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Summary
Acids, bases, and salts are three important types of chemical compounds. Elements combine in fixed ways to make these compounds. Acids usually taste sour, like a lemon. Bases often feel soapy and can taste bitter. It is not safe to taste or touch all chemicals, as some can be harmful. Scientists use special substances called indicators to tell if something is an acid or a base. Indicators change color when they meet an acid or a base. For example, litmus paper turns red in an acid and blue in a base. Early scientists found that all acids contain hydrogen. When acids dissolve in water, they release hydrogen ions, which then combine with water to form hydronium ions. These hydronium ions are what make a solution acidic. Bases are often metal oxides or metal hydroxides. When bases that dissolve in water, called alkalis, are put in water, they release hydroxide ions. These hydroxide ions make the solution alkaline.
Acids can be classified in different ways. Some come from living things, like plants, and are called organic acids. Acetic acid in vinegar is an example. These are usually weak. Other acids come from minerals and are called inorganic or mineral acids, like hydrochloric acid. These are often strong. Acids can also be grouped by whether they contain oxygen (oxy-acids) or not (hydracids). The basicity of an acid tells us how many hydronium ions one molecule of that acid can produce in water. For example, hydrochloric acid is monobasic because it produces one hydronium ion. Sulphuric acid is dibasic. The strength of an acid depends on how much it breaks into ions in water. Strong acids break apart completely, while weak acids only break apart a little. When diluting an acid, always add the acid slowly to water, not water to acid, because this process releases a lot of heat.
Bases are substances that react with acids to form salt and water. This reaction is called neutralization. If a base dissolves in water, it is called an alkali. Sodium hydroxide is a strong alkali. Ammonium hydroxide is a weak alkali. Just like acids, bases can be classified by their acidity, which is the number of hydroxide ions they can produce or the number of hydrogen ions they can neutralize. For example, sodium hydroxide is monoacidic. Calcium hydroxide is diacidic. Bases can be prepared in several ways, such as metals reacting with oxygen, or reactive metals reacting with water.
Salts are compounds formed when an acid reacts with a base. They consist of a positive ion from the base and a negative ion from the acid. Common table salt, sodium chloride, is an example. Salts can be normal, acid, or basic. Normal salts are formed by the complete neutralization of an acid by a base. Acid salts still have some replaceable hydrogen from the acid. Basic salts still have some replaceable hydroxide from the base. Salts can be prepared by methods like direct combination of elements, an acid reacting with a metal, an acid reacting with a carbonate, or neutralization. Some salts are soluble in water, while others are insoluble. Insoluble salts are often prepared by mixing solutions of two soluble salts, causing the insoluble salt to form as a solid, called a precipitate.
The pH scale measures how acidic or alkaline a solution is. It ranges from 0 to 14. A pH of 7 is neutral, like pure water. A pH less than 7 is acidic, and a pH greater than 7 is alkaline. The pH is very important in daily life, affecting our bodies, farming, and even causing acid rain. Some salts, when they dissolve in water, can change the pH of the water. This is called hydrolysis.
Many crystalline salts contain water molecules locked into their structure, known as water of crystallization. For example, blue copper sulphate crystals contain five water molecules. If heated, they lose this water and turn white. Some salts lose their water of crystallization when exposed to air; this is called efflorescence. Other substances, called hygroscopic substances, absorb moisture from the air. If they absorb so much moisture that they dissolve in it, they are called deliquescent. Drying agents are used to remove moisture, while dehydrating agents can remove chemically combined water from compounds.
Workbook solutions (Concise/Selina)
Intext Questions and Answers I
1. (a) What do you understand by the term acid?
Answer: Acids are defined as compounds which contain one or more hydrogen atoms and when dissolved in water, produce hydronium ions (H3O+) the only positively charged ions.
(b) Name the positive ion formed when an acid is dissolved in water.
Answer: The positive ion formed when an acid is dissolved in water is the hydronium ion (H3O+).
(c) Draw the structure of this ion.
Answer:
2. Write the ionisation of sulphuric acid showing the formation of hydronium ion.
Answer: The ionisation of sulphuric acid showing the formation of hydronium ions is:
H2SO4 + 2H2O → 2H3O+ + SO42-
3. Water is never added to acid in order to dilute it. Why?
Answer: Water is not added to acid as it is an exothermic process and in this process so much heat is produced that splashing of acidic solution may occur, also the container may break which can be fatal to the person.
4. Define the term ‘basicity’ of an acid. Give the basicity of nitric acid, sulphuric acid and phosphoric acid.
Answer: The basicity of an acid is defined as the number of hydronium ions (H3O)+ that can be produced by the ionisation of one molecule of that acid in aqueous solution.
The basicity of nitric acid (HNO3) is one, as it is a monobasic acid.
The basicity of sulphuric acid (H2SO4) is two, as it is a dibasic acid.
The basicity of phosphoric acid (H3PO4) is three, as it is a tribasic acid.
5. Give two examples of each of the following:
(a) oxy-acid
Answer: Two examples of oxy-acids are nitric acid (HNO3) and sulphuric acid (H2SO4).
(b) hydracid
Answer: Two examples of hydracids are hydrochloric acid (HCl) and hydrobromic acid (HBr).
(c) tribasic acid
Answer: Two examples of tribasic acids are phosphoric acid (H3PO4) and citric acid C3H5O(COOH)3.
(d) dibasic acid
Answer: Two examples of dibasic acids are sulphuric acid (H2SO4) and oxalic acid (COOH)2.
6. Name the:
(a) acidic anhydride of the following acids:
(i) sulphurous acid
Answer: The acidic anhydride of sulphurous acid (H2SO3) is sulphur dioxide (SO2).
(ii) nitric acid
Answer: The acidic anhydride of nitric acid (HNO3) is dinitrogen pentoxide (N2O5).
(iii) phosphoric acid
Answer: The acidic anhydride of phosphoric acid (H3PO4) is phosphorus pentoxide (P2O5).
(iv) carbonic acid
Answer: The acidic anhydride of carbonic acid (H2CO3) is carbon dioxide (CO2).
(b) acids present in vinegar, grapes and lemon.
Answer: The acid present in vinegar is acetic acid.
The acid present in grapes is tartaric acid.
The acid present in lemon is citric acid.
7. What do you understand by the statement ‘acetic acid is a monobasic acid’?
Answer: Acetic acid (CH3COOH or C2H4O2) contains four hydrogen atoms in its molecule, but it is a monobasic acid because its molecule ionises by liberating only one hydronium ion.
8. Give a balanced equation for
(i) reaction of nitrogen dioxide with water
(ii) preparation of a non volatile acid from a volatile acid.
(i) reaction of nitrogen dioxide with water
Answer: Nitrogen dioxide (NO2) is a mixed or double acid anhydride because two acids, nitrous acid and nitric acid, are formed when it reacts with water. The balanced equation is:
2NO2 + H2O → HNO2 + HNO3
(ii) preparation of a non volatile acid from a volatile acid.
Answer: A non-volatile acid like sulphuric acid can be prepared by the oxidation of a non-metal like sulphur using concentrated nitric acid, which is more volatile. The balanced equation is:
S + 6HNO3 → H2SO4 + 2H2O + 6NO2
Similarly, phosphoric acid (less volatile) can be prepared from phosphorus using nitric acid:
P + 5HNO3 → H3PO4 + H2O + 5NO2
9. What do you understand by the strength of an acid? On which factor does the strength of an acid depend?
Answer: The strength of an acid is the measure of concentration of hydronium ions it produces in its aqueous solution. Or simply put, the more the number of H+ ions released by an acid in water, the stronger is the acid.
The strength of an acid depends on the degree of ionisation [α] and concentration of hydronium ions [H3O]+ produced by that acid in aqueous solution.
10. Explain the following:
(a) Carbonic acid gives an acid salt but hydrochloric acid does not.
Answer: Carbonic acid (H2CO3) is a dibasic acid, meaning it has two replaceable hydrogen ions. Therefore, it can form an acid salt by partial replacement of its hydrogen ions. For example, dibasic acids like sulphuric acid form acid salts such as NaHSO4.
Hydrochloric acid (HCl) is a monobasic acid, meaning it has only one ionisable hydrogen atom. Monobasic acids ionise in one step and so form only one normal salt, not an acid salt.
(b) Dil. HCl acid is stronger than highly concentrated acetic acid.
Answer: The strength of an acid is the measure of the concentration of hydronium ions it produces in its aqueous solution. Dilute HCl is a stronger acid than highly concentrated acetic acid because dil. HCl releases more H+ ions (or hydronium ions) in water compared to concentrated acetic acid. Acetic acid is a weak acid and does not ionise completely, whereas HCl is a strong acid and ionises almost completely even when dilute.
(c) H3PO3 is not a tribasic acid.
Answer: H3PO3 is a dibasic acid because in oxyacids of phosphorus, hydrogen atoms which are attached to oxygen atoms are replaceable. Hydrogen atoms directly bonded to phosphorus atoms are not replaceable. The structure of H3PO3 shows two -OH groups and one P-H bond, making only two hydrogens ionisable.
(d) Lead carbonate does not react with dilute HCl.
Answer: If the salt produced in a reaction between an acid and a carbonate is insoluble, then the reaction does not proceed. So, we do not expect lead carbonate to react with hydrochloric acid because lead chloride, which would be a product, can be insoluble or sparingly soluble, potentially forming a protective layer.
(e) Nitrogen dioxide is a double acid anhydride.
Answer: NO2 is called a mixed or double acid anhydride because two acids, nitrous acid (HNO2) and nitric acid (HNO3), are formed when it reacts with water. The reaction is: 2NO2 + H2O → HNO2 + HNO3.
11. How is an acid prepared from (a) Non-metal (b) salt? Give an equation for each.
Answer: (a) An acid can be prepared from a non-metal by the oxidation of the non-metal. For example, sulphur (a non-metal) is oxidised by concentrated nitric acid to form sulphuric acid:
S + 6HNO3 → H2SO4 + 2H2O + 6NO2
Another method is by the action of water on non-metallic oxides (acidic anhydrides). For example, sulphur trioxide (a non-metallic oxide) dissolves in water to give sulphuric acid:
SO3 + H2O → H2SO4 (Sulphuric acid)
(b) An acid can be prepared from a salt by displacement, where normal salts of more volatile acids are displaced by a less or non-volatile acid. For example, hydrochloric acid (a volatile acid) is prepared by the reaction of sodium chloride (a salt) with concentrated sulphuric acid (a less volatile acid):
NaCl + H2SO4 (conc.) → NaHSO4 + HCl (Hydrochloric acid)
NaNO3 + H2SO4 (conc.) → NaHSO4 + HNO3 (Nitric acid)
12. Give equations to show how the following are made from their corresponding anhydrides.
Answer: (a) sulphurous acid
SO2 + H2O → H2SO3 (Sulphurous acid)
(b) phosphoric acid
P2O5 + 3H2O → 2H3PO4 (Phosphoric acid)
(c) carbonic acid
CO2 + H2O → H2CO3 (Carbonic acid)
(d) sulphuric acid
SO3 + H2O → H2SO4 (Sulphuric acid)
13. Name an acid used:
(a) to flavour and preserve food
Answer: Acetic acid is used as table vinegar (flavour) and in cooking. Benzoic acid is used for the preservation of food. Citric acid is used for food preservation.
(b) in a drink
Answer: Carbonic acid is used in flavoured drinks.
(c) to remove ink spots
Answer: Oxalic acid is used as an ink stain remover.
(d) as an eyewash
Answer: Boric acid is used as an eye-wash/antiseptic.
14. Give reaction of acids with
(a) chlorides
Answer: Few chlorides react with dilute sulphuric acid. For example:
BaCl₂ + H₂SO₄ (dilute) → 2HCl + BaSO₄↓
Otherwise, chlorides generally react with concentrated H₂SO₄ on warming to liberate HCl.
NaCl + H₂SO₄ (conc.) –(below 200°C)–> NaHSO₄ + HCl
2NaCl + H₂SO₄ (conc.) –(above 200°C)–> Na₂SO₄ + 2HCl
The conditions are either using dilute acid for specific reactive chlorides or using concentrated acid with warming for most other chlorides.
(b) nitrates
Few nitrates react with dilute sulphuric acid. For example:
Pb(NO₃)₂ + H₂SO₄ (dilute) → 2HNO₃ + PbSO₄↓
Otherwise, nitrates generally react with concentrated H₂SO₄ on warming to liberate HNO₃.
KNO₃ + H₂SO₄ (conc.) –(below 200°C)–> KHSO₄ + HNO₃
2KNO₃ + H₂SO₄ (conc.) –(above 200°C)–> K₂SO₄ + 2HNO₃
The conditions are either using dilute acid for specific reactive nitrates or using concentrated acid with warming for most other nitrates.
Intext Questions and Answers II
1. What do you understand by an alkali? Give two examples of: (a) strong alkalis, (b) weak alkalis.
Answer: An alkali is a basic hydroxide which when dissolved in water produces hydroxyl (OH-) ions as the only negatively charged ions.
(a) Examples of strong alkalis are: Sodium hydroxide (NaOH) and Potassium hydroxide (KOH).
(b) Examples of weak alkalis are: Ammonium hydroxide (NH₄OH) and Calcium hydroxide (Ca(OH)₂).
2. What is the difference between: (a) an alkali and a base, (b) the chemical nature of an aqueous solution of HCl and an aqueous solution of NH₃.
Answer: (a) All alkalis are bases but all bases are not alkalis. For example: Ferric hydroxide [Fe(OH)₃] and Cupric hydroxide [Cu(OH)₂] are bases, but not alkalis because they are insoluble in water. An alkali is a base soluble in water.
(b) An aqueous solution of HCl is acidic because HCl is an acid that furnishes H⁺ ions (or H₃O⁺ ions) in aqueous solution. An aqueous solution of NH₃ (which forms NH₄OH) is basic because NH₄OH is an alkali that furnishes OH⁻ ions in aqueous solution.
3. Name the ions furnished by: (a) bases in solution, (b) an acid.
Answer: (a) Bases in solution, specifically alkalis, furnish hydroxyl (OH⁻) ions as the only negatively charged ions.
(b) An acid in aqueous solution furnishes hydrogen ion (or proton), i.e. H⁺ ion, which combines with a water molecule to form hydronium ion (H₃O⁺), the only positively charged ions.
4. 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.
Answer: (a) Sodium oxide (Na₂O) is a basic oxide which is soluble in water.
(b) Sodium hydroxide (NaOH) is a hydroxide which is highly soluble in water.
(c) Copper oxide (CuO) is a basic oxide which is insoluble in water.
(d) Ferric hydroxide [Fe(OH)₃] is a hydroxide which is insoluble in water.
(e) Carbonic acid (H₂CO₃) is a weak mineral acid.
(f) Cupric hydroxide [Cu(OH)₂] is a base which is not an alkali.
(g) Copper oxide (CuO) is an oxide which is a base.
(h) Water (H₂O) is a hydrogen containing compound which is not an acid.
(i) Ammonium hydroxide (NH₄OH) is a base which does not contain a metal ion.
5. 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?
Answer: To identify the contents of each test tube using only red litmus paper:
- Dip the red litmus paper into each solution.
- The solution that turns the red litmus paper blue is the basic solution.
- The solution that shows no change in the red litmus paper could be either acidic or distilled water.
- Now, take the litmus paper that turned blue (from testing the basic solution). Dip this blue litmus paper into the remaining two solutions.
- The solution that turns this blue litmus paper red is the acidic solution.
- The solution that causes no change to this blue litmus paper (and also caused no change to the initial red litmus paper) is distilled water.
6. HCl, HNO₃, C₂H₅OH, C₆H₁₂O₆ all contain H atoms but only HCl and HNO₃ show acidic character. Why?
Answer: HCl and HNO₃ show acidic character because when dissolved in water, they produce hydronium ions (H₃O⁺) as the only positively charged ions. C₂H₅OH (ethanol) and C₆H₁₂O₆ (glucose), although containing hydrogen atoms, do not ionise in water to produce H₃O⁺ ions and hence do not show acidic character. The acidic properties of an acid are actually the properties of hydronium ions present in it.
7. (a) Dry HCl gas does not change the colour or dry litmus paper. Why?
(b) Is PbO₂ a base or not? Comment.
(c) Do basic solutions also have H⁺(aq)? Explain why are they basic by taking an example?
Answer: (a) Dry HCl gas does not change the colour of dry litmus paper because the acidic properties of an acid are due to the presence of hydronium ions (H₃O⁺) which are formed when the acid dissolves in water. In the absence of water, dry HCl gas does not furnish H⁺ ions and hence does not show acidic character.
(b) Lead (IV) oxide (PbO₂) is a metallic oxide which reacts with hydrochloric acid to produce lead (II) chloride (a salt) and water, but the word ‘only’ excludes it from the class of bases, because chlorine is also produced. The reaction is: PbO₂(s) + 4HCl(aq) → PbCl₂(aq) + Cl₂(g) + 2H₂O. Thus Lead (IV) oxide PbO₂ is not a base.
(c) Yes, basic solutions also have H⁺(aq) ions. Water ionises to a very small extent, producing both [H₃O⁺] (aq.) and OH⁻ (aq.) ions. In pure water, the concentration of [H₃O⁺] (aq.) and OH⁻ (aq.) ions is equal (1 × 10⁻⁷ mol litre⁻¹). When a base is added to water, it increases the concentration of OH⁻ ions. Although the concentration of H₃O⁺ ions decreases, they are still present. A solution is basic because the concentration of OH⁻ ions is greater than the concentration of H₃O⁺ ions. For example, in a solution of NaOH, NaOH dissociates to give Na⁺ and OH⁻ ions, increasing the OH⁻ ion concentration, making the solution basic, even though H₃O⁺ ions from the ionisation of water are still present.
8. How would you obtain :
(a) a base from another base,
(b) an alkali from a base.
(c) salt from another salt?
Answer:(a) A base can be obtained from another base by double decomposition. For example, aqueous solutions of salts with a base (alkali) precipitate the respective metallic hydroxide.
FeCl₃ (salt solution) + 3NaOH (base/alkali) → Fe(OH)₃↓ (base) + 3NaCl (salt)
(b) An alkali can be obtained from a base (metallic oxide) by the action of water on soluble metallic oxides.
Na₂O (basic oxide) + H₂O → 2NaOH (alkali)
(c) A salt can be obtained from another salt by double decomposition (precipitation).
AgNO₃ (salt solution) + NaCl (salt solution) → AgCl↓ (salt) + NaNO₃ (salt)
9. 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
Answer: (a) Zn + 2HCl → ZnCl₂ + H₂↑
(b) H₂SO₄ + CuO → CuSO₄ + H₂O
(c) CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂↑
(d) CaSO₃ + 2HCl → CaCl₂ + H₂O + SO₂↑
(e) ZnS + 2HCl → ZnCl₂ + H₂S↑
10. The skin has and needs natural oils. Why is it advisable to wear gloves while working with strong alkalis?
Answer: Caustic soda and caustic potash are strong alkalis. We know that alkalis react with oil to form soap. Since our skin contains oil (in the form of fat), hence when we touch caustic soda or caustic potash, a reaction takes place and soapy solutions are formed. This can cause a mild corrosive action (slight burn) on the skin. Therefore, it is advisable to wear gloves while working with strong alkalis.
11. Complete the table :
Answer:
12. What do you understand by pH value? Two solutions X and Y have pH values of 4 and 10 respectively. Which one of these two will give a pink colour with phenolphthalein indicator?
Answer: The pH of a solution is the negative logarithm to the base 10 of the hydrogen ion concentration expressed in moles per litre. It represents the [H₃O⁺] ion concentration of the given aqueous solution.
Solution Y with a pH value of 10 will give a pink colour with phenolphthalein indicator because phenolphthalein turns pink in alkaline solutions (pH > 7). Solution X with pH 4 is acidic and phenolphthalein remains colourless in acidic solutions.
13. 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.
(a) If M is a metal, then the litmus will turn ………………..
(b) If M is a non-metal, then the litmus will turn ………………..
(c) If M is a reactive metal, then ……………….. will be evolved when M reacts with dilute sulphuric acid.
(d) If M is a metal, it will form ……………….. oxide, which will form ……………….. solution with water.
(e) If M is a non-metal, it will not conduct electricity in the form of ………………..
Answer:
(a) If M is a metal, then the litmus will turn blue.
(b) If M is a non-metal, then the litmus will turn red.
(c) If M is a reactive metal, then hydrogen will be evolved when M reacts with dilute sulphuric acid.
(d) If M is a metal, it will form basic oxide, which will form alkaline solution with water.
(e) If M is a non-metal, it will not conduct electricity in the form of ions.
14. 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).
Answer: (a) A common acid-base indicator like litmus changes colour to indicate whether a solution is acidic or basic (e.g., litmus turns red in acid, blue in alkali). A universal indicator is a mixture of indicator dyes that gives a spectrum of colours depending on how acidic or alkaline a solution is, providing a more precise pH value by showing different colours at different concentrations of hydrogen ions.
(b) Acidity of a base is the number of hydroxyl ions [OH]⁻ which can be produced per molecule of the base in aqueous solution or the number of hydrogen ions (of an acid) with which a molecule of that base will react to produce salt and water only. Basicity of an acid is defined as the number of hydronium ions (H₃O)⁺ that can be produced by the ionisation of one molecule of that acid in aqueous solution.
(c) An acid is a compound which contains one or more hydrogen atoms and when dissolved in water, produces hydronium ions (H₃O⁺) as the only positively charged ions. An alkali is a basic hydroxide which when dissolved in water produces hydroxyl (OH⁻) ions as the only negatively charged ions. Acids generally have a sour taste, while alkalis have a bitter taste and soapy touch. Acids turn blue litmus red, while alkalis turn red litmus blue.
15. What should be added to neutral solution in order to :
(a) increase the concentration of hydronium ions,
(b) decrease the concentration of hydronium ions
Answer: (a) To increase the concentration of hydronium ions in a neutral solution, an acid should be added.
(b) To decrease the concentration of hydronium ions in a neutral solution, an alkali (or base) should be added.
16. How does tooth enamel get damaged? What should be done to prevent it?
Answer: Substances like chocolates and sweets are degraded by bacteria present in our mouth. When the pH falls to 5.5 tooth decay starts. Tooth enamel (calcium phosphate) is the hardest substance in our body and it gets corroded. To prevent it, tooth paste, which is generally basic, is used to neutralise excess acid in the mouth. The saliva produced by salivary glands is slightly alkaline and helps to increase the pH to some extent.
17. When you use universal indicator, you see that solutions of different acids produce different colours. Indeed, solutions of the same acid with different concentrations will also give different colours. Why?
Answer: Universal indicators give different colours at different concentrations of hydrogen ions in a solution. Different acids, or the same acid at different concentrations, will produce different concentrations of hydrogen ions [H₃O⁺] when dissolved in water. Since the colour produced by a universal indicator depends on the concentration of hydrogen ions, different colours are observed.
18. 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.
Answer: 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
Liberation of hydrogen:
(a) With magnesium powder: Solution A (pH 1.8) would be most likely to liberate hydrogen, followed by Solution E (pH 5). Stronger acids react more vigorously.
Word equation: Magnesium + Acid (Solution A or E) → Magnesium salt + Hydrogen
(b) With powdered zinc metal: Solution A (pH 1.8) would be most likely to liberate hydrogen, followed by Solution E (pH 5).
Word equation: Zinc + Acid (Solution A or E) → Zinc salt + Hydrogen
19. Solution P has a pH of 13, solution Q has 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?
Answer:
(a) Solution P (pH 13) will liberate ammonia from ammonium sulphate on heating because it is a strong alkali.
(b) Solution R (pH 2) is a strong acid.
(c) Solution Q (pH 6) is a weak acid and will contain molecules as well as ions. (Weak acids do not ionise completely in solution).
20. Select the word/s given which are required to correctly complete the blanks
[ammonia, ammonium carbonate, carbon dioxide, hydrogen, hydronium, hydroxide, precipitate, salt, water]
(i) A solution M turns blue litmus red, so it must contain (1) ……………….. ions; another solution O turns red litmus blue and hence it must contain (2) ……………….. ions.
(ii) When solution M and O are mixed together, the products will be (3) ……………….. and (4) ………………..
(iii) If a piece of magnesium was put into solution M, (5) ……………….. gas would be evolved.
Answer: (i) A solution M turns blue litmus red, so it must contain (1) hydronium ions; another solution O turns red litmus blue and hence it must contain (2) hydroxide ions.
(ii) When solution M and O are mixed together, the products will be (3) salt and (4) water.
(iii) If a piece of magnesium was put into solution M, (5) hydrogen gas would be evolved.
Intext Questions and Answers III
1. Define the following and give two examples in each case: (a) a normal salt, (b) an acid salt, (c) a basic salt.
Answer: (a) A normal salt is a salt formed by the complete replacement of the ionisable hydrogen atoms of an acid by a metallic or an ammonium ion.
Examples of normal salts are:
- Sodium chloride (NaCl)
- Sodium carbonate (Na₂CO₃)
(b) An acid salt is a salt formed by the partial replacement of the ionisable hydrogen atoms of a polybasic acid by a metal or an ammonium ion.
Examples of acid salts are:
- Sodium hydrogen sulphate (NaHSO₄)
- Sodium hydrogen sulphite (NaHSO₃)
(c) A basic salt is a salt formed by the partial replacement of the hydroxyl group of a di- or a tri-acidic base by an acid radical.
Examples of basic salts are:
- Basic lead chloride [Pb(OH)Cl]
- Basic magnesium chloride [Mg(OH)Cl]
2. Answer the following questions related to salts and their preparations:
(a) What is a ‘salt’?
Answer: A salt is a compound formed by the partial or total replacement of the ionisable hydrogen atoms of an acid by a metallic ion or an ammonium ion. Ionically, a salt is an ionic compound, which dissociates in water to yield a positive ion other than hydrogen ion (H⁺) and a negative ion other than hydroxyl ion (OH⁻).
(b) What kind of salt is prepared by precipitation?
Answer: Insoluble salts are prepared by precipitation, which is a double decomposition reaction where a solid precipitate is formed.
(c) Name a salt prepared by direct combination. Write an equation for the reaction that takes place in preparing the salt you have named.
Answer: A salt prepared by direct combination is sodium chloride.
The equation for the reaction is: 2Na(molten) + Cl₂ → 2NaCl
(d) Name the procedure used to prepare a sodium salt such as sodium sulphate.
Answer: The procedure used to prepare a sodium salt such as sodium sulphate involves neutralisation of an alkali (like sodium hydroxide or sodium carbonate) with an acid (dilute sulphuric acid), often by titration to determine the correct ratio of reactants, followed by evaporation of the resulting solution to obtain crystals.
3. Explain the following methods with examples.
(a) Direct combination
Answer: Direct combination is a method of preparing salts by heating two elements together.
For example, sodium chloride can be prepared by the direct combination of molten sodium and chlorine gas:
2Na(molten) + Cl₂ → 2NaCl
(b) Displacement
Answer: Displacement, in the context of preparing soluble salts of active metals, involves the action of dilute acids on these active metals. It is a chemical change in which a more active element displaces a less active element (in this case, hydrogen from the acid).
For example, zinc sulphate is prepared by the displacement reaction between zinc metal and dilute sulphuric acid:
Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂↑
(c) Double decomposition (precipitation)
Answer: Double decomposition, also known as precipitation, is a chemical change in which two compounds in solution react to form two other compounds by the mutual exchange of radicals. A solid precipitate (an insoluble salt) is formed as a result of this reaction.
For example, barium sulphate (an insoluble salt) is prepared by the double decomposition reaction between barium chloride solution and sulphuric acid:
BaCl₂ (aq) + H₂SO₄ (aq) → BaSO₄↓ (s) + 2HCl (aq)
(d) Neutralisation of insoluble base
Answer: Neutralisation of an insoluble base is a method to prepare salts where an acid reacts with an insoluble base (like a metallic oxide or hydroxide) to form salt and water. Neutralisation is the process by which H⁺ ions of an acid react completely with the [OH⁻] ions of a base (or O²⁻ from an oxide) to give salt and water only.
For example, copper (II) sulphate can be prepared by neutralising the insoluble base copper (II) oxide with dilute sulphuric acid:
CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l)
(e) Neutralisation of an alkali (titration)
Answer: Neutralisation of an alkali, often carried out by titration, is a method to prepare soluble salts where an acid reacts with a soluble base (alkali) to form salt and water. Titration helps to determine the exact volumes of acid and alkali required for complete neutralisation.
For example, sodium nitrate can be prepared by the neutralisation of sodium hydroxide (an alkali) with nitric acid:
NaOH(aq) + HNO₃(aq) → NaNO₃(aq) + H₂O(l)
4. How would you prepare :
(a) copper sulphate crystals from a mixture of charcoal and black copper oxide,
Answer: To prepare copper sulphate crystals from a mixture of charcoal and black copper oxide:
- Take dilute sulphuric acid in a beaker and heat it on a wire gauze.
- Add the mixture of charcoal and black copper oxide, in small quantities at a time, and keep on stirring the solution. The black copper oxide will react with the hot dilute sulphuric acid to form copper (II) sulphate, while charcoal will not react. Continue adding the mixture until no more copper oxide dissolves and some excess solid (copper oxide and charcoal) settles at the bottom. This indicates that all the acid has been used up.
- Filter the hot solution to remove the unreacted charcoal and excess copper oxide. Collect the filtrate, which is copper (II) sulphate solution, in a china dish.
- Evaporate the filtrate by heating to the point of crystallisation (when a thin crust or small crystals start to form on the surface or on a glass rod dipped into the solution).
- Allow the concentrated solution to cool slowly. Bright blue crystals of copper (II) sulphate pentahydrate (CuSO₄·5H₂O) will form.
- Collect the crystals, wash them with a little cold water if necessary, and dry them between the folds of filter paper or in a desiccator.
(b) zinc sulphate crystals from zinc dust (powdered zinc and zinc oxide),
Answer: To prepare zinc sulphate crystals from zinc dust (which contains powdered zinc and zinc oxide):
- Take dilute sulphuric acid (e.g., 1 volume of acid : 5 volumes of water) in a beaker and heat it gently on a wire gauze.
- Add the zinc dust (containing zinc metal and zinc oxide) in small portions, a little at a time, with constant stirring. The zinc metal will react with the acid to produce hydrogen gas (effervescence will be observed), and the zinc oxide (a basic oxide) will react with the acid in a neutralisation reaction.
Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂↑
ZnO(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂O(l) - Continue adding zinc dust until it settles at the base of the beaker and no more effervescence (if primarily zinc metal was reacting) or dissolution occurs. This indicates that all the acid has been used up.
- Filter off the excess unreacted zinc dust.
- Collect the filtrate (zinc sulphate solution) in a china dish.
- Evaporate the solution gently to concentrate it until the point of crystallisation is reached.
- Allow the hot concentrated solution to cool. White, needle-shaped crystals of hydrated zinc sulphate (ZnSO₄·7H₂O, white vitriol) will separate out.
- Filter the crystals, wash them with a small amount of cold water, and dry them between the folds of a filter paper.
(c) sodium hydrogen carbonate crystals.
Answer: To prepare sodium hydrogen carbonate crystals:
The principle is to pass carbon dioxide gas into a cold solution of sodium carbonate.
Reaction: Na₂CO₃ + CO₂ + H₂O → 2NaHCO₃
Procedure:
- Dissolve nearly 5 grams of anhydrous sodium carbonate in about 25 cm³ of distilled water in a flask.
- Cool the solution by keeping the flask in a freezing mixture (e.g., ice and salt).
- Pass carbon dioxide gas slowly through the cold sodium carbonate solution.
- Crystals of sodium bicarbonate (sodium hydrogen carbonate) will precipitate out after sometime as it is less soluble than sodium carbonate under these conditions.
- Filter the crystals of sodium hydrogen carbonate, wash them with a small amount of cold water, and dry them in the folds of filter paper or in a desiccator
(d) Calcium sulphate from calcium carbonate.
Answer: To prepare calcium sulphate from calcium carbonate, a two-step process is generally used because direct reaction of calcium carbonate with sulphuric acid can form an insoluble layer of calcium sulphate that prevents further reaction.
- First, convert insoluble calcium carbonate into a soluble calcium salt. This can be done by reacting calcium carbonate with dilute hydrochloric acid or dilute nitric acid. For example, with dilute hydrochloric acid:
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂↑(g)
Filter the solution if there are any unreacted solids to get a clear solution of calcium chloride. - Next, precipitate calcium sulphate by adding a solution of a soluble sulphate (like sodium sulphate solution) or dilute sulphuric acid to the calcium chloride solution obtained in step 1.
Using sodium sulphate solution:
CaCl₂(aq) + Na₂SO₄(aq) → CaSO₄(s)↓ + 2NaCl(aq)
Or using dilute sulphuric acid:
CaCl₂(aq) + H₂SO₄(aq) → CaSO₄(s)↓ + 2HCl(aq)
The white precipitate of calcium sulphate is then filtered, washed with water to remove any soluble impurities, and dried.
5. 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 for preparing the following salts by giving a suitable equation in each case:
Answer: 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 for preparing the following salts by giving a suitable equation in each case:
(1) Anhydrous ferric chloride,
Answer: The best method is A – direct combination of two elements.
Suitable equation: 2Fe(s) + 3Cl₂(g) → 2FeCl₃(s) (Iron heated with dry chlorine gas)
(2) Lead chloride.
Answer: The best method is E – reaction of two solutions of salts to form a precipitate.
Suitable equation: Pb(NO₃)₂(aq) + 2NaCl(aq) → PbCl₂↓(s) + 2NaNO₃(aq)
(Alternatively, Pb(NO₃)₂(aq) + 2HCl(aq) → PbCl₂↓(s) + 2HNO₃(aq))
(3) Sodium sulphate.
Answer: The best method is D – titration of a dilute acid with a solution of soluble base.
Suitable equation: 2NaOH(aq) + H₂SO₄(dil) → Na₂SO₄(aq) + 2H₂O(l)
(Alternatively, Na₂CO₃(aq) + H₂SO₄(dil) → Na₂SO₄(aq) + H₂O(l) + CO₂↑(g), though titration with NaOH is more typical for method D)
(4) Copper sulphate.
Answer: The best method is C – reaction of a dilute acid with an insoluble base.
Suitable equation: CuO(s) + H₂SO₄(dil) → CuSO₄(aq) + H₂O(l)
(Alternatively, Cu(OH)₂(s) + H₂SO₄(dil) → CuSO₄(aq) + 2H₂O(l) or CuCO₃(s) + H₂SO₄(dil) → CuSO₄(aq) + H₂O(l) + CO₂↑(g))
6. Name:
(a) a chloride which is insoluble in cold water but dissolves in hot water,
Answer: Lead chloride (PbCl₂)
(b) a chloride which is insoluble,
Answer: Silver chloride (AgCl)
(c) two sulphates which are insoluble.
Answer: Barium sulphate (BaSO₄) and Lead sulphate (PbSO₄). (Also Calcium sulphate CaSO₄ and Silver sulphate Ag₂SO₄ are listed as exceptions for soluble sulphates).
(d) a basic salt,
Answer: Basic lead chloride [Pb(OH)Cl] (Other examples: Basic magnesium chloride [Mg(OH)Cl], Basic copper chloride [Cu(OH)Cl], Basic copper nitrate [Cu(OH)NO₃])
(e) an acidic salt,
Answer: Sodium hydrogen sulphate (NaHSO₄) (Other examples: Sodium hydrogen sulphite (NaHSO₃), Disodium hydrogen phosphate (Na₂HPO₄), Sodium dihydrogen phosphate (NaH₂PO₄))
(f) a mixed salt,
Answer: Sodium potassium carbonate (NaKCO₃) (Another example: Bleaching powder CaOCl₂)
(g) a complex salt,
Answer: Sodium argentocyanide (Na[Ag(CN)₂]) (Other examples: Silver amino chloride [Ag(NH₃)₂]Cl, Tetraammine copper (II) sulphate [Cu(NH₃)₄]SO₄, Potassium mercuric iodide K₂[HgI₄], Sodium zincate Na₂ZnO₂)
(h) a double salt
Answer: Potash alum (K₂SO₄·Al₂(SO₄)₃·24H₂O) (Another example: Mohr’s salt FeSO₄·(NH₄)₂SO₄·6H₂O, Dolomite CaCO₃·MgCO₃)
7. 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 contains ………………… ions. A base reacts with an acid to form ………….. and water only. This type of reaction is known as …………………
Answers: An acid is a compound which when dissolved in water forms hydronium ions as the only positively charged ions. A base is a compound which is soluble in water contains hydroxyl (OH⁻) ions. A base reacts with an acid to form salt and water only. This type of reaction is known as neutralisation.
8. What would you observe when:
(a) blue litmus is introduced into a solution of hydrogen chloride gas,
Answer: If hydrogen chloride gas is dissolved in an organic solvent (like toluene) and is perfectly dry, there is no change in the colour of blue litmus. However, if it is an aqueous solution of hydrogen chloride gas (hydrochloric acid), or if moisture is present, the blue litmus will turn red.
(b) red litmus paper is introduced into a solution of ammonia in water,
Answer: Red litmus paper turns blue when introduced into a solution of ammonia in water (ammonium hydroxide).
(c) red litmus paper is introduced in caustic soda solution?
Answer: Red litmus paper turns blue when introduced in caustic soda (sodium hydroxide) solution.
9. Explain why:
(a) it is necessary to find out the ratio of reactants required in the preparation of sodium sulphate,
Answer: It is necessary to find out the ratio of reactants required in the preparation of sodium sulphate because both reactants (sodium hydroxide/sodium carbonate and sulphuric acid) as well as the product (sodium sulphate) are soluble in water. An excess of either of the reactants cannot be removed by filtration. Therefore, it is necessary to find out on a small scale, the ratio of the solutions of the two reactants required for complete neutralisation before preparation, usually by titration.
(b) fused calcium chloride is used in the preparation of FeCl₃?
Answer: Fused calcium chloride is used in the preparation of FeCl₃ because Iron (III) chloride is highly deliquescent. Fused calcium chloride acts as a drying agent, keeping the FeCl₃ dry.
(c) Anhydrous FeCl₃ cannot be prepared by heating hydrated Iron (III) chloride.
Answer: Anhydrous FeCl₃ cannot be prepared by simply heating the hydrated ferric chloride (FeCl₃·6H₂O) because on heating, FeCl₃·6H₂O produces Fe₂O₃, H₂O and HCl, instead of anhydrous FeCl₃.
2[FeCl₃·6H₂O]
Fe₂O₃ + 9H₂O + 6HCl
10. Match the salts given in column A to their methods of preparation in column B. Write a balanced equation for each preparation.
Answer:
11. (a) Give the pH value of pure water. Does it change if common salt is added to it?
Answer: The pH value of pure water is 7. The pH of pure water does not change if common salt (NaCl) is added to it, because NaCl is a salt of a strong acid (HCl) and a strong base (NaOH) and forms a neutral solution.
(b) Classify the following solutions as acids, bases or salts.
Ammonium hydroxide, barium chloride, sodium chloride, sodium hydroxide, H₂SO₄ and HNO₃.
Answer:
- Ammonium hydroxide: Base (weak alkali)
- Barium chloride: Salt
- Sodium chloride: Salt
- Sodium hydroxide: Base (strong alkali)
- H₂SO₄: Acid
- HNO₃: Acid
12. Complete the following table and write one equation for each to justify the statement:
Answer:
Answers:
13. 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,
Answer: CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l) (Copper (II) sulphate)
(b) An insoluble salt by the action of an acid on another salt,
Answer: Pb(NO₃)₂(aq) + H₂SO₄(aq) → PbSO₄(s)↓ + 2HNO₃(aq) (Lead (II) sulphate from lead nitrate)
Alternatively, if it means converting an insoluble salt to another insoluble salt via a soluble intermediate:
PbCO₃(s) + 2HNO₃(aq) → Pb(NO₃)₂(aq) + H₂O(l) + CO₂(g)↑
Then, Pb(NO₃)₂(aq) + H₂SO₄(aq) → PbSO₄(s)↓ + 2HNO₃(aq)
(c) An insoluble base by the action of a soluble base on a soluble salt,
Answer: CuSO₄(aq) + 2NaOH(aq) → Cu(OH)₂(s)↓ + Na₂SO₄(aq) (Copper (II) hydroxide)
(d) A soluble sulphate by the action of an acid on a metal.
Answer: Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)↑ (Zinc sulphate)
14. You are provided with the following chemicals: NaOH, Na₂CO₃, H₂O, Zn(OH)₂, CO₂, HCl, Fe, H₂SO₄, Cl₂, Zn. Using the suitable chemicals from the given list only, state briefly how you would prepare:
(a) iron (III) chloride
Answer: Method: Direct combination (Synthesis).
Briefly: Pass dry chlorine gas (Cl₂) over heated iron (Fe).
Equation: 2Fe(s) + 3Cl₂(g) 2FeCl₃(s)
(b) sodium sulphate
Answer: Method: Neutralisation (Titration).
Briefly: Titrate sodium hydroxide solution (NaOH) with dilute sulphuric acid (H₂SO₄) using a suitable indicator. Then evaporate the neutralised solution to get crystals.
Equation: 2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)
(Alternatively, using Na₂CO₃: Na₂CO₃(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + H₂O(l) + CO₂(g)↑)
(c) sodium zincate
Answer: Method: Action of alkali on certain metals/oxides/hydroxides.
Briefly: React zinc hydroxide (Zn(OH)₂) with sodium hydroxide solution (NaOH).
Equation: Zn(OH)₂(s) + 2NaOH(aq) → Na₂ZnO₂(aq) + 2H₂O(l)
(Alternatively, using Zn metal: Zn(s) + 2NaOH(aq) → Na₂ZnO₂(aq) + H₂(g)↑
(d) iron (II) sulphate
Answer: Method: Displacement (Action of dilute acid on active metal).
Briefly: React iron filings (Fe) with dilute sulphuric acid (H₂SO₄). Filter and crystallise.
Equation: Fe(s) + H₂SO₄(aq) → FeSO₄(aq) + H₂(g)↑
(e) sodium chloride.
Answer: Method: Neutralisation (Titration).
Briefly: Titrate sodium hydroxide solution (NaOH) with dilute hydrochloric acid (HCl) using a suitable indicator. Then evaporate the neutralised solution to get crystals.
Equation: NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)
(Alternatively, using Na₂CO₃: Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g)↑)
15. For each of the salt A, B, C and D, suggest a suitable method for its preparation.
(a) A is a sodium salt.
Answer: Method D: titration of a dilute acid with a solution of soluble base (e.g., NaOH or Na₂CO₃).
(b) B is an insoluble salt.
Answer: Method E: reaction of two solutions of salts to form a precipitate (Precipitation/Double Decomposition).
(c) C is a soluble salt of copper.
Answer: Method C: reaction of a dilute acid with an insoluble base (e.g., CuO or Cu(OH)₂).
(d) D is a soluble salt of zinc.
Answer: Method B: reaction of a dilute acid with a metal (Zinc metal). Or Method C: reaction of a dilute acid with an insoluble base (e.g., ZnO or Zn(OH)₂).
16. Choosing only the substances from the list given in the box below, write equations for the reactions which you would use in the laboratory to obtain:
[Dilute sulphuric acid, Copper, Copper carbonate, Iron, Sodium carbonate, Sodium, Zinc]
(a) Sodium sulphate,
Answer: Na₂CO₃(s) + H₂SO₄(dilute aq) → Na₂SO₄(aq) + H₂O(l) + CO₂(g)↑
(Alternatively, using Sodium metal, though it’s very reactive and not a typical lab prep for the salt directly with acid for beginners: 2Na(s) + H₂SO₄(dilute aq) → Na₂SO₄(aq) + H₂(g)↑)
(b) Copper sulphate,
Answer: CuCO₃(s) + H₂SO₄(dilute aq) → CuSO₄(aq) + H₂O(l) + CO₂(g)↑
(Copper metal will not react with dilute sulphuric acid as it is below hydrogen in reactivity series. Copper oxide would be better if available, but from the list, copper carbonate is suitable.)
(c) Iron(II) sulphate,
Answer: Fe(s) + H₂SO₄(dilute aq) → FeSO₄(aq) + H₂(g)↑
(d) Zinc carbonate.
Answer: Step 1: Prepare a soluble zinc salt first, e.g., zinc sulphate.
Zn(s) + H₂SO₄(dilute aq) → ZnSO₄(aq) + H₂(g)↑
Step 2: React the soluble zinc salt with sodium carbonate.
ZnSO₄(aq) + Na₂CO₃(aq) → ZnCO₃(s)↓ + Na₂SO₄(aq)
17. From the formula listed below, choose one in each case corresponding to the salt having the description given ahead : AgCl, CuCO₃, CuSO₄.5H₂O, KNO₃, NaCl, NaHSO₄, Pb(NO₃)₂, ZnCO₃, ZnSO₄.7H₂O
(a) an acid salt.
Answer: NaHSO₄
(b) an insoluble chloride.
Answer: AgCl
(c) on treating with concentrated sulphuric acid, this salt changes from blue to white.
Answer: CuSO₄.5H₂O
(d) on heating, this salt changes from green to black.
Answer: CuCO₃ (Copper carbonate is typically green, and on heating decomposes to black CuO and CO₂)
(e) this salt gives nitrogen dioxide on heating.
Answer: Pb(NO₃)₂ (Lead nitrate on heating gives PbO, NO₂ and O₂)
18. (a) Ca₃(PO₄)₂ is an example of a compound called ……………….. (acid salt / basic salt / normal salt).
Answer: normal salt
(b) Write the balanced equation for the reaction of : A named acid and a named alkali.
Answer: Named acid: Hydrochloric acid (HCl)
Named alkali: Sodium hydroxide (NaOH)
Balanced equation: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
19. State the terms defined by the following sentences :
(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 ionisation of the solute molecules.
Answer: Weak acid solution (or solution of a weak acid)
20. 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 an acid on a 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
Answer: Copper (II) chloride – B (e.g., CuO + 2HCl → CuCl₂ + H₂O)
(ii) Iron (II) chloride
Answer: Iron (II) chloride – A (e.g., Fe + 2HCl → FeCl₂ + H₂)
(iii) Iron (III) chloride
Answer: Iron (III) chloride – C (e.g., 2Fe + 3Cl₂ → 2FeCl₃)
(iv) Lead (II) chloride
Answer: Lead (II) chloride – E (e.g., Pb(NO₃)₂ + 2NaCl → PbCl₂↓ + 2NaNO₃)
(v) Sodium chloride
Answer: Sodium chloride – D (e.g., NaOH + HCl → NaCl + H₂O)
21. From the list given below, which one is : [SO₂, SiO₂, Al₂O₃, CO, MgO, Na₂O ]
(a) a covalent oxide of a metalloid.
Answer: SiO₂ (Silicon dioxide; silicon is a metalloid)
(b) an oxide which when dissolved in water form acid.
Answer: SO₂ (Sulphur dioxide; SO₂ + H₂O → H₂SO₃)
(c) a basic oxide.
Answer: MgO (Magnesium oxide) or Na₂O (Sodium oxide)
(d) an amphoteric oxide.
Answer: Al₂O₃ (Aluminium oxide)
Intext Question and Answers IV
1. What do you understand by water of crystallisation? Give four substances which contain water of crystallisation and write their common names.
Answer: Some salts unite with a definite quantity of water which is known as the water of crystallisation. Water of crystallisation is in loose chemical combination with the salts, and it can be driven out by heating the crystals of these salts above 100°C.
Four substances which contain water of crystallisation and their common names are:
- Washing soda crystals (Sodium carbonate decahydrate)
- Epsom salt (Magnesium sulphate heptahydrate)
- Glauber’s salt (Sodium sulphate decahydrate)
- Blue vitriol (Copper (II) sulphate pentahydrate)
2. (a) Define efflorescence. Give examples.
Answer: Efflorescence is the property of some salts to lose wholly, or partly their water of crystallisation when their crystals are exposed to dry air even for a short time. They become powdery. Such substances are called efflorescent substances.
Examples of efflorescent substances:
(a) Washing soda [hydrated sodium carbonate], when exposed to dry air, becomes a monohydrate.
(b) Glauber’s salt [hydrated sodium sulphate, Na₂SO₄·10H₂O] becomes a powdery anhydrous sodium sulphate when exposed to air.
(c) Epsom salt [magnesium sulphate heptahydrate], when exposed to dry air, becomes a monohydrate.
(b) Define deliquescence. Give examples.
Answer: Certain water-soluble substances, when exposed to the atmosphere at ordinary temperature, absorb moisture from the atmospheric air to become moist, and ultimately dissolve in the absorbed water, forming a saturated solution. Such a substance is called a deliquescent substance, and the phenomenon is called deliquescence.
Examples of deliquescent substances:
- Caustic soda (NaOH).
- Caustic potash (KOH).
- Magnesium chloride (MgCl₂).
- Zinc chloride (ZnCl₂).
- Calcium chloride (CaCl₂).
- Ferric chloride (FeCl₃).
- Zinc nitrate [Zn(NO₃)₂], and
- Copper nitrate [Cu(NO₃)₂].
3. Answer the questions below:
(a) What name is given to the water in the compound copper sulphate-5-water ?
Answer: The water in the compound copper sulphate-5-water is known as the water of crystallisation.
(b) If copper sulphate-5-water is heated, anhydrous copper sulphate is formed. What is its colour ?
Answer: If copper sulphate-5-water (blue crystal) is heated, anhydrous copper sulphate (white powder) is formed. Its colour is white.
(c) By what means, other than heating, could you dehydrate copper sulphate-5-water and obtain anhydrous copper sulphate ?
Answer: Concentrated sulphuric acid can remove water molecules from blue vitriol (CuSO₄·5H₂O), so it is a dehydrating agent as well. Thus, by using concentrated sulphuric acid, one could dehydrate copper sulphate-5-water and obtain anhydrous copper sulphate.
(d) Name a deliquescent salt.
Answer: A deliquescent salt is Ferric chloride (FeCl₃). Other examples include Caustic soda (NaOH), Caustic potash (KOH), Magnesium chloride (MgCl₂), Zinc chloride (ZnCl₂), and Calcium chloride (CaCl₂).
(e) Why does hydrated copper sulphate turn white on heating?
Answer: Hydrated copper sulphate (CuSO₄·5H₂O), which is blue, turns white on heating because it loses its water of crystallisation and forms anhydrous copper sulphate (CuSO₄), which is a white powder.
4. State your observations when the following are exposed to atmosphere
(a) Washing soda crystals
Answer: When washing soda crystals (hydrated sodium carbonate, Na₂CO₃·10H₂O) are exposed to dry air, they lose their water of crystallisation and become a monohydrate (Na₂CO₃·H₂O), turning powdery. This phenomenon is efflorescence.
(b) Iron (III) chloride salt.
Answer: When Iron (III) chloride salt (FeCl₃) is exposed to the atmosphere, it absorbs moisture from the atmospheric air to become moist, and ultimately dissolves in the absorbed water, forming a saturated solution. This is because it is a deliquescent substance.
5. 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⁻.
Answer: Sodium hydrogen sulphate is an acid salt. Acid salts ionise in water solution to give hydronium ions, and therefore, they show all the properties of an acid. Although it is a salt, its dissociation in water produces H⁺ ions, which is characteristic of acids.
(b) Anhydrous calcium chloride is used in a desiccator.
Answer: Anhydrous calcium chloride is used in a desiccator because it is a hygroscopic substance, meaning it absorbs moisture (water vapour) from the atmosphere without dissolving in it. Hygroscopic substances are desiccating agents used to keep other substances dry.
6. Explain clearly how conc. H₂SO₄ is used as dehydrating as well as drying agent.
Answer: Concentrated sulphuric acid (H₂SO₄) is a hygroscopic substance, meaning it absorbs moisture from other substances. This property makes it a drying agent, used to dry gases like chlorine, sulphur dioxide, hydrogen chloride, etc.
Concentrated sulphuric acid is also a dehydrating agent because it can remove water molecules even from compounds. For example, it can remove water molecules from blue vitriol (CuSO₄·5H₂O). Dehydrating agents remove chemically combined elements of water in the ratio of 2:1 (hydrogen:oxygen) from a compound and represent a chemical change.
7. Distinguish between drying and dehydrating agent.
Answer: The differences between a drying agent and a dehydrating agent are:
8. State whether a sample of each of the following would increase or decrease in mass if exposed to air.
(a) Solid NaOH,
Answer: Solid NaOH (Caustic soda) is a deliquescent substance. If exposed to air, it will absorb moisture from the atmospheric air, become moist, and ultimately dissolve in the absorbed water, forming a saturated solution. Therefore, its mass would increase.
(b) Solid CaCl₂,
Answer: Solid CaCl₂ (Calcium chloride) is a deliquescent substance. If exposed to air, it will absorb moisture from the atmospheric air, become moist, and ultimately dissolve in the absorbed water, forming a saturated solution. Therefore, its mass would increase.
(c) Solid Na₂CO₃·10H₂O,
Answer: Solid Na₂CO₃·10H₂O (Washing soda) is an efflorescent substance. If exposed to dry air, it will lose wholly, or partly its water of crystallisation and become powdery. Therefore, its mass would decrease.
(d) Conc. Sulphuric acid,
Answer: Conc. Sulphuric acid (H₂SO₄) is a hygroscopic substance. If exposed to air, it will absorb moisture (water vapour) from the atmosphere. Therefore, its mass would increase.
(e) Iron (III) Chloride
Answer: Iron (III) Chloride (FeCl₃) is a deliquescent substance. If exposed to air, it will absorb moisture from the atmospheric air, become moist, and ultimately dissolve in the absorbed water, forming a saturated solution. Therefore, its mass would increase.
9. (a) Why does common salt get wet during the rainy season?
Answer: Common salt [sodium chloride] turns moist and ultimately forms a solution, on exposure to air [especially during the rainy season]. Though pure sodium chloride is not deliquescent, the commercial version of the salt contains impurities, like magnesium chloride and calcium chloride, which are deliquescent substances. These impurities absorb moisture from the air, causing the common salt to get wet.
(b) How can this impurity be removed?
Answer: This kind of impurity (like magnesium chloride and calcium chloride in common salt) can be removed by passing a current of dry hydrogen chloride gas through a saturated solution of the affected salt. Pure sodium chloride is produced as a precipitate, which can be recovered by filtering and washing first with a little water and finally with alcohol.
(c) Name a substance which changes the blue colour of copper sulphate crystals to white.
Answer: Concentrated sulphuric acid is a substance which changes the blue colour of copper sulphate crystals (CuSO₄·5H₂O) to white (CuSO₄) because it is a dehydrating agent and removes the water of crystallisation.
(d) Name two crystalline substances which do not contain water of crystallisation.
Answer: Two crystalline substances which do not contain water of crystallisation are:
- Common salt (NaCl)
- Nitre (KNO₃)
Other examples include Sugar (C₁₂H₂₂O₁₁), Potassium permanganate (KMnO₄), and Ammonium chloride (NH₄Cl).
10. Name the salt which on hydrolysis forms (a) acidic (b) basic and (c) neutral solution. Give a balanced equation for each reaction
Answer: (a) Salts of strong acids and weak bases give an acidic solution (pH less than 7) on hydrolysis. Examples include Iron (III) chloride (FeCl₃), Copper sulphate (CuSO₄), Aluminium chloride (AlCl₃), and Ammonium sulphate [(NH₄)₂SO₄].
An equation for the hydrolysis of Copper sulphate forming an acidic solution is:
CuSO₄ + 2H₂O ⇌ H₂SO₄ (strong acid) + Cu(OH)₂ (weak base)
(b) Salts formed from strong bases like NaOH and KOH and weak acids like H₂CO₃, CH₃COOH, etc., give alkaline solutions on hydrolysis. They have pH more than 7. Examples include Sodium carbonate (Na₂CO₃) and potassium acetate (CH₃COOK).
An equation for the hydrolysis of Sodium carbonate forming a basic solution is:
Na₂CO₃ + 2H₂O ⇌ H₂CO₃ (weak acid) + 2NaOH (strong alkali)
(c) Salts derived from strong acids and strong bases, such as NaCl, Na₂SO₄, K₂SO₄ and KNO₃, give a neutral solution in water on hydrolysis.
An example is Sodium chloride (NaCl). Since it is formed from a strong acid (HCl) and a strong base (NaOH), its hydrolysis does not significantly alter the H⁺ or OH⁻ concentration, resulting in a neutral solution.
The ions of the salt do not react with water to produce H⁺ or OH⁻ ions, so the solution remains neutral. For NaCl:
NaCl (aq) → Na⁺(aq) + Cl⁻(aq)
Neither Na⁺ nor Cl⁻ hydrolyses significantly.
11. State the change noticed when blue litmus and red litmus are introduced in the following solutions:
(a) Na₂CO₃ solution
Answer: Na₂CO₃ solution is alkaline because it is a salt of a strong base (NaOH) and a weak acid (H₂CO₃).
When blue litmus is introduced, it will remain blue.
When red litmus is introduced, it will turn blue.
(b) NaCl solution
Answer: NaCl solution is neutral because it is a salt of a strong acid (HCl) and a strong base (NaOH).
When blue litmus is introduced, it will remain blue.
When red litmus is introduced, it will remain red.
(c) NH₄NO₃ solution
Answer: NH₄NO₃ solution is acidic because it is a salt of a strong acid (HNO₃) and a weak base (NH₄OH).
When blue litmus is introduced, it will turn red.
When red litmus is introduced, it will remain red.
(d) MgCl₂ solution
Answer: MgCl₂ solution is slightly acidic. It is formed from a strong acid (HCl) and a relatively weaker base (Mg(OH)₂).
When blue litmus is introduced, it will turn slightly red or pink.
When red litmus is introduced, it will remain red.
Exercise
MCQs
1. The acid which contains four hydrogen atoms is :
(a) Formic acid
(b) Sulphuric acid
(c) Nitric acid
(d) Acetic acid
Answer: (d) Acetic acid
2. A black coloured solid which on reaction with dilute sulphuric acid forms a blue coloured solution is :
(a) Carbon
(b) Manganese [IV] oxide
(c) Lead [II] oxide
(d) Copper [II] oxide
Answer: (d) Copper [II] oxide
3. A weak acid is :
(a) Formic acid
(b) Sulphuric acid
(c) Nitric acid
(d) Hydrochloric acid
Answer: (a) Formic acid
4. A complex salt is:
(a) Zinc sulphate
(b) Sodium hydrogen sulphate
(c) Iron [ammonium sulphate]
(d) Tetrammine copper [II] sulphate
Answer: (d) Tetrammine copper [II] sulphate
5. Which one of the following will not produce acid with water?
(a) CO
(b) CO₂
(c) NO₂
(d) SO₃
Answer: (a) CO
6. The basicity of acetic acid is :
(a) 3
(b) 1
(c) 4
(d) 2
Answer: (b) 1
7. Lead sulphate can be prepared by the following methods.
P Reacting lead with sulphuric acid.
Q Reacting lead nitrate with sodium sulphate solution.
R Reacting lead oxide with sodium sulphate solution.
(a) Only P
(b) Only Q
(c) Only R
(d) Both Q and R
Answer: (b) Only Q
8. Assertion (A): All acids form hydronium ions when dissolved in water.
Reason (R): Every acid has hydrogen as the necessary element.
(a) (1)
(b) (2)
(c) (3)
(d) (4)
Answer: (a) (1)
9. Assertion (A): All bases react with acids to form salt and water only.
Reason (R): All alkalies are bases but all bases are not alkalies.
(a) (1)
(b) (2)
(c) (3)
(d) (4)
Answer: (b) (2)
10. Assertion (A): Acids which contain carbon are necessarily organic acids.
Reason (R): All organic acids are derived from plants and animals.
(a) (1)
(b) (2)
(c) (3)
(d) (4)
Answer: (d) (4)
11. Assertion (A): Acids have a pH of more than 7.
Reason (R): Acids produce hydronium ions as the only positively charged ion in solution.
(a) (1)
(b) (2)
(c) (3)
(d) (4)
Answer: (d) (4)
12. Assertion (A): Iron (III) chloride is stored in a moisture proof container.
Reason (R): Iron (III) chloride is hygroscopic in nature.
(a) (1)
(b) (2)
(c) (3)
(d) (4)
Answer: (c) (3)
13. Assertion (A): Water is added to acids to dilute it.
Reason (R): The addition of water to acids is a highly exothermic reaction. Splashes of acid solution may come in contact with skin and container may also break.
(a) (1)
(b) (2)
(c) (3)
(d) (4)
Answer: (d) (4)
Very Short Answer Type
1. (a) A solution has a pH of 7. Explain how you would : (i) increase its pH; (ii) decrease its pH;
Answer:
(i) To increase the pH of a solution that has a pH of 7, an alkali should be added. When an alkali is added to water, the concentration of OH- ions increases, and the alkaline nature increases, leading to an increase in pH above 7.
(ii) To decrease the pH of a solution that has a pH of 7, an acid should be added. When an acid is added to water, the concentration of H3O+ ions increases, and the acidic nature increases, leading to a decrease in pH below 7.
(b) If a solution changes the colour of litmus from red to blue, what can you say about its pH ?
Answer: If a solution changes the colour of litmus from red to blue, it indicates that the solution is alkaline. Alkaline solutions have a pH greater than 7.
(c) What can you say about the pH of a solution, that liberates carbon dioxide from sodium carbonate ?
Answer: Acids liberate carbon dioxide on reaction with metallic carbonates like sodium carbonate. Therefore, a solution that liberates carbon dioxide from sodium carbonate is acidic, and its pH would be less than 7.
2. Match the following:
Answer:
3. Match the salts given in Column I with their method of preparation given in Column II.
Answer:
4. From the following list, Na2O, SO2, SiO2, Al2O3, MgO, CO, select an oxide which dissolves in water forming an acid.
Answer: From the given list, SO2 (sulphur dioxide) is an acidic oxide that dissolves in water to form an acid (sulphurous acid, H2SO3).
5. From the list given below, which one is :
(i) a deliquescent compound
(ii) a compound soluble in hot water but insoluble in cold water
(iii) a compound which in the aqueous solution state, is neutral in nature.
A. Nitroso iron [II] sulphate,
B. Iron [III] chloride,
C. Chromium sulphate,
D. Lead chloride,
E. Sodium chloride
Answer: (i) A deliquescent compound is B. Iron [III] chloride.
(ii) A compound soluble in hot water but insoluble in cold water is D. Lead chloride.
(iii) A compound which in the aqueous solution state, is neutral in nature is E. Sodium chloride.
6. From the following list of salts, choose the salt that most appropriately fits the description given ahead : [AgCl, MgCl2, NaHSO4, PbCO3, ZnCO3, KNO3, Ca(NO3)2].
Answer:(i) A deliquescent salt from the list is MgCl2.
(ii) An insoluble chloride from the list is AgCl.
7. Give a suitable word or phrase for:
Answer: (i) A salt formed by incomplete neutralisation of an acid by a base is an Acid salt.
(ii) The definite number of water molecules bound to some salts is Water of crystallisation.
(iii) The process in which a substance absorbs moisture from the atmospheric air to become moist, and ultimately dissolves in the absorbed water is Deliquescence.
(iv) A salt containing a metal ion surrounded by other ions or molecules is a Complex salt.
(v) A base which is soluble in water is an Alkali.
8. Name the method used for the preparation of the following salts from the list given below:
(i) Sodium nitrate
(ii) Iron (III) chloride
(iii) Lead chloride
(iv) Zinc sulphate
(v) Sodium hydrogen sulphate.
A. Simple displacement,
B. Neutralisation,
C. Decomposition by acid,
D. Double decomposition,
E. Direct synthesis
Answer: (i) Sodium nitrate can be prepared by B. Neutralisation (titration of NaOH with HNO3).
(ii) Iron (III) chloride can be prepared by E. Direct synthesis (Direct combination of iron and chlorine).
(iii) Lead chloride can be prepared by D. Double decomposition (Precipitation reaction between a soluble lead salt and a soluble chloride).
(iv) Zinc sulphate can be prepared by A. Simple displacement (action of dilute sulphuric acid on zinc metal).
(v) Sodium hydrogen sulphate can be prepared by C. Decomposition by acid (action of concentrated sulphuric acid on sodium chloride at a temperature below 200°C).
Short Answer Type
1. Define ‘Neutralization’.
Answer: Neutralisation is the process by which H+ ions of an acid react completely with the [OH-] ions of a base to give salt and water only.
2. State your observation when zinc granules are added to copper sulphate solution.
Answer: When zinc reacts with copper sulphate, zinc replaces the copper, producing zinc sulphate and depositing copper. As a result, the blue solution gradually turns colourless.
Reaction: Zn + CuSO₄ ⟶ Cu + ZnSO₄
3. Salt S is prepared by reacting dilute sulphuric acid with copper oxide. Identify S.
Answer: When dilute sulphuric acid reacts with copper oxide (CuO), the reaction is CuO + H2SO4 → CuSO4 + H2O. Therefore, salt S is Copper (II) sulphate (CuSO4).
4. From the substances given in the list below, write the equations for the reactions by which you could obtain the compounds given ahead:
[Dilute sulphuric acid, Zinc, Lead, sodium carbonate, sodium sulphite, calcium carbonate]
- hydrogen,
- sulphur dioxide,
- carbon dioxide,
- zinc carbonate (two steps required).
Answer: To obtain hydrogen:
Zn + H₂SO₄ (dil) → ZnSO₄ + H₂
To obtain sulphur dioxide:
Na₂SO₃ + H₂SO₄ (dil) → Na₂SO₄ + H₂O + SO₂
To obtain carbon dioxide:
CaCO₃ + H₂SO₄ (dil) → CaSO₄ + H₂O + CO₂
To obtain zinc carbonate (two steps):
Step 1: Zn + H₂SO₄ (dil) → ZnSO₄ + H₂
Step 2: ZnSO₄ + Na₂CO₃ → ZnCO₃ + Na₂SO₄
5. Choosing the substances from the given list : dil. sulphuric acid, copper, iron, sodium, copper [II] carbonate, sodium carbonate, sodium chloride, zinc nitrate.
Write balanced equations for the reactions which would be used in the laboratory to obtain the salts given ahead:
- Sodium sulphate
- Zinc carbonate
- Copper (II) sulphate
- Iron (II) sulphate
Answer: (a) Sodium sulphate:
Na₂CO₃ + H₂SO₄ (dil) → Na₂SO₄ + H₂O + CO₂↑
(b) Zinc carbonate:
Zn(NO₃)₂ + Na₂CO₃ → ZnCO₃↓ + 2NaNO₃
(c) Copper(II) sulphate:
CuCO₃ + H₂SO₄ (dil) → CuSO₄ + H₂O + CO₂↑
(d) Iron(II) sulphate:
Fe + H₂SO₄ (dil) → FeSO₄ + H₂↑
6. You are given three solutions A, B and C. A is a strong acid, B is a weak acid and C is a strong alkali.
(a) Which solution contains solute molecules in addition to water molecules?
(b) Which solution gives gelatinous white precipitate with zinc sulphate which disappears in excess solution?
Answer: (a) Solution B, the weak acid, contains solute molecules in addition to water molecules because weak acids do not ionise completely in solution, so they contain ions as well as molecules.
(b) Solution C, the strong alkali, will give a gelatinous white precipitate with zinc sulphate which disappears in excess solution. For example, if C is NaOH:
ZnSO4 + 2NaOH → Zn(OH)2↓ (gelatinous white) + Na2SO4
Zn(OH)2 + 2NaOH (excess) → Na2ZnO2 (sodium zincate, soluble) + 2H2O
7. Write balanced chemical equations for the following:
(a) Reaction of silver nitrate solution and sodium chloride solution.
(b) Lead nitrate solution is added to sodium chloride solution.
(c) State what happens to the crystals of washing soda when they are exposed to air. Name the phenomenon exhibited.
Answer: (a) Reaction of silver nitrate solution and sodium chloride solution:
AgNO₃ (aq) + NaCl (aq) → AgCl↓ (s) + NaNO₃ (aq)
(b) Lead nitrate solution is added to sodium chloride solution:
Pb(NO₃)₂ (aq) + 2NaCl (aq) → PbCl₂↓ (s) + 2NaNO₃ (aq)
(c) When crystals of washing soda (Na₂CO₃·10H₂O) are exposed to air, they lose wholly or partly their water of crystallisation and become powdery. This phenomenon is called efflorescence.
Na₂CO₃·10H₂O (dry air) → Na₂CO₃·H₂O + 9H₂O
8. Give a balanced equation for the preparation of the following salts:
(a) Copper [II] sulphate from copper [II] oxide.
(b) Iron [III] chloride from the metal iron.
(c) Lead sulphate from lead carbonate.
(d) Potassium sulphate from KOH solution.
(e) Lead [II] chloride from lead carbonate [give two equations].
(f) Sodium sulphate — using dilute sulphuric acid.
(g) Copper chloride — using copper carbonate.
Answer: (a) Copper [II] sulphate from copper [II] oxide:
CuO + H2SO4 (dil) → CuSO4 + H2O
(b) Iron [III] chloride from the metal iron:
2Fe (heated) + 3Cl2 (dry) → 2FeCl3
(c) Lead sulphate from lead carbonate:
First step: PbCO3 + 2HNO3 (dil) → Pb(NO3)2 + H2O + CO2↑
Second step: Pb(NO3)2 + H2SO4 (dil) → PbSO4↓ + 2HNO3
(d) Potassium sulphate from KOH solution:
2KOH + H2SO4 (dil) → K2SO4 + 2H2O
(e) Lead [II] chloride from lead carbonate [give two equations]:
First step: PbCO3 + 2HNO3 (dil) → Pb(NO3)2 + H2O + CO2↑
Second step: Pb(NO3)2 + 2HCl (dil) → PbCl2↓ + 2HNO3
(f) Sodium sulphate – using dilute sulphuric acid:
2NaOH + H2SO4 (dil) → Na2SO4 + 2H2O
(or Na2CO3 + H2SO4 (dil) → Na2SO4 + H2O + CO2↑)
(g) Copper chloride using copper carbonate:
CuCO3 + 2HCl (dil) → CuCl2 + H2O + CO2↑
9. Write the balanced equations for the preparation of the following compounds (as the major product) starting from iron and using only one other substance :
Answer: (a) Iron(II) chloride from iron (using hydrochloric acid):
Fe + 2HCl (dil) → FeCl₂ + H₂↑
(b) Iron(III) chloride from iron (using chlorine):
2Fe + 3Cl₂ (heat) → 2FeCl₃
(c) Iron(II) sulphate from iron (using dilute sulphuric acid):
Fe + H₂SO₄ (dil) → FeSO₄ + H₂↑
(d) Iron(II) sulphide from iron (using sulphur):
Fe + S (heat) → FeS
Long Answer Type
1. Draw the structure of the stable positive ion formed when an acid dissolves in water.
Answer: The hydronium ion (H₃O⁺) is the positively charged ion formed when an acid dissolves in water. It results from a hydrogen ion (proton, H⁺) attaching itself to a water molecule (H₂O).
2. Write balanced reactions for the following conversions (A, B, C, D).
Fe ──A→ FeCl₂ ──B→ FeCO₃ ──C→ Fe(NO₃)₂ ──D→ Fe(OH)₂
Answer: A: Fe + 2HCl (dil) → FeCl₂ + H₂↑
B: FeCl₂ (aq) + Na₂CO₃ (aq) → FeCO₃↓ + 2NaCl (aq)
C: FeCO₃ (s) + 2HNO₃ (aq) → Fe(NO₃)₂ (aq) + H₂O (l) + CO₂↑ (g)
D: Fe(NO₃)₂ (aq) + 2NaOH (aq) → Fe(OH)₂↓ + 2NaNO₃ (aq)
3. Give equations for the following conversions A to E :
Answer: (i) ZnSO₄ + Na₂CO₃ → Na₂SO₄ + ZnCO₃↓
(ii) ZnCO₃ + 2HNO₃ → Zn(NO₃)₂ + H₂O + CO₂↑
(iii) Zn(NO₃)₂ + 2NaOH → Zn(OH)₂ + 2NaNO₃
(iv) Zn(OH)₂ —Δ→ ZnO + H₂O
(v) ZnO + H₂SO₄ → ZnSO₄ + H₂O
4. 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.)
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?
Answer: (a) The first step required to prepare lead sulphate from lead carbonate is to convert the insoluble lead carbonate into a soluble lead salt, specifically soluble lead nitrate, by reacting it with dilute nitric acid.
(b) The equation for the reaction in the first step is:
PbCO3 (s) + 2HNO3 (aq) → Pb(NO3)2 (aq) + H2O (l) + CO2↑ (g)
(c) The direct addition of dilute sulphuric acid to lead carbonate is an impractical method of preparing lead sulphate because an insoluble crust of lead sulphate is formed on the lead carbonate which prevents further reaction.
5. The diagram given below is to prepare Iron [III] chloride in the laboratory.
(i) What is substance B?
(ii) What is the purpose of B?
(iii) Why is iron[III] chloride to be stored in a closed container?
(iv) Write the equation for the reaction between iron and chlorine.
Answer: (i) Substance B, through which chlorine gas is passed before it reaches the heated iron wire, is concentrated sulphuric acid.
(ii) The purpose of B (concentrated sulphuric acid) is to dry the chlorine gas.
(iii) Iron [III] chloride is to be stored in a closed container because Iron (III) chloride is highly deliquescent and will absorb moisture from the air if left exposed.
(iv) The equation for the reaction between iron and chlorine is:
2Fe (heated) + 3Cl2 (dry) → 2FeCl3
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