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Lesson 4 · 25 min

Potentiometric titration: end points from potentials

Coloured or cloudy samples defeat indicators, but an electrode works regardless. Find end points from the first and second derivatives of the E–V curve and from Gran plots, and see how an autotitrator doses cleverly.

In this lesson

  • Explain what potentiometric titration offers over indicators and sketch an E–V curve
  • Locate the end point from the maximum of the first derivative and the zero of the second derivative, interpolating linearly
  • Derive the Gran function for a weak-acid titration and extrapolate a straight line to the end point and KaK_\mathrm{a}
  • Choose a sensible dosing strategy for automatic potentiometric titration

Soy sauce is dark brown, red wine is purple and orange juice is full of pulp — in such samples you simply cannot see phenolphthalein change. Yet standard methods for the total acidity of soy sauce or the titratable acidity of wine are still titrations: the indicator has just been replaced by a pH electrode, and the instrument logs the potential after each addition. Potentiometric titration ignores colour and turbidity, resolves several breaks, and can be left to an autotitrator overnight.

Predict first

25.00 mL of 0.1000 M acetic acid is titrated with 0.1000 M NaOH in 0.50 mL steps. The largest ΔE/ΔV falls between 24.50 and 25.00 mL (midpoint 24.75 mL). Where does interpolating the zero of the second derivative put the end point?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

From the E–V curve to the end point

Put a combination pH electrode and the burette tip into the same beaker; after each addition, stir, wait for a stable reading and log (V, E). Potential is linear in pH, so the E–V curve is the titration curve upside down: the potential changes sharply near equivalence, and the end point is where the curve is steepest — the inflection.

There are three classic ways to find it:

  1. first derivative: compute ΔE/ΔV between neighbouring readings, plot it at the midpoint volume, and take the peak;
  2. second derivative: difference the first derivative again; Δ²E/ΔV² changes sign at the end point, and linear interpolation finds the zero;
  3. Gran plot: transform the data before the end point into a straight line and extrapolate it to the axis.

Derivatives need dense, accurate points near the end point, so autotitrators use dynamic dosing: 0.5–1 mL at a time while the potential barely moves, shrinking automatically to 0.01–0.05 mL near the break.

Ve=V1+D1D1−D2 (V2−V1)D=Δ2EΔV2V_\mathrm{e} = V_1 + \frac{D_1}{D_1 - D_2}\,(V_2 - V_1)\qquad D = \frac{\Delta^2 E}{\Delta V^2}
With second-derivative values D₁ and D₂ of opposite sign at V₁ and V₂, linear interpolation gives the end-point volume.
V1,V2V_1, V_2
the two volumes where the second derivative changes sign
D1,D2D_1, D_2
the corresponding second-derivative values

Worked example

An end point from a derivative table

Titrating 25.00 mL of acetic acid with 0.1000 M NaOH gives near the end point: 24.60 mL, +26.9 mV; 24.80 mL, +8.9 mV; 25.00 mL, −102.2 mV; 25.20 mL, −213.0 mV; 25.40 mL, −230.7 mV. Find the end point.

Gran plots: using data far from the end point

Derivatives rely on a handful of points near the end point — exactly where the potential changes fastest, the electrode responds slowest and noise matters most. In the 1950s the Swedish chemist Gran took another route: transform the data before the end point into a straight line.

When strong base titrates a weak acid HA, after VbV_\mathrm{b} mL we have [AX−]∝Vb[\ce{A-}] \propto V_\mathrm{b} and [HA]∝Ve−Vb[\mathrm{HA}] \propto V_\mathrm{e} - V_\mathrm{b}. Substituting into Ka=[HX+][AX−]/[HA]K_\mathrm{a} = [\ce{H+}][\ce{A-}]/[\mathrm{HA}] and rearranging gives the Gran equation: Vb⋅10−pHV_\mathrm{b}\cdot10^{-\mathrm{pH}} is a straight line in VbV_\mathrm{b} that meets the axis at VeV_\mathrm{e}, with slope −Ka-K_\mathrm{a} (times a ratio of activity coefficients). It uses the steady, accurate readings of the buffer region and still gives an end point when the break is small because the acid is very weak or very dilute.

Vb⋅10−pH=γHAγAX− Ka (Ve−Vb)V_\mathrm{b}\cdot 10^{-\mathrm{pH}} = \frac{\gamma_{\mathrm{HA}}}{\gamma_{\ce{A-}}}\,K_\mathrm{a}\,(V_\mathrm{e} - V_\mathrm{b})
The Gran function for a weak acid titrated with strong base. Plot V_b·10⁻ᵖᴴ against V_b and extrapolate the line to zero to find V_e.
VbV_\mathrm{b}
volume of strong base added
VeV_\mathrm{e}
end-point volume
γ\gamma
activity coefficients (their ratio is nearly constant at fixed ionic strength)
A combination pH electrode logs E after each addition · pH from the exact charge balance

Potentiometric titration: end points from the E–V curve

Each addition
Equivalence 05101520253035Titrant volume V / mL−300−200−1000100200E / mV
05101520253035Titrant volume V / mL01.0 × 10⁻⁴2.0 × 10⁻⁴3.0 × 10⁻⁴4.0 × 10⁻⁴V·10⁻ᵖᴴ

Ka from the Gran slope ≈ 1.75 × 10⁻⁵

MethodEnd point / mLMinus equivalence
Largest ΔE/ΔV24.500−0.500
Zero of Δ²E/ΔV²24.990−0.010
Gran line extrapolated25.028+0.028
  • E–V curve
  • Equivalence
The pH after each addition is solved exactly by the acid–base kernel (ideal solution, 25 °C) and the electrode reads E = −59.16 mV·(pH − 7), optionally with deterministic ±0.3 mV noise. “Dynamic” dosing imitates an autotitrator, shrinking the additions where the potential changes fast. Derivatives are difference quotients of neighbouring readings; the zero of the second derivative is found by linear interpolation; the Gran function uses 10⁻ᵖᴴ for [H⁺] (constant activity coefficients) and is fitted by least squares to the readings between 20 % and 90 % of the way to the end point.
With 1 mL steps the first-derivative end point is only good to ±0.5 mL, but the Gran line extrapolates almost exactly to 25.00 mL and its slope gives Ka ≈ 1.75 × 10⁻⁵. Then switch to phosphoric acid: both breaks can be found by derivatives; try turning the noise off to see the difference.

Concept check

In a Gran plot of Vb⋅10−pHV_\mathrm{b}\cdot10^{-\mathrm{pH}} against VbV_\mathrm{b} for a weak acid titrated with strong base, what do the x-intercept and slope give?

Try the calculation

In the same acetic acid titration the Gran function Vb⋅10−pHV_\mathrm{b}\cdot10^{-\mathrm{pH}} is 2.627×10−42.627\times10^{-4} after 10.00 mL of NaOH and 8.765×10−58.765\times10^{-5} after 20.00 mL (units mL·M). Extrapolating the line through these points, what is the end-point volume?

mL

Key takeaways

  • Potentiometric titration replaces the indicator with an electrode, ignores colour and turbidity, and resolves several breaks.
  • The first-derivative peak is good only to half a step; interpolating the second-derivative zero is far finer.
  • The weak-acid Gran function Vb⋅10−pHV_\mathrm{b}\cdot10^{-\mathrm{pH}} is linear in VbV_\mathrm{b}: the intercept is VeV_\mathrm{e} and the slope −Ka-K_\mathrm{a}.
  • Dynamic dosing shrinks the steps near the break and gets a better end point from fewer readings.
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Data and references

  1. D. C. Harris, Quantitative Chemical Analysis, 9th ed. (2016), Ch. 11 (finding the end point with a pH electrode; Gran plots).
  2. G. Gran, “Determination of the equivalence point in potentiometric titrations. Part II”, Analyst 77, 661–671 (1952).
  3. D. A. Skoog, D. M. West, F. J. Holler, S. R. Crouch, Fundamentals of Analytical Chemistry, 9th ed. (2014), Ch. 21 (potentiometric titrations).
  4. J. Mendham, R. C. Denney, J. D. Barnes, M. Thomas, Vogel’s Textbook of Quantitative Chemical Analysis, 6th ed. (Prentice Hall, 2000), Ch. 15.