The equivalence point in an acid–base titration is the stoichiometric point where the reacting amounts are exactly in the mole ratio given by the balanced neutralization reaction. For a monoprotic strong acid titrated by a strong base, the equivalence point occurs when \(n(\mathrm{H^+}) = n(\mathrm{OH^-})\).
1) Equivalence point vs endpoint
| Term | Definition | How it is detected |
|---|---|---|
| Equivalence point | Stoichiometric completion of the reaction (exact mole balance). | Calculated from moles and reaction stoichiometry; can be located on a titration curve where the steep change centers. |
| Endpoint | Experimental signal chosen to indicate completion (often an indicator color change). | Observed in the lab; ideally very close to the equivalence point but not guaranteed to be identical. |
2) Strong acid–strong base case (HCl with NaOH)
Balanced reaction: \[ \mathrm{HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)}. \] The mole ratio is \(1:1\), so at the equivalence point: \[ n(\mathrm{HCl}) = n(\mathrm{NaOH}). \]
Given data (example).
- Initial acid: \(V_a = 25.00\ \text{mL} = 0.02500\ \text{L}\), \(C_a = 0.1000\ \text{M}\) (HCl).
- Titrant base: \(C_b = 0.1250\ \text{M}\) (NaOH).
- Assume \(25^\circ\mathrm{C}\) and ideal strong electrolyte behavior.
3) Volume of titrant at the equivalence point
Step 1: compute initial moles of acid. \[ n_a = C_aV_a = (0.1000\ \mathrm{mol/L})(0.02500\ \mathrm{L}) = 0.002500\ \mathrm{mol}. \]
Step 2: at equivalence, \(n_b = n_a\), so \(C_bV_{\mathrm{eq}} = n_a\). \[ V_{\mathrm{eq}}=\frac{n_a}{C_b}=\frac{0.002500\ \mathrm{mol}}{0.1250\ \mathrm{mol/L}}=0.02000\ \mathrm{L}=20.00\ \mathrm{mL}. \]
4) pH at the equivalence point (strong acid–strong base)
At the equivalence point, the strong acid and strong base have neutralized completely. The solution contains primarily spectator ions (\(\mathrm{Na^+}\), \(\mathrm{Cl^-}\)) and water. For a strong acid–strong base titration, the pH is governed by water autoionization: \[ \mathrm{H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)},\quad K_w = 1.0\times 10^{-14}\ (\text{at }25^\circ\mathrm{C}). \] Therefore, \[ [\mathrm{H^+}] = [\mathrm{OH^-}] = \sqrt{K_w} = 1.0\times 10^{-7}\ \mathrm{M}, \qquad \mathrm{pH} = 7.00. \]
5) Where the equivalence point appears on a titration curve
6) General calculation pattern around the equivalence point
For a 1:1 strong acid–strong base titration, the equivalence condition is: \[ C_aV_a = C_bV_{\mathrm{eq}}. \] Away from equivalence, excess strong acid or strong base controls pH through the remaining moles after reaction and dilution by total volume.
| Region | Mole accounting | pH expression (strong electrolytes) |
|---|---|---|
| Before equivalence (\(V_b < V_{\mathrm{eq}}\)) | \(n_{\text{excess}}(\mathrm{H^+}) = C_aV_a - C_bV_b\) | \[ [\mathrm{H^+}] = \frac{C_aV_a - C_bV_b}{V_a + V_b},\quad \mathrm{pH} = -\log([\mathrm{H^+}]). \] |
| At equivalence (\(V_b = V_{\mathrm{eq}}\)) | \(n(\mathrm{H^+})=n(\mathrm{OH^-})\) | \[ \mathrm{pH}\approx 7.00\ \text{at }25^\circ\mathrm{C}. \] |
| After equivalence (\(V_b > V_{\mathrm{eq}}\)) | \(n_{\text{excess}}(\mathrm{OH^-}) = C_bV_b - C_aV_a\) | \[ [\mathrm{OH^-}] = \frac{C_bV_b - C_aV_a}{V_a + V_b},\quad \mathrm{pOH} = -\log([\mathrm{OH^-}]),\quad \mathrm{pH}=14-\mathrm{pOH}. \] |
7) Key takeaways for the equivalence point
- The equivalence point is defined by stoichiometry, not by the indicator.
- For monoprotic strong acid–strong base, \(V_{\mathrm{eq}}=\dfrac{C_aV_a}{C_b}\) and \(\mathrm{pH}\approx 7.00\) at \(25^\circ\mathrm{C}\).
- For weak acid or weak base titrations, the equivalence-point pH is not 7 because the conjugate species hydrolyzes; the equivalence point remains stoichiometric, but the pH shifts.