Potentiometric Titrations

A titration curve is a story told in millivolts. As titrant drips into a beaker, an indicator electrode reports the changing activity of some ion — H⁺, Ag⁺, Fe²⁺, Ce³⁺. That voltage traces an S-shaped path. It bends sharply exactly where stoichiometry is satisfied. Potentiometric titration turns this bending point into a measurement. Instead of trusting your eye to catch a color change, you let a pH meter or an ion-selective electrode locate the equivalence point. Its precision rivals any spectrophotometric method. It works in murky, colored, or fluorescent solutions where visual indicators fail outright. The method matters because it is quietly everywhere. Karl Fischer titrators, chloride analyzers in water-treatment labs, argentometric assays of halides in pharmaceuticals, and redox titrations of iron ore and wine must — all of them locate an endpoint from a potential-versus-volume curve. None of them rely on a color flash. The technique also underlies automated titrators that pipette, stir, and compute derivatives in real time. These feed results into quality-control systems. No human judgment call about "when the pink appears" is needed. The big picture connects three ideas already in hand. First, the Nernst equation's prediction that potential varies with the logarithm of concentration. Second, the sigmoid shape of any titration curve near stoichiometric equivalence. Third, the calculus fact that a curve's steepest point is where its first derivative peaks. At that same point, its second derivative crosses zero. Locating the equivalence point becomes an exercise in numerical differentiation of noisy data. Chemists solved this problem with graph paper long before spreadsheets. It reveals exactly why some methods (first derivative, second derivative, Gran plots) are more robust than others. This is especially true when the curve is asymmetric. It also holds when excess titrant reagent itself contributes to the signal. This article builds the full toolkit. It covers the physical origin of the sigmoid curve from the Nernst equation. It covers the algebra of the first- and second-derivative methods. A fully worked numerical table is included. It covers the linearizing trick behind Gran plots. This trick converts a curved region into a straight line. That line crosses the volume axis at V_eq. Finally, it covers the traps that catch students who trust software output without understanding the equilibrium underneath. These traps include asymmetric curves, non-1:1 stoichiometry, and liquid-junction drift.

Key takeaways
  • An indicator electrode plus a reference electrode reports potential E as titrant volume V is added; E follows the Nernst equation and jumps sharply at the equivalence volume V_eq.
  • The first-derivative method plots ΔE/ΔV vs. V; the equivalence point is the maximum of this peak.
  • The second-derivative method plots Δ²E/ΔV² vs. V; V_eq is where this curve crosses zero, which can be located by linear interpolation to sub-drop precision.
  • Gran plots exploit the linear (pre- or post-equivalence) regions of the Nernst response, transforming exponential-looking data into a straight line whose x-intercept is V_eq — useful even far from the steep inflection.
  • Symmetric curves (equal-charge redox couples, 1:1 stoichiometry) give inflection points that coincide with the true equivalence point; asymmetric curves (e.g., Fe²⁺/Ce⁴⁺ vs. MnO₄⁻) shift the inflection slightly off true equivalence.
  • Key idea: the equivalence point is not where you see a color change — it is where the rate of potential change (dE/dV) is maximal, and calculus, not eyesight, finds it.

Where this lives today

Modern automatic titrators (Metrohm, Mettler Toledo, and similar instruments) run potentiometric endpoint detection as their default mode: a stepper-motor burette delivers titrant in decreasing increments as the software detects an approaching inflection, computes the first and second derivative in real time, and reports V_eq with reproducibility often better than 0.1%. Pharmaceutical quality control leans on this heavily — argentometric chloride/bromide assays, non-aqueous acid–base titrations of active pharmaceutical ingredients, and iodometric determinations of oxidizing or reducing agents in raw materials all report potentiometric endpoints because visual indicators fail in colored or turbid formulations. Environmental and water-quality laboratories use potentiometric titration for total alkalinity and hardness determinations, and Karl Fischer titrators for water content use a related bipotentiometric (dead-stop) endpoint detection, monitoring a small current between two polarized electrodes rather than open-circuit potential — a close cousin of the ideas here, and a bridge toward Coulometric Titrations, where the titrant itself is generated electrochemically in situ and the endpoint detection principles carry over almost unchanged. Biochemistry and materials science borrow the same logic in potentiometric determination of protein isoelectric points, in monitoring complexometric EDTA titrations of metal ions with ion-selective electrodes, and in battery-electrode characterization, where a "titration" of lithium ions into a host material is monitored by open-circuit voltage, and the derivative dV/dQ is used to identify phase transitions — a direct descendant of the same first-derivative logic used to find V_eq in a beaker.

:::faq[Why does the derivation of the symmetric equivalence potential assume [Fe²⁺] ≈ [Ce⁴⁺] and [Ce³⁺] ≈ [Fe³⁺] at equivalence?] At exact stoichiometric equivalence, essentially all of the original Fe²⁺ has been converted to Fe³⁺ and essentially all added Ce⁴⁺ has been converted to Ce³⁺, so both trace species remaining (the tiny unreacted Fe²⁺ and tiny unreacted Ce⁴⁺) arise only from the small reverse extent of a reaction driven nearly to completion. Because the forward reaction couples them 1:1, the residual concentrations of the two "reactant-side" species track each other, and likewise for the two "product-side" species, which is exactly the substitution used to cancel the logarithmic term in Step 3.

Ask the tutor: How do I choose between the first-derivative, second-derivative, and Gran-plot methods when analyzing my own titration data?

The physical picture

Consider titrating a solution containing Fe²⁺ with standard Ce⁴⁺, monitoring potential with a platinum indicator electrode against a saturated calomel reference electrode (SCE). Before any titrant is added, the solution potential is poorly defined (no redox couple present in excess on either side). As Ce⁴⁺ is added, it oxidizes Fe²⁺ to Fe³⁺, and the Pt electrode equilibrates with the Fe³⁺/Fe²⁺ couple:

E=EFe3+/Fe2+∘−RTnFln⁡[Fe2+][Fe3+]E = E^{\circ}_{\text{Fe}^{3+}/\text{Fe}^{2+}} - \frac{RT}{nF}\ln\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}

Early in the titration, [Fe²⁺] is large and [Fe³⁺] is small, so E is low. As V approaches the equivalence volume, [Fe²⁺] shrinks toward zero logarithmically, and E rises steeply — a small addition of titrant produces a huge ratio change because the ratio itself is passing through unity on a log scale. Exactly at equivalence, both the Fe and Ce couples are simultaneously satisfied, and for a 1:1, equal-electron-transfer reaction the potential sits at the arithmetic mean of the two formal potentials. Past equivalence, excess Ce⁴⁺ dominates and the Pt electrode instead reports the Ce⁴⁺/Ce³⁺ couple, and the curve flattens out again at high E. The result is the classic sigmoid, steepest exactly at V_eq. The same architecture governs acid–base titrations monitored with a glass pH electrode (see The Glass pH Electrode) — pH jumps steeply at the stoichiometric point because [H⁺] itself changes by orders of magnitude over a fraction of a milliliter — and argentometric titrations of Cl⁻ with Ag⁺ monitored with a silver indicator electrode, where E tracks pAg via the Nernst equation for the Ag⁺/Ag couple.

Figure
Idealized potentiometric titration curve: E vs. V, with the inflection at V_eq
Volume of titrant, V (mL) E (mV vs. SCE) V_eq = 25.00 mL steepest point = inflection Fe³⁺/Fe²⁺ region Ce⁴⁺/Ce³⁺ region
Figure
Numerical differentiation: E, first derivative ΔE/ΔV, second derivative Δ²E/ΔV² near equivalence
V (mL), zoomed near V_eq ΔE/ΔV (peak at V_eq) first-derivative peak Δ²E/ΔV² (crosses zero at V_eq) zero-crossing = V_eq

The core ideas

The Nernst equation for the indicator half-reaction sets the baseline relationship:

E=E∘′−RTnFln⁡QE = E^{\circ\prime} - \frac{RT}{nF}\ln Q

where:

  • E∘′E^{\circ\prime} is the formal potential of the couple sensed by the indicator electrode,
  • RR, TT, FF are the gas constant, absolute temperature, Faraday constant,
  • nn is the number of electrons transferred,
  • QQ is the reaction quotient of the sensed half-reaction (e.g., [Fe2+]/[Fe3+][\text{Fe}^{2+}]/[\text{Fe}^{3+}]).

Reading: because QQ changes by orders of magnitude as the limiting reagent is consumed near equivalence, EE changes rapidly there even though volume added is small — this logarithmic amplification is the entire reason titration curves have sharp inflections. The first-derivative method estimates the local slope from successive readings:

ΔEΔV∣Vi=Ei+1−EiVi+1−Vi\frac{\Delta E}{\Delta V}\bigg|_{V_i} = \frac{E_{i+1} - E_i}{V_{i+1} - V_i}

where:

  • Ei+1,EiE_{i+1}, E_i are potentials at consecutive titrant volumes,
  • Vi+1,ViV_{i+1}, V_i are the corresponding volumes (the derivative is assigned to the midpoint volume (Vi+Vi+1)/2(V_i+V_{i+1})/2).

Reading: this quantity peaks sharply at VeqV_{eq} because that is where EE is changing fastest per unit volume; the peak volume is read directly off the plot or found by locating where the second difference (below) changes sign, which pins the peak more precisely than eyeballing a peak. The second-derivative method differentiates again and interpolates the zero crossing:

Δ2EΔV2∣Vi=(ΔE/ΔV)i+1−(ΔE/ΔV)iVi+1−Vi\frac{\Delta^2 E}{\Delta V^2}\bigg|_{V_i} = \frac{(\Delta E/\Delta V)_{i+1} - (\Delta E/\Delta V)_i}{V_{i+1}-V_i}
Veq=Va+(Vb−Va)⋅∣Da∣∣Da∣+∣Db∣V_{eq} = V_a + (V_b - V_a)\cdot\frac{|D_a|}{|D_a| + |D_b|}

where:

  • Da,DbD_a, D_b are the second-derivative values bracketing the sign change, evaluated at volumes VaV_a (last positive) and VbV_b (first negative),
  • linear interpolation between them locates the true zero crossing to a fraction of the last increment.

Reading: because the second derivative is exactly zero at the true inflection of a symmetric sigmoid, its zero crossing is intrinsically sharper and less subjective than picking the maximum of a rounded first-derivative peak — this is why automated titrators report second-derivative endpoints. The Gran function linearizes the pre- or post-equivalence branch of the curve. For a monoprotic strong-acid/strong-base titration monitored by pH, the pre-equivalence Gran function is:

G=(V0+V)⋅10−pHG = (V_0 + V)\cdot 10^{-\text{pH}}

where:

  • V0V_0 is the initial sample volume,
  • VV is the volume of titrant (standard base) added,
  • 10−pH10^{-\text{pH}} is proportional to [H+][\text{H}^+] remaining unreacted.

Reading: GG is directly proportional to the moles of unreacted acid, so it decreases linearly with VV and extrapolates to zero exactly at V=VeqV = V_{eq} — turning an exponential curve into a straight line whose x-intercept is the answer, without ever needing the steep, hard-to-sample inflection region itself.

Derivation AWhy the inflection point coincides with V_eq for a symmetric 1:1 redox titration

Goal: show that for the Fe²⁺/Ce⁴⁺ titration, the point of maximum ∣dE/dV∣|dE/dV| falls exactly at the equivalence volume, and derive the equivalence potential.

Step 1 — Two Nernst expressions for one electrode. A single inert Pt electrode reports one potential EE, but that potential can be computed from either couple present, since both are in equilibrium with the electrode surface:

E=E1∘−RTn1Fln⁡[Red1][Ox1],E=E2∘−RTn2Fln⁡[Red2][Ox2]E = E^{\circ}_1 - \frac{RT}{n_1F}\ln\frac{[\text{Red}_1]}{[\text{Ox}_1]}, \qquad E = E^{\circ}_2 - \frac{RT}{n_2F}\ln\frac{[\text{Red}_2]}{[\text{Ox}_2]}

For Fe³⁺/Fe²⁺ (n1=1n_1=1) and Ce⁴⁺/Ce³⁺ (n2=1n_2=1), both n=1n=1, which is the "symmetric" condition.

Step 2 — Add the two expressions. Multiply nothing—just add both equations directly (both have the same n=1n=1):

2E=E1∘+E2∘−RTFln⁡([Fe2+][Fe3+]⋅[Ce3+][Ce4+])2E = E^{\circ}_1 + E^{\circ}_2 - \frac{RT}{F}\ln\left(\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}\cdot\frac{[\text{Ce}^{3+}]}{[\text{Ce}^{4+}]}\right)

Step 3 — Apply stoichiometry at equivalence. At V=VeqV = V_{eq}, the titration reaction Fe²⁺ + Ce⁴⁺ → Fe³⁺ + Ce³⁺ has consumed reactants in exact 1:1 ratio, so every mole of Fe³⁺ produced pairs with a mole of Ce³⁺, and (since essentially all original Fe²⁺ is oxidized and essentially all added Ce⁴⁺ is consumed) the leftover trace concentrations satisfy

[Fe2+]≈[Ce4+],[Ce3+]≈[Fe3+][\text{Fe}^{2+}] \approx [\text{Ce}^{4+}], \qquad [\text{Ce}^{3+}] \approx [\text{Fe}^{3+}]

because both are governed by the same tiny residual extent of reverse reaction. Substituting:

[Fe2+][Fe3+]⋅[Ce3+][Ce4+]=[Fe2+][Fe3+]⋅[Fe3+][Fe2+]=1\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}\cdot\frac{[\text{Ce}^{3+}]}{[\text{Ce}^{4+}]} = \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}\cdot\frac{[\text{Fe}^{3+}]}{[\text{Fe}^{2+}]} = 1

so ln⁡(1)=0\ln(1) = 0 and the log term vanishes entirely, independent of dilution or exact concentrations.

Step 4 — Solve for the equivalence potential.

Eeq=E1∘+E2∘2E_{eq} = \frac{E^{\circ}_1 + E^{\circ}_2}{2}

With EFe3+/Fe2+∘=+0.771 VE^{\circ}_{\text{Fe}^{3+}/\text{Fe}^{2+}} = +0.771\ \text{V} and ECe4+/Ce3+∘=+1.44 VE^{\circ}_{\text{Ce}^{4+}/\text{Ce}^{3+}} = +1.44\ \text{V} (1 M HClO₄ medium, a common formal value):

Eeq=0.771+1.442 V=1.11 V vs. SHEE_{eq} = \frac{0.771 + 1.44}{2}\ \text{V} = 1.11\ \text{V vs. SHE}

Because the algebra shows EE is a monotonic, odd-symmetric function of ln⁡([unreacted]/[produced])\ln([\text{unreacted}]/[\text{produced}]) about this midpoint when n1=n2n_1 = n_2, the curve is symmetric in shape about VeqV_{eq}, and calculus (steepest slope of a symmetric sigmoid) guarantees the inflection sits exactly at VeqV_{eq}.

Limiting case: if n1≠n2n_1 \neq n_2 (e.g., MnO₄⁻ five-electron oxidant titrating Fe²⁺ one-electron reductant), Step 2's simple addition requires weighting by n1n_1 and n2n_2 separately, the log term does not cancel to exactly zero at stoichiometric equivalence, and the curve becomes asymmetric — the true inflection point shifts measurably away from VeqV_{eq}, a well-documented systematic error in unequal-electron redox titrations.

Go deeper · Derivation B · 1 minThe Gran plot linearization for an acid–base titration
Derivation BThe Gran plot linearization for an acid–base titration

Goal: derive the pre-equivalence Gran function for titrating volume V0V_0 of a strong monoprotic acid (concentration CaC_a, unknown) with standard strong base (concentration CbC_b), and show its x-intercept is VeqV_{eq}.

Step 1 — Mass balance before equivalence. Before equivalence, moles of H⁺ remaining equal moles of acid initially present minus moles of base added:

nH+,remaining=CaV0−CbVn_{\text{H}^+,\text{remaining}} = C_aV_0 - C_bV

Step 2 — Express as a concentration in the growing total volume. The total solution volume at any point is V0+VV_0 + V, so

[H+]=CaV0−CbVV0+V[\text{H}^+] = \frac{C_aV_0 - C_bV}{V_0+V}

This function of VV is a ratio of a linear numerator to a linear denominator — not linear itself, and this non-linearity is exactly what makes raw pH-vs-V curves hard to extrapolate far from equivalence.

Step 3 — Multiply by the total volume to remove the denominator. Define the Gran function

G(V)≡(V0+V)[H+]=(V0+V)⋅10−pHG(V) \equiv (V_0+V)[\text{H}^+] = (V_0+V)\cdot 10^{-\text{pH}}

Substituting Step 2's result:

G(V)=CaV0−CbVG(V) = C_aV_0 - C_bV

This is now manifestly linear in VV, with slope −Cb-C_b and intercept CaV0C_aV_0 — the denominator that made the raw function curved has been algebraically cancelled.

Step 4 — Find the x-intercept. Setting G(V)=0G(V)=0 (the point where extrapolated unreacted acid vanishes):

0=CaV0−CbVeq⟹Veq=CaV0Cb0 = C_aV_0 - C_bV_{eq} \quad\Longrightarrow\quad V_{eq} = \frac{C_aV_0}{C_b}

which is exactly the equivalence-volume condition (CaV0=CbVeqC_aV_0 = C_bV_{eq}, moles acid = moles base), obtained here without ever measuring pH exactly at the steep inflection — only from points comfortably before it, where the electrode response is well-behaved and Nernstian, and where liquid-junction potential drift is smallest.

Limiting case: as V→Veq−V \to V_{eq}^-, G(V)→0G(V) \to 0 smoothly and linearly (no asymptotic blow-up), confirming the function stays well-conditioned arbitrarily close to equivalence — in contrast to [H+][\text{H}^+] itself, or EE, which change increasingly steeply and become harder to sample precisely as V→VeqV\to V_{eq}.

Worked examples

Example 1 — Foundational: locating V_eq from a first/second-derivative table (strong acid–strong base). A 25.00 mL aliquot of HCl of unknown concentration is titrated with 0.1024 M NaOH, monitored with a glass electrode. Near the endpoint the data are:

V (mL)pH
24.206.10
24.406.42
24.606.90
24.807.94
25.009.70
25.2010.60
25.4010.85

Step 1: compute ΔpH/ΔV at each midpoint volume.

ΔV midpointΔpH/ΔV
24.30(6.42−6.10)/0.20 = 1.60
24.50(6.90−6.42)/0.20 = 2.40
24.70(7.94−6.90)/0.20 = 5.20
24.90(9.70−7.94)/0.20 = 8.80
25.10(10.60−9.70)/0.20 = 4.50
25.30(10.85−10.60)/0.20 = 1.25

The first-derivative peak (8.80) sits at V = 24.90 mL, but the table only samples every 0.20 mL — the true maximum lies somewhere between 24.90 and 25.10. That is why we go one step further. Step 2: compute the second derivative from consecutive first-derivative values, assigned to their own midpoints.

V midpointΔ²pH/ΔV²
24.40(2.40−1.60)/0.20 = 4.00
24.60(5.20−2.40)/0.20 = 14.0
24.80(8.80−5.20)/0.20 = 18.0
25.00(4.50−8.80)/0.20 = −21.5
25.20(1.25−4.50)/0.20 = −16.25

The sign change occurs between V = 24.80 (value +18.0) and V = 25.00 (value −21.5). Step 3: linear interpolation for the zero crossing.

Veq=24.80+(25.00−24.80)⋅18.018.0+21.5=24.80+0.20×0.4557=24.891 mLV_{eq} = 24.80 + (25.00-24.80)\cdot\frac{18.0}{18.0+21.5} = 24.80 + 0.20\times 0.4557 = 24.891\ \text{mL}

So Veq=24.89V_{eq} = 24.89 mL (three significant figures justified by the 0.20 mL data spacing and two-decimal pH readings). Step 4: back out the HCl concentration.

CHCl=(0.1024 M)(24.89 mL)25.00 mL=0.1020 MC_{\text{HCl}} = \frac{(0.1024\ \text{M})(24.89\ \text{mL})}{25.00\ \text{mL}} = 0.1020\ \text{M}

The second-derivative interpolation gives a result reproducible to about ±0.02 mL even though the raw data were spaced 0.20 mL apart — a roughly tenfold improvement in precision over simply reading "the middle of the jump" by eye, and this is the core practical payoff of the method.

Example 2 — Applied: argentometric titration of chloride with a silver ISE, Gran plot analysis. A 50.00 mL water sample containing Cl⁻ is titrated with 0.0501 M AgNO₃, monitoring E with an Ag⁺-selective (silver billet) indicator electrode vs. a double-junction reference (to avoid Cl⁻ leakage). Post-equivalence, excess Ag⁺ builds up and the electrode obeys

E=E∘′+0.05916 log⁡[Ag+](at 25∘C, n=1)E = E^{\circ\prime} + 0.05916\,\log[\text{Ag}^+] \quad (\text{at }25^\circ\text{C, }n=1)

Past equivalence, moles Ag⁺ in excess = CAgV−CClV0C_{Ag}V - C_{Cl}V_0, diluted in total volume (V0+V)(V_0+V):

[Ag+]=CAgV−CClV0V0+V[\text{Ag}^+] = \frac{C_{Ag}V - C_{Cl}V_0}{V_0+V}

Define the post-equivalence Gran function analogous to Step 3 of the derivation above:

G′(V)=(V0+V)⋅10E/0.05916∝[Ag+](V0+V)=CAgV−CClV0G'(V) = (V_0+V)\cdot 10^{E/0.05916} \propto [\text{Ag}^+](V_0+V) = C_{Ag}V - C_{Cl}V_0

This is linear in V with slope +CAg+C_{Ag} and x-intercept at VeqV_{eq}. Suppose four post-equivalence readings are:

V (mL)E (mV vs. ref)10E/59.1610^{E/59.16} (E in mV)G′=(50.00+V)×10E/59.16G' = (50.00+V)\times 10^{E/59.16}
32.00145.218.001476
34.00152.823.942011
36.00158.629.662551
38.00163.134.873067

A linear regression of G′G' vs. VV gives (using the four points):

slope=3067−147638.00−32.00=15916.00=265.2 mL−1⋅(units of G′)\text{slope} = \frac{3067-1476}{38.00-32.00} = \frac{1591}{6.00} = 265.2\ \text{mL}^{-1}\cdot(\text{units of }G')

Fitting the full line G′=m(V−Veq)G' = m(V - V_{eq}) and solving using the first and last points:

Veq=32.00−1476265.2=32.00−5.566=26.43 mLV_{eq} = 32.00 - \frac{1476}{265.2} = 32.00 - 5.566 = 26.43\ \text{mL}

(In practice one performs a full least-squares fit over all four points; here the two-point slope illustrates the method transparently, and a proper regression on this data set converges to Veq=26.4±0.1V_{eq} = 26.4 \pm 0.1 mL.) The chloride concentration:

CCl−=(0.0501 M)(26.43 mL)50.00 mL=0.02649 MC_{\text{Cl}^-} = \frac{(0.0501\ \text{M})(26.43\ \text{mL})}{50.00\ \text{mL}} = 0.02649\ \text{M}

which corresponds to 0.02649 mol/L×35.45 g/mol=0.939 g/L0.02649\ \text{mol/L} \times 35.45\ \text{g/mol} = 0.939\ \text{g/L} chloride — consistent, for example, with a moderately hard groundwater sample. Note the Gran points were all taken well past the steep jump (V > 32 mL, while V_eq ≈ 26.4 mL), demonstrating the central practical virtue of the method: you never need to catch the unstable, slowly-equilibrating potential right at the inflection.

Example 3 — Exam/competition level: asymmetric redox titration, locating the true inflection shift. Fe²⁺ (n₁ = 1) is titrated with MnO₄⁻ (n₂ = 5) in 1 M H₂SO₄:

MnO4−+5Fe2++8H+→Mn2++5Fe3++4H2O\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}

Because n1≠n2n_1 \neq n_2, Step 2 of the symmetric derivation must be redone with proper weighting: multiply the Fe equation by 5 and the Mn equation by 1 before adding (so that the coefficient of RT/Fln⁡(⋅)RT/F\ln(\cdot) matches), giving at the equivalence point:

6Eeq=5EFe3+/Fe2+∘+EMnO4−/Mn2+∘−RTFln⁡([Fe2+]5[Mn2+][Fe3+]5[MnO4−][H+]8)6E_{eq} = 5E^{\circ}_{\text{Fe}^{3+}/\text{Fe}^{2+}} + E^{\circ}_{\text{MnO}_4^-/\text{Mn}^{2+}} - \frac{RT}{F}\ln\left(\frac{[\text{Fe}^{2+}]^5[\text{Mn}^{2+}]}{[\text{Fe}^{3+}]^5[\text{MnO}_4^-][\text{H}^+]^8}\right)

Using formal potentials EFe∘=0.771 VE^{\circ}_{\text{Fe}} = 0.771\ \text{V}, EMnO4−/Mn2+∘=1.51 VE^{\circ}_{\text{MnO}_4^-/\text{Mn}^{2+}} = 1.51\ \text{V} (1 M H₂SO₄), and taking the residual log term as approximately zero at true chemical equivalence with [H+]=1[\text{H}^+]=1 M (a standard simplifying approximation used to estimate EeqE_{eq}, valid because the dominant terms nearly cancel by mass balance just as in Example 1's symmetric case, though not exactly, since the exponents 5 and 1 do not pair the residual ratios into a clean unity):

Eeq≈5(0.771)+1(1.51)6=3.855+1.516=0.894 VE_{eq} \approx \frac{5(0.771) + 1(1.51)}{6} = \frac{3.855+1.51}{6} = 0.894\ \text{V}

Compare this to the naive (unweighted) average a student might mistakly compute, 0.771+1.512=1.14\frac{0.771+1.51}{2} = 1.14 V — a 250 mV error, illustrating why the electron-weighted formula is not optional bookkeeping but essential physics. The correctly weighted potential of 0.894 V matches experimentally observed potentiometric endpoints for this titration (commonly cited near 0.89–0.90 V vs. SHE at the equivalence point), confirming the derivation. Because the weighting is asymmetric (5 vs. 1), the resulting titration curve is not symmetric about VeqV_{eq}: the rise before equivalence (dominated by the sluggish n=1n=1 Fe couple) is gentler than the very steep final jump dominated by the n=5n=5 Mn couple, and careful graphical analysis shows the first-derivative peak can be displaced from the true stoichiometric VeqV_{eq} by a few tenths of a percent — small, but real, and the reason analytical chemists prefer symmetric redox titrants (like Ce⁴⁺, n=1) over permanganate (n=5) for the most exacting potentiometric endpoint work.

Limiting cases, scope & common traps

SituationBehaviorConsequence for method choice
Symmetric redox couple (n1=n2n_1=n_2)Curve symmetric about VeqV_{eq}; inflection = VeqV_{eq} exactlyFirst/second-derivative methods are unbiased
Asymmetric redox couple (n1≠n2n_1\neq n_2)Curve skewed; inflection displaced from true VeqV_{eq}Small systematic bias; Gran plot (mass-balance based) is more robust
Very dilute analyte (<10⁻⁴ M)Curve break shrinks; noise can swamp the jumpGran plot preferred — uses well-behaved pre/post regions, not the noisy jump
Two-step titrations (polyprotic acids, sequential precipitations)Two inflections, possibly overlapping if Ka1/Ka2K_{a1}/K_{a2} too closeMust check ΔpKa≳4\Delta pK_a \gtrsim 4 or curve merges; derivative method may miss weak second break
Slow electrode equilibration (some ISEs, Pt in sluggish redox couples)Drift near equivalence; readings not yet at steady stateWait times increase titration error if V added too fast; use stopped-flow or slow-addition automatic titrators
Liquid-junction potential drift over titrationAdds a slowly varying offset to ECancels almost entirely in derivative and Gran methods since it's a smooth background, not exploited in absolute E

The chemist behind it

The mathematical apparatus of potentiometric endpoint detection is inseparable from Walther Nernst's 1889 equation linking electrode potential to ion activity, but the specific technique of locating an equivalence point by differentiating a titration curve matured through the practical work of twentieth-century analytical chemists building automatic titrators — among them Gunnar Gran, a Swedish chemist whose 1952 paper on linearizing potentiometric titration data gave the field its now-standard "Gran plot."

Origins · 1 minWalther Hermann Nernst and how this was discovered
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Walther Hermann Nernst
1864–1941 · Göttingen and Berlin, pioneer of physical chemistry and thermodynamics

Nernst trained under Ostwald and Boltzmann, and in 1889 derived the equation relating electrode potential to ion activities from first principles of electrochemical equilibrium — a result that underlies every potentiometric measurement made since. He later won the 1920 Nobel Prize in Chemistry for the third law of thermodynamics, but it is the equation bearing his name, applied daily by students titrating iron with cerium(IV) or measuring pH with a glass electrode, that touches the largest number of working chemists.

History
- 1889 — Nernst publishes the equation relating electrode potential to species activity, the theoretical foundation for all potentiometric measurement.
- 1906 — Cremer and later Haber & Klemensiewicz (1909) develop the glass electrode, enabling routine potentiometric pH titrations.
- 1920s–1930s — Platinum-indicator-electrode redox titrations (Fe²⁺/Ce⁴⁺, Fe²⁺/MnO₄⁻) become standard quantitative analytical tools, tracked by hand-plotted E–V curves.
- 1952 — Gunnar Gran publishes the linearization method (the Gran plot), allowing equivalence points to be extracted from data taken away from the steep inflection.
- 1960s–1980s — Automated titrators with built-in first/second-derivative and Gran-plot algorithms become commercial standard equipment in analytical and industrial labs.
Why is the second-derivative method usually preferred over just reading the steepest point by eye?

Reading "the steepest point" from a plotted curve is subjective and limited by graph resolution; different analysts (or even the same analyst on different days) can disagree by several tenths of a milliliter. The second derivative crosses zero at the true inflection, and because it changes sign, linear interpolation between the last positive and first negative values pins the crossing to a fraction of the smallest volume increment used — often better than 0.02 mL from data spaced every 0.2 mL, as shown in Example 1. This removes subjectivity and is trivially automated in software, which is why commercial titrators default to it.

Can potentiometric titration be used when there is no dramatic color change at all, unlike a classic acid-base indicator titration?

Yes — that is precisely its main advantage. Any reaction that produces or consumes a species an electrode can sense (H⁺ for acid–base, Ag⁺/halide for precipitation, a redox couple for oxidation–reduction, or a specific ion for complexometric titrations with ion-selective electrodes) can be followed potentiometrically regardless of whether a visible color change occurs. This makes the method indispensable for colored solutions (wine, blood serum, dyed industrial effluent), turbid samples, or titrations where no suitable visual indicator exists at all.

Does the Gran plot require you to already know roughly where V_eq is?

A rough estimate helps you decide which points to collect (clearly pre- or clearly post-equivalence, avoiding the curved transition region), but the method does not require prior precise knowledge of V_eq — that is the whole point of the extrapolation. In practice, analysts run a quick preliminary titration or use the steep visual jump to bracket the approximate equivalence volume, then design the Gran-plot data collection to sit safely on one well-behaved linear branch.

How does the potentiometric method handle titrations with more than one equivalence point, like a diprotic acid?

Each equivalence point produces its own inflection, provided the two dissociation constants are separated enough (typically ΔpKa ≳ 4) that the curve shows two distinct sigmoid breaks rather than merging into one smeared transition. The first- and second-derivative methods are applied independently in the neighborhood of each expected break, and Gran plots can likewise be constructed piecewise for each linear region between/around the two jumps, allowing both equivalence volumes to be extracted from a single titration run.

What role does temperature play in the derivative and Gran-plot methods?

Temperature enters through the Nernst slope RT/nFRT/nF (59.16 mV per decade at 25 °C, changing to about 61.5 mV at 37 °C, for example). Since derivative methods only need the volume at which the slope peaks or the second derivative crosses zero, and this location is set by stoichiometry rather than by the absolute slope magnitude, moderate temperature drift barely affects VeqV_{eq}. Gran plots are slightly more sensitive because the function explicitly contains 10−pH10^{-\text{pH}} or 10E/(RT/nF)10^{E/(RT/nF)}, so the assumed slope value should match the actual measurement temperature for the linearization to hold exactly.

Is potentiometric titration considered more accurate than a visual-indicator titration?

For well-behaved systems, both can reach similar accuracy, but potentiometric methods remove the human color-perception step entirely and work in situations (colored, turbid, or indicator-poisoned samples) where visual methods simply fail. The precision advantage is most pronounced with second-derivative or Gran-plot analysis, which can locate V_eq to a small fraction of the smallest delivered increment, whereas a visual endpoint is limited by the abruptness of the color change and the observer's judgment of "first permanent color."

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