---
title: "Titration Methods and Curves"
book: "University Chemistry — Year 1"
subject: chemistry
language: en
chapter: 15
exercises: 12
source: https://one-course.com/books/chemistry/2/en/chapter/15-titration-methods-and-curves
license: CC-BY-NC-SA-4.0
credit: "One Chemistry Book, One Course (one-course.com)"
---

# Chapter 15 — Titration Methods and Curves

A pharmacist receives a batch of vitamin C tablets labelled $500\,\mathrm{mg}$ and must certify, before they are sold, that each holds what the label says. She will not weigh the vitamin directly: it is mixed with binders and fillers. She will titrate it — let it react with a reagent of known concentration, measure the volume needed, and compute the amount. Which reaction, how to see its end, and how sure the result is: these are the questions of this chapter, which brings the equilibria of the last five chapters to the laboratory bench.

**You already know.**

The school volume (grades 11 and 12) titrated acids by bases with a colour change, followed [titrations](#def-b1-titration-methods-titration) with a pH meter and a conductivity meter, and used the [equivalence](#def-b1-titration-methods-titration) relation $C_AV_A = C_BV_B$. The predominant-reaction method is in [Chapter 10](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#ch-b1-predominant-reaction); precipitation, complexation and redox equilibria in Chapters [11](https://one-course.com/books/chemistry/2/en/chapter/11-precipitation-and-solubility#ch-b1-precipitation), [12](https://one-course.com/books/chemistry/2/en/chapter/12-complexation#ch-b1-complexation) and [13](https://one-course.com/books/chemistry/2/en/chapter/13-redox-equilibria-and-the-nernst-equation#ch-b1-nernst).

![A teaching laboratory: a burette, a conical flask on a magnetic stirrer, gloves and goggles. A titration is the most common quantitative measurement in chemistry.](https://one-course.com/images/onecourse/chapters/chemistry-2/b1-titration-methods/img-1a8d5bb3fec2.jpg)

*A teaching laboratory: a burette, a conical flask on a magnetic stirrer, gloves and goggles. A [titration](#def-b1-titration-methods-titration) is the most common quantitative measurement in chemistry.*

## 15.1 Titration and equivalence

**Definition 15.1 (Titration, equivalence).**

A *titration* determines the amount of a species, the *analyte*, by a reaction with a reagent of known concentration, the *titrant*, added progressively. The reaction must be single, [quantitative](https://one-course.com/books/chemistry/2/en/chapter/7-describing-a-chemical-system-extent-activities-q-and-k#def-b1-extent-q-and-k-quantitative) ($K$ large) and fast. The *equivalence* is reached when titrant and analyte have been brought together in the stoichiometric proportions of the reaction; the volume added then is the equivalence volume, and the point of the titration curve at that volume is the *equivalence point*. The volume at which the experimenter stops, seeing a colour change or a break in a curve, is the *end point*; a good method makes it coincide with the equivalence.

**Proposition 15.2 (Equivalence relation).**

For a [titration](#def-b1-titration-methods-titration) reaction $a\,\mathrm{A} + b\,\mathrm{B} \to \cdots$, with [analyte](#def-b1-titration-methods-titration) A ($n_A$ mol) and [titrant](#def-b1-titration-methods-titration) B at concentration $C_B$, the [equivalence](#def-b1-titration-methods-titration) volume satisfies

$$
\frac{n_A}{a} = \frac{C_B V_{\mathrm{eq}}}{b} .
$$

**Proof.** Let $x$ be the extent after adding $V$ of [titrant](#def-b1-titration-methods-titration). Before [equivalence](#def-b1-titration-methods-titration) B is limiting and, the reaction being [quantitative](https://one-course.com/books/chemistry/2/en/chapter/7-describing-a-chemical-system-extent-activities-q-and-k#def-b1-extent-q-and-k-quantitative), $x = C_BV/b$; after it, A is limiting and $x = n_A/a$. The [equivalence](#def-b1-titration-methods-titration) is the volume where both are used up together, $n_A - ax = 0$ and $C_BV - bx = 0$: eliminating $x$ gives the relation. ∎

## 15.2 Direct and indirect titrations

**Definition 15.3 (Kinds of titration).**

In a *direct titration* the [titrant](#def-b1-titration-methods-titration) reacts with the [analyte](#def-b1-titration-methods-titration) itself. In a *back titration* a known excess of a reagent is added to the [analyte](#def-b1-titration-methods-titration), and the excess left is titrated. When a solution contains several [analytes](#def-b1-titration-methods-titration) that react in turn with the [titrant](#def-b1-titration-methods-titration), they are measured by *successive titrations*, one [equivalence](#def-b1-titration-methods-titration) after another; when they react together and give a single [equivalence](#def-b1-titration-methods-titration), by a *simultaneous titration*, which gives only their sum.

**Method 15.4 (Back titration).**

Use it when the direct reaction is slow, when the [analyte](#def-b1-titration-methods-titration) is unstable or volatile, or when no indicator fits the direct reaction.

1. Add a known amount $n_R$ of reagent R, in excess, and let it react completely with the [analyte](#def-b1-titration-methods-titration) A ( $\alpha\,\mathrm{A} + \rho\,\mathrm{R}$ ).
2. Titrate the excess R with a [titrant](#def-b1-titration-methods-titration) T ( $\rho'\,\mathrm{R} +  \tau\,\mathrm{T}$ ) to the [equivalence](#def-b1-titration-methods-titration) volume $V_T$ .
3. Compute $n_{R,\text{excess}} = \frac{\rho'}{\tau}C_TV_T$ , then $n_A = \frac{\alpha}{\rho}(n_R - n_{R,\text{excess}})$ .

The weekend problem titrates vitamin C this way: ascorbic acid reduces an excess of iodine, and the iodine left is titrated by thiosulfate.

## 15.3 pH-metric titrations and indicators

![A pH-metric titration. The combined glass electrode measures the pH after each addition from the burette; the stirrer mixes the solution. The electrode is calibrated beforehand with two buffer solutions.](https://one-course.com/images/onecourse/chapters/chemistry-2/b1-titration-methods/fig-e18af2ca437c.svg)

*A pH-metric [titration](#def-b1-titration-methods-titration). The combined glass electrode measures the pH after each addition from the burette; the stirrer mixes the solution. The electrode is calibrated beforehand with two [buffer solutions](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#def-b1-predominant-reaction-buffer).*

![Left: 10.0\, mL of hydrochloric acid and of ethanoic acid, both at 0.100\, mol/ L, titrated by sodium hydroxide at 0.100\, mol/ L. The weak acid has a buffer region with pH = pK_a at half-equivalence and an equivalence point at pH 8.73. Right: a mixture, 0.050\, mol/ L of each acid: the strong acid is titrated first (equivalence at 5\, mL), then the weak one (10\, mL).](https://one-course.com/images/onecourse/chapters/chemistry-2/b1-titration-methods/fig-2f9207dde1ba.svg)

*Left: $10.0\,\mathrm{mL}$ of hydrochloric acid and of ethanoic acid, both at $0.100\,\mathrm{mol}/\mathrm{L}$, titrated by sodium hydroxide at $0.100\,\mathrm{mol}/\mathrm{L}$. The [weak acid](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#def-b1-predominant-reaction-strong-acid) has a buffer region with pH $=
\mathrm{p}K_a$ at [half-equivalence](#def-b1-titration-methods-titration) and an [equivalence point](#def-b1-titration-methods-titration) at pH 8.73. Right: a mixture, $0.050\,\mathrm{mol}/\mathrm{L}$ of each acid: the [strong acid](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#def-b1-predominant-reaction-strong-acid) is titrated first ([equivalence](#def-b1-titration-methods-titration) at $5\,\mathrm{mL}$), then the weak one ($10\,\mathrm{mL}$).*

**Proposition 15.5 (Half-equivalence of a weak acid).**

When a [weak acid](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#def-b1-predominant-reaction-strong-acid) HA is titrated by a [strong base](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#def-b1-predominant-reaction-strong-acid), at [half-equivalence](#def-b1-titration-methods-titration) $\mathrm{pH} = \mathrm{p}K_a$, provided that the acid is not too strong nor too dilute ($\mathrm{p}K_a$ roughly between 4 and 10 at $0.1\,\mathrm{mol}/\mathrm{L}$).

**Proof.** At [half-equivalence](#def-b1-titration-methods-titration) the [quantitative reaction](https://one-course.com/books/chemistry/2/en/chapter/7-describing-a-chemical-system-extent-activities-q-and-k#def-b1-extent-q-and-k-quantitative) $\ce{HA + OH- -> A- + H2O}$ has converted half of HA: $[\mathrm{HA}] = [\mathrm{A^-}]$, and the Henderson relation gives $\mathrm{pH} = \mathrm{p}K_a$. The condition ensures that the reaction of HA with water and that of $\mathrm{A^-}$ with water do not change those amounts appreciably. ∎

**Proposition 15.6 (pH at the equivalence of a weak acid).**

At the [equivalence](#def-b1-titration-methods-titration) of a [weak acid](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#def-b1-predominant-reaction-strong-acid) ($\mathrm{p}K_a$) by a [strong base](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#def-b1-predominant-reaction-strong-acid), the solution is the conjugate base at concentration $C'$ (after dilution), and

$$
\mathrm{pH} = \tfrac12(\mathrm{p}K_e + \mathrm{p}K_a + \log C') .
$$

**Proof.** The [predominant reaction](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#def-b1-predominant-reaction-predominant-reaction) of the base with water, $\ce{A- + H2O <=> HA + OH-}$, has $K = K_e/K_a$; with little base converted, $[\ce{OH-}] = \sqrt{KC'}$ ([Proposition 10.12](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#prop-b1-predominant-reaction-ph-formulas)), whence the result. For ethanoic acid, $C' = 0.050\,\mathrm{mol}/\mathrm{L}$: $\frac12(14.00 + 4.76 - 1.30) = 8.73$. ∎

**Proposition 15.7 (Successive titrations).**

Two acids titrated by the same base give separate [equivalences](#def-b1-titration-methods-titration), each to better than $1\,\%$, if their $\mathrm{p}K_a$ differ by at least 4.

**Proof.** At the first [equivalence](#def-b1-titration-methods-titration), the reaction of the base $\mathrm{A_1^-}$ with the second acid $\mathrm{HA_2}$, $\mathrm{A_1^-} + \mathrm{HA_2}
\rightleftharpoons \mathrm{HA_1} + \mathrm{A_2^-}$, has $K = 10^{\mathrm{p}K_1 - \mathrm{p}K_2}$. Starting from equal amounts, the fraction $x$ of the second acid already titrated satisfies $x^2/(1 - x)^2 = K$; for $x \leqslant 0.01$, $K
\leqslant 10^{-4}$, that is $\mathrm{p}K_2 - \mathrm{p}K_1 \geqslant 4$. ∎

Phosphoric acid (2.15, 7.21, 12.34) gives two clear [equivalences](#def-b1-titration-methods-titration); the third is lost in the water of the base. Ethanoic acid mixed with hydrochloric acid (right of the figure) is titrated after it, the [strong acid](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#def-b1-predominant-reaction-strong-acid) being in effect a $\mathrm{p}K_a$ below 0.

**Method 15.8 (Locating an equivalence on a curve).**

1. Derivative: compute $\Delta\mathrm{pH}/\Delta V$ between successive points; the [equivalence](#def-b1-titration-methods-titration) is at its maximum.
2. Tangents: draw two parallel tangents to the curve on each side of the jump, and the parallel equidistant from them; it cuts the curve at the [equivalence point](#def-b1-titration-methods-titration) .

**Method 15.9 (Tangents, by computation).**

For a curve recorded by computer, fit the points of each side of the jump with a smooth function and take the inflection point; for a symmetric jump ([strong acid](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#def-b1-predominant-reaction-strong-acid), [strong base](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#def-b1-predominant-reaction-strong-acid)) the midpoint of the jump at pH 7 is the [equivalence point](#def-b1-titration-methods-titration).

**Definition 15.10 (Colour indicator, turning zone).**

A *colour indicator* is a weak acid–base pair HIn/In$^-$ whose two forms have different colours. Its colour changes over the *turning zone*, the pH range where neither form clearly dominates, about $\mathrm{p}K_{a}(\mathrm{HIn}) \pm 1$.

| indicator | $\mathrm{p}K_a$ | [turning zone](#def-b1-titration-methods-indicator) | colours (acid / base) |
| --- | --- | --- | --- |
| methyl orange | 3.5 | 2.5–4.5 | red / yellow |
| methyl red | 4.8 | 3.8–5.8 | red / yellow |
| phenol red | 7.9 | 6.9–8.9 | yellow / red |
| thymol blue (second change) | 9.2 | 8.2–10.2 | yellow / blue |

*Four indicators, $\mathrm{p}K_a$ from a critical compilation of [dissociation constants](https://one-course.com/books/chemistry/2/en/chapter/12-complexation#def-b1-complexation-formation-constant); the [turning zones](#def-b1-titration-methods-indicator) are $\mathrm{p}K_a \pm 1$.*

**Method 15.11 (Choosing an indicator).**

Choose an indicator whose [turning zone](#def-b1-titration-methods-indicator) lies inside the vertical jump of the curve, as close as possible to the pH at [equivalence](#def-b1-titration-methods-titration). [Strong acid](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#def-b1-predominant-reaction-strong-acid) by [strong base](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#def-b1-predominant-reaction-strong-acid) (jump 3.3 to 10.7 for $\pm0.1\,\mathrm{mL}$ here): any of the four. Ethanoic acid ([equivalence](#def-b1-titration-methods-titration) 8.73): phenol red or thymol blue; methyl orange would change colour long before the [equivalence](#def-b1-titration-methods-titration).

## 15.4 Potentiometric titrations

**Definition 15.12 (Potentiometric titration).**

A *potentiometric titration* follows the potential of an electrode in the solution against a [reference electrode](https://one-course.com/books/chemistry/2/en/chapter/13-redox-equilibria-and-the-nernst-equation#def-b1-nernst-reference-electrode), for a redox (platinum wire), precipitation (silver wire) or complexation [titration](#def-b1-titration-methods-titration).

**Proposition 15.13 (Remarkable points of a redox titration).**

For the [titration](#def-b1-titration-methods-titration) of $\mathrm{Red_1}$ by $\mathrm{Ox_2}$ with $n_2\,\mathrm{Red_1} + n_1\,\mathrm{Ox_2} \to n_2\,\mathrm{Ox_1} +
n_1\,\mathrm{Red_2}$ (couples of $n_1$ and $n_2$ electrons), $E(V_{\mathrm{eq}}/2) = E^\circ_1$, $E(2V_{\mathrm{eq}}) =
E^\circ_2$, and at [equivalence](#def-b1-titration-methods-titration)

$$
E_{\mathrm{eq}} = \frac{n_1E^\circ_1 + n_2E^\circ_2}{n_1 + n_2},
$$

for couples whose [half-equations](https://one-course.com/books/chemistry/2/en/chapter/13-redox-equilibria-and-the-nernst-equation#def-b1-nernst-couple) involve no other species (or at a pH that keeps the extra terms fixed).

**Proof.** At [half-equivalence](#def-b1-titration-methods-titration) half of $\mathrm{Red_1}$ is oxidised: $[\mathrm{Ox_1}] = [\mathrm{Red_1}]$ and the [Nernst equation](https://one-course.com/books/chemistry/2/en/chapter/13-redox-equilibria-and-the-nernst-equation#thm-b1-nernst-nernst) of couple 1 gives $E^\circ_1$. At $2V_{\mathrm{eq}}$ as much $\mathrm{Ox_2}$ is in excess as was reduced: $[\mathrm{Ox_2}] = [\mathrm{Red_2}]$, $E = E^\circ_2$. At [equivalence](#def-b1-titration-methods-titration), multiply the [Nernst equation](https://one-course.com/books/chemistry/2/en/chapter/13-redox-equilibria-and-the-nernst-equation#thm-b1-nernst-nernst) of couple 1 by $n_1$ and that of couple 2 by $n_2$ and add; the stoichiometry makes $[\mathrm{Ox_1}]/[\mathrm{Red_1}] \cdot [\mathrm{Ox_2}]/[\mathrm{Red_2}] = 1$ (the amounts of $\mathrm{Red_1}$ and $\mathrm{Ox_2}$ left are in the same ratio as the products), so the logarithms cancel. ∎

For iron(II) by permanganate in $1\,\mathrm{mol}/\mathrm{L}$ acid, $n_1 = 1$ and $n_2 =
5$: $E_{\mathrm{eq}} = (0.77 + 5 \times 1.51)/6 = 1.39$ V. The jump, from about 0.8 to 1.5 V, is so large that the [end point](#def-b1-titration-methods-titration) is sharp; in practice permanganate is its own indicator, its violet colour persisting at the first drop in excess.

![Left: potentiometric titration of 10.0\, mL of iron(II) at 0.100\, mol/ L by permanganate at 0.0200\, mol/ L in 1\, mol/ L acid (platinum electrode, potential against the hydrogen electrode). Right: conductimetric titration of 100\, mL of hydrochloric acid at 0.0100\, mol/ L by sodium hydroxide at 0.100\, mol/ L; the equivalence is at the intersection of two straight lines.](https://one-course.com/images/onecourse/chapters/chemistry-2/b1-titration-methods/fig-dfb910190bbe.svg)

*Left: [potentiometric titration](#def-b1-titration-methods-potentiometric) of $10.0\,\mathrm{mL}$ of iron(II) at $0.100\,\mathrm{mol}/\mathrm{L}$ by permanganate at $0.0200\,\mathrm{mol}/\mathrm{L}$ in $1\,\mathrm{mol}/\mathrm{L}$ acid (platinum electrode, potential against the hydrogen electrode). Right: conductimetric [titration](#def-b1-titration-methods-titration) of $100\,\mathrm{mL}$ of hydrochloric acid at $0.0100\,\mathrm{mol}/\mathrm{L}$ by sodium hydroxide at $0.100\,\mathrm{mol}/\mathrm{L}$; the [equivalence](#def-b1-titration-methods-titration) is at the intersection of two straight lines.*

## 15.5 Conductimetric, complexometric and precipitation titrations

**Definition 15.14 (Molar ionic conductivity).**

The conductivity $\kappa$ of a solution (in $\mathrm{S}/\mathrm{m}$) measures how well it carries current through the motion of its ions. The *molar ionic conductivity* $\lambda_i$ of an ion is its contribution per mole per unit volume; its value at infinite dilution, $\lambda_i^\circ$, is a constant of the ion and the solvent at a given temperature.

**Proposition 15.15 (Kohlrausch’s law).**

In dilute solution, the conductivity is the sum of the contributions of the ions,

$$
\kappa = \sum_i \lambda_i^\circ c_i ,
$$

*Kohlrausch’s law*; the molar conductivity of a salt is the sum of those of its ions.

**Proof.** *Admitted at this level.* ∎

[Kohlrausch’s law](#prop-b1-titration-methods-kohlrausch) expresses that dilute ions move independently; its limits at higher concentration are studied in the Year 2 volume. We use it here as an experimental law.

**Proposition 15.16 (Slopes of a conductimetric titration).**

When a [strong acid](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#def-b1-predominant-reaction-strong-acid) H$^+$X$^-$ is titrated by a [strong base](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#def-b1-predominant-reaction-strong-acid) Na$^+$OH$^-$ in a large volume (dilution neglected), the conductivity varies linearly with the volume added, with slope proportional to $\lambda^\circ(\ce{Na+}) - \lambda^\circ(\ce{H+}) < 0$ before [equivalence](#def-b1-titration-methods-titration) and to $\lambda^\circ(\ce{Na+}) + \lambda^\circ(\ce{OH-}) > 0$ after.

**Proof.** Before [equivalence](#def-b1-titration-methods-titration) each mole of hydroxide added removes one mole of $\ce{H+}$ ($\ce{H+ + OH- -> H2O}$) and brings one mole of $\ce{Na+}$; after [equivalence](#def-b1-titration-methods-titration) it adds one $\ce{Na+}$ and one $\ce{OH-}$. $\ce{X-}$ is unchanged. By [Kohlrausch’s law](#prop-b1-titration-methods-kohlrausch) the conductivity changes by those combinations of $\lambda^\circ$ times the amount added divided by the (fixed) volume. ∎

From the limiting conductivities, $\lambda^\circ(\ce{H+}) =
349.8\,\mathrm{S}\,\mathrm{cm}^{2}/\mathrm{mol}$ and $\lambda^\circ(\ce{OH-}) = 198.3$, and those of the salts $\ce{HCl}$ (426) and $\ce{NaCl}$ (126.5), [Kohlrausch’s law](#prop-b1-titration-methods-kohlrausch) gives $\lambda^\circ(\ce{Cl-}) = 426 - 349.8 = 76.2$ and $\lambda^\circ(\ce{Na+}) =
126.5 - 76.2 = 50.3$ $\mathrm{S}\,\mathrm{cm}^{2}/\mathrm{mol}$. The hydrogen and hydroxide ions conduct far better than the others: the curve is a sharp V.

Complexometric [titrations](#def-b1-titration-methods-titration) use edta: at pH 10 (ammonia buffer) calcium and magnesium react with it [quantitatively](https://one-course.com/books/chemistry/2/en/chapter/7-describing-a-chemical-system-extent-activities-q-and-k#def-b1-extent-q-and-k-quantitative) (conditional constants of [Chapter 12](https://one-course.com/books/chemistry/2/en/chapter/12-complexation#ch-b1-complexation)), and an indicator that is itself a weak complexing agent of the metal changes colour when edta takes the last metal ions from it — the [titration](#def-b1-titration-methods-titration) of water hardness. Precipitation [titrations](#def-b1-titration-methods-titration) use silver ions: chloride is titrated by silver nitrate with chromate as the indicator ([Exercise 11.8](https://one-course.com/books/chemistry/2/en/chapter/11-precipitation-and-solubility#exo-b1-precipitation-8)), or followed with a silver electrode ([Chapter 13](https://one-course.com/books/chemistry/2/en/chapter/13-redox-equilibria-and-the-nernst-equation#ch-b1-nernst)). The precision of all these results, set by the glassware and the [end point](#def-b1-titration-methods-titration), is treated in [Chapter 29](https://one-course.com/books/chemistry/2/en/chapter/29-lab-techniques-i-safety-measurement-and-separation#ch-b1-lab-techniques-1).

## 15.6 Exercises

**Exercise 15.1 ★.**

$20.0\,\mathrm{mL}$ of a sodium hydroxide solution need $12.4\,\mathrm{mL}$ of hydrochloric acid at $0.100\,\mathrm{mol}/\mathrm{L}$. Write the reaction, its constant, and compute the concentration of the base.

**Solution of Exercise 15.1.**

$\ce{H3O+ + OH- -> 2H2O}$, $K = 1/K_e = 10^{14.00}$. $C = 12.4 \times
0.100/20.0 = 0.0620\,\mathrm{mol}/\mathrm{L}$.

**Exercise 15.2 ★.**

Which indicator of the table suits the [titration](#def-b1-titration-methods-titration) of ammonia ($0.10\,\mathrm{mol}/\mathrm{L}$, $\mathrm{p}K_a = 9.25$) by hydrochloric acid at $0.10\,\mathrm{mol}/\mathrm{L}$? Compute the pH at [equivalence](#def-b1-titration-methods-titration) first.

**Solution of Exercise 15.2.**

At [equivalence](#def-b1-titration-methods-titration) the solution is ammonium chloride at $0.050\,\mathrm{mol}/\mathrm{L}$: $\mathrm{pH} = \frac12(9.25 - \log 0.050) = 5.28$. Methyl red (3.8–5.8) contains it; methyl orange, whose zone (2.5–4.5) is reached only after [equivalence](#def-b1-titration-methods-titration) as the pH falls, would change too late.

**Exercise 15.3 ★.**

$10.0\,\mathrm{mL}$ of iron(II) are titrated by permanganate at $0.0200\,\mathrm{mol}/\mathrm{L}$; the violet colour persists from $12.6\,\mathrm{mL}$. Write the reaction and compute the concentration of iron(II).

**Solution of Exercise 15.3.**

$\ce{MnO4- + 5Fe^2+ + 8H+ -> Mn^2+ + 5Fe^3+ + 4H2O}$; $C = 5 \times 0.0200 \times 12.6/10.0 = 0.126\,\mathrm{mol}/\mathrm{L}$.

**Exercise 15.4 ★.**

A water sample of $50.0\,\mathrm{mL}$ is titrated by edta at $0.0100\,\mathrm{mol}/\mathrm{L}$ at pH 10; the [equivalence](#def-b1-titration-methods-titration) is at $14.2\,\mathrm{mL}$. Edta reacts one to one with calcium and magnesium. Compute the total concentration of these two ions. Is this a simultaneous or a successive [titration](#def-b1-titration-methods-titration)?

**Solution of Exercise 15.4.**

$C = 0.0100 \times 14.2/50.0 = 2.84 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$ of calcium plus magnesium. Both react before the [end point](#def-b1-titration-methods-titration), which gives only their sum: a [simultaneous titration](#def-b1-titration-methods-kinds).

**Exercise 15.5 ★★.**

For the [titration](#def-b1-titration-methods-titration) of $10.0\,\mathrm{mL}$ of ethanoic acid at $0.100\,\mathrm{mol}/\mathrm{L}$ by sodium hydroxide at $0.100\,\mathrm{mol}/\mathrm{L}$, compute the pH at 0, 5.0, 10.0 and $15.0\,\mathrm{mL}$ with the methods of [Chapter 10](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#ch-b1-predominant-reaction), and compare with the curve of this chapter.

**Solution of Exercise 15.5.**

0 mL: [weak acid](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#def-b1-predominant-reaction-strong-acid), $\mathrm{pH} = \frac12(4.76 + 1.00) = 2.88$. 5.0 mL: [half-equivalence](#def-b1-titration-methods-titration), 4.76. 10.0 mL: 8.73 ([Proposition 15.6](#prop-b1-titration-methods-ph-at-equivalence)). 15.0 mL: excess hydroxide $0.50\,\mathrm{mmol}$ in $25.0\,\mathrm{mL}$, $[\ce{OH-}] = 0.020$, $\mathrm{pH} = 12.30$. All agree with the computed curve to 0.01.

**Exercise 15.6 ★★.**

Show that a pH-metric [titration](#def-b1-titration-methods-titration) of phosphoric acid by sodium hydroxide gives two jumps, and compute the pH at each [equivalence](#def-b1-titration-methods-titration) for an acid at $0.100\,\mathrm{mol}/\mathrm{L}$ ($10.0\,\mathrm{mL}$, base $0.100\,\mathrm{mol}/\mathrm{L}$). Why is there no third jump?

**Solution of Exercise 15.6.**

The $\mathrm{p}K_a$ 2.15 and 7.21 differ by more than 4, as do 7.21 and 12.34: two [equivalences](#def-b1-titration-methods-titration), at 10.0 and $20.0\,\mathrm{mL}$. First: solution of $\ce{H2PO4-}$, $\mathrm{pH} \approx \frac12(2.15 + 7.21) = 4.68$; second: $\ce{HPO4^2-}$, $\frac12(7.21 + 12.34) = 9.78$ (the exact curve gives 4.71 and 9.66, the formulas neglecting dilution and water). The third acidity ($\mathrm{p}K_a$ 12.34) is too close to that of water: the hydroxide added reacts only partly and the pH rises without a jump.

**Exercise 15.7 ★★.**

A mixture of hydrochloric acid and ethanoic acid ($10.0\,\mathrm{mL}$) is titrated by sodium hydroxide at $0.100\,\mathrm{mol}/\mathrm{L}$: two [equivalences](#def-b1-titration-methods-titration), at $6.2\,\mathrm{mL}$ and $14.6\,\mathrm{mL}$. Compute the two concentrations. What is the pH halfway between the two [equivalences](#def-b1-titration-methods-titration)?

**Solution of Exercise 15.7.**

Hydrochloric acid: $6.2 \times 0.100/10.0 = 0.062\,\mathrm{mol}/\mathrm{L}$; ethanoic acid: $(14.6 - 6.2) \times 0.100/10.0 = 0.084\,\mathrm{mol}/\mathrm{L}$. Halfway, half of the ethanoic acid is titrated: $\mathrm{pH} = \mathrm{p}K_a = 4.76$.

**Exercise 15.8 ★★.**

Compute the potential of the platinum electrode in the [titration](#def-b1-titration-methods-titration) of iron(II) by permanganate (data of the chapter) at 2.0, 9.0, 11.0 and $20.0\,\mathrm{mL}$.

**Solution of Exercise 15.8.**

$V_{\mathrm{eq}} = 10.0\,\mathrm{mL}$. 2.0 mL: $E = 0.77 + 0.059\log(2/8) =
0.73$ V; 9.0 mL: $0.77 + 0.059\log 9 = 0.83$ V; 11.0 mL: excess permanganate $0.020\,\mathrm{mmol}$, $\ce{Mn^2+}$ $0.200\,\mathrm{mmol}$, $E = 1.51 +
\frac{0.059}{5}\log 0.10 = 1.50$ V; 20.0 mL: 1.51 V.

**Exercise 15.9 ★★.**

Using [Kohlrausch’s law](#prop-b1-titration-methods-kohlrausch) and the values of the chapter, compute the conductivity of the titrated hydrochloric acid at 0, 10.0 and $20.0\,\mathrm{mL}$ of base ($100\,\mathrm{mL}$ of acid at $0.0100\,\mathrm{mol}/\mathrm{L}$, base $0.100\,\mathrm{mol}/\mathrm{L}$, dilution taken into account), and the two slopes.

**Solution of Exercise 15.9.**

0 mL: $\kappa = (349.8 + 76.2) \times 0.0100 = 4.26\,\mathrm{mS}/\mathrm{cm}$. 10.0 mL: $\ce{Na+}$ and $\ce{Cl-}$ at $1.00/110 = 9.09 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$, $\kappa = (50.3 + 76.2) \times 9.09 \times 10^{-3} = 1.15$. 20.0 mL: $\ce{Na+}$ $0.0167$, $\ce{Cl-}$ and $\ce{OH-}$ $0.00833$ mol/L, $\kappa = 0.84 + 0.635 + 1.65
= 3.13$ mS/cm. Slopes (dilution neglected, $1\,\mathrm{mL}$ adds $1.0 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$): $(50.3 - 349.8) \times 10^{-3} = -0.30$ and $(50.3 + 198.3) \times 10^{-3} = +0.25$ mS/cm per mL.

**Exercise 15.10 ★★★.**

Prove that the pH-metric curve of a [strong acid](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#def-b1-predominant-reaction-strong-acid) titrated by a [strong base](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#def-b1-predominant-reaction-strong-acid) is symmetric about the [equivalence point](#def-b1-titration-methods-titration) when dilution is neglected, and compute the size of the jump between $V_{\mathrm{eq}} \pm 0.1$ mL for the [titration](#def-b1-titration-methods-titration) of the chapter. How does it change if all concentrations are divided by 100?

**Solution of Exercise 15.10.**

At $V_{\mathrm{eq}} - v$ the excess acid is $C_Bv$, at $V_{\mathrm{eq}} + v$ the excess base is $C_Bv$; in the same volume, $[\ce{H3O+}]$ before equals $[\ce{OH-}]$ after, so $\mathrm{pH}(V_{\mathrm{eq}} - v) +
\mathrm{pH}(V_{\mathrm{eq}} + v) = 14$: symmetry about (V$_{\mathrm{eq}}$, 7). Jump: $1.0 \times 10^{-5}\,\mathrm{mol}$ excess in $20\,\mathrm{mL}$, pH 3.30 to 10.70, 7.4 units. Divided by 100: $1.0 \times 10^{-7}$ mol in $20\,\mathrm{mL}$, pH 5.3 to 8.7, 3.4 units: a much less sharp [end point](#def-b1-titration-methods-titration).

**Exercise 15.11 ★★★.**

Derive the equation of the potentiometric curve of iron(II) by permanganate before and after the [equivalence](#def-b1-titration-methods-titration), and check the three remarkable points of [Proposition 15.13](#prop-b1-titration-methods-potentiometric-points). How would the [equivalence](#def-b1-titration-methods-titration) potential change at pH 1?

**Solution of Exercise 15.11.**

Before: $\ce{Fe^3+}$ formed $= 5C_{\mathrm{Mn}}V$, $\ce{Fe^2+}$ left $=
5C_{\mathrm{Mn}}(V_{\mathrm{eq}} - V)$, so $E = 0.77 + 0.059\log\frac{V}{V_{\mathrm{eq}}
- V}$: at $V_{\mathrm{eq}}/2$, $E = 0.77$. After: excess $\ce{MnO4-}$ $\propto
V - V_{\mathrm{eq}}$, $\ce{Mn^2+}$ $\propto V_{\mathrm{eq}}$: $E = 1.51 +
\frac{0.059}{5}\log\frac{V - V_{\mathrm{eq}}}{V_{\mathrm{eq}}}$ (pH 0); at $2V_{\mathrm{eq}}$, 1.51 V. At [equivalence](#def-b1-titration-methods-titration) $(0.77 + 5 \times 1.51)/6 =
1.39$ V. At pH 1 the permanganate couple is lower by $0.059 \times 8/5 =
0.094$ V: $E_{\mathrm{eq}} = (0.77 + 5 \times 1.416)/6 = 1.31$ V.

**Exercise 15.12 ★★★.**

Ammonium chloride ($0.100\,\mathrm{mol}/\mathrm{L}$, $\mathrm{p}K_a = 9.25$) cannot be titrated accurately by sodium hydroxide with an indicator. Explain from the curve, then propose a [back titration](#def-b1-titration-methods-kinds): excess sodium hydroxide, boil off the ammonia, titrate the remaining hydroxide with hydrochloric acid. Write the computation.

**Solution of Exercise 15.12.**

$\ce{NH4+ + OH- -> NH3 + H2O}$ has $K = 10^{14.00 - 9.25} = 10^{4.75}$: the reaction is not complete near the [equivalence](#def-b1-titration-methods-titration) and the jump is small, around pH 11 ([equivalence](#def-b1-titration-methods-titration) pH $14 - \frac12(4.75 + 1.30) = 11.0$). [Back titration](#def-b1-titration-methods-kinds): add $n_1$ of sodium hydroxide (excess), boil to drive off the ammonia formed, then titrate the hydroxide left with hydrochloric acid ($n_2$, sharp strong–strong jump): $n(\ce{NH4+}) = n_1 - n_2$.

## 15.7 Problem: How Much Vitamin C in the Tablet?

**Problem 15.1.**

Weekend problem — the reaction of iodine with ascorbic acid, a back titration by thiosulfate, the computation of the result, and its uncertainty, ending on the mass of ascorbic acid per tablet

A tablet is crushed and dissolved in water in a $100.0\,\mathrm{mL}$ volumetric flask. A $10.00\,\mathrm{mL}$ aliquot is taken with a pipette, and $20.00\,\mathrm{mL}$ of iodine solution at $0.0250\,\mathrm{mol}/\mathrm{L}$ are added (excess). The iodine left is titrated by sodium thiosulfate at $0.0500\,\mathrm{mol}/\mathrm{L}$ with starch as the indicator: the blue colour vanishes at $8.90\,\mathrm{mL}$. Ascorbic acid is $\ce{C6H8O6}$ ($M = 176.0\,\mathrm{g}/\mathrm{mol}$); iodine oxidises it to $\ce{C6H6O6}$. $E^\circ(\ce{I2}/\ce{I-}) = 0.53$ V, $E^\circ(\ce{S4O6^2-}/\ce{S2O3^2-}) = 0.02$ V.

**Part I — The reactions.**

1. Write the [half-equation](https://one-course.com/books/chemistry/2/en/chapter/13-redox-equilibria-and-the-nernst-equation#def-b1-nernst-couple) of the couple $\ce{C6H6O6}$ / $\ce{C6H8O6}$ .
2. Write the reaction of iodine with ascorbic acid.
3. Give the [oxidation number](https://one-course.com/books/chemistry/2/en/chapter/13-redox-equilibria-and-the-nernst-equation#def-b1-nernst-oxidation-number) of iodine in $\ce{I2}$ and $\ce{I-}$ .
4. Write the reaction of iodine with thiosulfate and compute its constant.
5. Starch gives a deep blue colour with iodine. Why does the colour vanish exactly at the [equivalence](#def-b1-titration-methods-titration) of the thiosulfate [titration](#def-b1-titration-methods-titration) ?
6. Ascorbic acid is slowly oxidised by the oxygen of air. Why does that argue for adding the iodine at once, in excess?

**Part II — Why a [back titration](#def-b1-titration-methods-kinds)?**

7. What would a [direct titration](#def-b1-titration-methods-kinds) of ascorbic acid by iodine require?
8. Describe the [back titration](#def-b1-titration-methods-kinds) as the three steps of [Method 15.4](#met-b1-titration-methods-back-titration) .
9. Why must the iodine be in excess? Check it with the result.
10. Why is the thiosulfate solution standardised shortly before use?
11. The aliquot is $10.00\,\mathrm{mL}$ of $100.0\,\mathrm{mL}$ : what is the dilution factor?
12. Which glassware delivers the $20.00\,\mathrm{mL}$ of iodine solution?

**Part III — The computation.**

13. Compute the amount of iodine introduced.
14. Compute the amount of thiosulfate used.
15. Deduce the amount of iodine in excess.
16. Deduce the amount of iodine that reacted with ascorbic acid.
17. Compute the amount of ascorbic acid in the aliquot.
18. Compute the amount, then the mass, of ascorbic acid in the tablet.
19. Compare with the label ( $500\,\mathrm{mg}$ ).

**Part IV — How sure?** The tolerances are: flask $(100.0 \pm 0.1)\,\mathrm{mL}$, pipettes $(10.00 \pm 0.02)\,\mathrm{mL}$ and $(20.00 \pm 0.03)\,\mathrm{mL}$, burette reading $\pm0.05\,\mathrm{mL}$; the two concentrations are known to $0.2\,\%$.

20. Express $n(\text{ascorbic acid})$ in the tablet as a function of the measured quantities.
21. Which relative uncertainties enter the amount of iodine introduced and the amount in excess?
22. Why does a difference of two amounts amplify the relative uncertainty?
23. Combining the contributions as in [Chapter 29](https://one-course.com/books/chemistry/2/en/chapter/29-lab-techniques-i-safety-measurement-and-separation#ch-b1-lab-techniques-1) (quadratic sum), estimate the relative uncertainty of the result.
24. Give the mass of ascorbic acid per tablet with its uncertainty, to two significant figures.

**Solution of Problem 15.1.**

**1.** $\ce{C6H6O6 + 2H+ + 2e- -> C6H8O6}$. **2.** $\ce{C6H8O6 + I2 -> C6H6O6 + 2I- + 2H+}$. **3.** 0 and $-$I. **4.** $\ce{I2 + 2S2O3^2- -> 2I- + S4O6^2-}$, $\log K = 2(0.53 - 0.02)/0.059
= 17$. **5.** The blue colour needs iodine; it disappears with the last iodine, at the [equivalence](#def-b1-titration-methods-titration). **6.** Losses to air would lower the result; added at once and in excess, the iodine oxidises the ascorbic acid quickly and completely. **7.** A fast, [quantitative reaction](https://one-course.com/books/chemistry/2/en/chapter/7-describing-a-chemical-system-extent-activities-q-and-k#def-b1-extent-q-and-k-quantitative) during the whole [titration](#def-b1-titration-methods-titration), without air oxidation while the [titrant](#def-b1-titration-methods-titration) is slowly added, and an [end point](#def-b1-titration-methods-titration) at the first excess of iodine. **8.** Excess iodine $n_R$; reaction with ascorbic acid; [titration](#def-b1-titration-methods-titration) of the iodine left by thiosulfate. **9.** Otherwise some ascorbic acid would remain and the iodine excess, zero, would not tell how much. Here $2.775 \times 10^{-4}\,\mathrm{mol}$ reacted out of $5.000 \times 10^{-4}\,\mathrm{mol}$: excess indeed. **10.** Thiosulfate solutions change slowly with time; its concentration enters the result directly. **11.** 10. **12.** A $20.00\,\mathrm{mL}$ volumetric pipette. **13.** $20.00 \times 10^{-3} \times 0.0250 = 5.000 \times 10^{-4}\,\mathrm{mol}$. **14.** $8.90 \times 10^{-3} \times 0.0500 = 4.450 \times 10^{-4}\,\mathrm{mol}$. **15.** Half: $2.225 \times 10^{-4}\,\mathrm{mol}$. **16.** $5.000 - 2.225 = 2.775 \times 10^{-4}\,\mathrm{mol}$. **17.** One to one: $2.775 \times 10^{-4}\,\mathrm{mol}$. **18.** $\times 10$: $2.775 \times 10^{-3}\,\mathrm{mol}$, $\times 176.0$: $0.488\,\mathrm{g}$. **19.** About $2\,\%$ below the label. **20.** $n = \bigl(C_IV_I - \frac12C_TV_T\bigr)\,V_{\mathrm{flask}}/V_{\mathrm{aliquot}}$. **21.** Iodine introduced: pipette $0.03/20.00 = 0.15\,\%$ and concentration $0.2\,\%$, about $0.25\,\%$, that is $1.3 \times 10^{-6}\,\mathrm{mol}$. Excess: burette $0.05/8.90 = 0.56\,\%$ and concentration $0.2\,\%$, about $0.60\,\%$, that is $1.3 \times 10^{-6}\,\mathrm{mol}$. **22.** The absolute uncertainties combine, while the difference is smaller than either amount: $\sqrt{1.3^2 + 1.3^2} \times 10^{-6} =
1.8 \times 10^{-6}\,\mathrm{mol}$, $0.66\,\%$ of $2.775 \times 10^{-4}\,\mathrm{mol}$, more than either term. **23.** With the flask ($0.1\,\%$) and the aliquot pipette ($0.2\,\%$): $\sqrt{0.66^2 + 0.1^2 + 0.2^2} = 0.70\,\%$. **24.** $0.488 \pm 0.003$ g: **$0.49\,\mathrm{g}$ of ascorbic acid per tablet**.