Selina Concise Solutions for ICSE Class 10 Chemistry Chapter 11 Sulphuric Acid

Question 1. Comment, sulphuric acid is referred to as:
(a) King of chemicals
(b) Oil of vitriol
Answer:
(a) Sulphuric acid is called King of Chemicals because there is no other manufactured compound which is used by such a large number of key industries.
(b) Sulphuric acid is referred to as Oil of vitriol as it was obtained as an oily viscous liquid by heating crystals of green vitriol.

📝 Teacher's Note: Use historical context to help students remember these names — show them how ancient chemists discovered these properties. Connect "vitriol" to the glass-like appearance of crystals.

🎯 Exam Tip: Remember both names and their reasons. Questions often ask for explanations, so don't just memorize the names.

Question 2. Sulphuric acid is manufactured by contact process
(a) Give two balanced equations to obtain \( SO_2 \) in this process.
(b) Give the conditions for the oxidation of \( SO_2 \)
Answer:
(a) Two balanced equations to obtain \( SO_2 \) is:
(i) \( 4FeS_2 + 11O_2 \rightarrow 2Fe_2O_3 + 8SO_2 \)
(ii) \( S + O_2 \rightarrow SO_2 \)
(b) The conditions for the oxidation of \( SO_2 \) are:
(i) The temperature should be as low as possible. The yield has been found to be maximum at about 410°C - 450°C
(ii) High pressure (2 atm) is favoured because the product formed has less volume than reactant.
(iii) Excess of oxygen increases the production of sulphur trioxide.
(iv) Vanadium pentoxide or platinised asbestos is used as catalyst.
(c) Sulphuric acid is not obtained directly by reacting \( SO_3 \) with water because the reaction is highly exothermic which produce the fine misty droplets of sulphuric acid that is not directly absorbed by water.
(d) The chemical used to dissolve \( SO_3 \) is concentrated sulphuric acid. The product formed is oleum.
(e) Main reactions of this process are:
\( S + O_2 \rightarrow SO_2 \)
\( 2SO_2 + O_2 \xrightarrow[450°C]{V_2O_5} 2SO_3 \)
\( SO_3 + H_2SO_4 \rightarrow H_2S_2O_7 \)
\( H_2S_2O_7 + H_2O \rightarrow 2H_2SO_4 \)

📝 Teacher's Note: Emphasize that contact process has multiple steps and optimal conditions are crucial for maximum yield. Use Le Chatelier's principle to explain why certain conditions are chosen.

🎯 Exam Tip: Learn all reaction equations and conditions. Questions often test the complete process, not just individual steps.

Question 3. Why is water not added to concentrated \( H_2SO_4 \) in order to dilute it?
Answer: Water is not added to concentrated acid since it is an exothermic reaction. If water is added to the acid, there is a sudden increase in temperature and the acid being in bulk tends to spurt out with serious consequences.
In simple words: Adding water to concentrated acid creates so much heat so quickly that the acid can splash and burn you badly.

📝 Teacher's Note: Always demonstrate the correct dilution method — add acid to water slowly while stirring. Use the memory aid "Do as you oughta, add acid to water."

🎯 Exam Tip: Mention both the exothermic nature and the safety hazard of spurting acid for full marks.

Question 4. Why the impurity of arsenic oxide must be removed before passing the mixture of \( SO_4 \) and air through the catalytic chamber?
Answer: Impurity of ARSENIC poisons the catalyst [i.e. deactivates the catalyst]. So, it must be removed before passing the mixture of \( SO_2 \) air through the catalytic chamber.
In simple words: Arsenic acts like poison for the catalyst, making it stop working properly, so it must be cleaned out first.

📝 Teacher's Note: Explain catalyst poisoning with everyday analogies — like how certain foods can make us sick. Emphasize that even small amounts of poison can ruin the entire process.

🎯 Exam Tip: Use the term "catalyst poisoning" and explain that arsenic deactivates the catalyst for full marks.

Question 5. Give two balanced reactions of each type to show the following properties of sulphuric acid:
(a) Acidic nature.
(b) Oxidising agent
(c) Hygroscopic nature
(d) Non-volatile nature
Answer:
Balanced reactions are:
(a) Acidic nature:
(i) Dilute \( H_2SO_4 \) reacts with basic oxides to form sulphate and water.
\( 2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O \)
(ii) \( CuO + H_2SO_4 \rightarrow CuSO_4 + H_2O \)
(iii) It reacts with carbonate to produce \( CO_2 \).
\( Na_2CO_3 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O + CO_2 \)
(b) Oxidising agent:
\( H_2SO_4 \rightarrow H_2O + SO_2 + [O] \)
Nascent oxygen oxidizes non-metals, metals and inorganic compounds.
For example,
Carbon to carbon dioxide
\( C + H_2SO_4 \rightarrow CO_2 + H_2O + 2SO_2 \)
Sulphur to sulphur dioxide
\( S + H_2SO_4 \rightarrow 3SO_2 + 2H_2O \)
(c) Hygroscopic nature:
It has great affinity for water. It readily absorbs moisture from atmospheric air.
\( HCOOH \xrightarrow{conc. H_2SO_4} CO + H_2O \)
\( C_6H_{12}O_6 \xrightarrow{conc. H_2SO_4} 6C + 6H_2O \)
(d) Non-volatile nature:
It has a high boiling point (356°C) so it is considered to be non-volatile. Therefore, it is used for preparing volatile acids like hydrochloric acid, nitric acid from their salts by double decomposition reaction.
\( NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl \)
\( KCl + H_2SO_4 \rightarrow KHSO_4 + HCl \)

📝 Teacher's Note: Group these properties logically and show how the same acid can behave differently under different conditions. Use demonstrations for hygroscopic and dehydrating properties.

🎯 Exam Tip: Learn at least two reactions for each property. Questions often ask for specific examples, and variety in answers shows thorough understanding.

Question 6. Give a chemical test to distinguish between:
(a) dilute sulphuric acid and dilute hydrochloric acid
(b) dilute sulphuric acid and conc. Sulphuric acid
Answer:
(a) Bring a glass rod dipped in Ammonia solution near the mouth of each test tubes containing dil. HCl and dil. \( H_2SO_4 \) each.

(b)
1. Dilute sulphuric acid treated with zinc gives Hydrogen gas which burns with pop sound. Concentrated \( H_2SO_4 \) gives \( SO_2 \) gas with zinc and the gas turns Acidified potassium dichromate paper green.
2. Barium chloride solution gives white ppt. with dilute \( H_2SO_4 \). This white ppt. is insoluble in all acids. Concentrated \( H_2SO_4 \) and NaCl mixture when heated gives dense white fumes if glass rod dipped in Ammonia solution is brought near it.

📝 Teacher's Note: Demonstrate these tests in the lab. The ammonia test is particularly striking and helps students remember the difference between these acids.

🎯 Exam Tip: Describe both the test procedure and expected results clearly. Mention specific observations like "white fumes" or "pop sound."

Question 7. Name the products formed when hot and concentrated sulphuric acid reacts with the following:
(a) Sulphur
(b) NaOH
(c) Sugar
(d) Carbon
(e) Copper
Answer:
(a) When sulphuric acid reacts with sulphur the product formed is Sulphur dioxide is formed.
\( S + 2H_2SO_4 \rightarrow 3SO_2 + 2H_2O \)
(b) When sulphuric acid reacts with sodium hydroxide it neutralizes base to form sodium sulphate.
\( 2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O \)
(c) When sulphuric acid reacts with sugar it forms carbon
\( C_{12}H_{22}O_{11} \xrightarrow{conc. H_2SO_4} 12C + 11H_2O \)
(d) When sulphuric acid reacts with carbon it forms carbon dioxide and sulphur dioxide gas.
\( C + 2H_2SO_4 \rightarrow CO_2 + 2H_2O + 2SO_2 \)
(e) When sulphuric acid reacts with copper it forms copper sulphate and sulphur dioxide.
\( Cu + H_2SO_4 \rightarrow CuSO_4 + 2H_2O + SO_2 \)

📝 Teacher's Note: The sugar-acid reaction is dramatic and memorable for students. Emphasize how the same acid shows different properties (acidic, oxidizing, dehydrating) with different substances.

🎯 Exam Tip: Write balanced equations along with product names. Questions often test both the products and the chemical equations.

Question 8. Why is:
(a) Concentrated sulphuric acid kept in air tight bottles?
(b) \( H_2SO_4 \) not a drying agent for \( H_2S \)?
(c) Sulphuric acid used in the preparation of HCl and \( HNO_3 \)? Give equations in both cases.
Answer:
(a) Concentrated sulphuric acid is hygroscopic substance that absorbs moisture when exposed to air. Hence, it is stored in air tight bottles.
(b) Sulphuric acid is not a drying agent for \( H_2S \) because it reacts with \( H_2S \) to form sulphur.
\( H_2SO_4 + H_2S \rightarrow 2H_2O + SO_2 + S \)
(c) Concentrated sulphuric acid has high boiling point (356°C). So, it is considered to be non-volatile. Hence, it is used for preparing volatile acids like Hydrochloric acid and Nitric acids from their salts by double decomposition.
\( NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl \)
\( NaNO_3 + H_2SO_4 \rightarrow NaHSO_4 + HNO_3 \)

📝 Teacher's Note: Connect the hygroscopic property to everyday observations — why car batteries need sealed caps, why concentrated acid bottles feel heavy with moisture.

🎯 Exam Tip: For part (c), always mention the "non-volatile nature" and provide both equations. This shows complete understanding of the principle.

Question 9. What property of conc. \( H_2SO_4 \) is made use of in each of in each of the following cases? Give an equation for the reaction on each case:
(a) in the production of HCl gas when it reacts with a chloride
(b) in the preparation of CO and HCOOH
(c) as a source of hydrogen by diluting it and adding a strip of magnesium
(d) in the preparation of sulphur dioxide by warming a mixture of conc. Sulphuric acid and copper – turnings
(e) Hydrogen sulphide gas is passed through concentrated sulphuric acid
Answer:
(a) Due to its reducing property. i.e, it is a non-volatile acid.
\( NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl \) (Conc.)
(b) It is a dehydrating agent.
\( HCOOH \xrightarrow{conc. H_2SO_4} CO + H_2O \)
(c) Magnesium is present above hydrogen in the reactivity series so sulphuric acid is able to liberate hydrogen gas by reacting with magnesium strip.
\( Mg + H_2SO_4 \rightarrow MgSO_4 + H_2 \)
(d) Due to its oxidizing character
\( Cu + H_2SO_4 \rightarrow CuSO_4 + 2H_2O + SO_2 \)
(e) Due to its oxidizing property Hydrogen sulphide gas is passed through concentrated sulphuric acid to liberate sulphur dioxide and sulphur is formed.
\( H_2S + H_2SO_4 \rightarrow S + 2H_2O + SO_2 \)

📝 Teacher's Note: This question beautifully shows how one compound can have multiple properties. Create a chart showing property vs. application to help students organize this information.

🎯 Exam Tip: Each part asks for both the property AND the equation. Don't forget either component — both are needed for full marks.

Question 10. What is the name given to the salts of:
(a) sulphurous acid
(b) sulphuric acid?
Answer:
The name of the salt of
(a) Hydrogen sulphites and Sulphites.
(b) Sulphate and bisulphate.
In simple words: Different acids form different types of salts — sulphurous acid makes sulphites, while sulphuric acid makes sulphates.

📝 Teacher's Note: Help students distinguish between sulphites (from \( H_2SO_3 \)) and sulphates (from \( H_2SO_4 \)). Use examples of common salts they might know.

🎯 Exam Tip: Remember both normal salts and acid salts (bisulphates/hydrogen sulphites) for each acid.

Question 11. Give reasons for the following:
(a) Sulphuric acid forms two types of salts with NaOH
(b) Red brown vapours are produced when concentrated sulphuric acid is added to hydrogen bromide
(c) A piece of wood becomes black when concentrated sulphuric acid is poured on it
(d) Brisk effervescence is seen when oil of vitriol is added to sodium carbonate
Answer:
(a) Sulphuric acid is dibasic acid (can release two \( H^+ \) ions), so it can form two types of salts: normal salt (\( Na_2SO_4 \)) when both hydrogen atoms are replaced, and acid salt (\( NaHSO_4 \)) when only one hydrogen is replaced.
(b) Concentrated sulphuric acid acts as an oxidizing agent and oxidizes hydrogen bromide to bromine, which appears as red-brown vapours.
(c) Wood contains cellulose and other organic compounds. Concentrated sulphuric acid acts as a dehydrating agent, removing water from these organic molecules and leaving behind black carbon.
(d) Oil of vitriol is another name for sulphuric acid. When it reacts with sodium carbonate, carbon dioxide gas is produced, which causes brisk effervescence: \( Na_2CO_3 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O + CO_2 \)

📝 Teacher's Note: These observations are perfect for demonstrations. The wood-blackening experiment particularly impresses students and shows the dehydrating power vividly.

🎯 Exam Tip: For each observation, identify which specific property of sulphuric acid is responsible. This shows analytical thinking beyond mere memorization.

Question. (a) Two types of salts are formed when sulphuric acid reacts with NaOH because sulphuric acid is dibasic. (b) When hydrogen bromide reacts with sulphuric acid the bromine gas is obtained which produce red brown vapours. (c) A piece of wood becomes black when concentrated sulphuric acid is poured on it because it gives a mass of carbon. (d) When sulphuric acid is added to sodium carbonate it liberates carbon dioxide which produces brisk effervescence.
Answer: (a) Two types of salts are formed when sulphuric acid reacts with NaOH because sulphuric acid is dibasic.
NaOH + \( H_2SO_4 \rightarrow NaHSO_4 + H_2O \)
2NaOH + \( H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O \)
(b) When hydrogen bromide reacts with sulphuric acid the bromine gas is obtained which produce red brown vapours.
2KBr + \( 3H_2SO_4 \rightarrow 2KHSO_4 + SO_2 + Br_2 + 2H_2O \)
(c) A piece of wood becomes black when concentrated sulphuric acid is poured on it because it gives a mass of carbon.
(d) When sulphuric acid is added to sodium carbonate it liberates carbon dioxide which produces brisk effervescence.
\( Na_2CO_3 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O + CO_2 \)
In simple words: Sulphuric acid can react in different ways - it can form different types of salts, remove water from substances like wood turning them to carbon, and release gases when mixed with carbonates.

📝 Teacher's Note: Demonstrate each property with simple experiments - show students how concentrated sulphuric acid turns sugar black, and how it fizzes with baking soda. This helps them understand the different behaviors of the same acid.

🎯 Exam Tip: Always write balanced chemical equations for each reaction described, and mention specific observations like "red brown vapours" or "brisk effervescence" to score full marks.

Question. Copy and complete the following table:

Answer:

In simple words: Different substances react with sulphuric acid to produce different gases - metals like zinc give hydrogen, carbonates give carbon dioxide, and bleaching powder gives chlorine gas.

📝 Teacher's Note: Use the activity series to explain why zinc produces hydrogen gas, and connect this to students' knowledge of acid-metal reactions from earlier chapters.

🎯 Exam Tip: Remember the pattern - metals produce hydrogen, carbonates produce carbon dioxide, and chlorides of active metals produce chlorine when treated with sulphuric acid.

Question. Name a gas that can be oxidized to sulphur.
Answer: Hydrogen sulphide (\( H_2S \)) can be oxidized to sulphur.
In simple words: When hydrogen sulphide gas is oxidized, it loses hydrogen and forms solid sulphur as one of the products.

📝 Teacher's Note: Show students the rotten egg smell of hydrogen sulphide and explain how oxidation removes the hydrogen, leaving behind yellow sulphur powder.

🎯 Exam Tip: Write the chemical formula \( H_2S \) along with the name "hydrogen sulphide" to show complete understanding.

Question. Give the odour of the gas evolved and name the gas produced when sodium sulphide is added to solution of HCI in water.
Answer: When sodium sulphide is added to solution of HCl, Hydrogen sulphide gas is produced. It has rotten egg like smell.
In simple words: When these two chemicals mix, they produce a gas that smells exactly like rotten eggs - this gas is hydrogen sulphide.

📝 Teacher's Note: This is a safe way to demonstrate hydrogen sulphide production in the lab. Always emphasize the characteristic smell as an identification method.

🎯 Exam Tip: Always mention both the name of the gas AND its characteristic odour when asked - examiners look for complete identification.

Question. (a) Name the catalyst which helps in the conversion of sulphur dioxide to sulphur trioxide in step C. (b) In the contact process for the manufacture of sulphuric acid, sulphur trioxide is not converted to sulphuric acid by reacting it with water. Instead a two-steps procedure is used. Write the equations for the two steps involved in D. (c) What type of substance will liberate sulphur dioxide from sodium sulphite in step E? (d) Write the equation for the reaction by which sulphure dioxide is converted to sodium sulphite in step F.
Answer: (a) The catalyst which helps in the conversion of sulphur dioxide to sulphur trioxide in step C is Vanadium pentoxide.
(b) The two steps for the conversion of sulphur trioxide to sulphuric acid is:
(i) \( SO_3 + H_2SO_4 \rightarrow H_2S_2O_7 \)
(ii) \( H_2S_2O_7 + H_2O \rightarrow 2H_2SO_4 \)
(c) The substance that will liberate sulphur dioxide in step E is dilute \( H_2SO_4 \).
(d) The equation for the reaction by which sulphur dioxide is converted to sodium sulphite in step F is:
\( SO_2 + 2NaOH \rightarrow Na_2SO_3 + H_2O \) Or
\( Na_2O + SO_2 \rightarrow Na_2SO_3 \)
In simple words: The contact process uses vanadium pentoxide as a catalyst and a special two-step method to make sulphuric acid safely, avoiding the dangerous direct reaction with water.

📝 Teacher's Note: Draw the contact process flowchart on the board and explain why direct reaction of \( SO_3 \) with water is avoided - it's too violent and produces acid mist.

🎯 Exam Tip: Remember "V2O5" as the catalyst and the two-step oleum method - these are frequently asked details in the contact process.

Question. (a) Name the process used for the large-scale manufacture of sulphuric acid. (b) Which property of sulphuric acid accounts for its use as a dehydrating agent? (c) Concentrated sulphuric acid is both an oxidizing agent and a non-volatile acid. Write one equation each to illustrate the above mentioned properties of sulphuric acid.
Answer: (a) The process used for the large-scale manufacture of sulphuric acid is Contact process.
(b) Sulphuric acid has great affinity for water. It readily removes element of water from other compound. Thus it acts as a dehydrating agent.
(c) Concentrated acid is non-volatile thus it is used for the preparation of volatile acids:
\( NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl \)
Concentrated acid act as an oxidizing agent:
\( C + 2H_2SO_4 \rightarrow CO_2 + 2H_2O + 2SO_2 \)
In simple words: Sulphuric acid is made by the contact process, it loves water so much that it removes water from other substances, and it can also give oxygen to other substances in reactions.

📝 Teacher's Note: Demonstrate dehydration with sugar turning black, and explain how non-volatile acids can displace volatile acids from their salts.

🎯 Exam Tip: For property-based questions, always write the specific equation that demonstrates that exact property mentioned in the question.

Question. Some properties of sulphuric acid are listed below. Choose the property A, B, C or D which is responsible for the reactions (i) to (v) some properties may be repeated. A. Acid B. Dehydrating agent C. Non-volatile acid D. Oxidizing agent (i) \( C_{12}H_{22}O_{11} + nH_2SO_4 \rightarrow 12C + 11H_2O + nH_2SO_4 \) (ii) \( S + 2H_2SO_4 \rightarrow 3SO_2 + 2H_2O \) (iii) \( NaCI + H_2SO_4 \rightarrow NaHSO_4 + HCI \) (iv) \( CuO + H_2SO_4 \rightarrow CuSO_4 + H_2O \) (v) \( Na_2CO_3 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O + CO_2 \)
Answer: (i) B
(ii) D
(iii) C
(iv) A
(v) A
In simple words: Each reaction shows a different property - removing water from sugar (dehydrating), giving oxygen to sulphur (oxidizing), displacing volatile acid (non-volatile), and neutralizing base and carbonate (acidic).

📝 Teacher's Note: Create a chart showing each property with 2-3 example reactions. This helps students quickly identify which property is being demonstrated.

🎯 Exam Tip: Look for key clues - water removal = dehydrating, element oxidation = oxidizing agent, volatile acid formation = non-volatile acid property, neutralization = acidic property.

Question. (a) Name the acid formed when sulphur dioxide dissolves in water (b) Name the gas released when sodium carbonate is added to a solution of sulphur dioxide.
Answer: (a) The acid formed when sulphur dioxide dissolves in water is sulphurous acid.
(b) Carbondioxide gas is released when sodium carbonate is added to solution of sulphur dioxide.
In simple words: When sulphur dioxide gas dissolves in water, it forms a weak acid called sulphurous acid. When this acidic solution meets sodium carbonate, it releases carbon dioxide gas.

📝 Teacher's Note: Show students that \( SO_2 \) forms a weaker acid (\( H_2SO_3 \)) compared to \( SO_3 \) which forms stronger \( H_2SO_4 \). This pattern helps in understanding acid strength.

🎯 Exam Tip: Don't confuse sulphurous acid (\( H_2SO_3 \)) with sulphuric acid (\( H_2SO_4 \)) - the first has one less oxygen atom and is much weaker.

Question. (a) What is the property of concentrated sulphuric acid which allows it to be used in the preparation of hydrogen chloride and nitric acid? (b) What property of concentrated sulphuric acid is in action when sugar turns black in its presence?
Answer: (a) Concentrated sulphuric acid is non-volatile; hence it is used for the preparation of higher volatile acids.
(b) Due to its dehydrating nature sugar turns black in the presence of concentrated sulphuric acid.
In simple words: Sulphuric acid doesn't evaporate easily, so it can push out other acids that do evaporate easily. It also removes water from sugar, leaving behind black carbon.

📝 Teacher's Note: Demonstrate with a drop of concentrated sulphuric acid on sugar. The dramatic blackening shows dehydration, while displacement reactions show non-volatile nature.

🎯 Exam Tip: "Non-volatile" is the key word for acid preparation, "dehydrating" is the key word for organic substance reactions like sugar turning black.