Volumetric Analysis

Volumetric analysis is the branch of the paper where chemistry becomes a measurement: a solution of known strength is run from a burette into a measured volume of the unknown until the reaction is just complete. It carries about one question (our estimate from the MEC topic list), and that question is nearly always arithmetic — an equivalent weight, a conversion between two concentration units, or a titration volume from N_1V_1 = N_2V_2. The definitions are few; the marks are lost on units, on the n-factor of the reagent, and on the difference between molality and molarity.

8.1 Equivalent weight and every concentration unit

What a titration is

Term	Meaning
titrant	the solution of known concentration, run from the burette
analyte (titrand)	the solution being measured, pipetted into the conical flask
standard solution	a solution whose concentration is accurately known
equivalence point	the point at which the reactants have combined in the exact stoichiometric ratio
end point	the point at which the indicator changes colour — the observed approximation to the equivalence point
titration error	the difference between the end point and the equivalence point

The whole method rests on one measurement: the volume delivered. Everything else is a conversion from that volume to an amount of substance.

Equivalent weight and the n-factor

The n-factor is not a property of the substance alone — it is a property of the substance in a particular reaction. That is why KMnO₄ has three different equivalent weights.

Substance	Reaction role	n-factor	E (g eq⁻¹)
HCl	acid (basicity 1)	1	36.5
CH₃COOH	acid (basicity 1)	1	60
H₂SO₄	acid (basicity 2)	2	49
H₃PO₄	acid (basicity 3)	3	32.7
NaOH	base (acidity 1)	1	40
Ca(OH)₂	base (acidity 2)	2	37
Na₂CO₃	salt, 2 H⁺ accepted	2	53
NaHCO₃	salt, 1 H⁺ accepted	1	84
KMnO₄ (acidic)	Mn⁷⁺ → Mn²⁺, 5 e⁻	5	31.6
KMnO₄ (neutral)	Mn⁷⁺ → Mn⁴⁺, 3 e⁻	3	52.7
KMnO₄ (alkaline)	Mn⁷⁺ → Mn⁶⁺, 1 e⁻	1	158
K₂Cr₂O₇ (acidic)	2 Cr⁶⁺ → 2 Cr³⁺, 6 e⁻	6	49
FeSO₄	Fe²⁺ → Fe³⁺, 1 e⁻	1	152
H₂C₂O₄·2H₂O	C₂O₄²⁻ → 2CO₂, 2 e⁻	2	63
Na₂S₂O₃	2S₂O₃²⁻ → S₄O₆²⁻, 2 e⁻ for two	1	158
I₂	I₂ → 2I⁻, 2 e⁻	2	127

The concentration units

Unit	Definition	Formula
% w/w	mass of solute per 100 g of solution	\dfrac{m_{\text{solute}}}{m_{\text{solution}}}\times 100
% w/v	mass of solute per 100 mL of solution	\dfrac{m_{\text{solute}}\text{ (g)}}{V\text{ (mL)}}\times 100
g L⁻¹	mass of solute per litre of solution	m/V
Molarity M	moles of solute per litre of solution	\dfrac{n}{V(\text{L})}
Molality m	moles of solute per kilogram of solvent	\dfrac{n}{m_{\text{solvent}}(\text{kg})}
Normality N	equivalents of solute per litre of solution	\dfrac{\text{eq}}{V(\text{L})} = M \times n\text{-factor}
Formality F	formula masses per litre of solution	\dfrac{\text{formula mass}}{\text{L}}
ppm	parts per million by mass	\dfrac{m_{\text{solute}}}{m_{\text{solution}}}\times 10^{6} ≈ mg kg⁻¹
ppb	parts per billion by mass	\dfrac{m_{\text{solute}}}{m_{\text{solution}}}\times 10^{9} = μg kg⁻¹
mole fraction x	moles of one component ÷ total moles	\dfrac{n_i}{n_1+n_2+\dots}

Molarity from percentage and density. This is the conversion the paper actually asks, and it needs the density because a percentage is by mass while molarity is by volume:

Molality of a glucose solution

Problem. A 10 % w/w aqueous glucose solution contains 10 g of glucose (M = 180) in 90 g of water. Find its molality.

Solution. n = 10/180 = 0.0556 mol; solvent = 90 g = 0.090 kg. m = 0.0556/0.090 = 0.617 mol kg⁻¹. Note that the denominator is the mass of water, not the mass of the solution (100 g). Using 0.100 kg gives 0.556, the classic wrong option.

Dilution

Diluting a solution adds solvent but not solute, so the number of moles is unchanged:

8.2 Primary and secondary standards; the law of equivalence

Primary and secondary standards

Quality of a primary standard	Why it matters
very high purity (99.9 % or better)	the mass weighed is the mass of the substance
stable in air — not hygroscopic, not efflorescent, not oxidised	the mass does not drift while you weigh it
high molar mass	a weighing error of 1 mg is a smaller fraction of a larger mass
readily soluble in water	it must dissolve completely and quickly
reacts rapidly and stoichiometrically	a sharp end point, a known equation
non-toxic and reasonably cheap	it is used in every laboratory

Primary standards	Used to standardise
oxalic acid H₂C₂O₄·2H₂O	NaOH, KMnO₄
anhydrous sodium carbonate Na₂CO₃	HCl, H₂SO₄
potassium hydrogen phthalate (KHP)	NaOH
borax Na₂B₄O₇·10H₂O	HCl
potassium dichromate K₂Cr₂O₇	Na₂S₂O₃, Fe²⁺ solutions
sodium chloride NaCl	AgNO₃
pure zinc or copper	EDTA, iodine
Mohr's salt FeSO₄·(NH₄)₂SO₄·6H₂O	KMnO₄, K₂Cr₂O₇

Secondary standard	Why it cannot be a primary standard
NaOH, KOH	hygroscopic; absorb CO₂ from the air, so the effective concentration drifts
HCl, H₂SO₄	HCl is volatile; both absorb water and their strength varies
KMnO₄	decomposes slowly, oxidises water and traces of organic matter; light-sensitive
Na₂S₂O₃	decomposes in acid and is attacked by CO₂ and bacteria; concentration falls with time
I₂	volatile and slightly soluble in water; solutions change strength on standing

The law of equivalence and the normality equation

Equivalents can be counted in three ways, and all three must agree:

That identity is what makes normality so convenient: it removes the coefficients from the calculation. Twenty millilitres of 0.1 N acid neutralises exactly twenty millilitres of 0.1 N base, whatever the acid and whatever the base, because "equivalent" already means "capable of reacting with one mole of the other partner".

Back titration

A back titration is used when the reaction is slow, when the analyte is insoluble, or when the end point is hard to see. A measured excess of a standard reagent is added, the mixture is left to react completely, and the unreacted excess is then titrated:

where N_1V_1 refers to the excess reagent added and N_2V_2 to the unreacted part.

Choosing the indicator

The indicator must change colour within the steep part of the titration curve, so that the end point falls as close as possible to the equivalence point.

Titration	pH at equivalence	Indicator
strong acid + strong base	7	phenolphthalein (8.3–10.0) or methyl orange (3.1–4.4)
weak acid + strong base	above 7 (about 8.7)	phenolphthalein only
strong acid + weak base	below 7 (about 5.1)	methyl orange only
weak acid + weak base	near 7, no sharp jump	none is satisfactory
KMnO₄ against a reductant	—	self-indicating (purple to colourless)
iodine against thiosulphate	—	starch (blue-black to colourless)

Formula and fact sheet

Quantity	Relation
Equivalent weight	E = M/n; equivalents = \text{mass}/E
n-factor: acid	basicity (replaceable H⁺): HCl 1, H₂SO₄ 2, H₃PO₄ 3
n-factor: base	acidity (OH⁻): NaOH 1, Ca(OH)₂ 2, Al(OH)₃ 3
n-factor: salt	total charge on the cation: NaCl 1, Na₂CO₃ 2, Al₂(SO₄)₃ 6
n-factor: redox	electrons gained or lost per formula unit
KMnO₄	5 (acid, Mn²⁺), 3 (neutral, MnO₂), 1 (alkaline, MnO₄²⁻)
K₂Cr₂O₇	6 in acid (2Cr³⁺); E = 49
Molarity	M = n/V(\text{L}); M = \dfrac{10 \times \%\text{w/w} \times d}{M_B}
Molality	m = n/\text{kg solvent}
Normality	N = M \times n\text{-factor}
Formality	formula masses per litre; used for ionic solids such as NaCl
ppm / ppb	mg kg⁻¹ / μg kg⁻¹ (≈ mg L⁻¹ for dilute aqueous)
Mole fraction	x_i = n_i/\sum n; \sum x_i = 1
Dilution	M_1V_1 = M_2V_2; N_1V_1 = N_2V_2
Law of equivalence	N_1V_1 = N_2V_2 at the equivalence point
Back titration	equivalents of sample = N_1V_1 - N_2V_2
Strength	g L⁻¹ = N \times E

Indicator	pH range	Colour change	Use
methyl orange	3.1–4.4	red to yellow	strong acid + weak base
methyl red	4.4–6.2	red to yellow	strong acid + weak base
bromothymol blue	6.0–7.6	yellow to blue	strong acid + strong base
phenolphthalein	8.3–10.0	colourless to pink	weak acid + strong base
starch	—	blue-black to colourless	iodometric titrations
KMnO₄ itself	—	purple to colourless	permanganate titrations

Primary standards: oxalic acid, Na₂CO₃, KHP, borax, K₂Cr₂O₇, NaCl, pure Zn/Cu, Mohr's salt. Secondary standards: NaOH, KOH, HCl, H₂SO₄, KMnO₄, Na₂S₂O₃, I₂.

Chapter summary

• Volumetric analysis measures the volume of a standard solution needed to react completely with a measured volume of analyte; the end point is the indicator's approximation to the equivalence point.
• Equivalent weight E = M/n; the n-factor is the basicity of an acid, the acidity of a base, the cation charge of a salt, or the electron change of an oxidant or reductant.
• KMnO₄ has three equivalent weights because it is reduced to Mn²⁺ (n = 5), MnO₂ (n = 3) or MnO₄²⁻ (n = 1); K₂Cr₂O₇ is always n = 6.
• Molarity is per litre of solution and changes with temperature; molality is per kilogram of solvent and does not. Normality = molarity × n-factor.
• M = 10 \times \%\text{w/w} \times d / M_B is the only conversion that needs the density; ppm and ppb are mass ratios, and mole fractions sum to 1.
• Dilution conserves moles: M_1V_1 = M_2V_2.
• A primary standard is pure, stable, non-hygroscopic, soluble, fast-reacting and of high molar mass; NaOH, HCl, KMnO₄, Na₂S₂O₃ and I₂ are secondary standards and must be standardised.
• At the equivalence point N_1V_1 = N_2V_2, and equivalents can be counted as NV, mass/E, or n-factor × moles.
• In a back titration the equivalents used by the sample are the excess added minus the excess titrated back.
• Choose the indicator whose transition range lies on the steep part of the curve: phenolphthalein for weak acid + strong base, methyl orange for strong acid + weak base.

Practice questions

Set A — Recall (23 questions)

1. The equivalent weight of an acid is its molar mass divided byaits basicitybits aciditycthe number of oxygen atomsdthe number of moles
2. The equivalent weight of H₂SO₄ (M = 98) isa98b24.5c49d32.7
3. The n-factor of K₂Cr₂O₇ in acidic medium isa3b6c2d7
4. The equivalent weight of KMnO₄ in acidic medium (M = 158) isa158b52.7c79d31.6
5. The equivalent weight of anhydrous Na₂CO₃ (M = 106) isa106b53c35.3d26.5
6. Molarity is the number of moles of solute peralitre of solutionbkilogram of solventclitre of solventdkilogram of solution
7. Molality is expressed inamol L⁻¹bmol kg⁻¹ceq L⁻¹dg L⁻¹
8. Normality is equal to molarity multiplied byamolar massbdensitycthe n-factordvolume in litres
9. One part per million is the same asa1 g per kgb1 mg per gc1 μg per kgd1 mg per kg
10. The sum of the mole fractions of all the components of a solution isa1b100c0dequal to the molarity
11. Formality is normally used foraweak acidsbgasescionic solids that do not exist as moleculesdnon-electrolytes
12. Which substance is a primary standard?aNaOHboxalic acidcHCldKMnO₄
13. Which substance is NOT a primary standard?aNaOHbNa₂CO₃cpotassium hydrogen phthalatedborax
14. The law of equivalence is written asaM_1V_1 = M_2V_2bN_1V_2 = N_2V_1cN_1 = N_2dN_1V_1 = N_2V_2
15. The indicator used for a strong acid–weak base titration isaphenolphthaleinbmethyl orangecstarchdlitmus
16. The indicator used for a weak acid–strong base titration isamethyl orangebstarchcphenolphthaleindmethyl red
17. The number of equivalents of a solute is equal toaN \times V in litresbM \times V in litrescmass × EdN/V
18. The equivalent weight of oxalic acid dihydrate (M = 126) isa126b45c90d63
19. The point at which the indicator changes colour in a titration is called theaequivalence pointbend pointcneutral pointdstoichiometric point
20. Potassium permanganate needs no external indicator because it isacolourlessbyellowcpurple and turns colourlessdwhite
21. A primary standard should haveaa high molar massblow solubilityca low molar massdhigh volatility
22. The burette should be rinsed withadistilled waterbthe analyte solutioncthe indicatordthe titrant
23. 1 M H₂SO₄ is the same asa0.5 Nb2 Nc1 Nd4 N

Answers and explanations — Set A

1. (a) — An acid's n-factor is its basicity, the number of replaceable H⁺ per molecule. For a base it would be the acidity, the number of OH⁻.

2. (c) — Sulphuric acid is dibasic, so E = 98/2 = 49. Option (a) forgets the n-factor and (d) is the value for H₃PO₄.

3. (b) — Two chromium atoms go from +6 to +3, so 2 × 3 = 6 electrons are gained per formula unit. Option (a) counts only one chromium atom.

4. (d) — Mn goes from +7 to +2, so n = 5 and E = 158/5 = 31.6. Option (b) is the value in neutral medium (n = 3).

5. (b) — The carbonate ion takes two protons, so n = 2 and E = 106/2 = 53. Option (a) uses n = 1, which is right only for NaHCO₃.

6. (a) — Molarity is moles per litre of solution, not of solvent. The volume includes the solute.

7. (b) — Molality is moles of solute per kilogram of solvent, so its unit is mol kg⁻¹ and it does not change with temperature.

8. (c) — N = M \times n-factor. The n-factor is 1 for HCl, 2 for H₂SO₄ and 5 for KMnO₄ in acid.

9. (d) — ppm is 1 mg per kg (or per litre of dilute aqueous solution). Option (c) is one part per billion.

10. (a) — Mole fractions are fractions of the total number of moles, so they must add up to exactly 1.

11. (c) — Formality counts formula masses per litre and is used for ionic solids such as NaCl, which contain no discrete molecules.

12. (b) — Oxalic acid is pure, stable and non-hygroscopic. NaOH absorbs CO₂ and water, HCl is volatile and KMnO₄ decomposes.

13. (a) — NaOH is hygroscopic and absorbs CO₂ from the air, so its concentration drifts. The other three are standard primary standards.

14. (d) — At the equivalence point the equivalents of the two reactants are equal: N_1V_1 = N_2V_2. Option (a) is true only for a 1 : 1 reaction in molar terms.

15. (b) — The salt of a weak base is acidic at equivalence (pH about 5), so the indicator must change in the acid region: methyl orange.

16. (c) — The salt of a weak acid is basic at equivalence (pH about 8.7), so phenolphthalein, which changes at 8.3–10.0, is the right choice.

17. (a) — Equivalents = N \times V in litres. Option (b) gives moles, not equivalents.

18. (d) — Oxalate loses two electrons, so n = 2 and E = 126/2 = 63. Option (b) is the anhydrous value, 90/2.

19. (b) — The end point is what you observe; the equivalence point is the true stoichiometric point. The small difference between them is the titration error.

20. (c) — The purple permanganate ion is reduced to almost colourless Mn²⁺, so the first permanent pink tint marks the end point.

21. (a) — A high molar mass makes the weighing error a smaller fraction of the mass taken. Low solubility and volatility would both be disqualifying.

22. (d) — Rinsing with the titrant removes water that would otherwise dilute it. Rinsing with water would make the delivered concentration lower than stated.

23. (b) — Sulphuric acid has two replaceable protons, so N = 2M = 2 N. Option (c) is the value for a monobasic acid.

Set B — Understanding (14 questions)

1. KMnO₄ has different equivalent weights in acidic, neutral and alkaline media becauseaits n-factor depends on the electrons gainedbits molar mass changescthe temperature changesdit decomposes
2. NaOH cannot be used as a primary standard because itais insoluble in waterbhas a low molar masscis a strong basedabsorbs water and CO₂ from the air
3. Molality is preferred to molarity in colligative-property work because molalityais easier to measurebdoes not change with temperaturecuses a smaller mass of solutedis always numerically smaller
4. 0.1 M H₂SO₄ is 0.2 N becauseaits molar mass is 98bit is a strong acidcit has two replaceable H⁺ per moleculedit contains two oxygen atoms
5. When a solution is diluted with water, the number of moles of soluteastays the samebincreasescdecreasesddoubles
6. Phenolphthalein is unsuitable for a strong acid–weak base titration because the pH at equivalence isaabove 10bexactly 7cbelow 3dbelow its colour-change range
7. A primary standard should have a high molar mass so thatait dissolves fasterbit reacts fasterca weighing error is a small fraction of the massdit is cheaper
8. A back titration is used whenathe reaction is very fastbthe analyte is insoluble or the reaction is slowcno indicator is availabledthe titrant is a primary standard
9. The difference between the end point and the equivalence point is called theatitration errorbindicator constantcpercentage puritydbuffer error
10. No external indicator is needed in a permanganate titration becauseathe analyte is already colouredbthe end point is a precipitatecKMnO₄ is its own indicatordthe pH changes sharply
11. Which solution has the greater normality: 0.1 M H₂SO₄ or 0.1 M HCl?aHClbboth are equalcit depends on the volumedH₂SO₄, which has two replaceable protons
12. A concentration of 1 ppm in water meansa1 g per litreb1 mg per kgc1 μg per kgd1 mol per litre
13. The pipette should be rinsed withadistilled waterbthe titrantcthe solution it will deliverdthe indicator
14. A few drops of distilled water left in the conical flask before the titrationado not change the number of moles of analytebincrease its concentrationcchange the end pointdinvalidate the titration

Answers and explanations — Set B

1. (a) — The manganese is reduced to Mn²⁺ (5 e⁻), MnO₂ (3 e⁻) or MnO₄²⁻ (1 e⁻), so E = 158/5, 158/3 or 158/1. The molar mass never changes.

2. (d) — Pellets of NaOH take up water and CO₂, forming carbonate, so the mass weighed no longer corresponds to the stated amount of NaOH.

3. (b) — Molality uses a mass of solvent, which does not expand or contract with temperature; a molarity would change as the solution's volume changed.

4. (c) — Normality counts replaceable H⁺, and H₂SO₄ has two. The strength of the acid is irrelevant to the n-factor.

5. (a) — Dilution adds solvent only, so the moles of solute are conserved. That is exactly why M_1V_1 = M_2V_2.

6. (d) — The equivalence pH of about 5 lies below phenolphthalein's range of 8.3–10.0, so the indicator would change far too late.

7. (c) — A 1 mg error on a 2 g mass is 0.05 %; the same error on a 0.05 g mass is 2 %. A high molar mass makes the relative error small.

8. (b) — When the analyte dissolves slowly, reacts slowly, or gives no sharp end point, an excess of reagent is added and the remainder is titrated back.

9. (a) — This difference is the titration error. A good indicator keeps it small by changing on the steep part of the curve.

10. (c) — The purple colour of MnO₄⁻ disappears as it is reduced, so the first permanent pink colour is the end point.

11. (d) — N = M \times n-factor: 0.1 M H₂SO₄ is 0.2 N while 0.1 M HCl is 0.1 N. Normality does not depend on the volume taken.

12. (b) — One part per million is 1 mg kg⁻¹ (≈ 1 mg L⁻¹ for dilute aqueous solutions). Option (c) is 1 ppb.

13. (c) — The pipette must deliver exactly the solution it holds; rinsing with water would dilute it and rinsing with titrant would add extra reagent.

14. (a) — The water adds volume but no analyte, and the titration measures moles, not concentration in the flask. The end point is unchanged.

Set C — Application (9 questions)

1. The molarity of 98 % w/w H₂SO₄ of density 1.84 g mL⁻¹ isa9.2 Mb1.84 Mc18.4 Md36.8 M
2. The normality of 0.05 M K₂Cr₂O₇ in acidic medium isa0.30 Nb0.05 Nc0.10 Nd0.60 N
3. 25 mL of 2.0 M HCl is diluted to 500 mL. The new molarity isa0.50 Mb0.05 Mc1.0 Md0.10 M
4. The volume of 0.200 M HCl needed to neutralise 25.0 mL of 0.100 M Na₂CO₃ completely isa12.5 mLb25.0 mLc50.0 mLd5.0 mL
5. The mass of Na₂CO₃ needed to prepare 250 mL of 0.100 N solution is (Na = 23)a1.325 gb2.650 gc0.662 gd5.300 g
6. 1.20 g of impure CaCO₃ is dissolved in 50.0 mL of 0.500 N HCl; the excess acid needs 20.0 mL of 0.250 N NaOH. The percentage of CaCO₃ (E = 50) isa20.0 %b96.0 %c66.7 %d83.3 %
7. The molality of a 10 % w/w glucose solution (M = 180) isa0.556 mb1.00 mc0.617 md0.100 m
8. 0.020 g of Ca²⁺ dissolved in 1.00 kg of water gives a concentration ofa2 ppmb20 ppmc200 ppmd0.02 ppm
9. 0.200 g of impure NaOH needs 48.0 mL of 0.100 M HCl for neutralisation. The percentage purity of the NaOH is (Na = 23, O = 16, H = 1)a48 %b80 %c96 %d100 %

Answers and explanations — Set C

1. (c) — M = \dfrac{10 \times 98 \times 1.84}{98} = 18.4 M. Option (d) is the normality, since H₂SO₄ has two replaceable protons.

2. (a) — N = M \times 6 = 0.05 \times 6 = 0.30 N. Option (c) wrongly uses n = 2, the n-factor of oxalic acid.

3. (d) — M_2 = M_1V_1/V_2 = 2.0 \times 25/500 = 0.10 M. Option (a) divides by 100 instead of 500.

4. (b) — N(\text{Na}_2\text{CO}_3) = 0.200 N and N(\text{HCl}) = 0.200 N, so N_1V_1 = N_2V_2 gives 25.0 mL. By moles: 0.00250 mol carbonate needs 0.00500 mol acid.

5. (a) — Equivalents = 0.250 \times 0.100 = 0.0250; mass = 0.0250 \times 53 = 1.325 g. Option (b) uses E = 106, i.e. n = 1.

6. (d) — Acid taken 0.0250 eq, acid left 0.00500 eq, so 0.0200 eq reacted with the carbonate; mass = 0.0200 \times 50 = 1.00 g, and 1.00/1.20 = 83.3 %.

7. (c) — 10 g glucose is 0.0556 mol and the solvent is 90 g = 0.090 kg, so m = 0.0556/0.090 = 0.617. Option (a) divides by the 100 g of solution.

8. (b) — 0.020 g = 20 mg in 1.00 kg, so 20 ppm. Option (a) forgets to convert grams to milligrams.

9. (c) — Moles of HCl = 0.0480 \times 0.100 = 0.00480; mass of NaOH = 0.00480 \times 40 = 0.192 g; purity = 0.192/0.200 = 96 %.