Acid-base titration curves: strong-strong, weak-strong, and polyprotic titrations

Intuition Beginner

A titration is a controlled reaction between a solution of known concentration (the titrant) and a solution of unknown concentration (the analyte). You add titrant incrementally, measure pH after each addition, and plot the result as a titration curve: pH on the vertical axis, volume of titrant added on the horizontal axis.

The curve has a characteristic shape. It starts flat, rises slowly through a buffer region, then shoots up steeply at the equivalence point -- the volume at which exactly enough titrant has been added to react with all the analyte. After the equivalence point the curve flattens again.

The steepness at the equivalence point makes the titration useful. A tiny addition of titrant causes a large pH jump, which you can detect with an indicator or a pH meter. The volume at which this jump occurs tells you how much analyte was present.

For a strong acid titrated with a strong base, the equivalence point occurs at pH 7. For a weak acid titrated with a strong base, the equivalence point is above pH 7 because the conjugate base of the weak acid makes the solution basic.

Visual Beginner

A strong acid-strong base titration curve is symmetric and S-shaped, crossing pH 7 at the equivalence point. A weak acid-strong base curve has a buffer plateau before the equivalence point and crosses above pH 7.

The half-equivalence point (at half the equivalence-point volume) is where exactly half the weak acid has been neutralised. Here $[\text{HA}]=[{\text{A}}^{-}]$, so pH = pKa. This is the standard experimental method for determining pKa from a titration curve.

Worked example Beginner

Titrate 25.00 mL of 0.100 M HCl with 0.100 M NaOH. Calculate the pH at four points.

Point 1: Before any NaOH added (V = 0 mL). HCl is a strong acid, so $[{\text{H}}^{+}]=0.100$ M. pH = 1.00.

Point 2: After adding 10.00 mL NaOH. The added NaOH neutralises some HCl. Moles HCl remaining = $0.100\times 25.00-0.100\times 10.00=1.50$ mmol. Total volume = 35.00 mL.

$$[{\text{H}}^{+}]=\frac{1.50}{35.00}=0.0429\text{M},\text{pH}=1.37$$

Point 3: Equivalence point (V = 25.00 mL). All HCl has been neutralised. The solution contains only NaCl and water. pH = 7.00.

Point 4: After adding 30.00 mL NaOH (5.00 mL past equivalence). Excess NaOH = $0.100\times 5.00=0.500$ mmol. Total volume = 55.00 mL.

$$[{\text{OH}}^{-}]=\frac{0.500}{55.00}=0.00909\text{M},\text{pOH}=2.04,\text{pH}=11.96$$

The pH jumps from 3.00 (at 24.00 mL NaOH) to 11.00 (at 26.00 mL NaOH) -- a change of 8 pH units for just 2 mL of added titrant.

Check your understanding Beginner

Formal definition Intermediate+

Regions of a titration curve

A titration curve for a monoprotic weak acid HA with a strong base has four distinct regions, each governed by different equilibrium mathematics.

Region 1: Before any titrant added. The solution contains only the weak acid. pH is determined by the weak-acid equilibrium:

$$[{\text{H}}^{+}]=\sqrt{K_a{c}_a}$$

where $c_a$ is the initial acid concentration. This approximation holds when $K_a{c}_a\gg K_w$ and $c_a\gg [{\text{H}}^{+}]$.

Region 2: Buffer region (after some base added, before equivalence point). The added strong base converts some HA to ${\text{A}}^{-}$, creating a buffer. The pH is given by the Henderson-Hasselbalch equation:

$$\text{pH}=\text{p}{K}_a+\log \frac{n_b}{n_a-n_b}$$

where $n_a$ is the initial moles of HA and $n_b$ is the moles of strong base added. The special case $n_b=n_a/2$ (the half-equivalence point) gives pH = pKa. The buffer region extends from roughly 10% to 90% neutralisation.

Region 3: Equivalence point. All HA has been converted to ${\text{A}}^{-}$. The pH is determined by the hydrolysis of the conjugate base:

$${\text{A}}^{-}+{\text{H}}_2\text{O}\rightleftharpoons \text{HA}+{\text{OH}}^{-}$$

The concentration of ${\text{A}}^{-}$ at the equivalence point is $c_{{\text{A}}^{-}}=n_a/(V_a+V_e)$, where $V_e$ is the equivalence-point volume. The pH calculation uses $K_b=K_w/K_a$:

$$[{\text{OH}}^{-}]=\sqrt{K_b\cdot c_{{\text{A}}^{-}}}$$

For a weak base titrated with a strong acid, the roles are reversed: the equivalence point is acidic due to hydrolysis of the conjugate acid.

Region 4: Post-equivalence region. Beyond the equivalence point, the pH is governed by the excess strong base (or strong acid, in a weak base-strong acid titration):

$$[{\text{OH}}^{-}]=\frac{c_b(V_b-V_e)}{V_a+V_b}$$

The contribution of the weak base ${\text{A}}^{-}$ is negligible compared to the excess strong base.

Strong acid-strong base titrations

When both titrant and analyte are strong, there is no buffer region. The pH is governed entirely by the excess of whichever strong species remains:

$$[{\text{H}}^{+}]=\frac{n_{\text{acid}}-n_{\text{base}}}{V_{\text{total}}}$$

The equivalence point is at pH 7.00 (at 25 C). The titration curve is symmetric about the equivalence point: the pH change from 1 mL before equivalence to the equivalence point equals the change from the equivalence point to 1 mL after equivalence.

Weak acid-strong base titrations

The weak acid-strong base curve differs from the strong-strong case in three ways. First, the initial pH is higher (HA is only partially dissociated). Second, a buffer plateau appears between about 10% and 90% neutralisation, where the Henderson-Hasselbalch equation applies. Third, the equivalence point is above pH 7 because the salt formed (${\text{A}}^{-}$) is a weak base.

The pH at the equivalence point depends on the concentration of the conjugate base and its $K_b$. More dilute solutions have lower $[{\text{A}}^{-}]$ at the equivalence point, and the pH approaches 7 from above as concentration decreases. For a $0.100\text{M}$ acetic acid titration with $0.100\text{M}$ NaOH, the equivalence-point pH is approximately 8.73. For a $0.00100\text{M}$ solution under the same conditions, the equivalence-point pH is approximately 7.87.

Weak base-strong acid titrations

The mirror case: a weak base B (such as ${\text{NH}}_3$) titrated with a strong acid (such as HCl). The reaction is $\text{B}+{\text{H}}^{+}\to {\text{BH}}^{+}$. The initial pH is basic, the buffer region involves the $\text{B}/{\text{BH}}^{+}$ pair, and the equivalence point is acidic because ${\text{BH}}^{+}$ is a weak acid. For $25.00\text{mL}$ of $0.100\text{M}{\text{NH}}_3$ ($K_b=1.8\times 10^{-5}$) titrated with $0.100\text{M}$ HCl, the equivalence-point pH is approximately 5.28.

Indicator selection

An acid-base indicator is a weak acid (or base) whose protonated and deprotonated forms have different colours. The indicator transition occurs over approximately 2 pH units centred on its $\text{p}{K}_{\text{In}}$:

$$\text{p}{K}_{\text{In}}-1<\text{pH}<\text{p}{K}_{\text{In}}+1$$

To select an indicator for a titration, the indicator transition range must overlap the steep portion of the titration curve at the equivalence point. For a strong acid-strong base titration (equivalence pH 7), phenolphthalein ($\text{p}{K}_{\text{In}}=9.3$, range 8.2-10.0) and bromothymol blue ($\text{p}{K}_{\text{In}}=7.0$, range 6.0-7.6) both work because the pH change is so steep that even indicators with $\text{p}{K}_{\text{In}}$ 2-3 units from 7 still fall within the sharp rise.

For a weak acid-strong base titration (equivalence pH > 7), phenolphthalein is the standard choice because its transition range (8.2-10.0) brackets the equivalence-point pH of most weak-acid titrations. Methyl red ($\text{p}{K}_{\text{In}}=5.0$, range 4.2-6.2) would give a false early endpoint because the colour change occurs well before the equivalence point.

For a weak base-strong acid titration (equivalence pH < 7), methyl red is appropriate because its transition range brackets the acidic equivalence point.

Counterexamples to common slips

• Equivalence point is not the same as endpoint. The equivalence point is the theoretical stoichiometric point. The endpoint is the observed indicator colour change. A good indicator makes the endpoint coincide with the equivalence point to within the experimental error.
• Equivalence point is not always pH 7. Only strong acid-strong base titrations have pH 7 at equivalence. The pKa of the weak species determines the equivalence-point pH.
• The half-equivalence point is not the equivalence point. The half-equivalence point is where pH = pKa. The equivalence point is where all analyte has been neutralised. These are different volumes and different pH values.

Key result Intermediate+

Theorem (Equivalence-point pH depends on the weak species). For the titration of a weak acid HA ($K_a$) at initial concentration $c_a$ with a strong base at concentration $c_b$, the pH at the equivalence point is determined by the hydrolysis of the conjugate base ${\text{A}}^{-}$ at concentration $c_{{\text{A}}^{-}}=c_a{V}_a/(V_a+V_e)$, where $V_e=c_a{V}_a/c_b$. When $c_{{\text{A}}^{-}}\gg [{\text{OH}}^{-}]$:

$$[{\text{OH}}^{-}]=\sqrt{\frac{K_w}{K_a}\cdot c_{{\text{A}}^{-}}},\text{pH}=7.00+\frac{1}{2}\text{p}{K}_a+\frac{1}{2}\log c_{{\text{A}}^{-}}$$

Proof. At the equivalence point, the solution contains only the salt NaA and water. The anion ${\text{A}}^{-}$ is a weak base with $K_b=K_w/K_a$. From the base hydrolysis equilibrium:

$$K_b=\frac{[\text{HA}][{\text{OH}}^{-}]}{[{\text{A}}^{-}]}$$

With $[\text{HA}]=[{\text{OH}}^{-}]$ (from the stoichiometry of the hydrolysis reaction) and $[{\text{A}}^{-}]\approx c_{{\text{A}}^{-}}$ (valid when the extent of hydrolysis is small):

$$[{\text{OH}}^{-}{]}^2=K_b\cdot c_{{\text{A}}^{-}}=\frac{K_w}{K_a}\cdot c_{{\text{A}}^{-}}$$

Taking $-\log$:

$$\text{pOH}=\frac{1}{2}\text{p}{K}_w-\frac{1}{2}\text{p}{K}_a-\frac{1}{2}\log c_{{\text{A}}^{-}}$$ $$\text{pH}=\text{p}{K}_w-\text{pOH}=7.00+\frac{1}{2}\text{p}{K}_a+\frac{1}{2}\log c_{{\text{A}}^{-}}\square$$

This result shows that the equivalence-point pH increases with pKa (weaker acids give more basic equivalence points) and with concentration (more dilute solutions approach pH 7). The analogous result for a weak base-strong acid titration is $\text{pH}=7.00-\frac{1}{2}\text{p}{K}_b-\frac{1}{2}\log c_{{\text{BH}}^{+}}$.

Corollary (Titration feasibility). A weak acid can be accurately titrated with a strong base only if $K_a\cdot c_a\ge 10^{-8}$. Below this threshold, the pH change at the equivalence point is too gradual for precise endpoint detection.

Worked example: complete titration curve

Titrate 25.00 mL of 0.100 M ${\text{CH}}_3\text{COOH}$ ($\text{p}{K}_a=4.76$) with 0.100 M NaOH. Calculate pH at 0, 5.00, 12.50, 20.00, 25.00, and 30.00 mL NaOH.

V = 0 mL. Weak acid alone: $[{\text{H}}^{+}]=\sqrt{1.74\times 10^{-5}\times 0.100}=1.32\times 10^{-3}$ M, pH = 2.88.

V = 5.00 mL. Buffer region. Moles ${\text{A}}^{-}=0.100\times 5.00=0.500$ mmol. Moles HA = $2.50-0.500=2.00$ mmol.

$$\text{pH}=4.76+\log \frac{0.500}{2.00}=4.76-0.602=4.16$$

V = 12.50 mL (half-equivalence). $[\text{HA}]=[{\text{A}}^{-}]$, so pH = 4.76.

V = 20.00 mL. Moles ${\text{A}}^{-}=2.00$ mmol, moles HA = 0.500 mmol.

$$\text{pH}=4.76+\log \frac{2.00}{0.50}=4.76+0.602=5.36$$

V = 25.00 mL (equivalence). $c_{{\text{A}}^{-}}=2.50/50.00=0.0500$ M.

$$\text{pH}=7.00+\frac{4.76}{2}+\frac{\log (0.0500)}{2}=7.00+2.38-0.65=8.73$$

V = 30.00 mL. Excess NaOH: $0.100\times 5.00=0.500$ mmol in 55.00 mL.

$$[{\text{OH}}^{-}]=0.500/55.00=0.00909\text{M},\text{pOH}=2.04,\text{pH}=11.96$$

Bridge. The four-region structure of the titration curve connects directly to 14.10.02 buffer chemistry: the buffer region of the curve is a dynamic Henderson-Hasselbalch system where each addition of titrant shifts the $[{\text{A}}^{-}]/[\text{HA}]$ ratio. The equivalence-point theorem quantifies how the pKa of the weak species controls the endpoint pH, which determines indicator selection. In 14.11.01 electrochemistry, the pH electrode converts each point on the titration curve into a voltage via the Nernst equation, enabling potentiometric endpoint detection that is more precise than visual indicators.

Exercises Intermediate+

Polyprotic acid titration curves Master

Sequential dissociation and multiple equivalence points

A polyprotic acid has two or more dissociable protons, each governed by a separate $K_a$. Phosphoric acid ${\text{H}}_3{\text{PO}}_4$ has $\text{p}{K}_{a1}=2.15$, $\text{p}{K}_{a2}=7.20$, $\text{p}{K}_{a3}=12.35$. Each dissociation produces a separate equivalence point in the titration curve.

$${\text{H}}_3{\text{PO}}_4\stackrel{K_{a1}}{\to }{\text{H}}_2{\text{PO}}_4^{-}\stackrel{K_{a2}}{\to }{\text{HPO}}_4^{2-}\stackrel{K_{a3}}{\to }{\text{PO}}_4^{3-}$$

The titration curve has three distinct buffer plateaus, one for each conjugate pair, separated by equivalence points where the pH rises steeply. The first equivalence point (all ${\text{H}}_3{\text{PO}}_4$ converted to ${\text{H}}_2{\text{PO}}_4^{-}$) occurs at:

$$\text{pH}=\frac{\text{p}{K}_{a1}+\text{p}{K}_{a2}}{2}=\frac{2.15+7.20}{2}=4.68$$

The second equivalence point (all ${\text{H}}_2{\text{PO}}_4^{-}$ converted to ${\text{HPO}}_4^{2-}$) occurs at:

$$\text{pH}=\frac{\text{p}{K}_{a2}+\text{p}{K}_{a3}}{2}=\frac{7.20+12.35}{2}=9.78$$

Whether the third equivalence point is observable depends on the separation of $\text{p}{K}_{a3}$ from the strong-base region. For ${\text{H}}_3{\text{PO}}_4$, $\text{p}{K}_{a3}=12.35$, so the third equivalence point is poorly defined because the pH change is too gradual to detect precisely.

Conditions for resolving multiple equivalence points

Two successive equivalence points are resolvable when $\text{p}{K}_{a(i+1)}-\text{p}{K}_{ai}>3$. When the pKa values are this far apart, the buffer regions do not overlap significantly, and there is a steep pH rise between them. For phosphoric acid, $\text{p}{K}_{a2}-\text{p}{K}_{a1}=5.05$ and $\text{p}{K}_{a3}-\text{p}{K}_{a2}=5.15$, so both the first and second equivalence points are well resolved.

When successive pKa values are closer together (separation less than about 3), the buffer regions merge and the individual equivalence points are not distinguishable. Citric acid ($\text{p}{K}_{a1}=3.13$, $\text{p}{K}_{a2}=4.76$, $\text{p}{K}_{a3}=6.40$) has separations of only 1.63 and 1.64 units. Its titration curve shows a single broad rise rather than three distinct steps.

Speciation at each equivalence point

At the first equivalence point of the ${\text{H}}_3{\text{PO}}_4$ titration, the dominant species is ${\text{H}}_2{\text{PO}}_4^{-}$. The fractional composition gives ${\alpha }_1>0.99$ at pH 4.68, meaning the solution is essentially pure ${\text{H}}_2{\text{PO}}_4^{-}$. The second equivalence point gives ${\alpha }_2>0.99$ at pH 9.78, dominated by ${\text{HPO}}_4^{2-}$.

The pH at each equivalence point is calculated using the amphoteric-intermediate formula:

$${\text{pH}}_{e,1}=\frac{\text{p}{K}_{a1}+\text{p}{K}_{a2}}{2},{\text{pH}}_{e,2}=\frac{\text{p}{K}_{a2}+\text{p}{K}_{a3}}{2}$$

These are exact in the limit of well-separated pKa values and moderate concentration. The corrections at finite concentration involve the concentration-dependent terms from the exact charge-balance equation.

Worked example: phosphoric acid titration

Titrate 25.00 mL of 0.100 M ${\text{H}}_3{\text{PO}}_4$ with 0.100 M NaOH.

First equivalence point: $V_{e1}=25.00$ mL. pH = 4.68. Use methyl orange (transition 3.1-4.4) or bromocresol green (3.8-5.4) as indicator. The colour change at the first equivalence point is gradual, making potentiometric detection preferable.

Second equivalence point: $V_{e2}=50.00$ mL. pH = 9.78. Use phenolphthalein (8.2-10.0) or thymolphthalein (9.3-10.5). The second equivalence point is sharper than the first because the pKa separation is the same but the pH range of the steep rise is narrower.

Third equivalence point: $V_{e3}=75.00$ mL. pH $\approx 12.35+\frac{1}{2}\log c_{{\text{PO}}_4^{3-}}$. With $c_{{\text{PO}}_4^{3-}}\approx 0.025$ M, pH $\approx 12.35-0.30=12.05$. This point is not practically useful because the titration curve is too gradual for reliable endpoint detection.

The Gran plot Master

Linearisation of the titration curve

The Gran plot is a method for locating the equivalence point by transforming the titration curve into a linear function. The standard Gran plot for a weak acid titrated with strong base plots $V_b\times 10^{-\text{pH}}$ (i.e., $V_b[{\text{H}}^{+}]$) against $V_b$ in the region before the equivalence point.

Before the equivalence point, the Henderson-Hasselbalch equation gives:

$$[{\text{H}}^{+}]=K_a\cdot \frac{[\text{HA}]}{[{\text{A}}^{-}]}=K_a\cdot \frac{n_a-c_b{V}_b}{c_b{V}_b}$$

Multiplying both sides by $V_b$ and approximating $[{\text{H}}^{+}]\approx 10^{-\text{pH}}$ (neglecting activity corrections):

$$V_b[{\text{H}}^{+}]=\frac{K_a}{c_b}(n_a-c_b{V}_b)=\frac{K_a{n}_a}{c_b}-K_a{V}_b$$

This is linear in $V_b$ with slope $-K_a$ and x-intercept $V_b=n_a/c_b=V_e$. The equivalence point is found by extrapolating the straight line to the x-axis.

Advantages over direct endpoint detection

The Gran plot has three advantages over simply reading the steepest point of the titration curve. First, it uses data from the buffer region (well before the equivalence point), where the pH measurements are most precise because the buffer resists disturbance from CO${}_2$ absorption and other perturbations. Second, it provides both the equivalence-point volume and $K_a$ simultaneously (from the slope). Third, it is less sensitive to systematic errors in the pH measurement because the linear extrapolation averages over many data points.

The Gran plot is particularly useful for weak acids with $K_a$ close to the feasibility limit ($K_a\cdot c_a\approx 10^{-8}$), where the titration curve has no sharp equivalence point but the Gran plot remains linear. It is also the standard method for determining the equivalence point in Gran-type acid-base titrations used to measure the alkalinity of natural waters.

Modified Gran functions

The standard Gran function works before the equivalence point. After the equivalence point, a second Gran function plots $(V_a+V_b)\times 10^{\text{pH}-14}$ (i.e., $(V_a+V_b)[{\text{OH}}^{-}]$) against $V_b$. This function is linear after the equivalence point with x-intercept $V_e$.

For a diprotic acid, two Gran plots can be constructed, one for each equivalence point. The first uses the function $V_b[{\text{H}}^{+}]$ linear in the region before $V_{e1}$, and the second uses the function $V_b[{\text{H}}^{+}]$ linear between $V_{e1}$ and $V_{e2}$, or $(V_a+V_b)[{\text{OH}}^{-}]$ after $V_{e2}$.

Derivation of the post-equivalence Gran function

After the equivalence point, the solution contains excess strong base. The concentration of excess hydroxide is:

$$[{\text{OH}}^{-}]=\frac{c_b(V_b-V_e)}{V_a+V_b}$$

Multiplying by $(V_a+V_b)$:

$$(V_a+V_b)[{\text{OH}}^{-}]=c_b(V_b-V_e)=c_b{V}_b-c_b{V}_e$$

This is linear in $V_b$ with slope $c_b$ and x-intercept $V_e$. Plotting $(V_a+V_b)\times 10^{\text{pH}-\text{p}{K}_w}$ versus $V_b$ gives a straight line that crosses the x-axis at the equivalence-point volume.

Derivative methods for endpoint detection Master

First and second derivative plots

The equivalence point can be located by finding the maximum of $d\text{pH}/d{V}_b$ (the first derivative) or the zero crossing of $d^2\text{pH}/d{V}_b^2$ (the second derivative). The first-derivative plot has a peak at the equivalence point. The second-derivative plot crosses zero at the equivalence point.

For $n$ equispaced data points with spacing $\Delta V$, the numerical first derivative at point $i$ is:

$${\left(\frac{d\text{pH}}{dV}\right)}_i=\frac{{\text{pH}}_{i+1}-{\text{pH}}_{i-1}}{2\Delta V}$$

The second derivative is:

$${\left(\frac{d^2\text{pH}}{d{V}^2}\right)}_i=\frac{{\text{pH}}_{i+1}-2{\text{pH}}_i+{\text{pH}}_{i-1}}{(\Delta V{)}^2}$$

The zero crossing of the second derivative is found by linear interpolation between the last positive and first negative values. This method is more precise than locating the peak of the first derivative because the zero crossing is sharper.

Comparison of methods

For a strong acid-strong base titration at $0.1\text{M}$ concentration, the equivalence point can be located to within $\pm 0.02\text{mL}$ by the second-derivative method, $\pm 0.05\text{mL}$ by the first-derivative method, and $\pm 0.1\text{mL}$ by visual indicator. The Gran plot achieves $\pm 0.02\text{mL}$ for strong acids and is the preferred method for weak acids near the feasibility limit.

For automatic titrators, the tangent method (locating the inflection point by fitting a tangent to the steepest portion) is the most common algorithm. The precision of automatic endpoint detection depends on the density of data points near the equivalence point. Modern titrators collect data at 0.01 mL intervals and use polynomial fitting to locate the inflection point to $\pm 0.005\text{mL}$.

Titration of mixtures Master

Mixture of a strong acid and a weak acid

When a solution contains both a strong acid (HCl) and a weak acid (HA), the titration curve shows two equivalence points. The first corresponds to the complete neutralisation of the strong acid (because the strong acid is neutralised first, being the stronger proton source). The second corresponds to the neutralisation of the weak acid.

The first equivalence point is detected as a change in the slope of the titration curve. Before it, the pH is governed by the strong acid alone (the weak acid contributes negligibly to $[{\text{H}}^{+}]$ in the presence of a much higher concentration of strong acid). Between the first and second equivalence points, the weak acid and its conjugate base form a buffer, and the Henderson-Hasselbalch equation applies.

The analysis of such mixtures requires that $K_a(\text{HA})\le 10^{-4}$ and that the two acids be present in comparable concentrations. If $K_a$ is too large, the weak acid begins to titrate before the strong acid is fully consumed, and the two equivalence points merge.

Mixture of two weak acids

Two weak acids with sufficiently different pKa values ($\Delta \text{p}{K}_a>4$) can be titrated sequentially. The stronger acid (lower pKa) is titrated first. The pH at the first equivalence point is determined by a combination of the first acid's conjugate base and the as-yet-untitrated second acid. The analysis is complex and requires solving the full charge-balance equation.

Systematic treatment: the master equation Master

Exact pH as a function of titrant volume

The exact pH at any point during the titration of a weak acid HA ($K_a$) at initial concentration $c_a$ and volume $V_a$ with a strong base MOH at concentration $c_b$ is obtained from the charge-balance equation:

$$[{\text{H}}^{+}]+\frac{c_b{V}_b}{V_a+V_b}=\frac{K_a}{K_a+[{\text{H}}^{+}]}\cdot \frac{c_a{V}_a}{V_a+V_b}+\frac{K_w}{[{\text{H}}^{+}]}$$

The left side is the total positive charge from protons and the added metal cation. The right side is the total negative charge from the conjugate base ${\text{A}}^{-}$ and hydroxide. Multiplying through by $(V_a+V_b)[{\text{H}}^{+}](K_a+[{\text{H}}^{+}])$ gives a polynomial in $[{\text{H}}^{+}]$ that can be solved numerically.

This single equation reduces to the four-region approximations used at the beginner and intermediate levels. Before any titrant is added ($V_b=0$), it reduces to the weak-acid equation. In the buffer region, the Henderson-Hasselbalch equation is recovered. At the equivalence point ($c_b{V}_b=c_a{V}_a$), the conjugate-base hydrolysis equation emerges. Past the equivalence point, the excess-strong-base approximation dominates.

Activity corrections in titration curves

The systematic treatment above uses concentrations. In rigorous work, activities replace concentrations. The activity coefficient corrections become significant near the equivalence point of dilute titrations and in the presence of inert electrolytes. The Debye-Huckel correction shifts the equivalence-point pH by 0.05-0.2 units at ionic strengths of 0.01-0.1 M.

For primary standard calibrations (NIST-traceable pH standards), the ionic strength is fixed by adding a background electrolyte (typically KCl at $I=0.1$ M), and the pH values are corrected using the Davies equation. The corrected titration curves agree with potentiometric measurements to within 0.01 pH units.

Historical notes Master

The first quantitative titration is attributed to Joseph Louis Gay-Lussac, who in 1828 described the titration of silver with chloride using potassium chromate as an indicator (the Mohr method). Gay-Lussac's method was developed for assaying the silver content of coinage and established the principle that a sharp colour change marks the stoichiometric endpoint.

Izaak Kolthoff systematised the theory of acid-base titrations in the 1930s and 1940s. His 1950 review in Analytical Chemistry ^{[Kolthoff 1950]} codified the four-region analysis of titration curves and established the feasibility criterion $K_a\cdot c_a\ge 10^{-8}$ for accurate endpoint detection. Kolthoff's treatment remains the standard framework taught in quantitative analysis courses.

Gunnar Gran introduced the Gran plot linearisation in 1950 ^{[Gran 1950]}. Gran's insight was that the titration curve, which is nonlinear in pH-versus-volume space, can be transformed into a linear function by multiplying the volume by the hydrogen-ion concentration (or hydroxide concentration, after the equivalence point). The linear extrapolation provides both the equivalence-point volume and the dissociation constant. The Gran plot is now the standard method for endpoint determination in water-alkalinity titrations and weak-acid analysis.

Sorensen's introduction of the pH scale in 1909 ^{[Sorensen 1909]} was the prerequisite for potentiometric titration. Before pH, acid-base titrations relied exclusively on visual indicators. The glass electrode, developed by Haber and Klemensiewicz in 1909 and refined by MacInnes and Dole in the 1930s, enabled the continuous measurement of pH throughout the titration and made titration curves experimentally accessible. Automatic titrators, combining a burette drive with a pH meter and a recording device, were commercialised in the 1950s and are now standard laboratory instruments.

Connections Master

• Buffer solutions (14.10.02): The buffer region of every weak-acid titration curve is a dynamic Henderson-Hasselbalch system. The titration curve is essentially a plot of how buffer capacity varies from near-zero (pure weak acid) through maximum (half-equivalence) back to near-zero (approaching equivalence).
• Electrochemistry (14.11): Potentiometric titration uses a pH electrode whose response is governed by the Nernst equation. The Gran plot and derivative methods for endpoint detection rely on the high precision of potentiometric pH measurement, which is fundamentally an electrochemical technique.
• Solubility equilibria (14.10.04): Precipitation titrations (Mohr, Volhard) share the same equivalence-point analysis framework developed here. The Ksp of the precipitate replaces Ka as the governing equilibrium constant, but the four-region curve structure and indicator selection principles are identical.

Bibliography Master

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  publisher = {W. H. Freeman},
  year = {2015},
  edition = {9th}
}

@book{Atkins2023,
  author = {Atkins, P. and de Paula, J.},
  title = {Physical Chemistry},
  publisher = {Oxford University Press},
  year = {2023},
  edition = {12th}
}

@book{Ringbom1963,
  author = {Ringbom, A.},
  title = {Complexation in Analytical Chemistry},
  publisher = {Wiley},
  year = {1963}
}