1. The Basics
Acids: Sour-tasting substances (like lemon juice) that release hydrogen ions (H⁺) in water. Examples: HCl, H₂SO₄.
Bases: Bitter-tasting, soapy-feeling substances that release hydroxide ions (OH⁻) in water.
Salts: Neutral compounds formed when an acid reacts with a base. The H⁺ ion from the acid is replaced by a metal ion. Example: NaCl (table salt).
2. Properties & Identification
pH Scale: Measures how acidic or basic a solution is (0-14). pH 7 is neutral. Below 7 is acidic, above 7 is basic.
Indicators: Dyes that change color to show pH. Litmus turns red in acid and blue in base.
Reactions: Acids react with metals (producing H₂ gas) and carbonates (producing CO₂ gas).
3. The Main Reaction: Neutralization
The core reaction is: Acid + Base → Salt + Water + Heat. This is how salts are typically made.
4. Types of Salts
Normal Salt: From complete neutralization (e.g., NaCl).
Acid Salt: Has replaceable H atoms left (e.g., Baking Soda, NaHCO₃).
Basic Salt: Has OH ions (e.g., Zn(OH)Cl).
5. Water of Crystallization
Some salts have water molecules trapped in their crystals, which gives them shape and color. For example, blue Copper Sulphate (CuSO₄·5H₂O) turns white when heated, losing its water.
In short: This chapter explains how acids and bases cancel each other out to form salts, how to identify them, and the different ways they are prepared and classified.
INTEXT – QUESTION- 1
1) What do you understand by the terms, acid and base?
Ans: Acids and bases are chemical opposites.Acids taste sour (like lemon juice), can sting, and turn blue litmus paper red. They work by releasing hydrogen ions (H⁺).
Bases feel slippery (like soap), taste bitter, and turn red litmus paper blue. They work by releasing hydroxide ions (OH⁻).When mixed, an acid and base neutralize each other, forming water and a salt
2) Explain the formation of hydronium ion. Write the ionization of sulphuric acid showing the formation of hydronium ion.
Ans: Formation of Hydronium Ion (H₃O⁺)
A hydronium ion is formed when a hydrogen ion (H⁺), which is essentially a proton, attaches itself to a water molecule (H₂O). A hydrogen ion is too reactive to exist alone in water. It is immediately attracted to the lone pair of electrons on the oxygen atom of a water molecule, forming a coordinate covalent bond and creating the hydronium ion (H₃O⁺).
Simple Equation: H⁺ + H₂O → H₃O⁺
Ionization of Sulphuric Acid with Hydronium Ion Formation
Sulphuric acid (H₂SO₄) is a strong acid. It ionizes in water by donating its protons (H⁺ ions) to water molecules, which then become hydronium ions.
The ionization occurs in two steps:
1. First Ionization (Complete):
H₂SO₄ + H₂O → H₃O⁺ + HSO₄⁻
(This step is complete, forming the first hydronium ion.)
2. Second Ionization (Partial):
HSO₄⁻ + H₂O ⇌ H₃O⁺ + SO₄²⁻
(This step is partial and reaches an equilibrium.)
Overall Simplified Reaction:
H₂SO₄ + 2H₂O → 2H₃O⁺ + SO₄²⁻
3) Water is never added to acid in order to dilute it. Why?
Ans: Water should never be added to acid because the reaction is highly exothermic (releases a lot of heat). Adding water to concentrated acid can cause the mixture to splatter or boil violently, potentially leading to acid burns.
To dilute acid safely, always add the acid slowly to water while stirring. This method allows the heat to be absorbed and dissipated safely by the larger volume of water.
4) Define the term ‘basicity’ of an acid. Give the basicity of: nitric acid, sulphuric acid and phosphoric acid.
Ans: Basicity of an acid is defined as the number of replaceable hydrogen ions (H⁺ ions) present in one molecule of that acid.
The basicity of the given acids is:
Nitric acid (HNO₃): 1 (Monobasic)
Sulphuric acid (H₂SO₄): 2 (Dibasic)
Phosphoric acid (H₃PO₄): 3 (Tribasic)
5) Give two examples of each of the following:
(a) oxy-acid
(b) hydracids
(c) monobasic acid
(d) dibasic acid
(e) tribasic acid
Ans: (a) Oxyacids: – HNO3, H2SO4
(b) Hydracid:- HCl, HBr
(c) Monobasic acid:- HCl, HBr
(d) Dibasic acid: – H2SO4 , H2CO3
(e) Tribasic acid:- H3PO4, H3PO
6) Name the: (a) acidic anhydride of the following acids:
(i) sulphurous acid
(ii) nitric acid
(iii) phosphoric acid
(iv) carbonic acid
(b) acids present in vinegar, grapes and lemon
(c) (i) ion that turns blue litmus red,
(ii) ion that turns red litmus blue.
Ans: (a) The anhydride of following acids are:
(i) Sulphurous acid: SO2
(ii) Nitric acid: N2O5
(iii) Phosphoric acid: P2O5
(iv) Carbonic acid : CO2
(b) Acids present in following are:
Vinegar: Acetic acid Grapes: Tartaric acid and Malic acid Lemon: Citric acid
(c) (i) H+ ion turn blue litmus red.
(ii) OH- ion turns red litmus blue.
7) What do you understand by the statement ‘acetic acid is a monobasic acid?
Ans: A molecule of acetic acid (CH₃COOH) can donate only one proton (H⁺ ion).This single proton comes specifically from the acidic -COOH group (the hydrogen attached to the oxygen). The other three hydrogen atoms attached to the carbon chain are not acidic and do not detach as protons.
8)(a) Give a balanced equation for reaction of nitrogen dioxide with water. (b) How many types of salts does dibasic acid produce when it reacts with caustic soda solution? Give equation(s)
Ans: (a) Balanced Equation for Nitrogen Dioxide with Water
Nitrogen dioxide (NO₂) gas is not very soluble in water and undergoes a disproportionation reaction. The balanced chemical equation is:
2NO₂ (g) + H₂O (l) → HNO₂ (aq) + HNO₃ (aq)
This forms a mixture of nitrous acid and nitric acid.
(b) Salts of a Dibasic Acid with Caustic Soda
A dibasic acid (H₂A) has two replaceable hydrogen atoms and can form two types of salts with sodium hydroxide (caustic soda):
Acid Salt (NaHA): Formed when one molecule of acid reacts with one molecule of NaOH.
H₂A + NaOH → NaHA + H₂O
Example: Carbonic acid forms sodium bicarbonate.
H₂CO₃ + NaOH → NaHCO₃ + H₂O
Normal Salt (Na₂A): Formed when one molecule of acid reacts with two molecules of NaOH.
H₂A + 2NaOH → Na₂A + 2H₂O
Example: Carbonic acid forms sodium carbonate.
H₂CO₃ + 2NaOH → Na₂CO₃ + 2H₂O
9)Carbonic acid gives an acid salt but hydrochloric acid does not. Explain.
Ans: Carbonic acid (H₂CO₃) is a weak diprotic acid (has two replaceable H⁺ ions). It can lose one proton to form a stable bicarbonate ion (HCO₃⁻), resulting in an acid salt like NaHCO₃ (sodium bicarbonate).Hydrochloric acid (HCl) is a strong monoprotic acid (has only one replaceable H⁺ ion). When it reacts, it forms only normal salts (like NaCl). There is no second proton to lose, so it cannot form an acid salt.
10) What do you understand by the strength of an acid? On which factor does the strength of an acid depend?
Ans: Strength of an Acid
The strength of an acid is its ability to donate a proton (H⁺ ion) in water. A strong acid fully ionizes, releasing a high concentration of H⁺ ions.
What It Depends On
Acid strength mainly depends on how easily the bond between hydrogen and the rest of the molecule breaks. This is influenced by the bond’s polarity and strength.
For example:
In HCl, the H–Cl bond is polar and weak, so it breaks easily — making it a strong acid.
In CH₃COOH (acetic acid), the O–H bond is harder to break — making it a weak acid.
11) Dil. HCl acid is stronger than highly concentrated acetic acid. Explain.
Ans: The strength of an acid is determined by its ability to dissociate (ionize) in water and donate protons (H⁺ ions).
HCl (Hydrochloric Acid) is a strong acid. This means it completely dissociates into H⁺ and Cl⁻ ions, even when it is dilute. This high degree of ionization provides a high concentration of H⁺ ions, making it a strong acid regardless of concentration.
CH₃COOH (Acetic Acid) is a weak acid. This means it only partially dissociates into H⁺ and CH₃COO⁻ ions, establishing an equilibrium where most molecules remain undissociated. This low degree of ionization results in a low concentration of H⁺ ions, even in a highly concentrated solution.
Therefore, a dilute solution of HCl will have a much higher concentration of H⁺ ions than a highly concentrated solution of acetic acid, making the dilute HCl stronger. Acid strength is an inherent property based on dissociation, not concentration.
12) How is an acid prepared from a (a) Non-metal (b) Salt? Give an equation for each.
Ans: (a) Acid from a Non-metal
Non-metal oxides dissolve in water to form acids (acidic anhydrides).
Example: Burning sulfur gives sulfur dioxide. Dissolving it in water forms sulfurous acid.
Equation:
SO₂(g) + H₂O(l) → H₂SO₃(aq)
(b) Acid from a Salt
A less volatile acid is prepared from its salt using a more volatile acid (e.g., sulfuric acid).
Example: Heating sodium chloride with concentrated sulfuric acid produces hydrogen chloride gas. Dissolving it in water gives hydrochloric acid.
Equation:
2NaCl(s) + H₂SO₄(l) → Na₂SO₄(s) + 2HCl(g)
HCl(g) dissolves: HCl(g) → H⁺(aq) + Cl⁻(aq)
13) Name an acid used:
(a) to flavor and preserve food,
(b) in a drink,
(c) to remove ink spots,
(d) as an eyewash
Ans: (a) SO2 +H2O —> H2SO3
(b) P2O5 +3H2O —> 2H3PO4
(c) CO2 + H2O —> H2CO3
(d) SO3 + H2O —>H2SO4
14) Give equations to show how the following are made from their corresponding anhydrides.
(a) sulphurous acid
(b) phosphoric acid,
(c) carbonic acid
(d) sulphuric acid
Ans: (a) Citric acid
(b) Carbonic acid
(c) Oxalic acid
(d) Boric acid
INTEXT – QUESTION – 2
1) What do you understand about alkali? Give two examples of:
(a) strong alkalis
(b) weak alkalis
Ans: (a) Strong Alkalis
Ionize completely in water, releasing a high concentration of OH⁻ ions.
Examples: NaOH, KOH.
(b) Weak Alkalis
Ionize only partially in water, producing fewer OH⁻ ions.
Examples: NH₃(aq), Ca(OH)₂.
2) What is the difference between: (a) an alkali and a base,
(b) an alkali and a metal hydroxide?
Ans: (a) Alkali vs. Base
Base: A base is any substance that can neutralize an acid. It is a broad term.
Alkali: An alkali is a specific type of base that is soluble in water.
In short: All alkalis are bases, but not all bases are alkalis. For example, copper oxide (CuO) is a base but not an alkali because it doesn’t dissolve in water. Sodium hydroxide (NaOH) is both a base and an alkali because it dissolves.
(b) Alkali vs. Metal Hydroxide
Metal Hydroxide: This is a compound containing a metal ion bonded to hydroxide ions (OH⁻). It is a definition based on chemical composition.
Alkali: This refers to the property of a substance (specifically, a soluble base).
In short: An alkali is almost always a metal hydroxide that dissolves in water. However, not all metal hydroxides are alkalis. For example, calcium hydroxide is slightly soluble and is considered an alkali, while iron hydroxide is insoluble and is therefore not an alkali. Ammonia (NH₃) is a rare example of an alkali that is not a metal hydroxide.
3) Define in terms of ionization: (a) an acid (b) an alkali
Ans: (a) Acid
An acid is a type of substance that, when it is mixed into water, will release hydrogen ions (H⁺) into the solution. This is what gives acids their common properties, such as their sharp, sour taste (like in lemon juice or vinegar) and their ability to react with certain metals. It’s these freely moving hydrogen ions that are responsible for the acidic behaviour.
(b) Alkali
An alkali is a specific category of base. While a base is any substance that can accept hydrogen ions, an alkali is special because it is a base that can dissolve in water. When it dissolves, it releases hydroxide ions (OH⁻) into the solution. Common examples include sodium hydroxide (found in many soaps) and calcium hydroxide. It’s the hydroxide ions that give alkalis their slippery feel and their bitter taste. In short, all alkalis are bases, but not all bases are soluble enough to be called alkalis.
4) Name the ions furnished by:
(a) bases in solution
(b) a weak alkali
(c) an acid
Ans: (a) Bases in solution give hydroxide ion.
(b) Weak alkali gives hydroxide ions.
(c) An acid gives a hydronium ion.
5) Give one example in each case:
(a) A basic oxide which is soluble in water,
(b) A hydroxide which is highly soluble in water,
(c) A basic oxide which is insoluble in water,
(d) a hydroxide which is insoluble in water,
(e) A weak mineral acid,
(f) a base which is not an alkali
(g) An oxide which is a base,
(h) A hydrogen containing compound which is not an acid,
(i) A base which does not contain a metal ion.
Ans:(a) Barium oxide
(b) Sodium hydroxide
(c) Manganese oxide
(d) Copper hydroxide
(e) Carbonic acid
(f) Ferric hydroxide
(g) Copper oxide
(h) Ammonia
(i) Ammonium hydroxide
6) You have been provided with three test tubes. One of them contains distilled water and the other two have an acidic solution and a basic solution respectively. If you are given only red litmus paper, how will you identify the contents of each test tube?
Ans: Since you only have red litmus paper, follow these steps:
1.Dip the red litmus paper into the first solution.
If it turns blue, the solution is basic.
If there is no change (it stays red), it is either acidic or distilled water.
2.Now, use the identified base to test the remaining two solutions.
Take a drop of the basic solution (from Step 1) and add it to a sample of one of the remaining unknown solutions.
Dip the red litmus paper into this mixture.
If the litmus turns blue, that unknown solution is distilled water (the base diluted it but it’s still basic).
If there is no change (it stays red), the unknown solution is acidic (it neutralized the base).
3.The last test tube will contain the final substance.
7) HCl, HNO3, C2H5OH, C6H12O6 all contain H atoms but only HCI and HNO3 show acidic character. Why?
Ans: HCl and HNO3 show acidic character because they can easily donate a proton (H⁺ ion) in an aqueous solution.In HCl and HNO3, the hydrogen is bonded to highly electronegative atoms (Cl, O). This polar bond weakens the H-atom’s hold, allowing it to split off as an H⁺ ion.In C₂H₅OH (ethanol) and C₆H₁₂O₆ (glucose), the hydrogen is bonded to carbon and oxygen in stable covalent bonds. These bonds are not easily broken, so they do not release H⁺ ions and thus do not show acidic behavior.
8) Dry HCI gas does not change the colour of dry litmus paper. Why?
Ans: Dry HCl gas does not change the colour of dry litmus paper because it cannot release hydrogen ions (H⁺) in the absence of water. Litmus paper requires moisture (water) to function, as the colour change is caused by a reaction with these H⁺ ions. Since both the gas and the paper are dry, no reaction occurs.
9)Is PbO2 a base or not? Comment.
Ans: No, PbO₂ (Lead(IV) oxide) is not a base. It is an acidic oxide.
Comment:
Bases are typically compounds that react with acids to form salt and water. PbO₂ does not display this behavior. Instead, it behaves as an oxidizing agent and reacts with strong bases to form salts, which is a characteristic property of acidic oxides. For example, it reacts with hot, concentrated sodium hydroxide (a base) to form sodium plumbate (Na₂PbO₃).
10) (a) what effect does the concentration of [H3O+] ions have on the solution?
(b) Do basic solutions also have H+(aq)? Why are they basic?
Ans: (a) Effect of [H₃O⁺] ion concentration:
The concentration of hydronium ions, [H₃O⁺], directly determines whether a solution is acidic or basic and defines its pH.
High [H₃O⁺]: The solution is acidic (pH < 7). This leads to properties like a sour taste and the ability to react with metals.
Low [H₃O⁺]: The solution is basic (pH > 7). This results in properties like a bitter taste and a slippery feel.
(b) Do basic solutions have H⁺(aq)? Why are they basic?
Yes, basic solutions do contain H⁺(aq) ions (which are the same as H₃O⁺ ions).
However, they are basic because they contain a much higher concentration of hydroxide ions (OH⁻). The defining feature of a base is that it reduces the concentration of H₃O⁺ ions by reacting with them, leaving an excess of OH⁻ ions, which gives the solution its basic properties.
11) How would you obtain: (a) a base from another base
(b) an alkali from a base,
(c) salt from another salt?
Ans: (a) Obtaining a base from another base
This is typically done through a double displacement reaction where two bases exchange their cations (positive ions) or anions (negative ions).
Example:
To get the base Sodium hydroxide (NaOH) from Calcium hydroxide [Ca(OH)₂], react it with Sodium carbonate (Na₂CO₃).
Ca(OH)₂ + Na₂CO₃ → 2NaOH + CaCO₃↓
The insoluble Calcium carbonate (CaCO₃) precipitates out, leaving Sodium hydroxide in solution.
(b) Obtaining an alkali from a base
An alkali is a base that is soluble in water. Many insoluble bases can be converted into a soluble alkali (like ammonia) through a simple reaction.
Example:
To get the alkali Ammonium hydroxide (NH₄OH) from the base Calcium hydroxide [Ca(OH)₂], react it with an ammonium salt.
2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2NH₄OH (or 2NH₃ + 2H₂O)
This reaction releases ammonia gas, which dissolves in water to form ammonium hydroxide.
(c) Obtaining a salt from another salt
This is commonly achieved through double displacement (precipitation) reactions or by reacting with an acid or a base.
Example (Double Displacement):
To get the salt Silver chloride (AgCl) from Sodium chloride (NaCl), react it with Silver nitrate (AgNO₃).
NaCl + AgNO₃ → AgCl↓ + NaNO₃
The insoluble Silver chloride forms as a precipitate.
Example (With Acid):
To get Sodium sulfate (Na₂SO₄) from Sodium chloride (NaCl), react it with Sulfuric acid (H₂SO₄) under specific conditions.
2NaCl + H₂SO₄ → Na₂SO₄ + 2HCl↑
12) Write balanced equations to satisfy each statement:
(a) Acid + Active metal ⟶ Salt + hydrogen
(b) Acid + Base ⟶ Salt + water
(c) Acid + carbonate Or bicarbonate ⟶ Salt + water + carbon dioxide
(d) Acid + sulphite Or bisulphite ⟶ Salt + water + sulphur dioxide
(e) Acid + sulphide ⟶ Salt + Hydrogen sulphide
Ans: (a) Mg +2HCl⟶MgCl2 + H2
(b) HCl + NaOH⟶NaCl + H2O
(c) CaCO3 +2HCl⟶CaCl2 +H2O + CO2
(d) CaSO3 + 2HCl⟶ CaCl2 + H2O+ SO2
(e) ZnS + 2HCl⟶ZnCl2 + H2S
13) The skin has and needs natural oils. Why is it advisable to wear gloves while working with strong allkalis?
Ans: Strong alkalis react with the natural oils in your skin in a process called saponification. This literally turns your skin’s protective oils into soap, stripping away its natural barrier. This leads to severe dryness, cracking, and chemical burns as the alkali can then penetrate deep into the skin. Gloves create an essential protective barrier to prevent this painful and damaging reaction.
14) Why are alkalis soapy to touch? What do you understand about PH value?
Ans: Alkalis feel soapy because they react with the natural oils and fats on your skin. This chemical reaction creates a thin layer of soap, which is what causes that slippery sensation.
What is pH value?
pH is a scale from 0 to 14 that measures how acidic or alkaline a substance is.
A pH of 7 is neutral.
A pH below 7 is acidic.
A pH above 7 is alkaline (a base).
15) Complete the table:
| Indicator | Neutral | Acidic | Alkaline |
| Litmus | Purple | ________ | ________ |
| Phenolphthalein | Colourless | ________ | ________ |
| Indicator | Neutral | Acidic | Alkaline |
| Litmus | Purple | Blue to red | Red to blue |
| Phenolphthalein | Colourless | Colourless | Pink |
16) Two solutions X and Y have Ph values of 4 and 10 respectively. Which one of these two will give a pink colour with a phenolphthalein indicator?
Ans: Solution Y (pH 10) will turn phenolphthalein pink.
Phenolphthalein changes colour in the basic pH range (above 8.3), and solution Y is basic (pH 10). Solution X is acidic (pH 4) and will remain colourless.
17) You are supplied with five solutions: A, B, C, D and E with Ph values as follows: A= 1.8, B = 7, C= 8.5, D = 13, and E=5 Classify these solutions as neutral, slightly or strongly acidic and slightly or strongly alkaline. Which solution would be most likely to liberate hydrogen with:
(a) magnesium powder,
(b) powdered zinc metal. Give a word equation for each reaction.
Ans: Classification of Solutions:
A (pH 1.8): Strongly acidic
B (pH 7): Neutral
C (pH 8.5): Slightly alkaline
D (pH 13): Strongly alkaline
E (pH 5): Slightly acidic
(a) Most likely to liberate hydrogen with magnesium powder:
Solution A (strongly acidic) reacts vigorously with magnesium.
Word Equation:
Magnesium + Acid → Magnesium salt + Hydrogen
(b) Most likely to liberate hydrogen with powdered zinc metal:
Solution A (strongly acidic) reacts with zinc to produce hydrogen.
Word Equation:
Zinc + Acid → Zinc salt + Hydrogen
Note: Zinc also reacts with strong alkalis (e.g., Solution D) to liberate hydrogen, but the question emphasizes “most likely,” and acidic solutions typically show faster and more vigorous reactions with metals like zinc. However, for completeness:
Zinc reacts with strong alkali (e.g., NaOH) as:
Zinc + Sodium hydroxide → Sodium zincate + Hydrogen
But since Solution A is strongly acidic, it is the primary choice for hydrogen liberation with zinc.
Final Answer:
For both (a) and (b), Solution A (pH 1.8, strongly acidic) is most likely to liberate hydrogen.
Word Equations:
(a) Magnesium + Hydrochloric acid → Magnesium chloride + Hydrogen
(b) Zinc + Hydrochloric acid → Zinc chloride + Hydrogen
(Assuming the acid in Solution A is hydrochloric acid for simplicity; the specific acid may vary, but the reaction type is general for acids with reactive metals.)
18) (a) what are the acidic range and the alkaline range in the Ph scale?
(b) State one advantage of using ‘Ph paper’ for measuring the Ph value of an unknown solution.
Ans: (a) Acidic pH is 0 to below 7; alkaline pH is above 7 to 14.
(b) pH paper is fast and simple, giving an instant, approximate reading via color change.
19) Distinguish between: (a) a common acid base indicator and a universal indicator,
(b) acidity of bases and basicity of acids,
(c) acid and alkali (other than indicators).
Ans: (a) Common Acid-Base Indicator vs. Universal Indicator
Common Indicator: A single substance (e.g., litmus, phenolphthalein) that changes color over a narrow pH range, indicating only whether a substance is an acid or a base.
Universal Indicator: A mixture of several different indicators. It changes through a spectrum of colors (red to violet) across the full pH scale, allowing it to show the approximate pH value (strength) of the solution.
(b) Acidity of Bases vs. Basicity of Acids
Acidity of Bases: Refers to the number of replaceable hydroxyl ions (OH⁻) present in one molecule of a base. For example, NaOH has an acidity of 1, while Ca(OH)₂ has an acidity of 2.
Basicity of Acids: Refers to the number of replaceable hydrogen ions (H⁺) present in one molecule of an acid. For example, HCl has a basicity of 1, while H₂SO₄ has a basicity of 2.
(c) Acid vs. Alkali
Acid: A substance that donates protons (H⁺ ions) when dissolved in water. It has a pH less than 7. All acids are not alkalis. (e.g., HCl, H₂SO₄, CH₃COOH).
Alkali: A subset of bases that is soluble in water. All alkalis produce hydroxide ions (OH⁻) in aqueous solution and have a pH greater than 7. All alkalis are bases, but not all bases are alkalis. (e.g., NaOH, KOH are soluble bases, hence alkalis; CuO is a base but not an alkali as it is insoluble).
20) How does tooth enamel get damaged? What should be done to prevent it?
Ans: (a) Common Indicator vs. Universal Indicator
Common Indicator: Shows one color change (e.g., red to blue) to tell if a substance is an acid or a base.
Universal Indicator: Shows a range of colors (a spectrum) to give an approximate pH value of the substance.
(b) Acidity of Bases vs. Basicity of Acids
Acidity of a Base: Refers to how many hydroxide (OH⁻) ions one molecule of the base can provide when dissolved.
Basicity of an Acid: Refers to how many hydrogen (H⁺) ions one molecule of the acid can donate.
(c) Acid vs. Alkali
Acid: Any substance that releases H⁺ ions in water. It has a pH less than 7.
Alkali: A special type of base that dissolves in water to release OH⁻ ions. It has a pH greater than 7. All alkalis are bases, but a base like copper oxide (insoluble in water) is not an alkali.
INTEXT – QUESTION – 3
1) Define an acidic salt, a normal salt and a mixed salt. Give two examples in each case of: (a) a normal salt, (b) an acid salt, (c) a mixed salt.
Ans:
Normal Salt: A salt formed by the complete replacement of all the replaceable hydrogen ions of an acid by a metal or ammonium ion. These salts are neutral.
Acidic Salt: A salt formed by the incomplete replacement of the replaceable hydrogen ions of an acid by a metal or ammonium ion. It still has acidic hydrogen and can produce H⁺ ions.
Mixed Salt: A salt formed from more than one acid or base, containing two different anions or cations.
Examples:
(a) Normal Salt:
Sodium chloride (NaCl)
Potassium sulfate (K₂SO₄)
(b) Acidic Salt:
Sodium bicarbonate (NaHCO₃)
Potassium hydrogen sulfate (KHSO₄)
(c) Mixed Salt:
Sodium potassium sulfate (NaKSO₄)
Bleaching Powder / Calcium oxychloride (Ca(OCl)Cl)
2) Answer the following questions related to salts and their preparations:
(a) What is a ‘salt’?
(b) What kind of salt is prepared by direct combination. Write an equation for the reaction that takes place in preparing the salt you have named.
(c) Name a salt prepared by direct combination. Write an equation for the equation for the reaction that takes place in preparing the salt you have named.
(d) Name the procedure used to prepare a sodium salt such as sodium sulphate.
Ans: (a) A salt is a compound formed when the hydrogen ion of an acid is wholly or partially replaced by a metal ion or an ammonium ion (NH₄⁺).
(b) & (c) A salt prepared by direct combination is iron(II) sulphide.
The equation for the reaction is:
Fe + S → FeS
(This is a direct combination of iron and sulphur.)
(d) The procedure used to prepare a sodium salt such as sodium sulphate is titration.
This is because the base sodium hydroxide (NaOH) is soluble, so the simple displacement method is not suitable. Titration ensures the acid (e.g., H₂SO₄) is exactly neutralized by the base to form the salt.
3)How are the following salts prepared:
(a) Calcium sulphate from calcium carbonate,
(b) Lead (II) oxide from lead,
(c) Lead carbonate from lead nitrate,
(d) Sodium nitrate from sodium hydroxide,
(e) Magnesium carbonate from magnesium chloride,
(f) Copper (II) sulphate from copper (II) oxide?
Ans: (a) Calcium Sulphate from Calcium Carbonate
Add dilute sulphuric acid to calcium carbonate. It will fizz as carbon dioxide is released. Gently evaporate the remaining solution to obtain calcium sulphate.
(b) Lead (II) Oxide from Lead
Strongly heat a piece of lead metal in air. The metal will react with oxygen to form a yellow powder of lead(II) oxide.
(c) Lead Carbonate from Lead Nitrate
Mix solutions of lead nitrate and sodium carbonate. A white precipitate of lead carbonate forms immediately. Filter it out, wash it with water, and let it dry.
(d) Sodium Nitrate from Sodium Hydroxide
Add dilute nitric acid to the sodium hydroxide solution until the mixture is neutral. Evaporate the resulting solution to crystallize out sodium nitrate.
(e) Magnesium Carbonate from Magnesium Chloride
Add sodium carbonate solution to magnesium chloride solution. A white, chalky precipitate of basic magnesium carbonate forms. Filter, wash, and dry the solid.
(f) Copper (II) Sulphate from Copper (II) Oxide
Warm black copper(II) oxide with dilute sulphuric acid until it dissolves, forming a blue solution. Filter out any leftover solid and evaporate the filtrate to get blue copper sulphate crystals.
4) (a) How is lead sulphate prepared in the laboratory?
(b) Why lead sulphate cannot be prepared by the action of diluting H2SO4 on lead oxide?
Ans: (a) Preparation of Lead Sulphate in the Laboratory
Lead sulphate is prepared by a double decomposition reaction. A soluble lead salt (like lead nitrate) is reacted with dilute sulphuric acid or a soluble sulphate (like sodium sulphate).
Procedure:
Dissolve lead nitrate in water in a beaker.
Add dilute sulphuric acid (or a solution of sodium sulphate) to it.
A white precipitate of insoluble lead sulphate is formed immediately.
Filter the precipitate, wash it with distilled water, and dry it.
Chemical Reaction:
Pb(NO₃)₂(aq) + H₂SO₄(aq) → PbSO₄(s) ↓ + 2HNO₃(aq)
or
Pb(NO₃)₂(aq) + Na₂SO₄(aq) → PbSO₄(s) ↓ + 2NaNO₃(aq)
(b) Why dil. H₂SO₄ cannot be used with PbO?
Lead sulphate cannot be prepared by the action of dilute sulphuric acid on lead oxide because the reaction does not proceed to completion.The initial reaction forms lead sulphate, but this compound is insoluble. It forms a protective layer or coating on the unreacted lead oxide particles underneath. This layer prevents further contact between the internal lead oxide and the external acid, stopping the reaction prematurely. Therefore, the reaction is incomplete and does not yield a good amount of the product.
5) Describe giving all practical details, how would you prepare:
(a) Copper sulphate crystals from a mixture of charcoal and black copper oxide,
(b) zinc sulphate crystals from zinc dust (powdered zinc and zinc oxide),
(c) lead sulphate from metallic lead,
(d) sodium hydrogen carbonate crystals.
Ans: (a) Copper Sulphate Crystals from Charcoal and Black Copper Oxide
Principle: Charcoal (carbon) reduces black copper oxide (CuO) to impure copper metal. This metal then reacts with sulphuric acid to form copper sulphate.
Steps:
Reduction: Mix the charcoal and CuO powder in a clay crucible. Heat strongly in a fume cupboard (to avoid CO gas). The mixture will glow and turn reddish-brown, forming copper metal.
2CuO + C → 2Cu + CO₂
Dissolving: Allow the residue to cool. Transfer it to a beaker and add dilute sulphuric acid. Warm the mixture with constant stirring. Copper reacts with the acid to form copper sulphate.
Cu + H₂SO₄ + [O] → CuSO₄ + H₂O (Oxygen from air)
Filtration: Filter the solution to remove unreacted charcoal and impurities.
Crystallization: Gently evaporate the filtrate until a saturated solution forms (test by dipping a glass rod; crystals should form on it). Allow the solution to cool slowly. Blue triclinic crystals of CuSO₄·5H₂O will form.
Drying: Filter out the crystals, wash them with a little cold water, and dry them between filter papers.
(b) Zinc Sulphate Crystals from Zinc Dust (Zn + ZnO)
Principle: Both zinc metal and zinc oxide react directly with dilute sulphuric acid to form zinc sulphate.
Steps:
Reaction: Place the zinc dust in a beaker. Add dilute sulphuric acid slowly (reaction is vigorous). Effervescence of hydrogen gas occurs from the zinc metal. Warm the mixture to ensure the zinc oxide also fully reacts.
Zn + H₂SO₄ → ZnSO₄ + H₂↑
ZnO + H₂SO₄ → ZnSO₄ + H₂O
Test for Completion: When effervescence stops and a little unreacted zinc dust remains (ensuring no excess acid is left), the reaction is complete.
Filtration: Filter the solution to remove the unreacted zinc dust and any impurities.
Crystallization: Evaporate the filtrate to saturation and then allow it to cool. Colourless monoclinic crystals of ZnSO₄·7H₂O will form.
Drying: Separate and dry the crystals as before.
(c) Lead Sulphate from Metallic Lead
Principle: Lead is first converted to lead nitrate (soluble) and then precipitated as insoluble lead sulphate using sulphuric acid.
Steps:
Dissolve Lead: Place metallic lead in a beaker and add moderate concentration nitric acid. Warm gently until all lead dissolves and brown fumes of NO₂ cease. You now have a solution of lead nitrate, Pb(NO₃)₂.
Pb + 4HNO₃ → Pb(NO₃)₂ + 2NO₂↑ + 2H₂O
Precipitation: Filter the lead nitrate solution. Add dilute sulphuric acid to the filtrate. A white precipitate of lead sulphate forms immediately.
Pb(NO₃)₂ + H₂SO₄ → PbSO₄↓ + 2HNO₃
Purification: Filter the mixture to collect the precipitate. Wash the precipitate thoroughly with cold distilled water to remove any nitric acid or other soluble impurities.
Drying: Since it is an insoluble precipitate, it is obtained as a powder. Spread it on a filter paper and dry it in an oven or a desiccator.
(d) Sodium Hydrogen Carbonate Crystals
Principle: A saturated solution of sodium carbonate is reacted with carbon dioxide to form the less soluble sodium hydrogen carbonate, which crystallizes out.
Steps:
Prepare Soda Solution: Prepare a saturated solution of sodium carbonate (Na₂CO₃) in water.
Carbonation: Pass a steady stream of carbon dioxide gas (from a Kipp’s apparatus or a gas cylinder) through this solution for about 30 minutes. The solution will become cloudy as crystals form.
Na₂CO₃ + CO₂ + H₂O → 2NaHCO₃
Crystallization: Continue passing CO₂ until no more crystals form. The reaction is complete.
Filtration: Filter under suction (using a Buchner funnel) to collect the white crystalline precipitate of NaHCO₃.
Drying: Wash the crystals with a small amount of ice-cold water to remove any mother liquor. Dry them in a desiccator over anhydrous calcium chloride. Do not heat, as it decomposes to sodium carbonate.
6) The following is a list of methods for the preparation of salts.
A – direct combination of two elements
B – reaction of a dilute acid with a metal.
C – reaction of a dilute acid with an insoluble base.
D – titration of a dilute acid with a solution of soluble base.
E – reaction of two solutions of salts to form a precipitate.
Choose from the above list A to E, the best method of preparing the following salts by giving a suitable equation in each case:
1. Anhydrous ferric chloride,
2. Lead chloride,
3. Sodium sulphate.
4. Copper sulphate.
Ans: 1. Anhydrous ferric chloride
Best Method: A – direct combination of two elements.
Reason: This method produces the pure, anhydrous (water-free) salt directly.
Equation: 2Fe(s) + 3Cl₂(g) → 2FeCl₃(s)
2. Lead chloride
Best Method: E – reaction of two solutions of salts to form a precipitate.
Reason: Lead chloride is insoluble in water and can be easily purified by filtration and washing.
Equation: Pb(NO₃)₂(aq) + 2NaCl(aq) → PbCl₂(s) + 2NaNO₃(aq)
3. Sodium sulphate
Best Method: D – titration of a dilute acid with a solution of soluble base.
Reason: Both reactants are soluble, and titration allows for the exact neutralization to produce a pure, soluble salt.
Equation: H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)
4. Copper sulphate
Best Method: C – reaction of a dilute acid with an insoluble base.
Reason: Copper(II) oxide is an insoluble base, so the reaction goes to completion and the excess solid can be filtered out, leaving a pure solution.
Equation: CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l)
7) Name:
(a) a chloride which is insoluble in cold water but dissolves in hot water,
(b) a chloride which is insoluble,
(c) two sulphates which are insoluble,
(d) a basic salt,
(e) an acidic salt,
(f) a mixed salt,
(g) a complex salt,
(h) a double salt,
(i) two salts whose solubility increases with temperature,
(j) a salt whose solubility decreases with temperature.
Ans: (a) Lead chloride (PbCl₂)
(b) Silver chloride (AgCl)
(c) Barium sulphate (BaSO₄), Lead sulphate (PbSO₄)
(d) Basic copper carbonate [CuCO₃·Cu(OH)₂]
(e) Sodium hydrogen sulphate (NaHSO₄)
(f) Sodium potassium sulphate (NaKSO₄)
(g) Sodium argentocyanide [Na[Ag(CN)₂]]
(h) Potash alum [K₂SO₄·Al₂(SO₄)₃·24H₂O]
(i) Potassium nitrate (KNO₃), Ammonium chloride (NH₄Cl)
(j) Calcium sulphate (CaSO₄) / Cerium(III) sulphate (Ce₂(SO₄)₃)
Best Methods for Preparing the Salts
1. Anhydrous Ferric Chloride
Method: Direct Combination (Synthesis)
Equation: 2Fe(𝑠) + 3Cl₂(𝑔) → 2FeCl₃(𝑠)
Reason: This method produces pure, anhydrous crystals. Crystallization from water would yield hydrated ferric chloride (FeCl₃·6H₂O).
2. Lead Chloride
Method: Precipitation (Double Decomposition)
Equation: Pb(NO₃)₂(𝑎𝑞) + 2NaCl(𝑎𝑞) → PbCl₂(𝑠) ↓ + 2NaNO₃(𝑎𝑞)
Reason: Lead chloride is insoluble in cold water and can be easily separated by filtration.
3. Sodium Sulphate
Method: Neutralization
Equation: 2NaOH(𝑎𝑞) + H₂SO₄(𝑎𝑞) → Na₂SO₄(𝑎𝑞) + 2H₂O(𝑙)
Reason: This is a simple acid-base reaction between a common, soluble acid and base, producing a soluble salt that can be crystallized.
4. Copper Sulphate
Method: Action of acid on metal oxide
Equation: CuO(𝑠) + H₂SO₄(𝑎𝑞) → CuSO₄(𝑎𝑞) + H₂O(𝑙)
Reason: Copper does not react directly with dilute sulphuric acid. Using its oxide ensures a complete and controlled reaction to form the salt solution.
8) Explain ‘salt hydrolysis’ name two salts which are:
(a) acidic
(b) basic
(c) neutral, when dissolved in water.
Ans: Salt Hydrolysis
Salt hydrolysis occurs when a salt’s ions react with water, changing the solution’s pH. This happens because the salt is made from either a weak acid or a weak base.
(a) Acidic Salts
These form from a strong acid and a weak base. The cation (positive ion) reacts with water, releasing H⁺ ions and making the solution acidic.
Examples: Ammonium chloride (NH₄Cl), Aluminium sulphate (Al₂(SO₄)₃)
(b) Basic Salts
These form from a weak acid and a strong base. The anion (negative ion) reacts with water, releasing OH⁻ ions and making the solution basic.
Examples: Sodium carbonate (Na₂CO₃), Potassium acetate (CH₃COOK)
(c) Neutral Salts
These form from a strong acid and a strong base. The ions do not react with water, so the solution’s pH remains neutral (pH 7).
Examples: Sodium chloride (NaCl), Potassium nitrate (KNO₃)
9) What would you observe when:
(a) blue litmus is introduced into a solution of ferric chloride,
(b) red litmus paper is introduced into a solution of sodium sulphate,
(c) red litmus paper is introduced in sodium carbonate solution?
Ans: (a) Blue litmus in ferric chloride solution:
The blue litmus paper turns red. This is because ferric chloride (FeCl₃) undergoes hydrolysis to produce hydrochloric acid, making the solution acidic.
(b) Red litmus in sodium sulphate solution:
No change is observed. The red litmus paper stays red. Sodium sulphate (Na₂SO₄) is a salt of a strong acid and a strong base, so its solution is neutral.
(c) Red litmus in sodium carbonate solution:
The red litmus paper turns blue. This is because sodium carbonate (Na₂CO₃) undergoes hydrolysis to produce a basic solution.
10) Write the balanced equations for the preparation of the following salts in the laboratory:
(a) A soluble sulphate by the action of an acid on an insoluble base,
(b) An insoluble salt by the action of an acid on another salt,
(c) An insoluble base by the action of a soluble base on a soluble salt,
(d) A soluble sulphate by the action of an acid on a metal.
Ans: (a) Soluble Sulphate (Acid + Insoluble Base)
Add an insoluble base (e.g., copper(II) oxide) to warm sulfuric acid until no more dissolves. The base neutralizes the acid, forming a soluble salt (e.g., copper sulfate) and water. Filter to remove any excess base.
(b) Insoluble Salt (Precipitation)
Mix two clear, soluble salt solutions (e.g., silver nitrate and sodium chloride). They instantly react to form a solid, insoluble salt (a precipitate, e.g., silver chloride) and a soluble byproduct. Filter, wash, and dry the precipitate.
(c) Insoluble Base (Soluble Base + Soluble Salt)
Add a soluble alkali (e.g., sodium hydroxide) to a solution of a soluble salt (e.g., copper sulfate). An insoluble metal hydroxide (e.g., copper(II) hydroxide) precipitates out. Filter, wash, and dry the solid.
(d) Soluble Salt (Acid + Metal)
Add a reactive metal (e.g., magnesium) to a dilute acid (e.g., sulfuric acid). The metal fizzes as it reacts, producing hydrogen gas and a soluble salt (e.g., magnesium sulfate). Once the fizzing stops, filter out any excess metal and evaporate the water to get crystals.
11) Give the preparation of the salt shown in the left column by matching with the methods given in the right column. Write a balanced equation for each preparation.
| Salt | Method of preparation |
| Zinc sulphate | Precipitation |
| Ferrous sulphide | Oxidation |
| Barium sulphate | Displacement |
| Ferric sulphate | Neutralisation |
| Sodium sulphate | synthesis |
Ans: Zinc Sulphate – Displacement
Zn(OH)2 + H2SO4 ⟶ ZnSO4 + 2H2O
Ferrous sulphide – synthesis
Fe + S ⟶ FeS
Barium sulphate – Precipitation
BaCI2+H2SO4 ⟶ BaSO4 + 2HCI
Ferric sulphate – Oxidation
Fe + H2SO4 ⟶ FeSO4 + H2
Sodium sulphate – Neutralisation
2NaOH + H2SO4 ⟶ Na2SO4 + 2H2O
12) You are provided with the following chemicals: NaOH, Na2CO3, H2O, Zn(OH)2, CO2, HCI, Fe, H2SO4, CI2, Zn. Using the suitable chemicals from the given list only, state briefly how you would prepare:
(a) iron (III) chloride,
(b) sodium sulphate,
(c) sodium zincate
(d) iron (II) sulphate,
(e) sodium chloride?
Ans: (a) Iron (III) Chloride
React Fe (iron) with Cl₂ (chlorine gas) to form iron (III) chloride.
Reaction: 2Fe + 3Cl₂ → 2FeCl₃
(b) Sodium Sulphate
Neutralize NaOH (sodium hydroxide) with H₂SO₄ (sulphuric acid).
Reaction: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
(c) Sodium Zincate
Dissolve the amphoteric hydroxide Zn(OH)₂ (zinc hydroxide) in NaOH (sodium hydroxide) solution.
Reaction: Zn(OH)₂ + 2NaOH → Na₂ZnO₂ + 2H₂O
(d) Iron (II) Sulphate
React Fe (iron) with H₂SO₄ (dilute sulphuric acid). Ensure the acid is dilute to prevent further oxidation to iron (III).
Reaction: Fe + H₂SO₄ → FeSO₄ + H₂↑
(e) Sodium Chloride
Neutralize NaOH (sodium hydroxide) with HCl (hydrochloric acid).
Reaction: NaOH + HCl → NaCl + H₂O
Alternatively, neutralize Na₂CO₃ (sodium carbonate) with HCl: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂↑
13) Define the term neutralization: (a) Give a reaction, mentioning clearly acid and base used in the reaction. (b) if one mole of a strong acid reacts with one mole of a strong base, the heat produced is always the same. Why?
Ans: (a) Definition and Reaction
Neutralization is a chemical reaction between an acid and a base, resulting in the formation of salt and water.
Reaction:
Hydrochloric acid (a strong acid) reacts with Sodium hydroxide (a strong base).
Acid: Hydrochloric Acid (HCl)
Base: Sodium Hydroxide (NaOH)
HCl (aq) + NaOH (aq) → NaCl (aq) + H₂O (l)
(b) Reason for Constant Heat
The heat released is always the same (approximately 57.1 kJ/mol) because the net reaction in all such cases is essentially the same:
H⁺ (aq) + OH⁻ (aq) → H₂O (l)
Strong acids and bases completely dissociate in water. Therefore, regardless of which strong acid and base are used, the actual reaction is between one mole of H⁺ ions and one mole of OH⁻ ions to form one mole of water. The energy change associated with forming this H–O bond is constant.
14) Explain why: (a) It is necessary to find out the ratio of reactants required in the preparation of sodium sulphate.
(b) fused calcium chloride is used in the preparation of FeCI3?
Ans: (a) Why it is necessary to find the ratio of reactants for sodium sulphate.
It is necessary to use the correct stoichiometric ratio of reactants (sulphuric acid and sodium hydroxide) to ensure the reaction produces only the desired neutral salt, sodium sulphate (Na₂SO₄), and water.
Incorrect Ratio (Excess Acid): If too much acid is used, the product will be sodium hydrogen sulphate (NaHSO₄), an acidic salt, instead of Na₂SO₄.
Incorrect Ratio (Excess Alkali): If too much alkali is used, it will remain unreacted and contaminate the final product.
Therefore, finding the exact ratio (which is 1:1 for H₂SO₄:NaOH in this neutralization reaction) is crucial to obtain a pure and neutral sample of sodium sulphate.
(b) Why fused calcium chloride is used in the preparation of FeCl₃.
Fused (anhydrous) calcium chloride (CaCl₂) is used as a drying agent in the preparation of anhydrous iron(III) chloride (FeCl₃).
Iron(III) chloride is highly deliquescent, meaning it absorbs moisture from the air and forms a sticky solution. To prevent this and keep the freshly prepared FeCl₃ powder dry, it is stored in a desiccator. Fused calcium chloride is placed in the desiccator because it is an excellent hygroscopic substance that aggressively absorbs atmospheric moisture, creating a dry environment and preserving the anhydrous state of FeCl₃.
15) Give the Ph value of pure water. Does it change if common salt is added to it?
Ans: The pH of pure water is 7 at 25°C (room temperature), as it is perfectly neutral.
When common salt (sodium chloride, NaCl) is added: No, the pH does not change. This is because common salt is a neutral salt formed from a strong acid (HCl) and a strong base (NaOH). Its ions (Na⁺ and Cl⁻) do not react with water to produce H⁺ or OH⁻ ions, so the solution remains neutral with a pH of approximately 7.
16) Classify the following solutions as acids, bases or salts ammonium hydroxide, barium chloride, sodium chloride, sodium hydroxide, H2SO4 and HNO3
Ans: Ammonium hydroxide – Base
Barium chloride – Salt
Sodium chloride – Salt
Sodium hydroxide – Base
H₂SO₄ (Sulfuric acid) – Acid
HNO₃ (Nitric acid) – Acid
INTEXT – QUESTION – 4
1) What do you understand about water crystallization? Give four substances which contain water of crystallization and write their common names.
Ans: Water of Crystallization is the fixed number of water molecules that are chemically attached to each formula unit of a salt in its crystalline form. These water molecules give the crystal its shape and structure. They can be removed by heating, which often causes a change in the substance’s color and form.
Four common substances containing water of crystallization are:
Copper(II) sulfate – CuSO₄·5H₂O
Common Name: Blue Vitriol
Calcium sulfate – CaSO₄·2H₂O
Common Name: Gypsum
Sodium carbonate – Na₂CO₃·10H₂O
Common Name: Washing Soda
Ferrous sulfate – FeSO₄·7H₂O
Common Name: Green Vitriol
2) (a) Define efflorescence. Give examples. (b) define deliquescence. Give examples.
Ans: (a) Efflorescence
Hydrated salts lose water of crystallization in dry air, forming a powdery coating.
Examples:
Washing soda (Na₂CO₃·10H₂O) becomes a white powder.
Gypsum (CaSO₄·2H₂O) may form a powdery film.
(b) Deliquescence
Substances absorb moisture from air and dissolve into a liquid.
Examples:
Calcium chloride (CaCl₂) turns into liquid.
Sodium hydroxide (NaOH) becomes wet and sticky.
Potassium hydroxide (KOH) absorbs moisture readily.
3) Distinguish between drying and dehydrating agent.
Ans: A drying agent removes moisture physically, such as silica gel absorbing water from the air.
A dehydrating agent removes water chemically, such as sulfuric acid reacting with sugar to leave behind carbon.
4)Explain clearly how conc, H2SO4 is used as a dehydrating as well as drying agent.
Ans: Concentrated H₂SO₄ as a Dehydrating and Drying Agent
Concentrated sulfuric acid (H₂SO₄) is a powerful hygroscopic substance, meaning it has a very strong affinity for water and readily absorbs it. This property allows it to function as both a drying agent and a dehydrating agent, though the mechanisms and outcomes differ.
1. As a Drying Agent
Purpose: To remove trace amounts of water vapor or moisture from an inert gas or an organic liquid.
Mechanism: It acts through simple physical absorption and chemical reaction. The acid greedily absorbs water molecules from the wet substance passing through it. It then reacts with the water to form stable hydrates like H₂SO₄·H₂O and H₂SO₄·2H₂O.
Example: Wet chlorine gas (Cl₂) is dried by bubbling it through concentrated H₂SO₄. The acid removes the water, and the dry Cl₂ gas passes through unchanged, as it does not react with H₂SO₄.
2. As a Dehydrating Agent
Purpose: To remove water that is chemically bound within a compound, often leading to a dramatic chemical change and the formation of a new product.
Mechanism: It acts through a chemical reaction. The acid doesn’t just absorb water; it provides a strong acidic environment and heat that catalyses the removal of hydrogen and oxygen atoms (in a 2:1 ratio) from the organic molecule itself.
Example:
When sugar (C₁₂H₂₂O₁₁) is treated with conc. H₂SO₄, the acid removes water, leaving behind a porous mass of carbon.
C₁₂H₂₂O₁₁ → 12C + 11H₂O
It dehydrates ethanol (CH₃CH₂OH) to form ethene (CH₂=CH₂).
CH₃CH₂OH → CH₂=CH₂ + H₂O
5) M is an element in the form of a powder. M burns in oxygen and the product obtained is soluble in water. The solution is tested with litmus. Write down only the word which will correctly complete each of the following sentences.
(i) If M is a metal, then the litmus will turn _____________.
(ii) If M is a non-metal, then the litmus will turn ______________.
(iii) If M is a reactive metal, then _________________ will be evolved when M reacts with dilute sulphuric acid.
(iv) If M is a metal, it will form ______________ oxide, which will form ____________ solution with water.
(v) If M is a non-metal, it will not conduct electricity in the form of __________.
Ans: (i) Blue
(ii) Red
(iii) Hydrogen
(iv) Basic ,Alkaline
(v) Graphite
6) Give reasons for the following:
(a) Sodium hydrogen sulphate is not an acid but it dissolves in water to give hydrogen ions, according to the equation NaHSO4 ⇆ H+ + Na+ + SO42-
(b) Anhydrous calcium chloride is used in a desiccator.
Ans:(a) Sodium hydrogen sulphate (NaHSO₄) is not an acid but it dissolves in water to give hydrogen ions.
Not an Acid: It is classified as a salt (specifically, an acid salt) because it is formed by the partial neutralization of a dibasic acid (H₂SO₄) by a base (NaOH). Its crystalline solid state does not exhibit typical acid properties.
Gives H⁺ Ions: In its aqueous solution, the HSO₄⁻ ion dissociates further to produce hydrogen ions (H⁺). This behavior makes its solution acidic, even though the compound itself is a salt.
(b) Anhydrous calcium chloride is used in a desiccator.
It is an excellent drying agent or desiccant because it is hygroscopic. This means it has a very strong tendency to absorb moisture from the air inside the desiccator, creating a dry environment to prevent stored substances from getting moist.
7) State whether a sample of each of the following would increase or decrease in a mass if exposed to air.
(a) Solid NaOH
(b) Solid CaCI2
(c) Solid Na2CO3 10H2O
(d) Conc, sulphuric acid
(e) Iron (III) Chloride
Ans: (a) Increases
(b) Increase
(c) Decrease
(d) Increases
(e) Increases
8) (a) why does common salt get wet during the rainy season?
(b) How can this impurity be removed?
(c) Name a substance which changes the blue colour of copper sulphate crystals to white.
(d) Name two crystalline substances which do not contain water of crystallization.
Ans: (a) Common salt gets wet during the rainy season because it contains impurities like magnesium chloride (MgCl₂), which is hygroscopic. This means it absorbs moisture from the humid air.
(b) This impurity can be removed by a process called crystallization. The impure salt is dissolved in water, filtered, and then the pure salt is obtained by evaporating the water.
(c) Concentrated Sulphuric Acid (H₂SO₄) changes the blue colour of copper sulphate crystals to white by removing their water of crystallization.
(d) Two crystalline substances that do not contain water of crystallization are:
Sodium Chloride (NaCl – Common salt)
Potassium Nitrate (KNO₃)
MISCELLANEOUS QUESTIONS BASED ON ICSE EXAMINATIONS
1) For each of the salt: A, B, C and D, suggest a suitable method of its preparation.
(a) A is a sodium salt.
(b) B is an insoluble salt
(c) C is a soluble salt of copper
(d) D is a soluble salt of zinc
Ans: (a) Sodium salt (soluble):
Titrate sodium hydroxide with a suitable acid. Evaporate the solution to obtain the salt.
(b) Insoluble salt:
Mix two soluble salt solutions to form a precipitate. Filter, wash, and dry the solid.
(c) Soluble copper salt:
Add excess copper(II) oxide to warm acid. Filter and crystallize the salt from the filtrate.
(d) Soluble zinc salt:
Add excess zinc metal to dilute acid. Filter and crystallize the salt from the solution.
2) (a) A solution has a Ph of 7. Explain how you would: (i) increase its Ph; (ii) decrease its pH.
(b) If a solution changes the colour of litmus from red to blue, what can you say about its Ph?
(c) What can you say about the Ph of a solution that liberates carbon dioxide from sodium carbonate?
Ans: (a) For a solution with a pH of 7 (neutral):
(i) To increase its pH (make it basic), add an alkaline substance, such as a small amount of sodium hydroxide (NaOH) solution.
(ii) To decrease its pH (make it acidic), add an acidic substance, such as a small amount of hydrochloric acid (HCl) or vinegar.
(b) If a solution changes litmus from red to blue, it indicates the solution is basic (alkaline). Therefore, its pH is greater than 7.
(c) A solution that liberates carbon dioxide from sodium carbonate must be acidic. Acids react with carbonates to produce CO₂ gas. Therefore, its pH is less than 7.
3) Answer the questions below, relating your answers only to salts in the following list: sodium chloride, anhydrous calcium chloride, copper sulphate – water. (a) What name is given to the water in the compound copper sulphate -5- water?
(b) If copper sulphate 5- water is heated, anhydrous copper sulphate is formed. What is its colour?
(c) By what means, other than heating, could you dehydrate copper sulphate -5- water and obtain anhydrous copper sulphate?
(d) Which one of the salts in the given list is deliquescent?
Ans: (a) Water of crystallization
(b) White
(c) Using a desiccator (or concentrated sulphuric acid)
(d) Anhydrous calcium chloride
4) Solution P has a PH of 13, solution Q had a PH OF 6 and solution R has a PH of 2. Which solution:
(a) will liberate ammonia from ammonium sulphate on heating?
(b) is a strong acid?
(c) contains molecules as well as ions?
Ans: (a) True. Solution P (pH 13) is a strong base and will liberate ammonia from ammonium sulphate on heating.
(b) True. Solution R (pH 2) is strongly acidic, indicating it is a strong acid.
(c) True. Solution Q (pH 6) is a weak acid, so it contains both molecules and ions.
5) (a) Outline the steps that would be necessary to convert insoluble lead (II) oxide into soluble lead chloride.
(b) write the balanced equations for the reactions, to convert insoluble lead (II) oxide into soluble lead chloride.
(c) A solution of iron (III) chloride has a PH less than 7. Is the solution acidic or alkaline?
Ans:(a) Steps to convert insoluble PbO into soluble PbCl₂:
Dissolve in acid: Add excess dilute nitric acid (HNO₃) to the lead(II) oxide. The oxide will react to form soluble lead(II) nitrate and water.
Precipitate the chloride: Add a solution of sodium chloride (NaCl) or hydrochloric acid (HCl) to the resulting lead(II) nitrate solution. This will form a white precipitate of insoluble lead(II) chloride.
Dissolve in hot water: Although lead(II) chloride is mostly insoluble in cold water, its solubility increases significantly with temperature. Filter the mixture and wash the precipitate.
Obtain soluble PbCl₂: Add hot (or boiling) distilled water to the precipitate and stir. The lead(II) chloride will dissolve. Upon cooling, crystals may form, but the solution will contain soluble PbCl₂.
(b) Balanced Equations:
Reaction with acid:
PbO(s) + 2HNO₃(aq) → Pb(NO₃)₂(aq) + H₂O(l)
Precipitation of PbCl₂:
Pb(NO₃)₂(aq) + 2NaCl(aq) → PbCl₂(s) + 2NaNO₃(aq)
(Alternatively, using HCl: Pb(NO₃)₂(aq) + 2HCl(aq) → PbCl₂(s) + 2HNO₃(aq))
(c) Acidity of Iron (III) Chloride Solution:
The solution is acidic.
This is because iron(III) ions (Fe³⁺) are strongly hydrolyzing. They react with water in a hydrolysis reaction to release hydrogen ions (H⁺), which lower the pH.
Fe³⁺(aq) + 3H₂O(l) ⇌ Fe(OH)₃(s) + 3H⁺(aq)
6) Choosing only substances from the list given in the box below, write equations which you would use in the laboratory to obtain:
(a) Sodium sulphate,
(b) Copper sulphate
(c) Iron (II) sulphate
(d) Zinc carbonate
Ans: (a) Sodium Sulphate
This is formed by reacting sodium carbonate with dilute sulphuric acid. Effervescence occurs.
Equation:
Na₂CO₃ + H₂SO₄ (dil.) → Na₂SO₄ + H₂O + CO₂↑
(b) Copper Sulphate
This is formed by reacting copper oxide with dilute sulphuric acid. The black oxide dissolves to form a blue solution.
Equation:
CuO + H₂SO₄ (dil.) → CuSO₄ + H₂O
(c) Iron (II) Sulphate
This is formed by reacting iron with dilute sulphuric acid. Effervescence of hydrogen gas is observed.
Equation:
Fe + H₂SO₄ (dil.) → FeSO₄ + H₂↑
(d) Zinc Carbonate
This is formed as a precipitate by reacting a soluble zinc salt (zinc sulphate) with sodium carbonate.
Equation:
ZnSO₄ + Na₂CO₃ → ZnCO₃↓ + Na₂SO₄
7) From the formula listed below, choose one, in each case corresponding to the salt having the given description:- AgCl, CuCO3, CuSO4, 5H2O, KNO3, NaCI, NaHSO4, Pb(NO3)2, ZnCO3, ZnSO4, 7H2O
(a) an acid salt
(b) an insoluble chloride
(c) on treatment with concentrated sulphuric acid, this salt changes from blue to white.
(d) On heating, this salt changes from green to black
(e) this salt gives nitrogen dioxide when heated.
Ans: (a) NaHSO4
(b) AgCl
(c) CuSO4.5H2O
(d) CuCO3
(e) Pb(NO3)2
8) Ca(H2PO4)2 is an example of a compound called _______________ (acid salt/ basic salt/ normal salt)
Ans: Acid salt
9) Write the balanced equation for the reaction of: A named acid and a named alkali.
Ans: 2NaOH + H2SO 4 ⟶ Na2SO4 + 2H2O
10) State the terms defined by the following sentences:
(a) A soluble base
(b) The insoluble solid formed when two solutions are mixed together.
(c) An acidic solution in which there is only partial ionization of the solute molecules.
Ans: (a) A soluble base
Answer: Alkali
(b) The insoluble solid formed when two solutions are mixed together.
Answer: Precipitate
(c) An acidic solution in which there is only partial ionization of the solute molecules.
Answer: Weak acid
11) Differentiate between the chemical nature of an aqueous solution of HCI and an aqueous solution of NH3.
Ans: An aqueous solution of HCl (hydrochloric acid) is a strong acid. When dissolved in water, it completely dissociates into ions (H⁺ and Cl⁻), resulting in a high concentration of H⁺ ions. This makes the solution highly acidic (pH < 7), a good conductor of electricity, and capable of reacting vigorously with bases and metals.An aqueous solution of NH₃ (ammonia) is a weak base. It does not dissociate completely but instead reacts with water to form OH⁻ ions (hydroxide ions) and NH₄⁺ ions. This results in a lower concentration of OH⁻ compared to a strong base, making the solution alkaline (pH > 7) but only mildly so. It is a poorer conductor of electricity and reacts with acids to form salts.
12) Write the balanced equations for the preparation of the following compounds (as the major product) starting from iron and using only one other substance:
(a) Iron (II) chloride
(b) Iron (III) chloride
(c) Iron (II) sulphate
(d) Iron (II) Sulphide.
Ans: (a) Iron (II) chloride
Iron reacts with dilute hydrochloric acid.
Fe(s) + 2HCl(aq) → FeCl₂(aq) + H₂(g)
(b) Iron (III) chloride
Iron reacts with chlorine gas.
2Fe(s) + 3Cl₂(g) → 2FeCl₃(s)
(c) Iron (II) sulphate
Iron reacts with dilute sulphuric acid.
Fe(s) + H₂SO₄(aq) → FeSO₄(aq) + H₂(g)
(d) Iron (II) Sulphide
Iron reacts directly with sulphur powder.
Fe(s) + S(s) → FeS(s)
Question 2004: Which of the following methods, A, B, C, D or E is generally used for preparing the chlorides listed below from (i) to (v), Answer by writing down the chloride and the letter pertaining to the corresponding method each letter is to be used only once.
A Action of acid on metal
B Action of an acid on an oxide or carbonate
C Direct combination
D Neutralization of an alkali by an acid
E Precipitation (double decomposition)
(i) Copper (II) chloride
(ii) Iron (II) chloride
(iii) Iron (III) chloride
(iv) Lead (II) chloride
(v) Sodium chloride
Ans: (i) Copper (II) chloride – (B) Action of an acid on an oxide or carbonate
(ii) Iron (II) chloride – (A) Action of acid on metal
(iii) Iron (III) chloride – (C) Direct combination
(iv) Lead (II) chloride – (E) Precipitation (double decomposition)
(v) Sodium chloride – (D) Neutralization of an alkali by an acid
Question 2005: The preparation of lead sulphate from lead carbonate is a two-step process. (lead sulphate cannot be prepared by adding dilute sulphuric acid to lead carbonate.)
(a) What is the first step that is required to prepare lead sulphate from lead carbonate?
(b) Write the equation for the reaction that will take place when this first step is carried out.
(c) Why is the direct addition of dilute sulphuric acid to lead carbonate an impractical method of preparing lead sulphate?
Ans: (a) Lead carbonate reacts with dilute nitric acid, forming soluble lead nitrate.
(b) The chemical reaction is:
PbCO₃ + 2HNO₃ → Pb(NO₃)₂ + H₂O + CO₂
(c) Sulfuric acid is not used because it creates an insoluble layer of lead sulfate. This coating protects the unreacted carbonate underneath from further acid contact, stopping the reaction.
Question 2(2005): Fill in the blanks with suitable words: An acid is a compound which when dissolved in water forms hydronium ions as the only …………………. Ions, a base is a compound which is soluble in water containing ………….. ions. A base reacts with an acid to form a ………… and water only. This type of reaction is known as ……………………..
Ans: Positively charged
Negatively charged
Salt
Neutralization reaction
Question 1(2007): From the list given below, select the word (s) required to correctly complete blanks (a) to (e) in the following passage:
Ammonia, ammonium, carbonate, carbon dioxide, hydrogen, hydronium, hydroxide, precipitate, salt, water. A solution X turns blue litmus red, so it must contain
(a) …………….. ions; another solution Y turns red litmus blue and therefore, must contain.
(b) ………………… ions, When solutions X and Y are mixed together, the products will be (c) ……………………. And
(d) ……………………….. if a piece of magnesium were put into solution X.
(e) …………………………. Gas would be evolved.
(Note: words chosen from the list are to be used only once. Write the answers as (1) (a), (b), (c) and so on. Do not copy the passage).
Ans: (a) hydronium
(b) hydroxide
(c) salt
(d) water
(e) Hydrogen
Question 2(2007): Match the following:
| Column A | Column B |
| (a) acid salt | A. Sodium potassium carbonate |
| (b) Mixed salt | B. Alum |
| (c) complex salt | C. Sodium carbonate |
| (d) Double salt | D. Sodium zincate |
| (e) Normal salt | E. Sodium hydrogen carbonate |
Ans:
| Column A | Column B |
| (a) acid salt | E. Sodium hydrogen carbonate |
| (b) Mixed salt | A. Sodium potassium carbonate |
| (c) complex salt | D. Sodium zincate |
| (d) Double salt | B. Alum |
| (e) Normal salt | C. Sodium carbonate |
Question 3(2007): Write balanced equations for the following reactions:
(a) Lead sulphate from lead nitrate solution and dilute sulphuric acid,
(b) Copper sulphate from copper and concentrated sulphuric acid.
(c) Lead chloride from lead nitrate solution and sodium chloride solution,
(d) Ammonium sulphate from ammonia and dilute sulphuric acid,
(e) Sodium chloride from sodium carbonate solution and dilute hydrochloric acid Ans: (a) Pb(NO3)2 + H2SO4 ⟶ PbSO4 + 2HNO3
(b) Cu + H2SO4 ⟶ CuSO4 + H2
(C) Pb(NO3)2 + 2NaCI ⟶ PbCI2 + 2NaNO3
(d) 2NH3 + H2SO4 ⟶ (NH4)2SO2
(e) Na2CO3 +2HCI ⟶ 2NaCI + H2O + CO2
Question 1(2008): What are the terms define by the following?
(i) A salt containing a metal ion surrounded by other ions or molecules,
(ii) A base which is soluble in water.
Ans: (i) Complex salts
(ii) Alkali
Question 2(2008): Making use only of substances chosen from those given below:
| Dilute sulphuric acid | sodium carbonate |
| Zinc | sodium sulphite |
| Lead | Calcium carbonate |
Give the equations for the reactions by which you could obtain:
(i) Hydrogen
(ii) sulphur dioxide
(iii) carbon dioxide,
(iv) zinc carbonate (two steps required).
Ans: (i) Nn + H2SO4 ⟶ ZnSO4 + H2
(ii) Na2SO3 ⟶ Na2O + SO2
(iii) Na2CO3 + H2SO4 ⟶ Na2SO4 + H2O + CO2
(iv) Zn + H2SO4 ⟶ ZnSO4 + H2 ZnSO4 + Na2CO3 ⟶ Na2SO4 + ZnCO3
