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Neutralisation Curve in Acid–Base Titration

What is a neutralisation curve, and how are pH values interpreted and calculated during a strong acid–strong base titration?

Subject: General Chemistry Chapter: Acid Base Equilibrium Topic: Titration of a Strong Acid with a Strong Base Answer included
neutralisation curve neutralization curve titration curve strong acid strong base titration equivalence point endpoint pH vs volume HCl NaOH titration
Accepted answer Answer included

Meaning of a neutralisation curve

A neutralisation curve (neutralization curve) is the graph of \(\mathrm{pH}\) on the vertical axis versus the volume of titrant added on the horizontal axis during an acid–base titration. The curve shape reflects the dominant species controlling \([\mathrm{H_3O^+}]\) and \([\mathrm{OH^-}]\) as neutralization proceeds.

\[ \mathrm{H_3O^+(aq) + OH^-(aq) \rightarrow 2H_2O(l)} \]

Reference relationships for pH and pOH

At \(25^\circ\text{C}\), water autoionization is described by

\[ K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}] = 1.0\times 10^{-14} \]

The pH and pOH definitions are

\[ \mathrm{pH} = -\log_{10}([\mathrm{H_3O^+}]),\qquad \mathrm{pOH} = -\log_{10}([\mathrm{OH^-}]) \]

\[ \mathrm{pH} + \mathrm{pOH} = 14.00 \quad (25^\circ\text{C}) \]

Strong acid–strong base neutralisation curve model

A standard general-chemistry model uses a strong monoprotic acid (such as \(\mathrm{HCl}\)) titrated by a strong base (such as \(\mathrm{NaOH}\)). Strong electrolytes are treated as fully dissociated in dilute aqueous solution.

The pH calculation is controlled by the excess strong species after the reaction stoichiometry is accounted for: excess \(\mathrm{H_3O^+}\) before equivalence and excess \(\mathrm{OH^-}\) after equivalence.

General pH expressions as a function of added volume

Let \(C_a\) and \(V_a\) be the initial acid concentration and volume, and let \(C_b\) and \(V_b\) be the base concentration and the added base volume. Moles of acid equivalents and base equivalents are

\[ n_a = C_aV_a,\qquad n_b = C_bV_b \]

Total solution volume (assuming volume additivity) is

\[ V_{\text{tot}} = V_a + V_b \]

Region before the equivalence point

Before equivalence, \(n_a > n_b\). Excess hydronium equivalents are \(n_a-n_b\), giving

\[ [\mathrm{H_3O^+}] = \frac{n_a-n_b}{V_{\text{tot}}} = \frac{C_aV_a - C_bV_b}{V_a+V_b} \]

\[ \mathrm{pH} = -\log_{10}\!\left(\frac{C_aV_a - C_bV_b}{V_a+V_b}\right) \]

Equivalence point

The equivalence point satisfies \(n_a=n_b\), so

\[ C_aV_a = C_bV_{b,\text{eq}} \quad\Rightarrow\quad V_{b,\text{eq}} = \frac{C_aV_a}{C_b} \]

For a strong acid–strong base titration at \(25^\circ\text{C}\), the solution at equivalence is dominated by spectator ions and water, so \(\mathrm{pH}\approx 7.00\).

Region after the equivalence point

After equivalence, \(n_b > n_a\). Excess hydroxide is \(n_b-n_a\), giving

\[ [\mathrm{OH^-}] = \frac{n_b-n_a}{V_{\text{tot}}} = \frac{C_bV_b - C_aV_a}{V_a+V_b} \]

\[ \mathrm{pOH} = -\log_{10}\!\left(\frac{C_bV_b - C_aV_a}{V_a+V_b}\right),\qquad \mathrm{pH} = 14.00 - \mathrm{pOH} \]

Worked values for a typical neutralisation curve

Consider \(V_a = 25.00\ \text{mL}\) of \(C_a = 0.100\ \text{M}\ \mathrm{HCl}\) titrated by \(C_b = 0.100\ \text{M}\ \mathrm{NaOH}\) at \(25^\circ\text{C}\). The equivalence volume is

\[ V_{b,\text{eq}} = \frac{(0.100)(0.02500)}{0.100} = 0.02500\ \text{L} = 25.00\ \text{mL} \]

Added base volume \(V_b\) (mL) Dominant excess species Key concentration pH
0.00 excess \(\mathrm{H_3O^+}\) \([\mathrm{H_3O^+}] = 0.100\ \text{M}\) \(\mathrm{pH}= -\log_{10}(0.100)=1.00\)
10.00 excess \(\mathrm{H_3O^+}\) \[ [\mathrm{H_3O^+}] = \frac{(0.100)(0.02500)-(0.100)(0.01000)}{0.02500+0.01000} = \frac{0.00150}{0.03500} = 4.29\times 10^{-2} \] \(\mathrm{pH}\approx -\log_{10}(4.29\times 10^{-2})=1.37\)
25.00 equivalence \(\mathrm{pH}\approx 7.00\) 7.00
30.00 excess \(\mathrm{OH^-}\) \[ [\mathrm{OH^-}] = \frac{(0.100)(0.03000)-(0.100)(0.02500)}{0.02500+0.03000} = \frac{0.00050}{0.05500} = 9.09\times 10^{-3} \] \[ \mathrm{pOH} = -\log_{10}(9.09\times 10^{-3}) = 2.04,\quad \mathrm{pH} = 14.00-2.04=11.96 \]

Visualization of a neutralisation curve

Neutralisation curve: strong acid–strong base titration A pH versus added base volume curve with low initial pH, a steep rise near the equivalence point, and high pH after equivalence. The equivalence point is marked near the inflection. added base volume (titrant) pH 0 7 14 equivalence point pH \(\approx 7\) excess \(\mathrm{H_3O^+}\) excess \(\mathrm{OH^-}\) rapid pH change
The steep central region marks the rapid pH change around the equivalence point that defines the practical endpoint range for suitable acid–base indicators in a strong acid–strong base titration.

Indicator choice and endpoint meaning

The endpoint is the observed signal change (often a color change) from an acid–base indicator. The equivalence point is the stoichiometric condition \(n_a=n_b\). For a strong acid–strong base neutralisation curve, the rapid pH change occurs near \(\mathrm{pH}\approx 7\), so many common indicators can work because the pH jump spans several pH units in a small volume range.

Comparison with other neutralisation curves

Curves involving weak acids or weak bases show buffer regions and equivalence-point pH values that differ from \(7.00\). The strong acid–strong base curve is distinguished by the absence of a buffer plateau and a near-neutral equivalence point at \(25^\circ\text{C}\).

Titration type Typical equivalence-point pH (25°C) Distinct curve feature
strong acid + strong base \(\approx 7\) very steep jump centered near neutral pH
weak acid + strong base \(> 7\) buffer region; equivalence in basic range
strong acid + weak base \(< 7\) smaller jump; equivalence in acidic range
weak acid + weak base depends on \(K_a\) and \(K_b\) limited pH jump; potentiometric methods common

Common pitfalls

Confusion between equivalence point and endpoint is common; the equivalence point is defined by stoichiometry, while the endpoint depends on the indicator transition range and experimental observation.

Dilution effects are significant away from the equivalence point; concentration expressions use \(V_{\text{tot}}=V_a+V_b\), not the initial volume alone.

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