---
title: "Titration"
book: "School Chemistry — Grades 1 to 12"
subject: chemistry
language: en
chapter: 36
exercises: 15
source: https://one-course.com/books/chemistry/1/en/chapter/36-titration
license: CC-BY-NC-SA-4.0
credit: "One Chemistry Book, One Course (one-course.com)"
---

# Chapter 36 — Titration

The box of an iron supplement promises $80\,\mathrm{mg}$ of iron in every tablet, as iron(II) [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion). Is it true? A chemist crushes a tablet, [dissolves](https://one-course.com/books/chemistry/1/en/chapter/2-mixing-and-dissolving#def-g2-mixing-and-dissolving-dissolve) it in acid, and adds a purple solution of potassium permanganate drop by drop from a long graduated tube. Each drop loses its colour at once, as the permanganate [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) are used up by the iron(II) [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion), until one drop, the last, leaves a faint pink that does not fade. The volume poured at that moment gives the amount of iron. This chapter turns a reaction into a measurement.

**You already know.**

An [oxidant](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidant) gains [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus), a [reductant](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidant) gives them; the permanganate [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) oxidises iron(II) [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) according to $\ce{MnO4^- + 8H+ + 5Fe^{2+} -> Mn^{2+} + 5Fe^{3+} + 4H2O}$ ([Chapter 35](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#ch-g11-redox)). [Molar concentration](https://one-course.com/books/chemistry/1/en/chapter/26-concentration-and-dilution#def-g10-concentration-and-dilution-molar-concentration) and [dilution](https://one-course.com/books/chemistry/1/en/chapter/26-concentration-and-dilution#def-g10-concentration-and-dilution-dilution) ([Chapter 26](https://one-course.com/books/chemistry/1/en/chapter/26-concentration-and-dilution#ch-g10-concentration-and-dilution)); the [progress table](https://one-course.com/books/chemistry/1/en/chapter/27-the-reaction-progress-table#def-g10-reaction-progress-table-extent) and the [limiting reactant](https://one-course.com/books/chemistry/1/en/chapter/27-the-reaction-progress-table#def-g10-reaction-progress-table-limiting) ([Chapter 27](https://one-course.com/books/chemistry/1/en/chapter/27-the-reaction-progress-table#ch-g10-reaction-progress-table)).

![Iron supplement tablets: how much iron does each really hold?](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-titration/img-f272ef2128a5.jpg)

*Iron supplement tablets: how much iron does each really hold?*

## 36.1 Titrating

**Definition 36.1 (Titration, titrant, titrated solution).**

A *titration* measures the amount of a dissolved species by making it react with another species whose concentration is accurately known. The solution of known concentration, added from a burette, is the *titrant*; the solution containing the species to be measured, placed in the flask, is the *titrated solution*.

**Proposition 36.2 (What a titration reaction must be).**

A reaction can be used for a [titration](#def-g11-titration-titration) if it is total (so that one [reactant](https://one-course.com/books/chemistry/1/en/chapter/12-chemical-reactions-reactants-and-products#def-g7-chemical-reactions-reactant) is entirely used up), fast (so that each drop reacts at once), and unique (the [titrant](#def-g11-titration-titration) reacts with nothing else in the flask).

**Proof.** If the reaction stopped before completion, or lagged behind the drops, or consumed [titrant](#def-g11-titration-titration) elsewhere, the volume poured would no longer measure the amount of the titrated species. ∎

![A titration set: the burette, held on a stand, delivers the titrant into the titrated solution, kept stirred.](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-titration/fig-829ac7b0e496.svg)

*A [titration](#def-g11-titration-titration) set: the burette, held on a stand, delivers the [titrant](#def-g11-titration-titration) into the [titrated solution](#def-g11-titration-titration), kept stirred.*

## 36.2 Equivalence

**Definition 36.3 (Equivalence, equivalent volume).**

*Equivalence* is the moment of a [titration](#def-g11-titration-titration) when the [titrant](#def-g11-titration-titration) added and the titrated species are in the proportions of the equation: both are used up. The volume of [titrant](#def-g11-titration-titration) poured at that moment is the *equivalent volume*, $V_E$.

**Proposition 36.4 (The relation at equivalence).**

For a [titration](#def-g11-titration-titration) reaction $a\,\ce{A} + b\,\ce{B} \longrightarrow$ products, with A titrated (amount $n_A$) by a [titrant](#def-g11-titration-titration) B of concentration $c_B$, at [equivalence](#def-g11-titration-equivalence)

$$
\frac{n_A}{a} = \frac{c_B\, V_E}{b} .
$$

**Proof.** At [equivalence](#def-g11-titration-equivalence) the [mixture](https://one-course.com/books/chemistry/1/en/chapter/8-pure-substances-and-mixtures#def-g6-pure-substances-and-mixtures-mixture) is stoichiometric ([Chapter 27](https://one-course.com/books/chemistry/1/en/chapter/27-the-reaction-progress-table#ch-g10-reaction-progress-table)): A and B, of initial amounts $n_A$ and $c_B V_E$, run out together, so $n_A / a = c_B V_E / b$. ∎

![Iron(II) titrated by permanganate at 0.0200\, mol/ L: the iron(II) ions run out exactly at the equivalent volume; after it, the permanganate ions are no longer consumed and accumulate.](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-titration/fig-f8ce89f3150b.svg)

*Iron(II) titrated by permanganate at $0.0200\,\mathrm{mol}/\mathrm{L}$: the iron(II) [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) run out exactly at the [equivalent volume](#def-g11-titration-equivalence); after it, the permanganate [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) are no longer consumed and accumulate.*

## 36.3 Locating equivalence by a colour change

**Proposition 36.5 (Seeing equivalence).**

When one of the species of the [titration](#def-g11-titration-titration) is coloured, [equivalence](#def-g11-titration-equivalence) is seen as a colour change of the flask: the colour of the [titrant](#def-g11-titration-titration) appears and persists as soon as the [titrant](#def-g11-titration-titration) is in excess, or the colour of the titrated species disappears when it runs out.

**Proof.** Before [equivalence](#def-g11-titration-equivalence), every [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) of [titrant](#def-g11-titration-titration) that falls into the flask is used up at once; after [equivalence](#def-g11-titration-equivalence), it remains. A coloured species is therefore present in the flask on one side of [equivalence](#def-g11-titration-equivalence) only. ∎

**Example 36.6 (Permanganate, its own indicator).**

The permanganate [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) $\ce{MnO4^-}$ is deep purple, and the manganese [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) $\ce{Mn^{2+}}$ it gives are almost colourless; the iron [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) give the solution only a pale yellow tint. Before [equivalence](#def-g11-titration-equivalence), each purple drop is decolourised as it falls; the first drop that is not marks [equivalence](#def-g11-titration-equivalence): the flask turns pale pink and stays so.

![The flask of an iron(II) titration by permanganate, before, at and after equivalence. The equivalent volume is read at the first lasting pink.](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-titration/fig-6aa67a176187.svg)

*The flask of an iron(II) [titration](#def-g11-titration-titration) by permanganate, before, at and after [equivalence](#def-g11-titration-equivalence). The [equivalent volume](#def-g11-titration-equivalence) is read at the first lasting pink.*

**Example 36.7 (Vitamin C and iodine).**

Vitamin C, ascorbic acid $\ce{C6H8O6}$ ($M = 176.0\,\mathrm{g}/\mathrm{mol}$), is a [reductant](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidant); iodine $\ce{I2}$ oxidises it:

$$
\ce{C6H8O6 + I2 -> C6H6O6 + 2H+ + 2I-} .
$$

$10.0\,\mathrm{mL}$ of orange juice, with a little starch, are titrated by an iodine solution at $5.00 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$. Before [equivalence](#def-g11-titration-equivalence) the iodine is used up; the first drop in excess turns the starch dark blue. If $V_E = 5.70\,\mathrm{mL}$, the juice held $n = 5.00 \times 10^{-3} \times 5.70 \times 10^{-3} = 2.85 \times 10^{-5}\,\mathrm{mol}$ of vitamin C, that is $5.0\,\mathrm{mg}$ in $10.0\,\mathrm{mL}$: $0.50\,\mathrm{g}/\mathrm{L}$.

## 36.4 Computing an unknown concentration

**Method 36.8 (Computing a concentration from a titration).**

1. Write the [balanced equation](https://one-course.com/books/chemistry/1/en/chapter/13-conservation-of-mass-and-balanced-equations#def-g8-balanced-equations-equation) of the [titration](#def-g11-titration-titration) reaction and read the coefficients $a$ (titrated species) and $b$ ( [titrant](#def-g11-titration-titration) ).
2. Read $V_E$ on the burette, and compute the amount of [titrant](#def-g11-titration-titration) poured, $n_B = c_B V_E$ .
3. At [equivalence](#def-g11-titration-equivalence) , $n_A = \dfrac{a}{b}\, c_B V_E$ .
4. Divide by the volume titrated: $c_A = n_A / V_A$ . If the sample was diluted before [titration](#def-g11-titration-titration) , multiply by the [dilution factor](https://one-course.com/books/chemistry/1/en/chapter/26-concentration-and-dilution#def-g10-concentration-and-dilution-dilution) .

**Example 36.9 (An iron(II) solution).**

$20.0\,\mathrm{mL}$ of an iron(II) solution, acidified, need $V_E = 12.6\,\mathrm{mL}$ of permanganate at $0.0200\,\mathrm{mol}/\mathrm{L}$. Here $a = 5$ ($\ce{Fe^{2+}}$) and $b = 1$ ($\ce{MnO4^-}$): $n(\ce{MnO4^-}) = 0.0200 \times 0.0126 = 2.52 \times 10^{-4}\,\mathrm{mol}$, so $n(\ce{Fe^{2+}}) = 5 \times 2.52 \times 10^{-4} = 1.26 \times 10^{-3}\,\mathrm{mol}$, and $c(\ce{Fe^{2+}}) = 1.26 \times 10^{-3} / 0.0200 = 0.0630\,\mathrm{mol}/\mathrm{L}$.

## 36.5 The titration as a measurement

**In the lab — Reading a burette.**

The burette is rinsed with the [titrant](#def-g11-titration-titration), filled, and its tip freed of [air](https://one-course.com/books/chemistry/1/en/chapter/4-air-a-mixture-of-gases#def-g4-air-a-mixture-of-gases-air) bubbles. The level is read at the bottom of the meniscus, with the eye at the same height, before and after the [titration](#def-g11-titration-titration): the volume poured is the difference. Graduations are every $0.1\,\mathrm{mL}$, and a reading can be estimated to $0.05\,\mathrm{mL}$. Near [equivalence](#def-g11-titration-equivalence) the [titrant](#def-g11-titration-titration) is added drop by drop, while the flask is swirled or stirred.

![A titration in progress: burette, stand and conical flask on a white tile.](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-titration/img-a41d4ab7c5fc.jpg)

*A [titration](#def-g11-titration-titration) in progress: burette, stand and conical flask on a white tile.*

**Remark 36.10 (How precise is a titration?).**

Each burette reading may be off by about $0.05\,\mathrm{mL}$, and the volume poured is the difference of two readings: off by up to $0.1\,\mathrm{mL}$. For $V_E = 14.3\,\mathrm{mL}$, that is $0.1 / 14.3 \approx
0.7\,\%$. A larger $V_E$ makes this relative error smaller, which is why [titrant](#def-g11-titration-titration) concentrations are chosen to give [equivalent volumes](#def-g11-titration-equivalence) of 10 to $20\,\mathrm{mL}$ with a $25\,\mathrm{mL}$ burette. The [titration](#def-g11-titration-titration) is repeated until two results agree within about $0.1\,\mathrm{mL}$. A finer treatment of the uncertainty of a measurement is given in the Year 1 volume.

**Safety.**

![](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-titration/fig-2b716bc1fa58.svg)

![](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-titration/fig-e4962b3c5e86.svg)

![](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-titration/fig-5780317c64dd.svg)

![](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-titration/fig-c10fa5cc5966.svg)

![](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-titration/fig-7df0ec0804a8.svg)

Solid potassium permanganate is an [oxidant](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidant) that can feed a fire, corrosive and harmful; its dilute solutions stain the skin and clothes brown. Iodine solutions are harmful. Goggles and gloves; the waste is collected, never poured down the sink.

## 36.6 Exercises

**Exercise 36.1 ★.**

In a [titration](#def-g11-titration-titration), what is the [titrant](#def-g11-titration-titration)? The [titrated solution](#def-g11-titration-titration)? What happens at [equivalence](#def-g11-titration-equivalence)?

**Solution of Exercise 36.1.**

The [titrant](#def-g11-titration-titration) is the solution of accurately known concentration, in the burette; the [titrated solution](#def-g11-titration-titration) contains the species to be measured, in the flask. At [equivalence](#def-g11-titration-equivalence) the [titrant](#def-g11-titration-titration) added and the titrated species are in the proportions of the equation: both are used up.

**Exercise 36.2 ★.**

What amount of permanganate [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) is contained in $12.5\,\mathrm{mL}$ of a solution at $0.0200\,\mathrm{mol}/\mathrm{L}$?

**Solution of Exercise 36.2.**

$n = 0.0200 \times 0.0125 = 2.50 \times 10^{-4}\,\mathrm{mol}$.

**Exercise 36.3 ★.**

On the figure of the amounts against the volume poured, read the [equivalent volume](#def-g11-titration-equivalence). What amount of iron(II) was in the flask at the start?

**Solution of Exercise 36.3.**

$V_E = 14.3\,\mathrm{mL}$; the iron(II) line starts at $14.3 \times 10^{-4}\,\mathrm{mol}$, that is $1.43 \times 10^{-3}\,\mathrm{mol}$.

**Exercise 36.4 ★.**

Why is no indicator needed for a [titration](#def-g11-titration-titration) by permanganate?

**Solution of Exercise 36.4.**

The [titrant](#def-g11-titration-titration) is itself coloured (purple) and its product almost colourless: the colour appears in the flask as soon as the permanganate is in excess.

**Exercise 36.5 ★.**

At the [equivalence](#def-g11-titration-equivalence) of the [titration](#def-g11-titration-titration) of iron(II) by permanganate, what is the ratio of the amount of iron(II) to the amount of permanganate poured? Why?

**Solution of Exercise 36.5.**

5: the equation consumes 5 iron(II) [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) per permanganate [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion), and at [equivalence](#def-g11-titration-equivalence) the [reactants](https://one-course.com/books/chemistry/1/en/chapter/12-chemical-reactions-reactants-and-products#def-g7-chemical-reactions-reactant) have been brought together in these proportions.

**Exercise 36.6 ★★.**

$10.0\,\mathrm{mL}$ of an acidified iron(II) solution need $8.40\,\mathrm{mL}$ of permanganate at $0.0200\,\mathrm{mol}/\mathrm{L}$. Compute the [molar concentration](https://one-course.com/books/chemistry/1/en/chapter/26-concentration-and-dilution#def-g10-concentration-and-dilution-molar-concentration) of iron(II), then its [mass concentration](https://one-course.com/books/chemistry/1/en/chapter/26-concentration-and-dilution#def-g10-concentration-and-dilution-mass-concentration).

**Solution of Exercise 36.6.**

$n(\ce{MnO4^-}) = 0.0200 \times 0.00840 = 1.68 \times 10^{-4}\,\mathrm{mol}$; $n(\ce{Fe^{2+}}) = 5 \times 1.68 \times 10^{-4} = 8.40 \times 10^{-4}\,\mathrm{mol}$; $c = 8.40 \times 10^{-4} / 0.0100 = 0.0840\,\mathrm{mol}/\mathrm{L}$; $C_m = 0.0840 \times 55.8 = 4.69\,\mathrm{g}/\mathrm{L}$.

**Exercise 36.7 ★★.**

A vitamin C tablet is dissolved to make $100.0\,\mathrm{mL}$ of solution. $10.0\,\mathrm{mL}$ of it, with starch, need $11.4\,\mathrm{mL}$ of iodine at $0.0100\,\mathrm{mol}/\mathrm{L}$. What mass of vitamin C does the tablet contain?

**Solution of Exercise 36.7.**

Ratio 1:1: $n = 0.0100 \times 0.0114 = 1.14 \times 10^{-4}\,\mathrm{mol}$ in $10.0\,\mathrm{mL}$, so $1.14 \times 10^{-3}\,\mathrm{mol}$ in the $100.0\,\mathrm{mL}$, that is $1.14 \times 10^{-3} \times 176.0 = 0.201\,\mathrm{g}$: about $200\,\mathrm{mg}$.

**Exercise 36.8 ★★.**

A reaction is total but takes several minutes. Why can it not be used as it is for a [titration](#def-g11-titration-titration) with a colour change?

**Solution of Exercise 36.8.**

Each drop would not react at once: the colour of the [titrant](#def-g11-titration-titration) would linger before [equivalence](#def-g11-titration-equivalence), and the end would be seen too early (or one would have to wait minutes after each drop).

**Exercise 36.9 ★★.**

Look at the three flasks. A student stops at a flask as purple as the third one. Is her [equivalent volume](#def-g11-titration-equivalence) too large or too small? What should she have done near [equivalence](#def-g11-titration-equivalence)?

**Solution of Exercise 36.9.**

Too large: she has gone past [equivalence](#def-g11-titration-equivalence), adding permanganate in excess. Near [equivalence](#def-g11-titration-equivalence) she should have added the [titrant](#def-g11-titration-titration) drop by drop, swirling, and stopped at the first lasting faint pink.

**Exercise 36.10 ★★.**

A concentrated iron(II) solution is first diluted ten times. Then $10.0\,\mathrm{mL}$ of the diluted solution need $13.0\,\mathrm{mL}$ of permanganate at $0.0200\,\mathrm{mol}/\mathrm{L}$. Find the concentration of the concentrated solution.

**Solution of Exercise 36.10.**

$n(\ce{Fe^{2+}}) = 5 \times 0.0200 \times 0.0130 = 1.30 \times 10^{-3}\,\mathrm{mol}$ in $10.0\,\mathrm{mL}$: $c = 0.130\,\mathrm{mol}/\mathrm{L}$ for the diluted solution, and $10 \times 0.130 = 1.30\,\mathrm{mol}/\mathrm{L}$ for the concentrated one.

**Exercise 36.11 ★★.**

The iron(II) solution of the chapter’s worked example has $c = 0.0630\,\mathrm{mol}/\mathrm{L}$. What mass of iron does $1.00\,\mathrm{L}$ of it contain?

**Solution of Exercise 36.11.**

$0.0630 \times 55.8 = 3.52\,\mathrm{g}$.

**Exercise 36.12 ★★★.**

A [titration](#def-g11-titration-titration) gives $V_E = 7.2\,\mathrm{mL}$. Estimate the relative error due to the burette readings, and compare with $V_E = 14.3\,\mathrm{mL}$. Suggest two ways of obtaining a larger [equivalent volume](#def-g11-titration-equivalence).

**Solution of Exercise 36.12.**

$0.1 / 7.2 \approx 1.4\,\%$, against $0.7\,\%$ for $14.3\,\mathrm{mL}$. Titrate a larger volume (or a more concentrated sample) of the solution, or use a more dilute [titrant](#def-g11-titration-titration).

**Exercise 36.13 ★★★.**

Hydrogen peroxide is titrated by permanganate in acid:

$$
\ce{2MnO4^- + 5H2O2 + 6H+ -> 2Mn^{2+} + 5O2 + 8H2O} .
$$

A disinfectant is diluted ten times; $10.0\,\mathrm{mL}$ of the diluted solution need $17.6\,\mathrm{mL}$ of permanganate at $0.0200\,\mathrm{mol}/\mathrm{L}$. Find the [molar concentration](https://one-course.com/books/chemistry/1/en/chapter/26-concentration-and-dilution#def-g10-concentration-and-dilution-molar-concentration) of hydrogen peroxide in the disinfectant, then its [mass concentration](https://one-course.com/books/chemistry/1/en/chapter/26-concentration-and-dilution#def-g10-concentration-and-dilution-mass-concentration) ($\ce{H2O2}$: $34.0\,\mathrm{g}/\mathrm{mol}$).

**Solution of Exercise 36.13.**

$n(\ce{MnO4^-}) = 0.0200 \times 0.0176 = 3.52 \times 10^{-4}\,\mathrm{mol}$; $n(\ce{H2O2}) = \frac{5}{2} \times 3.52 \times 10^{-4} = 8.80 \times 10^{-4}\,\mathrm{mol}$ in $10.0\,\mathrm{mL}$: $0.0880\,\mathrm{mol}/\mathrm{L}$ in the diluted solution, $0.880\,\mathrm{mol}/\mathrm{L}$ in the disinfectant, that is $0.880 \times 34.0 = 29.9\,\mathrm{g}/\mathrm{L}$ (about a 3 % solution).

**Exercise 36.14 ★★★.**

An iron tablet should contain about $1.4 \times 10^{-3}\,\mathrm{mol}$ of iron(II). What concentration of permanganate gives an [equivalent volume](#def-g11-titration-equivalence) near $15\,\mathrm{mL}$ with a $25\,\mathrm{mL}$ burette? What would go wrong with a solution ten times more concentrated?

**Solution of Exercise 36.14.**

$n(\ce{MnO4^-}) = 1.4 \times 10^{-3} / 5 = 2.8 \times 10^{-4}\,\mathrm{mol}$, to be delivered in about $0.015\,\mathrm{L}$: $c \approx 0.019\,\mathrm{mol}/\mathrm{L}$, so a $0.0200\,\mathrm{mol}/\mathrm{L}$ solution. Ten times more concentrated, $V_E$ would be about $1.4\,\mathrm{mL}$: the burette error would reach some $7\,\%$.

**Exercise 36.15 ★★★.**

The equation of the [titration](#def-g11-titration-titration) shows $\ce{8H+}$ for each $\ce{MnO4^-}$. For an [equivalent volume](#def-g11-titration-equivalence) of $14.3\,\mathrm{mL}$ at $0.0200\,\mathrm{mol}/\mathrm{L}$, what amount of hydrogen [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) is used up? Is $10\,\mathrm{mL}$ of an acid containing $1.0\,\mathrm{mol}/\mathrm{L}$ of hydrogen [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) enough? Why must the flask be acidified?

**Solution of Exercise 36.15.**

$8 \times 0.0200 \times 0.0143 = 2.29 \times 10^{-3}\,\mathrm{mol}$ of hydrogen [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion). $10\,\mathrm{mL}$ at $1.0\,\mathrm{mol}/\mathrm{L}$ hold $1.0 \times 10^{-2}\,\mathrm{mol}$: enough, more than four times over. Without acid the reaction of the chapter could not take place as written (hydrogen [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) are a [reactant](https://one-course.com/books/chemistry/1/en/chapter/12-chemical-reactions-reactants-and-products#def-g7-chemical-reactions-reactant)), and permanganate would react differently.

## 36.7 Problem: Is the Iron Tablet Honest?

**Problem 36.1.**

Weekend problem — an iron supplement claims 80 mg of iron per tablet: does a titration agree?

The box of an iron supplement claims $80\,\mathrm{mg}$ of iron, as iron(II), per tablet. A tablet is crushed and dissolved in about $50\,\mathrm{mL}$ of dilute sulfuric acid, then titrated by potassium permanganate at $0.0200\,\mathrm{mol}/\mathrm{L}$. The faint pink persists at $V_E = 14.3\,\mathrm{mL}$.

**Part I — The reaction.**

1. Name the two [redox couples](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-couple) involved. Which species is the [oxidant](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidant) , which the [reductant](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidant) ?
2. Write the two [half-equations](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-couple) .
3. Combine them into the equation of the [titration](#def-g11-titration-titration) , and check that it is balanced in charge.
4. Why is the tablet dissolved in acid?
5. Show that the reaction meets the conditions of a [titration](#def-g11-titration-titration) , and explain how [equivalence](#def-g11-titration-equivalence) is seen.

**Part II — The [titration](#def-g11-titration-titration).**

6. In which piece of glassware is the permanganate solution? How is its volume read?
7. What colour is the flask before [equivalence](#def-g11-titration-equivalence) , and why does each drop lose its colour?
8. Compute the amount of permanganate poured at [equivalence](#def-g11-titration-equivalence) .

**Part III — The iron.**

9. Deduce the amount of iron(II) in the tablet.
10. Compute its mass in milligrams.
11. Compare with the label: relative difference?
12. The volume poured may be off by $0.1\,\mathrm{mL}$ . By how many milligrams may the mass of iron be off? Is the label confirmed?

**Part IV — Several tablets.**

13. Two more tablets give $V_E = 14.1\,\mathrm{mL}$ and $14.4\,\mathrm{mL}$ . Compute their masses of iron.
14. Compute the mean of the three tablets.
15. The three results differ by more than the burette error. What does this say about the tablets?
16. Would permanganate at $0.100\,\mathrm{mol}/\mathrm{L}$ have been a better choice? Explain.
17. Many supplements also contain vitamin C, a [reductant](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidant) . What would it do to the [titration](#def-g11-titration-titration) , and to the mass of iron found?
18. The iron is present as iron(II) sulfate, $\ce{FeSO4}$ . What mass of it does one tablet contain?
19. How many iron(II) [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) is that?
20. State the final answer: what mass of iron does one tablet contain?

**Solution of Problem 36.1.**

**1.** $\ce{MnO4^-}$/$\ce{Mn^{2+}}$ and $\ce{Fe^{3+}}$/$\ce{Fe^{2+}}$. The [oxidant](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidant) is the permanganate [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion), the [reductant](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidant) the iron(II) [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion).

**2.** $\ce{MnO4^- + 8H+ + 5e- -> Mn^{2+} + 4H2O}$; $\ce{Fe^{2+} -> Fe^{3+} + e-}$.

**3.** The second is multiplied by 5: $\ce{MnO4^- + 8H+ + 5Fe^{2+} -> Mn^{2+} + 5Fe^{3+} + 4H2O}$. Charge on the left $-1 + 8 + 10 = +17$; on the right $+2 + 15 = +17$.

**4.** Hydrogen [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) are a [reactant](https://one-course.com/books/chemistry/1/en/chapter/12-chemical-reactions-reactants-and-products#def-g7-chemical-reactions-reactant): the reaction needs an [acidic solution](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-acidic).

**5.** It is total, fast and the only reaction of the permanganate in the flask; the permanganate, purple, is used up before [equivalence](#def-g11-titration-equivalence) and colours the flask pink after it.

**6.** In the burette; read at the bottom of the meniscus, eye level, before and after, to $0.05\,\mathrm{mL}$.

**7.** Pale yellow (iron [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion)); each purple drop is used up at once by the iron(II) [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) still present.

**8.** $0.0200 \times 0.0143 = 2.86 \times 10^{-4}\,\mathrm{mol}$.

**9.** $n(\ce{Fe^{2+}}) = 5 \times 2.86 \times 10^{-4} =
1.43 \times 10^{-3}\,\mathrm{mol}$.

**10.** $1.43 \times 10^{-3} \times 55.8 = 0.0798\,\mathrm{g} =
79.8\,\mathrm{mg}$.

**11.** $(80 - 79.8)/80 \approx 0.3\,\%$.

**12.** $0.1\,\mathrm{mL}$ out of $14.3\,\mathrm{mL}$ is $0.7\,\%$, about $0.6\,\mathrm{mg}$: the result, $79.8 \pm 0.6$ mg, agrees with the label.

**13.** $14.1$: $5 \times 0.0200 \times 0.0141 \times 55.8 =
78.7\,\mathrm{mg}$; $14.4$: $80.4\,\mathrm{mg}$.

**14.** $(79.8 + 78.7 + 80.4)/3 = 79.6\,\mathrm{mg}$.

**15.** The tablets themselves differ slightly in their iron content, by about $1\,\%$.

**16.** No: $V_E$ would be about $2.9\,\mathrm{mL}$, and the burette error $3.5\,\%$, five times worse.

**17.** Vitamin C would also be oxidised by the permanganate: the reaction would no longer be unique, $V_E$ would be too large, and the iron would be overestimated.

**18.** $1.43 \times 10^{-3} \times 151.9 = 0.217\,\mathrm{g}$, about $217\,\mathrm{mg}$.

**19.** $1.43 \times 10^{-3} \times 6.02 \times 10^{23} = 8.6 \times 10^{20}$ [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion).

**20.** About $80\,\mathrm{mg}$ of iron per tablet: the label is honest.