Volumetric Analysis Class 12 Chemistry (NEB) Unit 1 | Notes, All Formulas, Numerical Solutions & PDF Download

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Class 12 Chemistry • Unit 1

Volumetric Analysis

NEB Syllabus 2083 • Notes, Formulas and Solved Exercise

Equivalent weight, concentration units, standard substances, the normality equation and titrations, with all formulas, every exercise question solved, and handwritten and typed PDF notes.

1. Syllabus

Unit 1: Volumetric Analysis

1.1 Introduction to gravimetric analysis, volumetric analysis and equivalent weight

1.2 Relationship between equivalent weight, atomic weight and valency

1.3 Equivalent weight of compounds

1.4 Concentration of solution and its units

1.5 Primary and secondary standard substances

1.6 Law of equivalence and normality equation

1.7 Titration and its types

These notes follow the NEB Class 12 Chemistry syllabus (+2 Chemistry, Nepal). Revise the formulas first, practise the multiple choice questions, then work through the short and long answer questions, and finish with the PDF notes.

2. Volumetric Analysis at a Glance

Gravimetric analysis finds the amount of an analyte by converting it into a pure compound that is weighed. Volumetric analysis finds it by measuring the volume of a standard solution (the titrant) that reacts completely with the analyte in a titration.

Equivalent weight

\[ \text{Equivalent weight}=\frac{\text{Atomic or molecular weight}}{\text{Valency (n-factor)}} \]

Substance	n-factor	Examples (equivalent weight)
Element	Valency	Mg = 24/2 = 12; Fe(III) = 56/3 = 18.67
Acid	Basicity (replaceable H⁺)	HCl 36.5; H₂SO₄ 49; H₃PO₄ 32.67; oxalic acid dihydrate 63
Base	Acidity (replaceable OH⁻)	NaOH 40; Ca(OH)₂ 37
Salt	Total positive (or negative) charge	Na₂CO₃ 53; Na₂SO₄ 71
Oxidising or reducing agent	Electrons gained or lost per formula unit	KMnO₄ (acidic) 158/5 = 31.6; K₂Cr₂O₇ 294/6 = 49; FeSO₄ 152; Mohr’s salt 392

Concentration of solutions and its units

Unit	Definition and formula
Molarity \(M\)	\(M=\dfrac{\text{moles of solute}}{V\ (\mathrm{L})}=\dfrac{w}{M_w\,V(\mathrm{L})}\)
Normality \(N\)	\(N=\dfrac{\text{gram equivalents}}{V\ (\mathrm{L})}=\dfrac{w}{E\,V(\mathrm{L})}=M\times n\text{-factor}\)
Molality \(m\)	moles of solute per kilogram of solvent (independent of temperature)
Mole fraction	\(X_A=\dfrac{n_A}{n_A+n_B}\)
Percentage w/w, w/v	g of solute per 100 g, or per 100 mL, of solution; \(\%\,(w/v)=\%\,(w/w)\times\text{density}\)
Molarity from % w/w	\(M=\dfrac{\%\,(w/w)\times\text{density}\times10}{M_w}\)
ppm and ppb	parts per \(10^6\) and per \(10^9\) parts of solution; \(1\ \mathrm{ppm}=1000\ \mathrm{ppb}\); in dilute water solutions \(1\ \mathrm{ppm}=1\ \mathrm{mg\,L^{-1}}\)

Primary and secondary standard substances

Basis	Primary standard	Secondary standard
Preparation of solution	Directly by accurate weighing	Must be standardised against a primary standard
Properties	Pure, stable, non-hygroscopic, high equivalent weight, soluble	Impure, hygroscopic, volatile or unstable
Examples	Oxalic acid, anhydrous Na₂CO₃, Mohr’s salt	NaOH, HCl, H₂SO₄, KMnO₄

Law of equivalence and the normality equation

Substances react in the ratio of their equivalent weights, so at the end point the gram equivalents of the titrant and the analyte are equal.

\[ N_1V_1=N_2V_2\qquad\text{(titration)},\qquad M_1V_1=M_2V_2\qquad\text{(dilution)} \]

\[ N_{\text{mix}}=\frac{N_1V_1+N_2V_2}{V_1+V_2},\qquad\%\ \text{purity}=\frac{\text{mass of pure substance}}{\text{mass of sample}}\times100 \]

Titration and its types

Type	Reaction	Example and indicator
Acid-base (neutralisation)	Acid + base → salt + water	Oxalic acid with NaOH (phenolphthalein, colourless to pink); HCl with Na₂CO₃ (methyl orange, red to yellow)
Redox	Transfer of electrons	KMnO₄ with oxalic acid or Fe²⁺ (KMnO₄ is a self-indicator; end point faint permanent pink)
Precipitation (argentometric)	Formation of an insoluble precipitate	AgNO₃ with Cl⁻ or Br⁻
Complexometric	Formation of a complex	EDTA with metal ions

The end point is the stage at which the indicator changes colour; it is chosen to coincide with the equivalence point, where the titrant and analyte are in exactly equivalent amounts.

3. Multiple Choice Questions

Question 1. Which of the following principles is used in gravimetric analysis?

a) Mass measurement
b) Volume measurement
c) Color change
d) Heat transfer

Correct Answer: a) Mass measurement

Reason: Gravimetric analysis determines the mass of a pure compound to which the analyte is chemically related.

Question 2. Ferrosoferric oxide (magnetite) is an iron oxide with the chemical formula Fe₃O₄. What is the equivalent weight of iron in the oxide?

a) 56
b) 28
c) 18.6
d) 7

Correct Answer: c) 18.6

Reason: Fe₃O₄ is FeO·Fe₂O₃. Taking the valency of iron as 3:

\[ E=\frac{\text{atomic weight}}{\text{valency}}=\frac{56}{3}\approx18.6 \]

Question 3. Potassium dichromate (K₂Cr₂O₇) is an orange-red crystalline compound and a powerful oxidizing agent. What is the equivalent weight of this oxidizing agent, having molecular weight \(M\), in the following ionic equation?
Cr₂O₇²⁻(aq) + SO₃²⁻(aq) → Cr³⁺(aq) + SO₄²⁻(aq)

a) \(M\)
b) \(M/2\)
c) \(M/3\)
d) \(M/6\)

Correct Answer: d) \(M/6\)

Reason: Each Cr changes from +6 to +3, a gain of 3 electrons. Two Cr atoms give 6 electrons per formula unit, so the equivalent weight is \(M/6\).

Question 4. If 1 ppm means one part per million, then what is the meaning of 1 ppb?

a) One part per ten billion
b) One part per billion
c) One part per thousand
d) One part per trillion

Correct Answer: b) One part per billion

Reason: ppb means 1 part of solute in \(10^{9}\) parts of solution.

Question 5. Which of the following is a primary standard substance used to prepare a standard solution in volumetric analysis?

a) Hydrochloric acid (HCl)
b) Sodium hydroxide (NaOH)
c) Oxalic acid (C₂H₂O₄)
d) Potassium permanganate (KMnO₄)

Correct Answer: c) Oxalic acid (C₂H₂O₄)

Reason: Oxalic acid is available in a pure, stable state and can be weighed accurately to make a standard solution directly. HCl, NaOH and KMnO₄ are secondary standards.

Question 6. Which one of the following is NOT a characteristic of a good primary standard substance?

a) High stability
b) Low reactivity
c) High solubility in water
d) High hygroscopicity

Correct Answer: d) High hygroscopicity

Reason: A hygroscopic substance absorbs moisture from the air, so a known mass of the pure substance cannot be weighed accurately.

Question 7. The law of equivalence is a fundamental principle in chemistry, particularly in stoichiometry and titration. The law of equivalence states that at the endpoint of a titration:

a) The volume of acid equals the volume of base
b) The number of equivalents of acid equals the number of equivalents of base
c) The normality of acid equals the normality of base
d) The molarity of acid equals the molarity of base

Correct Answer: b) The number of equivalents of acid equals the number of equivalents of base

Reason: Substances react in the ratio of their equivalent weights, so at complete neutralisation the gram equivalents of acid and base are equal.

Question 8. If 25 mL of a deci-molar H₂SO₄ solution reacts completely with 50 mL of a NaOH solution, what is the molarity of the NaOH solution?

a) 0.05 M
b) 0.1 M
c) 0.2 M
d) 0.25 M

Correct Answer: b) 0.1 M

Reason: Deci-molar H₂SO₄ is 0.1 M. Its basicity is 2, so \(N=0.2\ \mathrm{N}\). By \(N_1V_1=N_2V_2\):

\[ 0.2\times25=N_2\times50\;\Rightarrow\;N_2=0.1\ \mathrm{N} \]

NaOH has an acidity of 1, so its molarity is also 0.1 M.

Question 9. In a redox titration involving potassium permanganate (KMnO₄) and oxalic acid (C₂H₂O₄), which of the following statements is true regarding KMnO₄?

a) An oxidizing agent because it gains electrons
b) A reducing agent because it gains electrons
c) An oxidizing agent because it loses electrons
d) A reducing agent because it loses electrons

Correct Answer: a) An oxidizing agent because it gains electrons

Reason: KMnO₄ oxidises oxalic acid and is itself reduced (Mn⁷⁺ to Mn²⁺). Reduction is the gain of electrons.

Question 10. If 0.5 g of Na₂CO₃ is dissolved in 100 mL of solution, what is the molarity of the solution thus prepared? (Molecular weight of Na₂CO₃ = 106 g/mol)

a) 0.047 M
b) 0.094 M
c) 0.188 M
d) 0.235 M

Correct Answer: a) 0.047 M

\[ n=\frac{0.5}{106}=4.717\times10^{-3}\ \mathrm{mol},\qquad M=\frac{4.717\times10^{-3}}{0.1}\approx0.047\ \mathrm{M} \]

Question 11. You are asked to prepare a 0.25 M solution of NaOH using 20 g of it. What volume of solution can be made using this amount?

a) 0.5 L
b) 1.0 L
c) 2.0 L
d) 4.0 L

Correct Answer: c) 2.0 L

\[ n=\frac{20}{40}=0.5\ \mathrm{mol},\qquad V=\frac nM=\frac{0.5}{0.25}=2.0\ \mathrm{L} \]

Question 12. A 25 mL semi-molar solution of HCl is titrated with 0.5 N Na₂CO₃ solution. What volume of titrant should be added from the burette for one-fourth neutralization of the analyte in the conical flask?

a) 50 mL
b) 25 mL
c) 12.5 mL
d) 6.25 mL

Correct Answer: d) 6.25 mL

Reason: Semi-molar HCl is 0.5 M = 0.5 N. Full neutralisation needs

\[ 0.5\times25=0.5\times V\;\Rightarrow\;V=25\ \mathrm{mL} \]

One-fourth neutralisation needs \(25/4=6.25\ \mathrm{mL}\).

Question 13. You are given a solution of 0.2 M sulfuric acid (H₂SO₄). How many grams of H₂SO₄ are present in 500 mL of this solution?

a) 9.8 g
b) 19.6 g
c) 4.9 g
d) 2.45 g

Correct Answer: a) 9.8 g

\[ n=0.2\times0.5=0.1\ \mathrm{mol},\qquad m=0.1\times98=9.8\ \mathrm{g} \]

Question 14. A solution contains 20 g of NaOH in 500 mL of water. If the solution is diluted 2 times, what is true for the diluted solution compared to the original solution?

a) Concentration of the solution remains the same
b) Moles of NaOH remain the same
c) Volume of the solution remains the same
d) Corrosiveness of the solution remains the same

Correct Answer: b) Moles of NaOH remain the same

Reason: Adding solvent changes the volume and the concentration, but the amount of solute (moles) does not change.

Question 15. A 0.7 g sample of an unknown monoprotic acid requires 35 mL of 0.1 M NaOH for complete neutralization. What is the equivalent weight of the acid?

a) 20
b) 40
c) 200
d) 400

Correct Answer: c) 200

\[ \text{eq of NaOH}=0.1\times0.035=0.0035,\qquad E=\frac{0.7}{0.0035}=200 \]

Question 16. You are tasked with preparing a solution of HCl. If you have 36.5 g of HCl and need to prepare 1 liter of a 0.5 N solution, what volume of water should be added to the HCl?

a) 1 L
b) 500 mL
c) 2 L
d) 250 mL

Correct Answer: c) 2 L

Reason: 36.5 g of HCl is 1 gram equivalent. For a 0.5 N solution:

\[ V=\frac{\text{equivalents}}{N}=\frac{1}{0.5}=2\ \mathrm{L} \]

The HCl is therefore diluted with water to a total volume of 2 L.

4. Short Answer Questions

Question 1. Define and differentiate the terms:

a) Molarity and normality

Molarity (M) is the number of moles of solute dissolved per litre of solution, \(M=\dfrac{\text{moles}}{V\ (\mathrm{L})}\). Normality (N) is the number of gram equivalents of solute dissolved per litre of solution, \(N=\dfrac{\text{gram equivalents}}{V\ (\mathrm{L})}\). The two are related by \(N=M\times n\text{-factor}\).

b) ppm and ppb

ppm (parts per million) is the mass of solute present in \(10^{6}\) parts by mass of solution. ppb (parts per billion) is the mass of solute present in \(10^{9}\) parts by mass of solution, so \(1\ \mathrm{ppm}=1000\ \mathrm{ppb}\).

c) Molarity and molality

Molarity is moles of solute per litre of solution and changes with temperature because the volume changes. Molality is moles of solute per kilogram of solvent and does not change with temperature because it depends only on masses.

d) No. of gram moles and no. of gram equivalents

The number of gram moles is the mass of a substance divided by its molar mass. The number of gram equivalents is the mass of the substance divided by its equivalent weight, where equivalent weight \(=\dfrac{\text{molar mass}}{n\text{-factor}}\).

e) Primary and secondary standard substances

A primary standard is a pure, stable, non-hygroscopic substance whose standard solution can be prepared directly by accurate weighing, for example oxalic acid. A secondary standard cannot be weighed accurately because it is impure, hygroscopic, volatile or unstable, so its solution must be standardised against a primary standard, for example NaOH and KMnO₄.

f) Acid-base and redox titrations

An acid-base titration is based on the neutralisation reaction between an acid and a base. A redox titration is based on an oxidation-reduction reaction in which electrons are transferred from a reducing agent to an oxidising agent.

Question 2. Volumetric and gravimetric analyses are types of quantitative analysis, both useful in the chemical and industrial worlds.

a) Define volumetric analysis. How is it different from gravimetric analysis?

Volumetric analysis is a quantitative method in which the volume of a solution of known concentration needed to react completely with the analyte is measured. In gravimetric analysis the analyte is precipitated as a pure compound, filtered, dried and weighed, so the quantity measured is mass instead of volume.

b) What is the application of volumetric analysis in chemistry?

It is used to find the unknown concentration of acids, bases, oxidising agents and reducing agents, and for quality control in industry, for example to determine acetic acid in vinegar.

c) Write any one limitation of volumetric analysis.

The volume of a solution changes with temperature, so the molarity and normality change and the results depend on the temperature at which they are measured.

Question 3. In gravimetric analysis, a sample of impure barium chloride weighing 0.500 g is dissolved in water and treated with excess sulfuric acid to precipitate barium sulfate. After filtration and drying, the precipitate weighs 0.434 g.

a) Write the balanced chemical equation for the reaction.

\[ \mathrm{BaCl_2(aq)+H_2SO_4(aq)\longrightarrow BaSO_4(s)\downarrow+2HCl(aq)} \]

b) Calculate the percentage of barium chloride in the original sample.

Molar mass of BaSO₄ = 137.33 + 32.06 + 4(16) = 233.39 g/mol, and of BaCl₂ = 137.33 + 2(35.45) = 208.23 g/mol.

\[ n(\mathrm{BaSO_4})=\frac{0.434}{233.39}=1.8595\times10^{-3}\ \mathrm{mol}=n(\mathrm{BaCl_2}) \]

\[ m(\mathrm{BaCl_2})=1.8595\times10^{-3}\times208.23=0.3872\ \mathrm{g},\qquad\%=\frac{0.3872}{0.500}\times100=77.4\% \]

c) Explain why excess sulfuric acid is used.

The excess sulfate ions shift the equilibrium towards BaSO₄ by the common-ion effect. This makes the precipitation complete and reduces the loss of barium sulfate by dissolving.

Question 4. Match the given list of secondary standard substances with their property so that they cannot be used as primary standard substances.

Substance	Property
HCl	It absorbs CO₂ and moisture from the air
I₂	It is reduced by sunlight
H₂SO₄	It is a gas
NaOH	It absorbs moisture from the air
KMnO₄	It sublimes

The correct matches are:

Substance	Property
HCl	It is a gas (its solutions lose HCl and change concentration)
I₂	It sublimes
H₂SO₄	It absorbs moisture from the air
NaOH	It absorbs CO₂ and moisture from the air
KMnO₄	It is reduced by sunlight

Question 5. (a) Define equivalent weight. Write the relationship between equivalent weight, atomic weight and valency. (b) Calculate the equivalent weight of the underlined species given:
H₃PO₄ + KOH → K₃PO₄ + H₂O
FeSO₄ + 7H₂SO₄ + K₂Cr₂O₇ → 3Fe₂(SO₄)₃ + K₂SO₄ + Cr₂(SO₄)₃ + 7H₂O

a) Equivalent weight and its relation with atomic weight and valency

The equivalent weight of a substance is the mass of it that combines with or displaces 1 g of hydrogen, 8 g of oxygen or 35.5 g of chlorine.

\[ \text{Equivalent weight}=\frac{\text{atomic weight}}{\text{valency}} \]

For a compound the molecular weight is divided by the n-factor.

b) Equivalent weights of H₃PO₄ and K₂Cr₂O₇

H₃PO₄: in this reaction all 3 H⁺ ions are replaced (K₃PO₄ is formed), so the basicity is 3:

\[ E=\frac{3(1)+31+4(16)}{3}=\frac{98}{3}=32.67 \]

K₂Cr₂O₇: each Cr changes from +6 to +3, so two Cr atoms gain 6 electrons:

\[ E=\frac{2(39)+2(52)+7(16)}{6}=\frac{294}{6}=49 \]

Question 6. Nitric acid is a highly important inorganic acid known for its strong oxidizing properties and wide range of applications. A bottle of nitric acid is labeled with 68% (w/w) and specific gravity 1.4 g/mL.

a) What do you mean by 68% (w/w)?

It means that 68 g of pure HNO₃ is present in 100 g of the solution.

b) Convert the given concentration into w/v.

\[ \%\ (w/v)=\%\ (w/w)\times\text{specific gravity}=68\times1.4=95.2\%\ (w/v) \]

c) Calculate the mass of HNO₃ dissolved in 1 liter of solution.

The mass of 1 L (1000 mL) of solution is \(1000\times1.4=1400\ \mathrm{g}\).

\[ m(\mathrm{HNO_3})=0.68\times1400=952\ \mathrm{g} \]

d) Which of the concentration units w/v or w/w depends upon temperature and why?

The w/v unit depends on temperature, because the volume of a solution expands or contracts as the temperature changes. The w/w unit uses only masses, which do not change with temperature.

Question 7. The law of equivalence is a fundamental concept in chemistry, particularly in stoichiometry and titration.

a) State the “law of equivalence”. How is this law useful in titration?

The law of equivalence states that chemical substances react together in the ratio of their equivalent weights, so equal numbers of gram equivalents of the reactants take part in a reaction. In a titration, the gram equivalents of the titrant and the analyte are equal at the end point, which gives the normality equation \(N_1V_1=N_2V_2\) and makes the calculation of an unknown concentration simple.

b) Use this law to calculate the mass of magnesium metal that is enough to react completely with 250 mL of decinormal sulfuric acid.

\[ \text{eq of H}_2\mathrm{SO_4}=0.1\times0.250=0.025,\qquad E(\mathrm{Mg})=\frac{24}{2}=12\ \mathrm{g\,eq^{-1}} \]

\[ m(\mathrm{Mg})=0.025\times12=0.30\ \mathrm{g} \]

Question 8. Mohr’s salt (NH₄)₂Fe(SO₄)₂·6H₂O is a double salt. It is widely used in analytical chemistry to prepare primary standard solutions.

a) What do you mean by a standard solution? Why is such a solution important in volumetric analysis?

A standard solution is a solution whose concentration is known accurately. It is important because it is the reference solution (titrant) used to find the unknown concentration of other solutions.

b) Write any three characteristics of Mohr’s salt so that it is used as a primary standard substance.

First, it is stable and is not easily oxidised by air, unlike simple FeSO₄. Second, it is non-hygroscopic and can be weighed accurately. Third, it has a high equivalent weight (392), which reduces the error in weighing.

c) Why is this salt called a double salt?

It crystallises as a single substance from a solution containing two salts, (NH₄)₂SO₄ and FeSO₄, but in water it gives the ions of both salts, NH₄⁺, Fe²⁺ and SO₄²⁻.

Question 9. The normal range for glucose concentration in blood (fasting) is around 70–100 milligrams per deciliter (mg/dL). If a person’s blood is found to contain 80 mg/dL, convert this glucose amount into: (a) % w/v, (b) g/L, (c) ppm and ppb, (d) molarity.

(a) % w/v: 80 mg/dL is 0.080 g in 100 mL, so the concentration is 0.080% (w/v).

(b) g/L:

\[ \frac{80\ \mathrm{mg}}{1\ \mathrm{dL}}=\frac{800\ \mathrm{mg}}{1\ \mathrm{L}}=0.8\ \mathrm{g\,L^{-1}} \]

(c) ppm and ppb: 1 L of blood weighs about 1000 g, so

\[ \mathrm{ppm}=\frac{0.8}{1000}\times10^{6}=800\ \mathrm{ppm},\qquad\mathrm{ppb}=800\times1000=8\times10^{5}\ \mathrm{ppb} \]

(d) Molarity: the molar mass of glucose (C₆H₁₂O₆) is 180 g/mol:

\[ M=\frac{0.8}{180}=4.44\times10^{-3}\ \mathrm{M} \]

Question 10. A student is asked to prepare 500 mL of a 0.100 M solution of HCl. However, the only available HCl solution is 12.0 M.

a) Calculate the number of moles present in 500 mL of a 0.100 M solution of HCl.

\[ n=MV=0.100\times0.500=0.050\ \mathrm{mol} \]

b) How much volume of HCl solution should the student use to prepare the desired solution?

\[ M_1V_1=M_2V_2\;\Rightarrow\;12.0\times V_1=0.100\times500\;\Rightarrow\;V_1=\frac{50}{12}=4.17\ \mathrm{mL} \]

c) What precautions should be followed during this work and why?

The concentrated acid should always be added slowly to water, with constant stirring, and never water to the acid, because the dilution is strongly exothermic and the acid can splash. Gloves, goggles and a lab coat should be worn, and the work should be done in a well-ventilated place because concentrated HCl gives off fumes.

Question 11. A 0.500 g sample of a mixture containing NaCl and NaBr is dissolved in water and titrated with 0.100 M AgNO₃.

a) Name the type of titration. Why is it called so?

It is a precipitation titration (argentometric titration). It is called so because the reaction between Ag⁺ and the halide ions forms insoluble precipitates of AgCl and AgBr.

b) If 25.00 mL of AgNO₃ is required to reach the endpoint, what is the percentage of NaCl in the mixture?

Let the mass of NaCl be \(x\) g, so the mass of NaBr is \((0.500-x)\) g. The moles of Ag⁺ used equal the moles of Cl⁻ plus Br⁻:

\[ \frac{x}{58.44}+\frac{0.500-x}{102.89}=0.100\times0.02500 \]

Solving for \(x\) gives the mass of NaCl, and the percentage of NaCl is \(\dfrac{x}{0.500}\times100\).

Question 12. In a redox reaction, potassium permanganate (KMnO₄) is used to oxidize oxalic acid (H₂C₂O₄) in an acidic medium.

a) Write the balanced redox equation for the reaction between KMnO₄ and H₂C₂O₄.

\[ \mathrm{2KMnO_4+5H_2C_2O_4+3H_2SO_4\longrightarrow K_2SO_4+2MnSO_4+10CO_2+8H_2O} \]

In ionic form: \(\mathrm{2MnO_4^-+5H_2C_2O_4+6H^+\rightarrow2Mn^{2+}+10CO_2+8H_2O}\).

b) If 25 mL of 0.1 M KMnO₄ is required to completely oxidize 50 mL of oxalic acid solution, calculate the molarity of the oxalic acid solution.

2 mol of KMnO₄ react with 5 mol of oxalic acid:

\[ \frac{M_1V_1}{2}=\frac{M_2V_2}{5}\;\Rightarrow\;\frac{0.1\times25}{2}=\frac{M_2\times50}{5}\;\Rightarrow\;M_2=0.125\ \mathrm{M} \]

5. Long Answer Questions

Question 1. An industrial chemist needs to prepare 100 liters of a 0.5 N H₂SO₄ solution for a manufacturing process using a bottle labeled 98% (w/w) and density 1.84 g/mL.

a) What do you mean by 0.5 N concentration of acid?

It means that each litre of the solution contains 0.5 gram equivalents of the acid.

b) Calculate the volume of concentrated sulfuric acid (H₂SO₄) required to prepare 100 liters of 0.5 N.

Molarity and normality of the concentrated acid:

\[ M=\frac{98\times1.84\times10}{98}=18.4\ \mathrm{M},\qquad N=18.4\times2=36.8\ \mathrm{N} \]

\[ N_1V_1=N_2V_2\;\Rightarrow\;36.8\times V_1=0.5\times100\;\Rightarrow\;V_1=\frac{50}{36.8}=1.36\ \mathrm{L} \]

c) Discuss any safety concerns and precautions that must be taken when preparing and handling such concentrated acid solutions in large quantities.

Concentrated H₂SO₄ is highly corrosive and dilution releases a large amount of heat. Gloves, goggles, a lab coat and an apron should be worn. The acid must be added slowly to a large volume of water, with constant stirring and cooling, never water to the acid, so that the heat is dissipated safely and the acid does not splash. The work should be done in a well-ventilated area with a safety shower and eyewash nearby, and spills should be neutralised with sodium bicarbonate.

Question 2. In biological research, a 2% (w/v) NaCl solution can be used as an isotonic solution to maintain osmotic balance for cells or tissues.

a) What do you mean by 2% (w/v)?

It means that 2 g of NaCl is dissolved in every 100 mL of the solution.

b) Calculate the molarity and normality of this solution if its density is 1.02 g/mL.

The concentration is given as w/v, so the density is not needed. The solution contains 2 g per 100 mL, that is 20 g/L:

\[ M=\frac{20}{58.5}=0.342\ \mathrm{M} \]

The n-factor of NaCl is 1, so the normality is also \(0.342\ \mathrm{N}\).

c) Why are different units needed for concentration measurement?

Different situations need different information. Molarity and normality are based on volume and are convenient in titrations, percentage units are used in everyday and medical preparations, ppm and ppb express very small amounts, and molality is used where temperature changes matter because it does not depend on volume.

Question 3. The given figure shows the titration between deci-molar caustic soda and oxalic acid solution in an acidic medium.

Class 12 Chemistry Volumetric Analysis titration of 0.1 M NaOH in a burette against 0.2 M oxalic acid in a conical flask

Figure: Standard NaOH solution in the burette (initial reading 0.00 mL) and oxalic acid solution (0.2 M) in the conical flask

a) Identify the type of titration, analyte and titrant in the figure.

It is an acid-base (neutralisation) titration. The analyte is the oxalic acid solution in the conical flask and the titrant is the standard NaOH solution in the burette.

b) Write the chemical reaction involved.

\[ \mathrm{H_2C_2O_4(aq)+2NaOH(aq)\longrightarrow Na_2C_2O_4(aq)+2H_2O(l)} \]

c) Calculate the volume of NaOH added from the burette for the complete reaction with 25 mL of oxalic acid.

Deci-molar NaOH is 0.1 M. The oxalic acid contains \(0.2\times25=5\ \mathrm{mmol}\), and each mole needs 2 mol of NaOH:

\[ n(\mathrm{NaOH})=2\times5=10\ \mathrm{mmol},\qquad V=\frac{10}{0.1}=100\ \mathrm{mL} \]

d) What is the concentration of the mixture in the conical flask when the oxalic acid is half neutralized?

Half of the oxalic acid (2.5 mmol) reacts with 5 mmol of NaOH, that is 50 mL of the titrant, and forms 2.5 mmol of Na₂C₂O₄. The other 2.5 mmol of H₂C₂O₄ remains. The total volume is \(25+50=75\ \mathrm{mL}\):

\[ [\mathrm{H_2C_2O_4}]=\frac{2.5}{75}=0.033\ \mathrm{M},\qquad[\mathrm{Na_2C_2O_4}]=\frac{2.5}{75}=0.033\ \mathrm{M} \]

Question 4. 100 mL of 0.1 M HCl is mixed with 200 mL of 0.2 M H₂SO₄ in a single container to make a mixture of acids.

a) What do you mean by 0.1 M HCl?

It means that each litre of the solution contains 0.1 mole of HCl.

b) Find the volume of sulfuric acid required to prepare 200 mL of 0.2 M from the bottle containing 98% (w/v) H₂SO₄ and having a density of 1.8 g/L.

98% (w/v) means 98 g in 100 mL, that is 980 g/L, so the density is not needed:

\[ M=\frac{980}{98}=10\ \mathrm{M},\qquad M_1V_1=M_2V_2\;\Rightarrow\;10\times V_1=0.2\times200\;\Rightarrow\;V_1=4\ \mathrm{mL} \]

c) Calculate the normality of the acid mixture in the container.

\[ \text{eq of HCl}=0.1\times1\times100=10\ \mathrm{meq},\qquad\text{eq of H}_2\mathrm{SO_4}=0.2\times2\times200=80\ \mathrm{meq} \]

\[ N=\frac{10+80}{100+200}=\frac{90}{300}=0.3\ \mathrm{N} \]

d) What happens to the concentration of the acid mixture when 700 mL of water is added? Calculate the concentration of the diluted solution.

The concentration decreases, because the same amount of acid is now present in a larger volume. The new volume is \(300+700=1000\ \mathrm{mL}\):

\[ N=\frac{90\ \mathrm{meq}}{1000\ \mathrm{mL}}=0.09\ \mathrm{N} \]

Question 5. A solution of KMnO₄ is standardized by titrating it against a known amount of oxalic acid in the presence of a mineral acid in a warm condition. The balanced equation for the reaction is: 2MnO₄⁻ + 5H₂C₂O₄ + 6H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

a) Why is it necessary to standardize a solution before using it in a titration?

KMnO₄ is a secondary standard, so a solution made by weighing it does not have an exactly known concentration. It is also slowly decomposed by light and by traces of organic matter. Standardisation against a primary standard gives its exact concentration.

b) What is the end point color change in this titration keeping KMnO₄ in the burette as the titrant?

The solution in the flask changes from colourless to a faint, permanent pink, because KMnO₄ acts as its own indicator.

c) Why is it necessary to warm the conical flask before the titration?

The reaction between permanganate and oxalic acid is slow at room temperature. Warming to about 60–70 °C speeds it up so that the end point is sharp.

d) If 25.00 mL of KMnO₄ solution is required to titrate 0.250 g of oxalic acid, what is the molarity of the KMnO₄ solution?

The molar mass of H₂C₂O₄ is 90 g/mol:

\[ n(\mathrm{H_2C_2O_4})=\frac{0.250}{90}=2.78\times10^{-3}\ \mathrm{mol} \]

2 mol of KMnO₄ react with 5 mol of oxalic acid:

\[ n(\mathrm{KMnO_4})=\frac25\times2.78\times10^{-3}=1.11\times10^{-3}\ \mathrm{mol},\qquad M=\frac{1.11\times10^{-3}}{0.02500}=0.0444\ \mathrm{M} \]

Question 6. Potassium permanganate (KMnO₄) is used in a titration to determine the concentration of an iron(II) sulfate (FeSO₄) solution. In acidic conditions, KMnO₄ reacts with Fe²⁺ ions according to the equation: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

a) What is the number of moles of KMnO₄ required to react with 1 mole of iron sulfate?

From the equation, 1 mol of MnO₄⁻ reacts with 5 mol of Fe²⁺, so 1 mol of FeSO₄ needs \(\dfrac15=0.2\) mol of KMnO₄.

b) Calculate the equivalent weight of KMnO₄ and FeSO₄ in the given equation (atomic mass of Mn = 55, Fe = 56).

KMnO₄ (M = 39 + 55 + 64 = 158) gains 5 electrons; FeSO₄ (M = 56 + 32 + 64 = 152) loses 1 electron:

\[ E(\mathrm{KMnO_4})=\frac{158}{5}=31.6,\qquad E(\mathrm{FeSO_4})=\frac{152}{1}=152 \]

c) What is the indicator in the given titration and why?

KMnO₄ itself acts as the indicator (self-indicator). Its purple colour disappears as it is reduced to colourless Mn²⁺, and one extra drop beyond the end point gives a permanent pink colour.

d) If a 25.0 mL sample of iron(II) sulfate requires 20.0 mL of 0.02 M KMnO₄ for a complete reaction, calculate the concentration of the iron(II) sulfate solution.

\[ n(\mathrm{KMnO_4})=0.02\times0.0200=4.0\times10^{-4}\ \mathrm{mol},\qquad n(\mathrm{Fe^{2+}})=5\times4.0\times10^{-4}=2.0\times10^{-3}\ \mathrm{mol} \]

\[ M=\frac{2.0\times10^{-3}}{0.0250}=0.08\ \mathrm{M} \]

Question 7. Solve the given numerical problems.

a) A bottle of a tile cleanser contains 135 g of HCl and 775 g of water. What is the percent by mass of HCl in this cleanser?

\[ \%=\frac{135}{135+775}\times100=14.8\% \]

b) How many grams of FeSO₄ are present in a solution marked 0.35 M FeSO₄?

A 0.35 M solution contains 0.35 mol of FeSO₄ in every litre. With a molar mass of 151.9 g/mol:

\[ m=0.35\times151.9=53.2\ \mathrm{g\ per\ litre} \]

c) Calculate the concentration in grams per liter of bench hydrochloric acid whose concentration is 12 mol/L.

\[ 12\ \mathrm{mol\,L^{-1}}\times36.5\ \mathrm{g\,mol^{-1}}=438\ \mathrm{g\,L^{-1}} \]

d) “Concentrated” hydrochloric acid is an aqueous solution of 37.2% HCl that is commonly used as a laboratory reagent. The density of this solution is 1.19 g/mL. What mass of HCl is contained in 0.500 L of this solution?

\[ \text{mass of solution}=500\times1.19=595\ \mathrm{g},\qquad m(\mathrm{HCl})=0.372\times595=221\ \mathrm{g} \]

e) What volume of 0.01 M KMnO₄ solution will contain 5 g of KMnO₄?

\[ n=\frac{5}{158}=0.0316\ \mathrm{mol},\qquad V=\frac{0.0316}{0.01}=3.16\ \mathrm{L} \]

f) 1 g of caustic soda is added to 2 liters of \(X\) M sulfuric acid so that the pH of the resulting solution is 7. Find the value of \(X\).

At pH 7 the acid is exactly neutralised. The moles of NaOH (and of OH⁻) are \(\dfrac{1}{40}=0.025\ \mathrm{mol}\). The acid supplies \(2\times2X=4X\) mol of H⁺:

\[ 4X=0.025\;\Rightarrow\;X=0.00625\ \mathrm{M} \]

g) What volumes of 12 M and 2 M NaOH solution should be mixed to prepare 2 liters of a 9 M solution?

Let the volume of 12 M NaOH be \(V\) litres, so the volume of 2 M NaOH is \((2-V)\) litres. The moles of NaOH are conserved:

\[ 12V+2(2-V)=9\times2\;\Rightarrow\;10V=14\;\Rightarrow\;V=1.4\ \mathrm{L} \]

So 1.4 L of 12 M NaOH and 0.6 L of 2 M NaOH are mixed.

h) Calculate the volume of 1 M NaOH required to neutralize 200 mL of 2 N HCl. What mass of NaCl is produced from the neutralization reaction?

\[ N_1V_1=N_2V_2\;\Rightarrow\;1\times V_1=2\times200\;\Rightarrow\;V_1=400\ \mathrm{mL} \]

\[ n(\mathrm{NaCl})=0.4\ \mathrm{mol},\qquad m=0.4\times58.5=23.4\ \mathrm{g} \]

i) Two liters of molar H₂SO₄ are mixed with one liter of normal NaOH solution. Calculate the strength of the mixture and the mass of salt produced from the reaction.

Molar H₂SO₄ is 2 N, so the acid has \(2\times2=4\) eq and the base has \(1\times1=1\) eq. The acid left over is \(4-1=3\) eq in a total volume of 3 L:

\[ N=\frac{3}{3}=1\ \mathrm{N}\ (\text{acidic}) \]

The salt formed is Na₂SO₄ (equivalent weight 71) and 1 eq of it is produced, so its mass is 71 g.

j) 50 mL of 2 N H₂SO₄ is mixed with 30 mL of 3 N NaOH. Predict whether the solution is acidic, basic or neutral. Also, calculate the resulting strength of the solution.

Acid: \(50\times2=100\ \mathrm{meq}\). Base: \(30\times3=90\ \mathrm{meq}\). The acid is in excess by 10 meq, so the solution is acidic:

\[ N=\frac{10}{50+30}=0.125\ \mathrm{N} \]

k) A 1.20 g sample of lye (sodium hydroxide is sold commercially as solid lye) required 45.0 mL of 0.5 M hydrochloric acid to neutralize. Calculate the percentage purity of NaOH in the sample.

\[ n(\mathrm{NaOH})=0.0450\times0.5=0.0225\ \mathrm{mol},\qquad m=0.0225\times40=0.90\ \mathrm{g} \]

\[ \%\ \text{purity}=\frac{0.90}{1.20}\times100=75\% \]

l) 5.125 g of washing soda crystals are dissolved and made up to 250 mL. 25.0 mL of this solution requires 35.8 mL of 0.05 M sulfuric acid for complete neutralization. Calculate the percentage of sodium carbonate in the washing soda crystals.

0.05 M H₂SO₄ is 0.1 N. In 25.0 mL the Na₂CO₃ equals \(35.8\times0.1=3.58\ \mathrm{meq}\), so in 250 mL it is 35.8 meq = 0.0358 eq. The equivalent weight of Na₂CO₃ is \(106/2=53\):

\[ m=0.0358\times53=1.897\ \mathrm{g},\qquad\%=\frac{1.897}{5.125}\times100=37.0\% \]

m) At a blood alcohol level above 500 mg per 100 mL, a person may die. What is the molar concentration of alcohol (ethanol) at the lethal level?

500 mg per 100 mL is 5 g/L, and the molar mass of ethanol (C₂H₅OH) is 46 g/mol:

\[ M=\frac{5}{46}=0.109\ \mathrm{M} \]

n) 8.492 g of salts having the formula (NH₄)₂SO₄·nH₂O were dissolved and made up to 250 mL using dilute acid and water. A 25.0 mL of this solution was titrated against a potassium permanganate solution of concentration 0.015 M, and 22.5 mL was consumed at the end point. Calculate the value of \(n\).

Only Fe²⁺ is oxidised by KMnO₄, so the salt that reacts is the iron(II) double salt (NH₄)₂SO₄·FeSO₄·\(n\)H₂O, whose anhydrous part, (NH₄)₂Fe(SO₄)₂, has a molar mass of 284 g/mol.

\[ n(\mathrm{MnO_4^-})=22.5\times0.015=0.3375\ \mathrm{mmol},\qquad n(\mathrm{Fe^{2+}})\ \text{in 25 mL}=5\times0.3375=1.6875\ \mathrm{mmol} \]

In the whole 250 mL there is \(1.6875\times10=16.875\ \mathrm{mmol}\), so

\[ M_{\text{salt}}=\frac{8.492}{0.016875}=503.2\ \mathrm{g\,mol^{-1}} \]

\[ 284+18n=503.2\;\Rightarrow\;n\approx12 \]

o) A commercial bleach solution contains 3.62% mass NaOCl in water. Calculate (a) the mole fraction and (b) the molality of NaOCl in the solution.

In 100 g of solution there are 3.62 g of NaOCl (molar mass 74.5 g/mol) and 96.38 g of water:

\[ n(\mathrm{NaOCl})=\frac{3.62}{74.5}=0.0486\ \mathrm{mol},\qquad n(\mathrm{H_2O})=\frac{96.38}{18}=5.354\ \mathrm{mol} \]

(a) Mole fraction:

\[ X(\mathrm{NaOCl})=\frac{0.0486}{0.0486+5.354}=0.00899 \]

(b) Molality:

\[ m=\frac{0.0486}{0.09638}=0.504\ \mathrm{mol\,kg^{-1}} \]

6. View PDF Notes

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7. Frequently Asked Questions

What is volumetric analysis in Class 12 Chemistry?

Volumetric analysis is a quantitative method in which the concentration of an unknown solution (the analyte) is found by measuring the volume of a solution of known concentration (the titrant) that reacts completely with it. It is carried out by titration, and the end point is shown by an indicator or by a self-indicating reagent such as KMnO₄.

What is the difference between gravimetric and volumetric analysis?

Gravimetric analysis finds the amount of an analyte by converting it into a pure, insoluble compound that is filtered, dried and weighed, so the measured quantity is mass. Volumetric analysis finds it by measuring the volume of a standard solution needed to react completely with the analyte, so the measured quantity is volume.

What is equivalent weight and how is it calculated?

Equivalent weight is the mass of a substance that combines with or displaces 1.008 g of hydrogen, 8 g of oxygen or 35.5 g of chlorine. Equivalent weight = atomic (or molecular) weight / valency (n-factor). For an acid it is molar mass divided by basicity, for a base molar mass divided by acidity, for a salt molar mass divided by the total positive charge, and for an oxidising or reducing agent molar mass divided by the number of electrons gained or lost.

What are the equivalent weights of common acids, bases and oxidising agents?

HCl 36.5; H₂SO₄ 98/2 = 49; H₃PO₄ in complete neutralisation 98/3 = 32.67; NaOH 40; Ca(OH)₂ 74/2 = 37; Na₂CO₃ 106/2 = 53; oxalic acid dihydrate 126/2 = 63; KMnO₄ in acidic medium 158/5 = 31.6; K₂Cr₂O₇ in acidic medium 294/6 = 49; FeSO₄ in oxidation to Fe³⁺ 152/1 = 152; Mohr’s salt 392.

What are molarity, normality and molality?

Molarity (M) is moles of solute per litre of solution. Normality (N) is gram equivalents of solute per litre of solution, so N = M × n-factor. Molality (m) is moles of solute per kilogram of solvent. Molarity and normality change with temperature because volume changes, while molality does not.

What are ppm and ppb?

ppm is parts per million, 1 part of solute in 10⁶ parts of solution (for dilute aqueous solutions 1 ppm = 1 mg per litre). ppb is parts per billion, 1 part in 10⁹ parts of solution, so 1 ppm = 1000 ppb. A blood glucose level of 80 mg/dL is 0.8 g/L, or about 800 ppm.

How do you convert between percentage by mass, percentage by volume and molarity?

% (w/v) = % (w/w) × density (g/mL). Molarity = % (w/w) × density × 10 / molar mass. For example, 68% (w/w) HNO₃ of density 1.4 g/mL is 95.2% (w/v), which contains 952 g of HNO₃ per litre, and 98% (w/w) H₂SO₄ of density 1.84 g/mL has M = 98 × 1.84 × 10 / 98 = 18.4 M, or 36.8 N.

What is a primary standard substance?

A primary standard is a substance pure and stable enough to prepare a standard solution directly by accurate weighing. It should be available in a pure state, be non-hygroscopic and stable in air, have a high equivalent weight, be highly soluble, and react quickly and completely. Examples are oxalic acid, anhydrous sodium carbonate and Mohr’s salt.

What is a secondary standard substance and why can NaOH, HCl and KMnO₄ not be primary standards?

A secondary standard is a substance whose solution must be standardised against a primary standard before use. NaOH absorbs CO₂ and moisture from air, HCl is a gas and its solutions change concentration, H₂SO₄ is hygroscopic, KMnO₄ is decomposed by light and organic matter, and I₂ sublimes. Examples of secondary standards are NaOH, HCl, H₂SO₄ and KMnO₄.

What is the law of equivalence and the normality equation?

The law of equivalence states that substances react in the ratio of their equivalent weights, so at the end point the gram equivalents of the reactants are equal. For a titration this gives the normality equation N₁V₁ = N₂V₂, where subscript 1 refers to the titrant and 2 to the analyte. For 25 mL of 0.2 N H₂SO₄ and 50 mL of NaOH, N₂ = 0.2 × 25 / 50 = 0.1 N.

What are the types of titration?

The main types are acid-base (neutralisation) titrations such as NaOH with oxalic acid; redox titrations such as KMnO₄ with oxalic acid or Fe²⁺; precipitation titrations such as AgNO₃ with chloride (argentometric titration); and complexometric titrations such as EDTA with metal ions.

Which indicator is used for which acid-base titration?

Phenolphthalein (colour change pH 8.2 to 10, colourless to pink) is used for a weak acid with a strong base, for example oxalic acid with NaOH. Methyl orange (pH 3.1 to 4.4, red to yellow) is used for a strong acid with a weak base, for example HCl with Na₂CO₃. Either indicator can be used for a strong acid with a strong base. In permanganate titrations KMnO₄ acts as a self-indicator, with the end point a faint permanent pink.

Why is the KMnO₄ and oxalic acid titration carried out warm and in acidic medium?

The reaction 2MnO₄⁻ + 5H₂C₂O₄ + 6H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O is slow at room temperature, so the flask is warmed to about 60 to 70 °C. Dilute sulfuric acid supplies the H⁺ needed for MnO₄⁻ to be reduced to colourless Mn²⁺. The end point is the first faint permanent pink colour, since KMnO₄ is its own indicator.

What is Mohr’s salt and why is it a double salt?

Mohr’s salt, (NH₄)₂Fe(SO₄)₂·6H₂O, is a double salt of ammonium sulfate and ferrous sulfate that crystallises as one substance but gives the ions NH₄⁺, Fe²⁺ and SO₄²⁻ in water. It is used as a primary standard because it is stable in air, non-hygroscopic and has a high equivalent weight (392), which reduces weighing errors.

What are the most important formulas of volumetric analysis?

Equivalent weight = molar mass / n-factor; N = M × n-factor; M = moles / V(L); N = gram equivalents / V(L) = mass / (equivalent weight × V in L); molality = moles of solute / kg of solvent; % (w/v) = % (w/w) × density; N₁V₁ = N₂V₂; M₁V₁ = M₂V₂ for dilution; N = (N₁V₁ + N₂V₂)/(V₁ + V₂) for mixing; % purity = (mass of pure substance / mass of sample) × 100.

What are the important questions from Volumetric Analysis for Class 12 NEB?

Frequently asked questions include definitions and differences (molarity and normality, primary and secondary standards, volumetric and gravimetric analysis), equivalent weight of acids, bases and oxidising agents, conversion of concentration units, the KMnO₄ titration with oxalic acid and ferrous salts, the law of equivalence problems with N₁V₁ = N₂V₂, dilution and mixing of solutions, and percentage purity. All are solved on this page.

What topics are included in Unit 1 Volumetric Analysis of NEB Class 12 Chemistry?

The unit covers: 1.1 introduction to gravimetric analysis, volumetric analysis and equivalent weight; 1.2 relationship between equivalent weight, atomic weight and valency; 1.3 equivalent weight of compounds; 1.4 concentration of solution and its units; 1.5 primary and secondary standard substances; 1.6 law of equivalence and normality equation; 1.7 titration and its types.

Where can I download Volumetric Analysis Class 12 notes in PDF for free?

This page includes handwritten notes and typed PDF notes for Volumetric Analysis that you can view online, together with all formulas and the full set of solved exercise questions and numericals.

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Volumetric Analysis

1. Syllabus

2. Volumetric Analysis at a Glance

3. Multiple Choice Questions

4. Short Answer Questions

5. Long Answer Questions

6. View PDF Notes

7. Frequently Asked Questions

Also Read: NEB Class 12 Notes

NEB (Grade 12)

1. Syllabus

2. Volumetric Analysis at a Glance

3. Multiple Choice Questions

4. Short Answer Questions

5. Long Answer Questions

6. View PDF Notes

7. Frequently Asked Questions

Also Read: NEB Class 12 Notes

NEB (Grade 12)

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