Volumetric Analysis

Quantitative analysis answers one question: how much of something is present. This unit teaches the volume-measuring route to that answer. You take a solution of accurately known concentration, run it into a measured volume of the unknown until the reaction is exactly complete, read the burette, and turn that reading into a mass or a percentage. Everything after this — the arithmetic of equivalents, the concentration units, the choice of indicator — exists only to make that one reading trustworthy.

1.1 Introduction to gravimetric analysis, volumetric analysis and equivalent weight

Analytical chemistry splits into two halves. Qualitative analysis asks what is present; quantitative analysis asks how much. Classical quantitative analysis then splits again, according to what you actually measure on the bench.

In a gravimetric estimation of sulphate, for example, the sulphate is precipitated as barium sulphate, and the precipitate is filtered, washed, ignited and weighed:

BaCl₂ + Na₂SO₄ → BaSO₄↓ + 2NaCl

From the mass of BaSO₄ (molar mass 233) the mass of SO₄²⁻ (96) follows by simple proportion. Chloride is estimated the same way as AgCl, and iron as Fe₂O₃.

Feature	Gravimetric analysis	Volumetric analysis
Quantity measured	Mass of a precipitate	Volume of a standard solution
Chief apparatus	Analytical balance, crucible, furnace	Burette, pipette, volumetric flask
Speed	Slow (hours to days)	Fast (minutes)
Accuracy	Very high (to 0.01 mg)	High (to about 0.1 %)
Indicator needed	No	Usually yes
Suits	Small numbers of very accurate results	Routine repeated analysis

Volumetric work needs a common currency in which "exactly enough" can be expressed. That currency is the equivalent.

Equivalent weight is a pure number (a relative mass), so it has no unit; the gram equivalent has the unit g.

1.2 Relationship between equivalent weight, atomic weight and valency

One atom of hydrogen has valency 1 and a relative atomic mass of 1.008, so by definition one equivalent of hydrogen is 1.008. Now take an element X of atomic weight A and valency v. One atom of X combines with v atoms of hydrogen, so A parts of X combine with 1.008v parts of hydrogen. Therefore the mass of X that combines with 1.008 parts of hydrogen is A/v:

Rearranged, this gives the two forms the board also asks for:

Element	Atomic weight	Valency	Equivalent weight
H	1.008	1	1.008
O	16	2	8
Cl	35.5	1	35.5
Mg	24	2	12
Al	27	3	9
Ca	40	2	20
Fe (ferrous)	56	2	28
Fe (ferric)	56	3	18.67

1.3 Equivalent weight of compounds: acid, base, salt, oxidizing and reducing agents

For a compound, replace "valency" by the n-factor — the number of reacting units the formula unit supplies. What counts as a unit depends on the class of compound.

Class	n-factor is	E =
Acid	basicity = number of replaceable H⁺	molar mass / basicity
Base	acidity = number of replaceable OH⁻	molar mass / acidity
Salt	total positive (or negative) charge per formula unit	molar mass / total charge
Oxidising agent	number of electrons gained per formula unit	molar mass / electrons gained
Reducing agent	number of electrons lost per formula unit	molar mass / electrons lost

Worked values you should know by heart:

Substance	Molar mass	n-factor	Equivalent weight
HCl	36.5	1	36.5
H₂SO₄	98	2	49
H₃PO₄ (full neutralisation)	98	3	32.67
NaOH	40	1	40
Ca(OH)₂	74	2	37
Na₂CO₃	106	2	53
CaCO₃	100	2	50
Al₂(SO₄)₃	342	6	57
K₂Cr₂O₇ (acid medium)	294	6	49
H₂C₂O₄·2H₂O (as reductant)	126	2	63
FeSO₄·7H₂O (as reductant)	278	1	278
Mohr's salt FeSO₄·(NH₄)₂SO₄·6H₂O	392	1	392
Na₂S₂O₃·5H₂O (with I₂)	248	1	248
I₂ (as oxidant)	254	2	127

The three equivalent weights of potassium permanganate

KMnO₄ (molar mass 158) is reduced to a different manganese species in each medium, so it gains a different number of electrons:

Medium	Half reaction	Electrons gained	E
Acidic (dil. H₂SO₄)	MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O	5	158/5 = 31.6
Neutral or faintly alkaline	MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻	3	158/3 = 52.67
Strongly alkaline	MnO₄⁻ + e⁻ → MnO₄²⁻	1	158/1 = 158

Each half reaction is balanced in both mass and charge — check the acidic one: charge on the left is (-1) + 8(+1) + 5(-1) = +2, and on the right +2.

1.4 Concentration units: percentage, g/L, molarity, molality, normality, formality, ppm and ppb

Concentration is the amount of solute in a stated amount of solvent or solution. Which unit you choose depends on what you can measure.

Unit	Symbol	Definition	Usual unit
Percentage by mass	% w/w	mass of solute per 100 g of solution	—
Percentage by mass/volume	% w/v	mass (g) of solute per 100 cm³ of solution	—
Percentage by volume	% v/v	volume of solute per 100 cm³ of solution	—
Strength	S	mass (g) of solute per litre of solution	g L⁻¹
Molarity	M	moles of solute per litre of solution	mol L⁻¹
Molality	m	moles of solute per kilogram of solvent	mol kg⁻¹
Normality	N	gram equivalents of solute per litre of solution	eq L⁻¹
Formality	F	formula masses of solute per litre of solution	mol L⁻¹
Parts per million	ppm	parts of solute per 10⁶ parts of solution	—
Parts per billion	ppb	parts of solute per 10⁹ parts of solution	—

The working formulas:

If a concentrated reagent bottle gives the percentage by mass p and the density d (in g cm⁻³), then one litre of it weighs 1000d g and contains 10pd g of solute, so

Formality is used for ionic solids, which contain no discrete molecules. A solution made by dissolving 58.5 g of NaCl in water and making it up to one litre is 1 F, because "one mole of NaCl molecules" does not exist in the solution — only Na⁺ and Cl⁻ ions.

ppm and ppb are for very dilute solutions, and for dilute aqueous solutions (density ≈ 1 g cm⁻³) 1 ppm is the same as 1 mg L⁻¹ and 1 ppb the same as 1 μg L⁻¹. The national standard for arsenic in drinking water in Nepal is 50 ppb, i.e. 0.05 mg of arsenic per litre — a limit set because many Terai tube-wells exceed the 10 ppb WHO guideline.

Worked example 1.2 — preparing a standard solution

Problem. What mass of anhydrous sodium carbonate is required to prepare 250 cm³ of 0.1 N Na₂CO₃ solution?

Solution. Equivalent weight of Na₂CO₃ = 106/2 = 53.

Gram equivalents required = N \times V_{\text{(L)}} = 0.1 \times 0.250 = 0.025 eq.

w = \text{eq} \times E = 0.025 \times 53 = 1.325\ \text{g}

So 1.325 g is weighed accurately, dissolved in distilled water, and the solution is made up to the 250 cm³ mark of a volumetric flask.

1.5 Primary and secondary standard substances

A titration is only as good as the solution whose concentration you claim to know.

A primary standard must satisfy all of the following:

1. Available in a state of high purity (99.9 % or better).
2. Stable in air — it must not absorb moisture (hygroscopic), lose water of crystallisation (efflorescent), or absorb CO₂.
3. Highly soluble in water, and the solution must be stable on storing.
4. A high equivalent weight, so that the weighing error is a small fraction of the mass taken.
5. It must react rapidly, completely and in one definite way (stoichiometrically).

Primary standards	Secondary standards
Oxalic acid, H₂C₂O₄·2H₂O	Hydrochloric acid, HCl
Anhydrous sodium carbonate, Na₂CO₃	Sodium hydroxide, NaOH
Potassium dichromate, K₂Cr₂O₇	Potassium permanganate, KMnO₄
Potassium hydrogen phthalate	Sodium thiosulphate, Na₂S₂O₃·5H₂O
Silver nitrate, AgNO₃; sodium chloride, NaCl	Iodine solution

Why the right-hand column fails: NaOH is deliquescent and absorbs CO₂ from the air, so the pellets you weigh are part Na₂CO₃ and part water. Concentrated HCl gives off fumes, so its concentration drifts. KMnO₄ is always slightly contaminated with MnO₂ and is slowly reduced by traces of organic matter in the water, so it is standardised freshly against oxalic acid.

1.6 Law of equivalence and normality equation

1.7 Titration and its types: acid-base titration, redox titration

The vocabulary is examined on its own:

Term	Meaning
Titrant	the standard solution in the burette
Titrate / analyte	the solution of unknown concentration in the flask
Aliquot	the measured portion of analyte delivered by the pipette
Equivalence point	the point at which chemically equivalent amounts have reacted
End point	the point at which the indicator changes colour — the observed equivalence point
Titration error	the (small) difference between end point and equivalence point
Concordant readings	two or more burette readings agreeing within 0.1 cm³

A titration is repeated until two concordant readings are obtained, and the mean of the concordant readings is used in the calculation. The first, rough titration is discarded.

Types of titration

Type	Reaction used	Typical example	Indicator
Acid-base (neutralisation)	H⁺ + OH⁻ → H₂O	NaOH vs HCl	phenolphthalein, methyl orange
Redox	electron transfer	KMnO₄ vs FeSO₄	self-indicating
Precipitation	formation of an insoluble salt	AgNO₃ vs NaCl	potassium chromate (Mohr)
Complexometric	formation of a complex ion	EDTA vs Ca²⁺, Mg²⁺	Eriochrome black T
Iodometric / iodimetric	I₂ ⇌ 2I⁻	Na₂S₂O₃ vs liberated I₂	starch (blue-black)

Acid-base titration

The four possible combinations behave differently near the equivalence point, and this decides the indicator (developed fully in Unit 2):

Acid vs base	pH at equivalence	Suitable indicator
Strong acid vs strong base	7	methyl orange or phenolphthalein
Weak acid vs strong base	> 7	phenolphthalein
Strong acid vs weak base	< 7	methyl orange
Weak acid vs weak base	≈ 7, no sharp jump	none reliable — use a pH meter

Sodium carbonate is a special case worth knowing, because it is neutralised in two stages and therefore has two end points:

Na₂CO₃ + HCl → NaHCO₃ + NaCl (first stage, pH ≈ 8.3, phenolphthalein)

NaHCO₃ + HCl → NaCl + H₂O + CO₂ (second stage, pH ≈ 4.0, methyl orange)

Phenolphthalein therefore measures half the carbonate and methyl orange measures all of it. This is the basis of the double-indicator method used to analyse mixtures of NaOH, Na₂CO₃ and NaHCO₃.

Redox titration

Here the titrant oxidises or reduces the analyte, and the n-factor is the number of electrons transferred. The two classic Nepali-syllabus systems are:

Permanganometry. KMnO₄ in dilute sulphuric acid titrated against oxalic acid or a ferrous salt. KMnO₄ is self-indicating: the purple MnO₄⁻ is decolourised as long as reductant remains, and the first permanent pale-pink tinge is the end point.

2KMnO₄ + 5H₂C₂O₄ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 10CO₂ + 8H₂O

2KMnO₄ + 10FeSO₄ + 8H₂SO₄ → K₂SO₄ + 2MnSO₄ + 5Fe₂(SO₄)₃ + 8H₂O

Dichrometry. K₂Cr₂O₇ is a primary standard and its solution keeps indefinitely, but it is not self-indicating; an external or internal redox indicator (diphenylamine) is used.

K₂Cr₂O₇ + 6FeSO₄ + 7H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 3Fe₂(SO₄)₃ + 7H₂O

Chapter summary

• Gravimetric analysis weighs a precipitate; volumetric analysis measures the volume of a standard solution. Volumetric work is faster, gravimetric more accurate.
• E = \text{molar mass}/n\text{-factor}, and for an element E = \text{atomic weight}/\text{valency}. The n-factor is basicity for an acid, acidity for a base, total ionic charge for a salt, and electrons transferred for a redox reagent.
• KMnO₄ has three equivalent weights: 31.6 (acidic, 5e⁻), 52.67 (neutral, 3e⁻) and 158 (strongly alkaline, 1e⁻).
• Key concentration relations: M = w \times 1000/(M_w V_{\text{cm}^3}), N = M \times n, S = N \times E, M = 10pd/M_w, m = 1000M/(1000d - MM_w). Molality and percentage by mass do not change with temperature; molarity and normality do.
• A primary standard (oxalic acid, Na₂CO₃, K₂Cr₂O₇) is pure, stable and of high equivalent weight, so it can be weighed directly; a secondary standard (HCl, NaOH, KMnO₄, Na₂S₂O₃) must be titrated against one.
• Law of equivalence \Rightarrow N_1V_1 = N_2V_2 (milliequivalents when V is in cm³), or n_1M_1V_1 = n_2M_2V_2 in molarities.
• In a permanganate titration, acidify with dilute H₂SO₄ only; KMnO₄ is self-indicating, while K₂Cr₂O₇ needs an indicator but is itself a primary standard.
• Na₂CO₃ titrated with HCl gives two end points: phenolphthalein at the bicarbonate stage (half the carbonate) and methyl orange at complete neutralisation.

Practice questions

Group A — Multiple choice (1 mark each)

1. The equivalent weight of KMnO₄ in acidic medium is [1]a158b79c52.67d31.6
2. Which of the following is a primary standard substance? [1]aNaOHbHClcK₂Cr₂O₇dKMnO₄
3. The concentration unit that does not change with temperature is [1]amolaritybnormalitycmolalitydformality
4. A 0.1 M solution of H₂SO₄ is the same as [1]a0.05 Nb0.1 Nc0.2 Nd0.4 N
5. In the titration of oxalic acid against KMnO₄ the indicator used is [1]aphenolphthaleinbmethyl orangecstarchdnone — KMnO₄ is self-indicating
6. 1 ppm of fluoride in drinking water is the same as [1]a1 g L⁻¹b1 mg L⁻¹c1 μg L⁻¹d1 mol L⁻¹

Answers to Group A

1. (d) — MnO₄⁻ gains 5 electrons, so E = 158/5 = 31.6.

2. (c) — K₂Cr₂O₇ is pure, non-hygroscopic, stable and of high equivalent weight.

3. (c) — molality uses the mass of solvent, and mass does not expand on heating.

4. (c) — N = M \times n = 0.1 \times 2 = 0.2 N.

5. (d) — the purple permanganate colour itself marks the end point.

6. (b) — for dilute aqueous solutions the density is 1 g cm⁻³, so 1 part in 10⁶ = 1 mg per 1000 g ≈ 1 mg L⁻¹.

Group B — Short answer (5 marks each)

1. Define equivalent weight. Calculate the equivalent weight of KMnO₄ in acidic, neutral and strongly alkaline media, writing the half reaction in each case. [5]
2. What mass of oxalic acid dihydrate is needed to prepare 500 cm³ of a 0.1 N solution? [5]
3. 25.0 cm³ of 0.1 N NaOH was exactly neutralised by 20.0 cm³ of sulphuric acid. Calculate the normality, molarity and strength in g L⁻¹ of the acid. [5]
4. Distinguish between a primary and a secondary standard substance. Give two examples of each and state why sodium hydroxide cannot be a primary standard. [5]
5. Concentrated nitric acid is 70 % by mass and has a density of 1.42 g cm⁻³. Calculate its molarity, and the volume of it needed to prepare 1.00 L of 0.5 M HNO₃. [5]
6. Why is only dilute sulphuric acid used to acidify a potassium permanganate titration, and why must an oxalic acid titration be carried out hot? [5]

Answers to Group B

1. Acidic: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, E = 158/5 = 31.6. Neutral or faintly alkaline: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻, E = 158/3 = 52.67. Strongly alkaline: MnO₄⁻ + e⁻ → MnO₄²⁻, E = 158/1 = 158.

2. E(\text{H}_2\text{C}_2\text{O}_4\cdot 2\text{H}_2\text{O}) = 126/2 = 63. Equivalents needed = 0.1 \times 0.500 = 0.05, so w = 0.05 \times 63 = 3.15\ \text{g}.

3. N_{\text{acid}} = (0.1 \times 25.0)/20.0 = 0.125\ \text{N}; M = N/n = 0.125/2 = 0.0625\ \text{M}; S = N \times E = 0.125 \times 49 = 6.125\ \text{g L}^{-1}.

4. A primary standard is pure, stable, highly soluble and of high equivalent weight, so a standard solution follows from one accurate weighing (oxalic acid, K₂Cr₂O₇). A secondary standard must be titrated against a primary standard (HCl, KMnO₄). NaOH is deliquescent and absorbs CO₂ from the air to form Na₂CO₃, so the mass weighed is not pure NaOH.

5. M = 10pd/M_w = (10 \times 70 \times 1.42)/63 = 994/63 = 15.78\ \text{M}. For the dilution, M_1V_1 = M_2V_2: V_1 = (0.5 \times 1000)/15.78 = 31.7\ \text{cm}^3.

6. HCl would itself be oxidised by permanganate (2MnO₄⁻ + 16H⁺ + 10Cl⁻ → 2Mn²⁺ + 5Cl₂ + 8H₂O), consuming extra KMnO₄ and giving a high result; HNO₃ is an oxidising agent and would oxidise the reductant in the flask. Only dilute H₂SO₄ merely supplies H⁺. The oxalate–permanganate reaction is very slow at room temperature (it is autocatalysed by the Mn²⁺ it produces), so the flask is warmed to 60–70 °C to get a sharp end point.

Group C — Long answer (8 marks each)

1. (a) State the law of equivalence and use it to derive the normality equation N_1V_1 = N_2V_2. [3](b) 1.06 g of anhydrous sodium carbonate was dissolved in water and the solution made up to 250 cm³. 25.0 cm³ of this solution required 20.0 cm³ of hydrochloric acid for complete neutralisation. Calculate the normality of the acid and its strength in g L⁻¹. [5]
2. (a) What is a redox titration? Write the balanced equation for the titration of ferrous sulphate with potassium permanganate in acidic medium and give the n-factor of each reagent. [3](b) 2.00 g of an impure sample of Mohr's salt was dissolved in dilute H₂SO₄ and made up to 250 cm³. 25.0 cm³ of this solution required 10.0 cm³ of 0.05 N KMnO₄. Calculate the percentage purity of the sample. [5]

Answers to Group C

1.(b) E(\text{Na}_2\text{CO}_3) = 106/2 = 53, so the equivalents taken are 1.06/53 = 0.0200 eq in 250 cm³, i.e. 20.0 meq.

The 25.0 cm³ aliquot contains 20.0 \times (25.0/250) = 2.00 meq. Therefore

N(\text{HCl}) = \frac{2.00}{20.0} = 0.100\ \text{N}

and S = N \times E = 0.100 \times 36.5 = 3.65\ \text{g L}^{-1}.

2.(a) A redox titration is one in which the titrant and analyte react by transfer of electrons, the end point being fixed by a redox indicator or by the titrant's own colour.

2KMnO₄ + 10FeSO₄ + 8H₂SO₄ → K₂SO₄ + 2MnSO₄ + 5Fe₂(SO₄)₃ + 8H₂O

n-factor of KMnO₄ = 5 (MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O); n-factor of FeSO₄ = 1 (Fe²⁺ → Fe³⁺ + e⁻).

2.(b) Milliequivalents of KMnO₄ used = 0.05 \times 10.0 = 0.500 meq, so the 25.0 cm³ aliquot contained 0.500 meq of Mohr's salt.

In the full 250 cm³, meq = 0.500 \times 10 = 5.00 meq = 5.00\times10^{-3} eq.

Since E = 392 (n-factor 1), the mass of pure Mohr's salt is

w = 5.00\times10^{-3} \times 392 = 1.96\ \text{g}

\text{purity} = \frac{1.96}{2.00} \times 100 = 98.0\ \%