
Strong acid–strong base titration permits a stoichiometric concentration model because the neutralisation reaction goes essentially to completion. Let \(C_a\) and \(V_a\) be the acid concentration and initial volume, and \(C_b\) and \(V_b\) be the base concentration and added volume. Moles are \(n_a = C_a V_a\) and \(n_b = C_b V_b\) with volumes in liters.
Before equivalence (acid in excess)
Excess moles of \(\text{H}^+\) are \(n_{\text{ex}} = n_a - n_b\), and the total volume is \(V_\text{tot} = V_a + V_b\). The hydrogen ion concentration is \([\text{H}^+] \approx n_{\text{ex}}/V_\text{tot}\), giving \[ \text{pH} = -\log\!\left(\frac{n_a - n_b}{V_a + V_b}\right) \]
At equivalence (stoichiometric equality)
For a monoprotic strong acid with a strong base at \(25^\circ\text{C}\), \(\text{pH} \approx 7\). The remaining solute is a neutral salt, and the pH is set primarily by water autoionization: \[ K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} \]
After equivalence (base in excess)
Excess moles of \(\text{OH}^-\) are \(n_{\text{ex}} = n_b - n_a\), and \[ \text{pOH} = -\log\!\left(\frac{n_b - n_a}{V_a + V_b}\right), \qquad \text{pH} = 14 - \text{pOH} \]
Worked values on a representative neutralisation curve
Consider \(25.0\ \text{mL}\) of \(0.100\ \text{M}\) HCl titrated by \(0.100\ \text{M}\) NaOH at \(25^\circ\text{C}\). The equivalence volume satisfies \(C_a V_a = C_b V_{b,\text{eq}}\), so \(V_{b,\text{eq}} = 25.0\ \text{mL}\).
| Added base \(V_b\) | Excess species | Key concentration | pH |
|---|---|---|---|
| 0.0 mL | \(\text{H}^+\) | \([\text{H}^+] = 0.100\) | \(\text{pH} = 1.00\) |
| 10.0 mL | \(\text{H}^+\) | \([\text{H}^+] = \dfrac{0.00250 - 0.00100}{0.0350} = 0.0429\) | \(\text{pH} = 1.37\) |
| 24.0 mL | \(\text{H}^+\) | \([\text{H}^+] = \dfrac{0.00250 - 0.00240}{0.0490} = 0.00204\) | \(\text{pH} = 2.69\) |
| 25.0 mL | none (stoichiometric) | water sets pH | \(\text{pH} \approx 7.00\) |
| 26.0 mL | \(\text{OH}^-\) | \([\text{OH}^-] = \dfrac{0.00260 - 0.00250}{0.0510} = 0.00196\) | \(\text{pH} = 14 - 2.71 = 11.29\) |
Visualization of typical neutralisation curves
Common pitfalls
Equivalence and end point are not synonyms; the neutralisation curve identifies the equivalence volume stoichiometrically, while an indicator end point depends on its transition range. Volume units are a frequent source of numerical error; mole calculations require liters.
Weak-acid and weak-base titrations require equilibrium chemistry around equivalence; the salt is not neutral, so pH is not constrained to 7. The half-equivalence statement \( \text{pH} = \text{p}K_a \) applies to weak acid–strong base titration when activities are approximated by concentrations.