Selina Concise Solutions for ICSE Class 10 Chemistry Chapter 3 Acids Bases and Salts

Intext - Question- 1

Question 1. What do you understand by the terms, acid and base?
Answer: (a) Acids are defined as compounds which contain one or more hydrogen atoms, and when dissolved in water, they produce hydronium ions (\( H_3O^+ \)), the only positively charged ions. (b) Hydronium ion (c) \( H_3O^+ \)

📝 Teacher's Note: Use everyday examples like lemon juice and vinegar to introduce acids. Show students how acids feel sour and change the color of indicators - this makes the abstract concept tangible.

🎯 Exam Tip: Always mention that acids produce \( H_3O^+ \) ions (not just \( H^+ \)) and emphasize "in aqueous solution" for full marks.

Question 2. Explain the formation of hydronium ion. Write the ionization of sulphuric acid showing the formation of hydronium ion.
Answer: Hydronium ions: They are formed by the reaction of \( H^+ \) (from acid) and water. It reacts with water to form \( H_3O^+ \) (Hydronium ion).
\( H^+ + H_2O \rightarrow H_3O^+ \)
Ionization of sulphuric acid showing the formation of hydronium ion:
\( H_2SO_4 + H_2O \rightleftharpoons H_3O^+ + HSO_4^- \)
\( HSO_4^- + H_2O \rightleftharpoons H_3O^+ + SO_4^{2-} \)

📝 Teacher's Note: Emphasize that free \( H^+ \) ions don't exist in water - they always bond with water molecules. Use the analogy of \( H^+ \) being like a "homeless proton" that needs a water molecule as shelter.

🎯 Exam Tip: Always show the step-wise ionization of dibasic acids like \( H_2SO_4 \) - this shows complete understanding and earns extra marks.

Question 3. Water is never added to acid in order to dilute it why?
Answer: If water is added to a concentrated acid, the heat generated causes the mixture to splash out and cause severe burns. Thus, water is never added to acid in order to dilute it. Always add acid to water slowly while stirring.

📝 Teacher's Note: Demonstrate this safety rule with a memorable phrase: "Do as you oughta, add acid to water." Show students a dilution demonstration using colored water to emphasize proper technique.

🎯 Exam Tip: Mention both the safety reason (heat generation and splashing) and the correct method (acid to water) for complete marks.

Question 4. Define the term 'basicity' of an acid. Give the basicity of: nitric acid, sulphuric acid and phosphoric acid.
Answer: Basicity: The basicity of an acid is defined as the number of hydronium ions (\( H_3O^+ \)) that can be produced by the ionization of one molecule of that acid in aqueous solution. The basicity of following compounds are:
Nitric acid: Basicity = 1
Sulphuric acid: Basicity = 2
Phosphoric acid: Basicity = 3

📝 Teacher's Note: Connect basicity to the number of replaceable hydrogen atoms. Use visual molecular structures to show students which hydrogens can be donated as ions.

🎯 Exam Tip: State the definition clearly first, then give the numerical values. Remember that basicity equals the number of replaceable H atoms in the molecule.

Question 5. Give two examples of each of the following: (a) oxy-acid (b) hydracids (c) monobasic acid (d) dibasic acid (e) tribasic acid
Answer: (a) Oxyacids: - \( HNO_3 \), \( H_2SO_4 \)
(b) Hydracid:- HCl, HBr
(c) Monobasic acid:- HCl, HBr
(d) Dibasic acid: - \( H_2SO_4 \), \( H_2CO_3 \)
(e) Tribasic acid:- \( H_3PO_4 \), \( H_3PO_3 \)

📝 Teacher's Note: Help students remember that oxy-acids contain oxygen while hydracids don't. Point out that the same acid can appear in multiple categories (like HCl being both hydracid and monobasic).

🎯 Exam Tip: Learn the common acids and their formulas - questions often test your knowledge of acid classifications through these examples.

Question 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.
Answer: (a) The anhydride of following acids are:
(i) Sulphurous acid: \( SO_2 \)
(ii) Nitric acid: \( N_2O_5 \)
(iii) Phosphoric acid: \( P_2O_5 \)
(iv) Carbonic acid: \( CO_2 \)
(b) Acids present in following are:
Vinegar: Acetic acid
Grapes: Tartaric acid and Malic acid
Lemon: Citric acid
(c) (i) \( H^+ \) ion turns blue litmus red.
(ii) \( OH^- \) ion turns red litmus blue.

📝 Teacher's Note: Connect acid anhydrides to everyday examples - like \( CO_2 \) in soda making it slightly acidic. Use taste tests (safely) to help students remember natural acids in foods.

🎯 Exam Tip: Remember the litmus test colors: "Blue becomes red in acid, red becomes blue in base" - this is frequently tested.

Question 7. What do you understand by the statement 'acetic acid is a monobasic acid?
Answer: Acetic acid is a monobasic acid which on ionization in water produce one hydronium ion per molecule of the acid. This means each molecule of acetic acid can donate only one hydrogen ion when dissolved in water.

📝 Teacher's Note: Show the molecular structure of acetic acid (\( CH_3COOH \)) and point out that only the hydrogen attached to oxygen is acidic - the three hydrogens on the methyl group are not replaceable.

🎯 Exam Tip: Always explain "one hydronium ion per molecule" to show you understand the relationship between molecular structure and basicity.

Question 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)
Answer: (a) \( 2NO_2 + H_2O \rightarrow HNO_2 + HNO_3 \)
(b) Two types of salts are produced when dibasic acid reacts with caustic soda. One is acidic salt and other normal salt.
Acid salts:
\( H_2SO_4 + NaOH \rightarrow NaHSO_4 + H_2O \)
Normal salts:
\( H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O \)

📝 Teacher's Note: Emphasize that dibasic acids have two replaceable hydrogens, so they can form either one normal salt (both H replaced) or one acid salt (one H replaced). Use mole ratios to explain this.

🎯 Exam Tip: Show both types of salts with balanced equations - this demonstrates complete understanding of neutralization reactions.

Question 9. Carbonic acid gives an acid salt but hydrochloric acid does not. Explain.
Answer: Carbonic acid is a dibasic acid with two replaceable hydrogen ions; therefore it forms one acid salt or one normal salt. Hydrochloric acid is a monobasic acid with one replaceable hydrogen ion and so forms only one normal salt. Since HCl has only one hydrogen to donate, it cannot form an acid salt because there would be no hydrogen left on the molecule after partial neutralization.

📝 Teacher's Note: Draw the molecular structures to show students why \( H_2CO_3 \) can lose one H (forming \( HCO_3^- \)) while HCl must lose its only H completely. This visual helps clarify the concept.

🎯 Exam Tip: Connect the answer to basicity - only polybasic acids can form acid salts because they have multiple hydrogens to donate partially.

Question 10. What do you understand by the strength of an acid? On which factor does the strength of an acid depend?
Answer: (a) Carbonic acid is a dibasic acid with two replaceable hydrogen ions; therefore it forms one acid salt or one normal salt. Hydrochloric acid is a monobasic acid with one replaceable hydrogen ion and so forms only one normal salt.
(b) Strength of an acid is the measure of concentration of hydronium ions it produces in its aqueous solution. Dil. HCl produces high concentration of hydronium ion compared to that of concentrated acetic acid. Thus, dil. HCl is stronger acid than highly concentrated acetic acid.
(c) \( H_3PO_3 \) is not a tribasic 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.
(d) The salt produced is insoluble in the solution so the reaction does not proceed. Hence, we do not expect lead carbonate to react with hydrochloric acid.
(e) \( NO_2 \) is called double acid anhydride because two acids - nitrous acid and nitric acid - are formed when it reacts with water.
\( 2NO_2 + H_2O \rightarrow HNO_2 + HNO_3 \)

📝 Teacher's Note: Distinguish between concentration and strength clearly - strength depends on degree of ionization, not amount of acid present. Use the analogy of "how completely the acid breaks apart" versus "how much acid you have."

🎯 Exam Tip: Remember that acid strength depends on ionization ability, not concentration. A dilute strong acid can be stronger than a concentrated weak acid.

Question 11. Dil. HCl acid is stronger than highly concentrated acetic acid. Explain.
Answer: Strength of an acid is the measure of concentration of hydronium ions it produces in its aqueous solution. Dil. HCl produces high concentration of hydronium ion compared to that of concentrated acetic acid. Thus, dil. HCl is stronger acid than highly concentrated acetic acid. This is because HCl ionizes almost completely even when dilute, while acetic acid ionizes only partially even when concentrated.

📝 Teacher's Note: Use conductivity experiments or pH measurements to demonstrate this concept practically. Show that dilute HCl has lower pH than concentrated acetic acid despite having less total acid.

🎯 Exam Tip: Emphasize the difference between "strong/weak" (degree of ionization) and "concentrated/dilute" (amount of solute) - this distinction is crucial for scoring marks.

Question 12. How is an acid prepared from a (a) Non-metal (b) Salt? Give an equation for each.
Answer: Acids are prepared from non-metals by their oxidation. For example:
Sulphur or phosphorus is oxidized by conc. Nitric acid to form sulphuric acid or phosphoric acid.
\( S + 6HNO_3 \rightarrow H_2SO_4 + 2H_2O + 6NO_2 \)
\( P + 5HNO_3 \rightarrow H_3PO_4 + H_2O + 5NO_2 \)
Acids are prepared from salt by displacement reaction. For example:
Nitric acid is prepared by using \( H_2SO_4 \) and sodium chloride.
\( NaCl + H_2SO_4 \rightarrow NaHSO_4 + HNO_3 \)

📝 Teacher's Note: Explain that stronger acids can displace weaker acids from their salts - this is the principle behind the salt displacement method. Connect this to the reactivity series concept.

🎯 Exam Tip: Learn the balanced equations for both methods. Remember that displacement works because \( H_2SO_4 \) is stronger than most other acids.

Question 13. Name an acid used: (a) to flavor and preserve food, (b) in a drink, (c) to remove ink spots, (d) as an eyewash
Answer: (a) Acetic acid - used to flavor and preserve food (vinegar)
(b) Carbonic acid - used in drinks (carbonated beverages)
(c) Oxalic acid - used to remove ink spots
(d) Boric acid - used as an eyewash

📝 Teacher's Note: Connect chemistry to daily life by showing students actual products containing these acids. Bring labels from food items, drinks, and household cleaners to make the lesson practical and memorable.

🎯 Exam Tip: Remember the specific uses - acetic acid for preservation, carbonic acid for fizzy drinks, oxalic acid for stain removal, and boric acid for antiseptic purposes.

Question 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
Answer: There appears to be a mismatch between the question and the provided answers. The question asks for equations showing how acids are made from anhydrides, but the answers list different acids. The correct equations should be:
(a) \( SO_2 + H_2O \rightarrow H_2SO_3 \) (sulphurous acid)
(b) \( P_2O_5 + 3H_2O \rightarrow 2H_3PO_4 \) (phosphoric acid)
(c) \( CO_2 + H_2O \rightarrow H_2CO_3 \) (carbonic acid)
(d) \( SO_3 + H_2O \rightarrow H_2SO_4 \) (sulphuric acid)
In simple words: Acid anhydrides are compounds that form acids when they react with water - like adding water to a powder to make a solution.

📝 Teacher's Note: Emphasize that anhydrides literally mean "without water" and demonstrate with simple examples like carbon dioxide dissolving in water to form carbonic acid. Show students the pattern: oxide + water = acid.

🎯 Exam Tip: Remember the pattern - non-metal oxides (anhydrides) + water = acids. Always balance the chemical equations properly.

Intext - Question - 2

Question. 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 ions (\( OH^- \)) as the only negatively charged ions.
(a) Strong alkalis: Sodium hydroxide (\( NaOH \)), Potassium hydroxide (\( KOH \))
(b) Weak alkalis: Calcium hydroxide (\( Ca(OH)_2 \)), Ammonium hydroxide (\( NH_4OH \))
In simple words: Alkalis are bases that dissolve in water and release hydroxide ions, making the solution basic or alkaline.

📝 Teacher's Note: Help students remember that all alkalis are bases but not all bases are alkalis - use the analogy that all cats are animals but not all animals are cats.

🎯 Exam Tip: Always mention that alkalis produce hydroxyl ions (\( OH^- \)) when dissolved in water - this is the key defining characteristic.

Question. What is the difference between:
(a) an alkali and a base,
(b) an alkali and a metal hydroxide?
Answer: (a) An alkali and a base:
1. Alkalis are soluble in water whereas bases may be or may not be soluble in water.
2. All alkalis are bases but all bases are not alkalis.
(b) An alkali and metal hydroxide:
1. Alkalis are soluble in water whereas metal hydroxides may be or may not be soluble in water
In simple words: Think of alkalis as the special group of bases that dissolve completely in water, like sugar dissolves in tea.

📝 Teacher's Note: Use examples like copper hydroxide (insoluble base) vs sodium hydroxide (soluble alkali) to make the distinction clear. Draw Venn diagrams to show the relationship.

🎯 Exam Tip: Remember the key word "soluble" - alkalis are always water-soluble bases. This is the main distinguishing factor.

Question. Define in terms of ionization:
(a) an acid
(b) an alkali
Answer: (a) An acid: Acids are defined as compounds which when dissolved in water produce hydronium ions (\( H_3O^+ \)).
(b) An alkali: Alkalis are compounds which when dissolved in water produces hydroxyl ions (\( OH^- \)).
In simple words: Acids release hydrogen ions while alkalis release hydroxide ions when they dissolve in water.

📝 Teacher's Note: Explain that ionization means breaking into charged particles. Use the analogy of salt dissolving into sodium and chloride ions to make this concept clearer.

🎯 Exam Tip: Always mention the specific ions produced - \( H_3O^+ \) for acids and \( OH^- \) for alkalis. Don't just say "hydrogen ions" for acids.

Question. Name the ions furnished by:
(a) bases in solution,
(b) a weak alkali
(c) an acid
Answer: (a) Bases in solution give hydroxide ion (\( OH^- \)).
(b) Weak alkali gives hydroxide ions (\( OH^- \)).
(c) An acid gives a hydronium ion (\( H_3O^+ \)).
In simple words: All basic substances produce hydroxide ions, while acidic substances produce hydronium ions in water.

📝 Teacher's Note: Emphasize that both strong and weak alkalis produce the same type of ion - the difference is only in the amount or concentration of ions produced.

🎯 Exam Tip: Write the complete ion symbols with charges - \( OH^- \) and \( H_3O^+ \) - to get full marks for ion identification questions.

Question. 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) Barium oxide (\( BaO \))
(b) Sodium hydroxide (\( NaOH \))
(c) Manganese oxide (\( MnO \))
(d) Copper hydroxide (\( Cu(OH)_2 \))
(e) Carbonic acid (\( H_2CO_3 \))
(f) Ferric hydroxide (\( Fe(OH)_3 \))
(g) Copper oxide (\( CuO \))
(h) Ammonia (\( NH_3 \))
(i) Ammonium hydroxide (\( NH_4OH \))
In simple words: These examples show the variety of basic and acidic compounds, some dissolve in water while others don't, and some contain metals while others don't.

📝 Teacher's Note: Use this question to reinforce the concept that solubility determines whether a base is an alkali. Point out that ammonium hydroxide is unique as a base without a metal.

🎯 Exam Tip: Learn these standard examples by heart - they frequently appear in various question formats. Pay special attention to the non-metal base examples.

Question. 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: The test tube containing distilled water does not affect the red litmus paper.
The test tube containing acidic solution does not change the red litmus paper.
But the test tube containing basic solution turns red litmus paper blue.
In simple words: Red litmus stays red in acid and neutral solutions but turns blue in basic solutions - like a color-changing indicator.

📝 Teacher's Note: Demonstrate this practically in class. Emphasize that neutral solutions don't change litmus color - this is often a point of confusion for students.

🎯 Exam Tip: Remember the key point - only basic solutions change red litmus to blue. Acids and neutral solutions keep red litmus red.

Question. HCl, HNO₃, C₂H₅OH, C₆H₁₂O₆ all contain H atoms but only HCl and HNO₃ show acidic character. Why?
Answer: It is because HCl and HNO₃ ionize in aqueous solution whereas ethanol and glucose do not ionize in aqueous solution.
In simple words: Just having hydrogen atoms doesn't make a compound acidic - the hydrogen must be able to separate as ions in water.

📝 Teacher's Note: Explain that ionization is the key - use the analogy of having money in your pocket vs having it available to spend. The hydrogen must be "available" as ions.

🎯 Exam Tip: Always mention "ionization in aqueous solution" as the reason. Don't just say "presence of hydrogen" - that's incomplete.

Question. Dry HCl gas does not change the colour of dry litmus paper. Why?
Answer: It is because HCl ionizes only in aqueous solution.
In simple words: HCl needs water to break into ions - without water, it can't show its acidic nature, like needing water to dissolve sugar.

📝 Teacher's Note: This demonstrates that acids need water to show acidic properties. Connect this to the previous question about ionization being key to acidic behavior.

🎯 Exam Tip: The key word is "aqueous" - always mention that acids need water to ionize and show acidic properties.

Question. Is PbO₂ a base or not? Comment.
Answer: Lead oxide is not a base because when it reacts with acid it forms chlorine along with salt and water. Thus, it is excluded from the class bases.
\( PbO_2 + 4HCl \rightarrow PbCl_2 + Cl_2 + 2H_2O \)
In simple words: PbO₂ behaves differently from normal bases - it produces chlorine gas instead of just salt and water when reacting with acids.

📝 Teacher's Note: Emphasize that this is an exception to normal base behavior. Use this to teach students that there are always exceptions in chemistry that need to be remembered.

🎯 Exam Tip: Remember this specific reaction and the fact that production of chlorine gas disqualifies PbO₂ from being classified as a base.

Question. (a) what effect does the concentration of [H₃O⁺] ion have on solution?
(b) Do basic solutions also have H⁺(aq)? Why are they basic?
Answer: (a) As the concentration of [H₃O⁺] increases in solution, the pH decreases. Consequently, the acidity of the solution increases.
(b) Yes, basic solutions also have H⁺(aq) ions. Basic solutions have lower concentration of H⁺(aq) in comparison to concentration of OH⁻(aq) ions.
In simple words: More H⁺ ions means more acidic, and basic solutions still have some H⁺ ions but much fewer than OH⁻ ions.

📝 Teacher's Note: Use the seesaw analogy - when H⁺ ions go up, pH goes down. Explain that even basic solutions have some H⁺ ions, just much fewer than OH⁻ ions.

🎯 Exam Tip: Remember the inverse relationship - higher H⁺ concentration means lower pH. Always mention the comparison between H⁺ and OH⁻ concentrations for basic solutions.

Question. How would you obtain:
(a) a base from other base
(b) an alkali from a base,
(c) salt from another salt?
Answer: (a) We can obtain a base from another base by double decomposition. The aqueous solution of salts with base precipitates the respective metallic hydroxide.
\( FeCl_3 + 3NaOH \rightarrow Fe(OH)_3 + 3NaCl \)
(b) An alkali from a base
\( Na_2CO_3 + Ca(OH)_2 \xrightarrow{\Delta} 2NaOH + CaCO_3 \)
(c) Salt from another salt
\( NH_4Cl + NaOH \rightarrow NaCl + H_2O + NH_3 \)
In simple words: These are chemical transformations where we use one compound to make another similar compound through different types of reactions.

📝 Teacher's Note: Emphasize the concept of double decomposition and precipitation. Show how heating can drive reactions forward by removing products as gases.

🎯 Exam Tip: Learn these specific equations as they represent important industrial processes. Note the heat symbol (Δ) in equation (b).

Question. 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) \( Mg + 2HCl \rightarrow MgCl_2 + H_2 \)
(b) \( HCl + NaOH \rightarrow NaCl + H_2O \)
(c) \( CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2 \)
(d) \( CaSO_3 + 2HCl \rightarrow CaCl_2 + H_2O + SO_2 \)
(e) \( ZnS + 2HCl \rightarrow ZnCl_2 + H_2S \)
In simple words: These are the main types of acid reactions - each produces different gases depending on what compound the acid reacts with.

📝 Teacher's Note: Teach students to recognize reaction patterns. Point out that the gas produced helps identify what type of compound reacted with the acid.

🎯 Exam Tip: Learn these reaction types as they form the basis of acid identification tests. Remember which gases are produced in each case.

Question. The skin has and needs natural oils. Why is it advisable to wear gloves while working with strong alkalis?
Answer: As we know that alkalis react with oil to form soap. As our skin contains oil so when we touch strong alkalis, a reaction takes place and soapy solution is formed. Hence we should wear gloves.
In simple words: Strong alkalis turn the oils in our skin into soap, which can damage and irritate the skin badly.

📝 Teacher's Note: This is a great safety lesson. Explain saponification in simple terms and emphasize laboratory safety protocols when handling chemicals.

🎯 Exam Tip: Connect this to the saponification reaction - alkali + oil = soap. This explains why alkalis are dangerous to skin.

Question. Why are alkalis soapy to touch? What do you understand by pH value?
Answer: Alkalis are soapy to touch as they react with oils of our skin to form soaps.
pH of a solution is the negative logarithm to the base 10 of hydrogen ion concentration expressed in moles per litre.
In simple words: Alkalis feel slippery because they make soap from skin oils, and pH is a scale that tells us how acidic or basic a solution is.

📝 Teacher's Note: Keep the pH definition simple initially - focus on the 0-14 scale before introducing the logarithmic definition. Connect the soapy feeling to everyday experience with soap.

🎯 Exam Tip: For pH definition, mention "negative logarithm of hydrogen ion concentration" for full marks. Connect the soapy feeling to saponification reaction.

Question. Complete the table:

Answer: The completed table shows how different indicators change color in different types of solutions.
In simple words: Indicators are like chemical mood rings that change color to show if a solution is acidic, basic, or neutral.

📝 Teacher's Note: Demonstrate these color changes practically. Emphasize that phenolphthalein is colorless in both neutral and acidic solutions - only turning pink in alkaline solutions.

🎯 Exam Tip: Remember that litmus shows purple in neutral, while phenolphthalein stays colorless in both neutral and acidic conditions. Pink phenolphthalein = definitely alkaline.

Question 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 phenolphthalein indicator?
Answer: The solution Y will give a pink colour with phenolphthalein. Solution Y has a pH of 10, which means it is basic (pH > 7). Phenolphthalein indicator turns pink in basic solutions and remains colorless in acidic solutions. Solution X with pH 4 is acidic and will not give any color change.
In simple words: Phenolphthalein is like a detective that only shows pink color when it meets a basic solution. Since Y is basic (pH 10), it will turn pink.

📝 Teacher's Note: Demonstrate this with actual phenolphthalein and solutions of different pH values. Students remember better when they see the color change happening. Emphasize that pH > 7 = basic = pink with phenolphthalein.

🎯 Exam Tip: Remember the simple rule: phenolphthalein turns pink only in basic solutions (pH > 7). Always check the pH value first before answering.

Question 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.
Answer:
A = Strongly acidic
B = Neutral
C = Weakly alkaline
D = Strongly alkaline
E = Weakly acidic

(a) Solution A (acidic solution) + Mg → H₂ + Mg salt
(b) Solution A (acidic solution) + Zn → H₂ + Zn salt

Both magnesium and zinc will react with the strongly acidic solution A to liberate hydrogen gas because metals react with acids to produce hydrogen gas and a salt.
In simple words: Only acidic solutions can make metals release hydrogen gas. Solution A is the most acidic, so it will work best with both metals.

📝 Teacher's Note: Use the pH scale diagram on the board. Show students that pH 0-3 = strongly acidic, 4-6 = weakly acidic, 7 = neutral, 8-10 = weakly basic, 11-14 = strongly basic.

🎯 Exam Tip: For hydrogen liberation, always choose the most acidic solution. Write the word equation as: Metal + Acid → Salt + Hydrogen gas.

Question 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.
Answer:
(a) The p H scale ranges from 0 to 14.
p H = 7, Solution is neutral
p H < 7, Solution is acidic
p H > 7, Solution is basic

(b) One advantage of measuring the pH of unknown solution by using pH paper is that we can come to know whether the solution is acidic, basic or neutral without wasting the solution.
In simple words: pH paper is like a quick test that tells you if something is sour (acidic), soapy (basic), or plain (neutral) without using up much of your solution.

📝 Teacher's Note: Bring actual pH paper to class and test common household items like lemon juice, soap water, and pure water. Students love hands-on activities.

🎯 Exam Tip: Remember the pH ranges: 0-6.9 = acidic, 7 = neutral, 7.1-14 = basic. For advantages of pH paper, mention "quick", "doesn't waste solution", and "easy to use".

Question 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).
Answer:
(a) A common acid base indicator and a universal indicator:
Acid base indicator like litmus tells us only whether a given substance is an acid or a base. Universal indicator gives an idea as to how acidic or basic a substance is. An universal indicator gives different colours with solutions of different p H values.

(b) Acidity of bases and basicity of acids:
Acidity of bases: The number of hydroxyl ions which can be produced per molecule of the base in aqueous solution.
Basicity of acid: The basicity of an acid is defined as the number of hydronium ions that can be produced by the ionization of one molecule of that acid in aqueous solution.

(c) Acid and alkali:
An acid is that substance which gives H⁺ ions when dissolved in water
An alkali is that substance which gives OH⁻ ions when dissolved in water.
In simple words: Common indicators are like yes/no answers, but universal indicators tell you exactly how much. Acids give H⁺ ions (like hydrogen pieces), while alkalis give OH⁻ ions (like hydroxide pieces).

📝 Teacher's Note: Use the analogy of a thermometer (universal indicator) vs a fever checker that just says hot/cold (common indicator). For ions, explain H⁺ makes things sour and OH⁻ makes things soapy.

🎯 Exam Tip: For distinguishing questions, always give clear definitions and mention the key difference. Universal indicator shows "degree" while common indicators show "type".

Question 20. 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 p H falls to 5.5 tooth decay starts. Tooth enamel is the hardest substance in our body and it gets corroded. The saliva produced by salivary glands is slightly alkaline, it helps to increase the p H, to some extent, but tooth paste which contain basic substance is used to neutralize excess acid in the mouth.
In simple words: Bacteria in our mouth eat sugar and make acid. When this acid becomes too strong (pH 5.5), it starts eating our tooth enamel. We use toothpaste because it's basic and fights the acid.

📝 Teacher's Note: Relate this to students' daily life. Explain why dentists say "brush after eating sweets" and how saliva is our body's natural defense system.

🎯 Exam Tip: Mention the critical pH value of 5.5, the role of bacteria, and the solution (basic toothpaste). Connect acid-base chemistry to real life.

Intext - Question - 3

Question 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.
Answer:
Acidic salt: Acid salts are formed by the partial replacement of the ionizable hydrogen atoms of a polybasic acid by a metal or an ammonium ion.

Normal salt: Normal salts are the salts formed by the complete replacement of the ionizable hydrogen atoms of an acid by a metallic or an ammonium ion.

Mixed salt: Mixed salts are those salts that contain more than one basic or acid radical.

Examples:
(a) A Normal salt: \( Na_2SO_4 \), NaCl
(b) An acid salt: \( NaHSO_4 \), \( Na_2HPO_4 \)
(c) A mixed salt: \( NaKCO_3 \), \( CaOCl_2 \)
In simple words: Normal salts are completely neutralized, acid salts still have some hydrogen left over, and mixed salts are like combining two different salts together.

📝 Teacher's Note: Use building blocks analogy - normal salt is like a complete house, acid salt is missing some parts, mixed salt is two houses joined together. Show the formulas clearly.

🎯 Exam Tip: Look for "H" in acid salt formulas (like NaHSO₄) and multiple metals in mixed salt formulas (like NaKCO₃). Always give the definition before examples.

Question 2. Answer the following questions related to salts and their preparations: (a) What is a 'salt'? (b) What kind of salt 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.
Answer:
(a) Salt is a compound formed by the partial or total replacement of the ionizable hydrogen atoms of an acid by a metallic ion or an ammonium ion.

(b) An insoluble salt can be prepared by precipitation.

(c) A salt prepared by direct combination is Iron (III) chloride.
Reaction: \( 2Fe + 3Cl_2 → 2FeCl_3 \)

(d) The name of the procedure used to prepare a sodium salt such as sodium sulphate is Neutralization of acid with base.
In simple words: A salt is what you get when an acid and a base react together. Some salts can be made by directly combining elements, like mixing iron and chlorine gas.

📝 Teacher's Note: Clarify that direct combination means element + element = compound, while neutralization means acid + base = salt + water. Show both types of reactions with examples.

🎯 Exam Tip: For salt definition, mention "replacement of hydrogen". For direct combination, always balance the chemical equation. Neutralization is the most common method for preparing salts.

Question 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?
Answer:
(a) Calcium sulphate from calcium carbonate: By decomposition of calcium carbonates by acids.

(b) Lead (II) oxide from lead: Lead oxide can be prepared from lead by Direct combination.

(c) Lead carbonate from lead nitrate: Lead carbonate is prepared from lead nitrate by precipitation (double decomposition) with Sodium carbonate.

(d) Sodium nitrate from sodium hydroxide: Sodium nitrate is prepared from sodium hydroxide by neutralizing it with nitric acid.

(e) Magnesium carbonate from Magnesium chloride: Magnesium carbonate from Magnesium chloride can be prepared by double decomposition with Sodium carbonate.

(f) Copper (II) sulphate from copper (II) oxide: Copper sulphate can be prepared from copper oxide by action with sulphuric acid.
In simple words: There are different ways to make salts - sometimes you mix with acids, sometimes you heat elements together, and sometimes you swap parts between two compounds.

📝 Teacher's Note: Group the methods: direct combination (element + element), neutralization (acid + base), precipitation (mixing two solutions), and acid + metal oxide. Students find patterns easier to remember.

🎯 Exam Tip: Identify the method first, then explain briefly. Common methods are: neutralization, precipitation, direct combination, and acid + metal oxide reaction.

Question 4. (a) How is lead sulphate prepared in the aboratory? (b) Why lead sulphate cannot be prepared by the action of dilute \( H_2SO_4 \) on lead oxide?
Answer:
(a) Lead sulphate is prepared from insoluble lead oxide, by first converting lead oxide into soluble lead nitrate with dilute nitric acid and then treating the resulting solution with sulphuric acid to obtain white ppt. of Lead sulphate.
\( PbO + 2HNO_3 → Pb(NO_3)_2 + H_2O \)
\( Pb(NO_3)_2 + H_2SO_4 → PbSO_4 + 2HNO_3 \)

(b) Lead sulfate is insoluble, when lead is added to sulfuric acid it only reacts on the surface. The lead becomes coated with insoluble lead sulfate and the lead in the interior can't react. Therefore lead sulfate cannot be prepared by adding dilute sulfuric acid.
In simple words: Lead sulphate forms a protective coating on lead, like paint on a wall. Once this coating forms, the acid cannot reach the lead underneath to react further.

📝 Teacher's Note: Use the analogy of trying to paint over already dried paint - the new paint can't stick properly. The insoluble coating prevents further reaction. This is called "passivation".

🎯 Exam Tip: For lead sulphate preparation, mention the two-step process: first make soluble lead nitrate, then precipitate lead sulphate. For part (b), explain the coating effect.

Question 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.
Answer:
(a) Copper sulphate crystals from mixture of charcoal and black copper oxide:
The carbon in the charcoal reduces the black copper oxide to reddish-brown copper. The lid must not be removed until the crucible is cool or the hot copper will be re-oxidized by air.
Take dilute sulphuric acid in a beaker and heat it on wire gauze. Add cupric oxide in small quantities at a time, with stirring till no more of it dissolves and the excess compound settles to the bottom.
Filter it hot and collect the filtrate in china dish. Evaporate the filtrate by heating to the point of crystallization and then allow it to cool and collect the crystals of copper sulphate pentahydrate.
Reaction: \( CuO + H_2SO_4 → CuSO_4 + H_2O \)
\( CuSO_4 + 5H_2O → CuSO_4 · 5H_2O \)

(b) Zinc sulphate crystals from Zinc dust:
Take dilute sulphuric acid in a beaker and heat it on wire gauze. Add some granulated zinc pieces with constant stirring. Add till the Zinc settles at base of the beaker. Effervescences take place because of liberation of hydrogen gas. When effervescence stops, it indicates that all the acid has been used up. The excess of zinc is filtered off. Collect the solution in china dish and evaporate the solution to get crystals. Filter, wash them with water and dry them between the folds of paper. The white needle crystals are of hydrated Zinc sulphate.
Reaction: \( Zn + H_2SO_4 → ZnSO_4 + H_2 \)
\( ZnSO_4 + 7 H_2O → ZnSO_4 · 7 H_2O \)

(c) Lead sulphate from metallic lead:
Metallic lead is converted to lead oxide by oxidation. Then lead sulphate is prepared from insoluble lead oxide, by first converting it into soluble lead nitrate. Then the lead nitrate solution is treated with sulphuric acid to obtain white ppt. of Lead sulphate.
Reaction: \( PbO + 2HNO_3 → Pb(NO_3)_2 + H_2O \)
In simple words: These are step-by-step recipes for making different salt crystals, like cooking but with chemicals. You mix, heat, filter, and let crystals form by cooling.

📝 Teacher's Note: Emphasize safety precautions - heating, handling acids, and proper ventilation. Show students actual crystals if possible. Explain why we filter hot and cool slowly for good crystal formation.

🎯 Exam Tip: For practical preparations, mention key steps: heating, stirring, filtering, evaporation, and crystallization. Always include the chemical equations and mention safety precautions.

Question. Sodium hydrogen carbonate crystals:
Dissolve 5 grams of anhydrous sodium carbonate in about 25 ml of distilled water in a flask. Cool the solution by keeping the flask in a freezing mixture. Pass carbon dioxide gas in the solution. Crystals of sodium bicarbonate will precipitate out after sometime. Filter the crystals and dry it in folds of filter paper.
Answer: Reaction: \( \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O} \rightarrow 2\text{NaHCO}_3 \)
In simple words: When you cool sodium carbonate solution and bubble carbon dioxide gas through it, it forms sodium bicarbonate crystals which you can filter out and dry.

📝 Teacher's Note: Demonstrate this with actual materials if possible. Students often forget that cooling is essential for crystal formation. Emphasize that CO2 gas must be passed slowly for best results.

🎯 Exam Tip: Always mention "filter and dry" in the final step when describing crystal preparation. Include the balanced equation with proper chemical formulas.

Question. 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:
(i) Anhydrous ferric chloride
(ii) Lead chloride
(iii) Sodium sulphate
(iv) Copper sulphate
Answer:
(i) Anhydrous ferric chloride: A (Direct combination of two elements)
\( 2\text{Fe} + 3\text{Cl}_2 \rightarrow 2\text{FeCl}_3 \)
(ii) Lead chloride: E (Reaction of two solutions of salts to form a precipitate)
\( \text{Pb(NO}_3)_2 + 2\text{HCl} \rightarrow \text{PbCl}_2 + 2\text{HNO}_3 \)
(iii) Sodium sulphate: D (Titration of dilute acid with a solution of soluble base)
\( 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \)
(iv) Copper sulphate: C (Reaction of dilute acid with an insoluble base)
\( \text{Cu(OH)}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + 2\text{H}_2\text{O} \)
In simple words: Different salts need different preparation methods - some are made by mixing elements directly, others by acid-base reactions, and some by mixing two salt solutions.

📝 Teacher's Note: Create a flowchart showing when to use each method. Students should memorize that method E (precipitation) is used for insoluble salts, while method D (titration) is for soluble salts from two soluble reactants.

🎯 Exam Tip: Remember the key rule - if the salt is insoluble, use precipitation method E. If both reactants are soluble, use titration method D to avoid excess reactants.

Question. 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
Answer:
(a) Lead chloride
(b) Silver chloride
(c) Barium sulphate and lead sulphate
(d) Basic lead chloride
(e) Sodium hydrogen sulphate
(f) Sodium potassium carbonate
(g) Sodium argentocyanide
(h) Potash alum
(i) Potassium bromide and potassium chloride
(j) Calcium sulphate
In simple words: Different salts have different properties - some dissolve only in hot water, some never dissolve, and some become less soluble when heated.

📝 Teacher's Note: Use actual examples with lead chloride dissolving in hot water to demonstrate temperature effects. Create memory aids like "Silver never dissolves" for AgCl.

🎯 Exam Tip: Remember the key exceptions - PbCl2 dissolves in hot water (unusual behavior), and CaSO4 becomes less soluble when heated (opposite to normal trend).

Question. Explain 'salt hydrolysis' name two salts which are:
(a) acidic
(b) basic
(c) neutral, when dissolved in water
Answer: The phenomenon, due to which salt formed by a weak acid and a strong base, or by a strong acid and a weak base, reacts with water to give an acidic or an alkaline solution, is known as salt hydrolysis.
(a) Acidic: Iron chloride, Copper sulphate
(b) Basic: Sodium carbonate, potassium acetate
(c) Neutral: Sodium chloride, sodium sulphate
In simple words: Salt hydrolysis is when a salt reacts with water to make the solution acidic or basic instead of neutral, depending on whether the salt came from strong or weak acids and bases.

📝 Teacher's Note: Explain using the rule: strong acid + weak base = acidic salt, weak acid + strong base = basic salt, strong acid + strong base = neutral salt. Use litmus paper demonstrations.

🎯 Exam Tip: Always give the definition first, then classify salts using the strong/weak acid-base rule. Remember FeCl3 and CuSO4 are classic examples of acidic salts.

Question. 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
Answer:
(a) Blue litmus will turn into red which will indicate the solution to be acidic.
(b) No change will be observed.
(c) Red litmus will turn into blue will indicate the solution to be basic
In simple words: Ferric chloride turns blue litmus red (acidic), sodium sulphate doesn't change litmus color (neutral), and sodium carbonate turns red litmus blue (basic).

📝 Teacher's Note: Demonstrate with actual litmus papers and salt solutions. Emphasize that "no change" is also a valid observation indicating neutrality.

🎯 Exam Tip: State the color change clearly and mention what it indicates (acidic/basic/neutral). Don't just say "turns red" - specify which litmus paper changes.

Question. 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
Answer:
(a) \( \text{MgCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{MgSO}_4 + \text{H}_2\text{O} + \text{CO}_2 \)
(b) \( \text{Pb(NO}_3)_2 + \text{H}_2\text{SO}_4 \rightarrow \text{PbSO}_4 + 2\text{HNO}_3 \)
(c) \( \text{Pb(NO}_3)_2 + \text{Na}_2\text{CO}_3 \rightarrow \text{PbCO}_3 + 2\text{NaNO}_3 \)
(d) \( \text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2 \)
In simple words: These equations show different ways to make salts - by mixing acids with bases, metals, or other salts, depending on what products you want.

📝 Teacher's Note: Point out the different types of reactions - neutralization, displacement, precipitation. Students should identify the type of reaction in each case.

🎯 Exam Tip: Balance equations carefully and choose appropriate examples that clearly show the required reaction type. Use common laboratory chemicals in your examples.

Question. 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: Zinc sulphate, Ferrous sulphide, Barium sulphate, Ferric sulphate, Sodium sulphate
Method of preparation: Precipitation, Oxidation, Displacement, Neutralisation, synthesis
Answer:
Zinc Sulphate - Displacement
\( \text{Zn(OH)}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + 2\text{H}_2\text{O} \)
Ferrous sulphide - synthesis
\( \text{Fe} + \text{S} \rightarrow \text{FeS} \)
Barium sulphate - Precipitation
\( \text{BaCl}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2\text{HCl} \)
Ferric sulphate - Oxidation
\( \text{Fe} + \text{H}_2\text{SO}_4 \rightarrow \text{FeSO}_4 + \text{H}_2 \)
Sodium sulphate - Neutralisation
\( 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \)
In simple words: Each salt can be prepared by a specific method - some by direct combination, others by precipitation or neutralization reactions.

📝 Teacher's Note: Emphasize that synthesis means direct combination of elements, while displacement involves replacing one element with another. Show the logic behind each matching.

🎯 Exam Tip: Think about the properties of each salt first - if it's insoluble, use precipitation; if it's from elements, use synthesis; if from acid-base, use neutralization.

Question. 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
(b) sodium sulphate
(c) sodium zincate
(d) iron (II) sulphate
(e) sodium chloride
Answer:
(a) Iron (III) Chloride: Iron chloride is formed by direct combination of elements.
\( 2\text{Fe} + 3\text{Cl}_2 \rightarrow 2\text{FeCl}_3 \)
(b) Sodium sulphate: By neutralization of caustic soda with dilute sulphuric acid
\( 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \)
(c) Sodium zincate: By the action of metals with alkalis
\( \text{Zn} + 2\text{NaOH} \rightarrow \text{Na}_2\text{ZnO}_2 + \text{H}_2 \)
(d) Iron (II) sulphate: Iron sulphate is prepared by the action of dilute acid on an active metal.
\( \text{Fe} + \text{H}_2\text{SO}_4 \rightarrow \text{FeSO}_4 + \text{H}_2 \)
(e) Sodium chloride: By the neutralization reaction of strong acid with strong base
\( \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \)
In simple words: Using only the given chemicals, you can make different salts by choosing the right combination and reaction type for each desired product.

📝 Teacher's Note: Stress that students must only use chemicals from the given list. Explain why sodium zincate formation requires heating and the unusual reaction of zinc with alkali.

🎯 Exam Tip: Read the question carefully and only use chemicals from the provided list. Show clear chemical equations with proper balancing for full marks.

Question. 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?
Answer: Neutralization 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.
(a) \( \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \)
(b) Neutralization is simply a reaction between H+ ions given by strong acid and OH- ions given by strong base. In case of all strong acids and strong bases, the number of H+ and OH- ions produced by one mole of a strong acid or strong base is always same. Hence the heat of neutralization of a strong acid with strong base is always same.
In simple words: Neutralization is when acid and base react to form salt and water, and the heat released is always the same for strong acids and bases because the basic reaction is always H+ + OH- = H2O.

📝 Teacher's Note: Emphasize that neutralization is fundamentally about H+ and OH- ions combining. Use thermometer demonstrations to show heat release during neutralization.

🎯 Exam Tip: Always define neutralization in terms of H+ and OH- ions, not just "acid + base". Explain the constant heat release using the ionic equation H+ + OH- → H2O.

Question. 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 FeCl3
Answer:
(a) Since sodium hydroxide and sulphuric acid are both soluble, an excess of either of them cannot be removed by filtration. Therefore it is necessary to find out on small scale, the ratio of solutions of the two reactants.
(b) As iron chloride is highly deliquescent, so it is kept dry with the help of fused calcium chloride.
In simple words: (a) You need the exact ratio because you can't filter out leftover acid or base if you use too much, and (b) calcium chloride absorbs moisture to keep iron chloride dry.

📝 Teacher's Note: Demonstrate titration to show how exact ratios are found. Explain deliquescence using examples like table salt becoming sticky in humid weather.

🎯 Exam Tip: For part (a), always mention that both reactants are soluble so excess cannot be removed by filtration. For part (b), use the term "deliquescent" and explain it absorbs moisture.

Question. Give the pH value of pure water. Does it change if common salt is added to it?
Answer: pH of pure water is 7 at 25°C. No, the pH does not change when common salt is added because sodium chloride is a neutral salt formed from strong acid (HCl) and strong base (NaOH), so it does not undergo hydrolysis.
In simple words: Pure water has pH 7 (neutral), and adding common salt doesn't change this because salt doesn't make the water acidic or basic.

📝 Teacher's Note: Use pH paper or digital pH meter to demonstrate this concept. Contrast with adding acidic or basic salts which do change pH.

🎯 Exam Tip: Always mention the temperature (25°C) when stating pH of pure water, and explain why NaCl doesn't change pH using the strong acid + strong base concept.

Question 16. Classify the following solutions as acids, bases or salts ammonium hydroxide, barium chloride, sodium chloride, sodium hydroxide, \( H_2SO_4 \) and \( HNO_3 \)
Answer: Acids: \( H_2SO_4 \) and \( HNO_3 \)
Bases: Ammonium hydroxide and sodium hydroxide.
Salts: Barium chloride and sodium chloride.
In simple words: We group these chemicals based on what they do - acids release hydrogen ions, bases release hydroxide ions, and salts are formed when acids and bases react together.

📝 Teacher's Note: Use the formula patterns to help students identify: acids usually contain H at the start, bases end with OH, and salts are combinations of metal and non-metal ions. Practice with more examples for reinforcement.

🎯 Exam Tip: Remember the key indicators - acids have H⁺ ions, bases have OH⁻ ions, and salts are ionic compounds formed from neutralization reactions.

Intext - Question - 4

Question. What do you understand by water of crystallization? Give four substances which contain water of crystallization and write their common names.
Answer: Some salts, while crystallizing out from their solutions, unite with definite quantity of water which is known as water of crystallization.

Four substances which contain water of crystallization:
\( Na_2CO_3 \cdot 10H_2O \) - Washing soda
\( MgSO_4 \cdot 7H_2O \) - Epsom salt
\( K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O \) - Potash alum
\( Na_2SO_4 \cdot 10H_2O \) - Glauber's salt
In simple words: Some salts trap water molecules inside their crystal structure when they form, like how a sponge holds water - this trapped water is called water of crystallization.

📝 Teacher's Note: Show students actual samples of hydrated salts like blue copper sulfate and explain how the water is chemically bound, not just absorbed. Heat some copper sulfate to demonstrate the color change.

🎯 Exam Tip: Always write the complete formula with the water molecules (like \( CuSO_4 \cdot 5H_2O \)) and include both chemical and common names for full marks.

Question. (a) Define efflorescence. Give examples. (b) Define deliquescence. Give examples.
Answer: (a) Efflorescence is the property of some substances to lose wholly, or partly their water of crystallization when their crystals are exposed to dry air even for a short time.
Examples are: Washing soda, Glauber's salt, Epsom salt

(b) 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.
For example: Caustic soda, Caustic potash
In simple words: Efflorescence means crystals lose their water and become powdery in dry air, while deliquescence means substances absorb so much water from air that they dissolve completely.

📝 Teacher's Note: Use washing soda and table salt as classroom demonstrations - leave washing soda in open air to show efflorescence, and show how impure salt becomes wet during monsoons due to deliquescent impurities.

🎯 Exam Tip: Remember the key difference - efflorescence loses water (crystals become powdery), deliquescence gains water (becomes liquid solution).

Question. Distinguish between drying and dehydrating agent.
Answer:

In simple words: Drying agents just remove surface moisture like a towel dries your hands, while dehydrating agents actually break chemical bonds and remove water from inside compounds.

📝 Teacher's Note: Use examples like silica gel (drying) vs concentrated sulfuric acid (dehydrating). Show how drying is reversible but dehydration often changes the compound permanently.

🎯 Exam Tip: Remember that drying is physical (reversible) while dehydrating is chemical (often irreversible) - this distinction often appears in exams.

Question. Explain clearly how conc. \( H_2SO_4 \) is used as dehydrating as well as drying agent.
Answer: Conc. \( H_2SO_4 \) removes the moisture from gases and it can also remove water molecules from blue vitriol. So conc. \( H_2SO_4 \) is used as dehydrating as well as drying agent.
In simple words: Concentrated sulfuric acid is like a super-powerful water remover - it can both soak up water vapor from air and also pull water molecules right out of chemical compounds.

📝 Teacher's Note: Demonstrate with blue copper sulfate - show how concentrated sulfuric acid turns it white by removing water of crystallization. Emphasize safety precautions when handling concentrated acid.

🎯 Exam Tip: Always mention both properties with examples - removes moisture (drying) and removes water from compounds like \( CuSO_4 \cdot 5H_2O \) (dehydrating).

Question. 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 __________.
Answer: (i) Blue
(ii) Red
(iii) Hydrogen
(iv) Basic, Alkaline
(v) Graphite
In simple words: Metals form basic oxides that turn litmus blue, while non-metals form acidic oxides that turn litmus red - it's like a simple test to identify what type of element we have.

📝 Teacher's Note: Use magnesium and sulfur as examples - burn both and test their oxide solutions with litmus. This visual demonstration helps students remember the pattern clearly.

🎯 Exam Tip: Remember the pattern: Metal → Basic oxide → Blue litmus; Non-metal → Acidic oxide → Red litmus. Exception: Graphite is the non-metal that conducts electricity.

Question. 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 \( NaHSO_4 \rightleftharpoons H^+ + Na^+ + SO_4^{2-} \) (c) Anhydrous calcium chloride is used in a desiccator.
Answer: (a) Sodium hydrogen sulphate is not an acid but undergoes partial replacement of the ionisable hydrogen atom and behaves as an acidic salt to give \( H^+ \) ions.

(b) As calcium chloride absorbs moisture and keeps the compound dry, so it is used in desiccators as a drying agent.
In simple words: Sodium hydrogen sulfate is a salt that still has some hydrogen left over, so it can release hydrogen ions like an acid. Calcium chloride is like a moisture sponge that keeps things dry in storage containers.

📝 Teacher's Note: Explain that acidic salts are formed from incomplete neutralization - they have both salt and acid properties. Show students actual desiccator setup with calcium chloride.

🎯 Exam Tip: For acidic salts, mention "partial replacement" and "acidic behavior" as key terms. For desiccators, emphasize the moisture-absorbing property of anhydrous calcium chloride.

Question. State whether a sample of each of the following would increase or decrease in mass if exposed to air. (a) Solid NaOH (b) Solid \( CaCl_2 \) (c) Solid \( Na_2CO_3 \cdot 10H_2O \) (d) Conc. sulphuric acid (e) Iron (III) Chloride
Answer: (a) Increases
(b) Increase
(c) Decrease
(d) Increases
(e) Increases
In simple words: Most of these substances absorb water from air and become heavier, except washing soda which loses its trapped water and becomes lighter.

📝 Teacher's Note: Relate to students' daily experience - show how washing soda powder forms from crystals, while substances like caustic soda become sticky when left open due to moisture absorption.

🎯 Exam Tip: Remember that hydrated salts (with water of crystallization) lose mass due to efflorescence, while hygroscopic substances gain mass by absorbing moisture.

Question. (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.
Answer: (a) Common salt contains impurities like magnesium chloride, which are deliquescent substances. So on exposure to air especially during the rainy season, table salt turns moist though sodium chloride is not deliquescent.

(b) This impurity 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 water and then with alcohol.

(c) Conc. \( H_2SO_4 \) can change the blue colour of copper sulphate to white.

(d) Two crystalline substances which do not contain water of crystallization are: Common salt, Nitre, Sugar.
In simple words: Salt gets wet because it has impurities that absorb water from humid air, like having a few water-loving substances mixed in with the salt crystals.

📝 Teacher's Note: Demonstrate with pure NaCl vs impure salt during humid weather. Show the dehydration of copper sulfate using concentrated sulfuric acid as a classroom demonstration.

🎯 Exam Tip: Always mention that pure NaCl is not deliquescent - it's the impurities (especially MgCl₂) that cause the problem. For dehydration, concentrated H₂SO₄ is the standard answer.

Miscellaneous Questions Based On Icse Examinations

Question. 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
Answer: (a) A is sodium salt: It is prepared by neutralization of a base with acids.

(b) B is an insoluble salt: An insoluble salt is obtained from another insoluble salt by double decomposition. The insoluble salt is first converted into a soluble salt which is then used to prepare the desired salt.

(c) C is soluble salt of copper: The soluble salt of copper can be prepared by the decomposition of carbonates by acids.

(d) D is soluble salt of Zinc: The soluble salt of Zinc can be prepared by decomposition of chlorides by conc. \( H_2SO_4 \).
In simple words: Different salts need different preparation methods - sodium salts by mixing acid and base, insoluble salts by swapping partners between two solutions, and metal salts by dissolving the metal compounds in acids.

📝 Teacher's Note: Emphasize that method selection depends on salt properties - solubility, reactivity of metal, and availability of starting materials. Use specific examples for each type.

🎯 Exam Tip: Match the method to salt type: Group 1 salts - neutralization; Insoluble salts - precipitation; Transition metal salts - acid decomposition of carbonates/oxides.

Question. (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?
Answer: (a)
(i) The pH increases by the addition of base.
(ii) The pH decreases by the addition of acid.

(b) If the solution changes the colour of litmus from red to blue, the pH indicates the presence of base.

(c) The solution that liberates carbon dioxide from sodium carbonate has pH less than 7.
In simple words: pH is like a scale of sourness - adding base makes it less sour (higher pH), adding acid makes it more sour (lower pH). Only acids can make fizzy reactions with carbonates.

📝 Teacher's Note: Use universal indicator to show pH changes visually. Demonstrate the carbonate test with vinegar and baking soda to show CO₂ evolution.

🎯 Exam Tip: Remember pH scale: Below 7 = acidic, 7 = neutral, above 7 = basic. Only acids (pH < 7) react with carbonates to release CO₂.

Question. 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?
Answer: (a) The name given to the water in the compound copper sulphate-5-water is water of crystallization.

(b) The anhydrous copper sulphate is white in colour.

(c) By adding dehydrating substances such as conc. sulphuric acid as they remove the water of crystallisation.

(d) Anhydrous calcium chloride
In simple words: Blue copper sulfate has water trapped in its crystals, and when this water is removed by heating or chemicals, it becomes white and dry.

📝 Teacher's Note: Use the blue to white color change of copper sulfate as a classic demonstration of dehydration. Students should observe this change both by heating and using concentrated acids.

🎯 Exam Tip: Remember the color change: Blue (hydrated) → White (anhydrous). Concentrated H₂SO₄ is the standard dehydrating agent besides heating.

Question. 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) P
(b) R
(c) Q
In simple words: Very basic solutions (pH 13) can release ammonia gas, very acidic ones (pH 2) are strong acids, and weak acids (pH 6) have both intact molecules and separated ions.

📝 Teacher's Note: Explain that strong bases can displace ammonia from ammonium salts, and weak acids only partially ionize. Use pH values to identify strength of acids and bases.

🎯 Exam Tip: pH 13 = very strong base, pH 2 = very strong acid, pH 6 = weak acid (partially ionized, so both molecules and ions present).

Question. (a) Outline the steps that would be necessary to convert insoluble lead (II) oxide into soluble lead chloride.
Answer: Step 1: Convert lead(II) oxide to a soluble lead salt by treating with nitric acid: \( PbO + 2HNO_3 \rightarrow Pb(NO_3)_2 + H_2O \)
Step 2: Precipitate lead chloride by adding sodium chloride solution: \( Pb(NO_3)_2 + 2NaCl \rightarrow PbCl_2 + 2NaNO_3 \)
Step 3: Filter and wash the lead chloride precipitate, then dissolve in hot water to get the soluble lead chloride solution.
In simple words: We first dissolve the insoluble oxide in acid to make it soluble, then add chloride to form the desired salt, like changing one puzzle piece into another through chemical reactions.

📝 Teacher's Note: Emphasize the two-step process: first make the insoluble compound soluble, then use precipitation to get the desired salt. This is a common strategy in salt preparation.

🎯 Exam Tip: For converting insoluble compounds, always think: Dissolve first (using acid), then precipitate the desired salt (using appropriate reagent).

Question. Write the balanced equations for the reactions, to convert insoluble lead (II) oxide into soluble lead choride.
Answer: Lead oxide is treated with dilute nitric acid to get soluble Lead nitrate. This Lead nitrate is treated with soluble Metallic chloride or dilute hydrochloric acid to get insoluble Lead chloride.
PbO + 2HNO₃ (dil) → Pb(NO₃)₂ + H₂O
Pb(NO₃)₂ + 2NaCl → PbCl₂ + 2NaNO₃
In simple words: We first make the lead oxide dissolve by treating it with acid, then add another chemical to form the chloride we want.

📝 Teacher's Note: Emphasize that this is a two-step process because direct reaction doesn't work well. Show students that we convert insoluble to soluble first, then back to insoluble but in the desired form.

🎯 Exam Tip: Always write both balanced equations clearly and mention that lead nitrate is the intermediate soluble compound formed.

Question. A solution of iron (III) chloride has a PH less than 7. Is the solution acidic or alkaline?
Answer: Acidic
In simple words: When pH is less than 7, the solution is always acidic. Iron chloride makes the water acidic when it dissolves.

📝 Teacher's Note: Use the pH scale diagram to show students that pH < 7 = acidic, pH = 7 = neutral, pH > 7 = basic. This is a fundamental concept they must memorize.

🎯 Exam Tip: Remember the pH scale: less than 7 is acidic, exactly 7 is neutral, more than 7 is basic/alkaline.

Question. 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

Dilute Sulphuric acid, Copper, Copper Carbonate, Iron, Sodium Carbonate, Sodium, Zinc

Answer:
(a) Sodium sulphate:
Na₂CO₃ + H₂SO₄ (dil) → Na₂SO₄ + H₂O + CO₂
(b) Copper sulphate:
CuCO₃ + H₂SO₄ (dil) → CuSO₄ + H₂O + CO₂
(c) Iron (II) sulphate:
Fe + H₂SO₄ (dil) → FeSO₄ + H₂
(d) Zinc Carbonate:
Zn + H₂SO₄(dil) → ZnSO₄ + H₂
ZnSO₄ + Na₂CO₃ → ZnCO₃ + Na₂SO₄
In simple words: We can make different salts by reacting the given substances with dilute sulfuric acid, and sometimes we need two steps like for zinc carbonate.

📝 Teacher's Note: Point out that metals react with acids to give hydrogen gas, while carbonates react with acids to give carbon dioxide gas. This pattern helps students predict products.

🎯 Exam Tip: For zinc carbonate, remember it needs two steps - first make zinc sulfate, then convert to carbonate using sodium carbonate.

Question. 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 treating 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 on heating.
Answer:
(a) NaHSO₄
(b) AgCl
(c) CuSO₄.5H₂O
(d) CuCO₃
(e) Pb(NO₃)₂
In simple words: Each salt has unique properties - some are acidic, some don't dissolve in water, and some change color when heated or treated with chemicals.

📝 Teacher's Note: Use actual samples or pictures to show color changes. Students remember better when they see the blue-to-white change of copper sulfate losing its water of crystallization.

🎯 Exam Tip: Learn the characteristic properties: AgCl is white and insoluble, copper compounds are often colored, and nitrates decompose to give brown NO₂ gas.

Question. Ca(H₂PO₄)₂ is an example of a compound called _______________ (acid salt/ basic salt/ normal salt)
Answer: Acid salt
In simple words: This compound still has hydrogen atoms that can be replaced, so it's called an acid salt.

📝 Teacher's Note: Explain that acid salts still have replaceable hydrogen atoms in their formula. Show how H₂PO₄⁻ can still donate a hydrogen ion.

🎯 Exam Tip: Look for hydrogen in the salt formula that can still act acidic - these are acid salts like NaHSO₄ or Ca(H₂PO₄)₂.

Question. Write the balanced equation for the reaction of: A names acid and a named alkali.
Answer: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
In simple words: When an acid meets a base, they neutralize each other to form a salt and water.

📝 Teacher's Note: This is the classic neutralization reaction. Emphasize that acid + base always gives salt + water, and make sure students can balance such equations.

🎯 Exam Tip: For neutralization reactions, always check that the equation is balanced and produces salt + water only.

Question. 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.
Answer:
(a) Alkali
(b) Precipitate
(c) Weak acid
In simple words: These are important chemistry terms - alkali is a base that dissolves in water, precipitate is a solid that forms when solutions mix, and weak acid doesn't fully break apart in water.

📝 Teacher's Note: These are fundamental definitions students must know. Use examples like NaOH for alkali, AgCl formation for precipitate, and acetic acid for weak acid.

🎯 Exam Tip: Learn these key terms by heart - they appear frequently in exams and are building blocks for understanding more complex concepts.

Question. Differentiate between the chemical nature of an aqueous solution of HCI and an aqueous solution of NH₃.
Answer:

In simple words: HCl makes the solution acidic (sour) while NH₃ makes it basic (bitter), and they have opposite effects on litmus paper.

📝 Teacher's Note: Use litmus paper demonstration to show the color changes. Students should understand that acids and bases have opposite properties.

🎯 Exam Tip: Remember the litmus test: acids turn blue litmus red, bases turn red litmus blue. This is a common exam question.

Question. 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.
Answer:
(a) Fe + 2HCl (dil) → FeCl₂ + H₂
(b) 2Fe (heated) + 3Cl₂ (dry) → 2FeCl₃
(c) Fe + H₂SO₄ (dil) → FeSO₄ + H₂
(d) Fe + S → FeS
In simple words: Iron can react with different substances to form different compounds - with acids it gives hydrogen gas, with chlorine gas it forms chloride, and with sulfur it forms sulfide.

📝 Teacher's Note: Show that Fe²⁺ forms with dilute acids while Fe³⁺ forms with dry chlorine gas. The oxidation states are important to distinguish.

🎯 Exam Tip: Note that dilute acids give Fe²⁺ compounds while direct combination with non-metals can give Fe³⁺ compounds.

Question. 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

Answer:
(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
In simple words: Different salts need different methods to prepare them - some from metals with acids, some by mixing solutions to form precipitates.

📝 Teacher's Note: Explain why each method is chosen - reactive metals use method A, unreactive metals use method B, and insoluble salts often use method E.

🎯 Exam Tip: Learn the general principle: reactive metals use direct acid reaction, unreactive metals need oxide/carbonate route, and insoluble salts use precipitation.

Question. 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 is to convert insoluble lead carbonate into soluble lead nitrate by treating lead carbonate with dilute nitric acid.
(b) PbCO₃ (s) + 2HNO₃(dil) → Pb(NO₃)₂ (aq) + H₂O (l) + CO₂ ↑
(c) When dilute sulphuric acid is added directly to lead carbonate, the lead sulphate thus formed will be deposited on solid lead carbonate disconnecting lead carbonate from sulphuric acid.
In simple words: We can't make lead sulfate directly because the product forms a coating that stops the reaction, so we use a two-step process instead.

📝 Teacher's Note: This is a great example of how insoluble products can stop reactions. Draw a diagram showing how PbSO₄ coating prevents further reaction.

🎯 Exam Tip: Remember that when the product is insoluble and coats the reactant, the reaction stops - that's why we need the two-step method.

Question. 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 a ………… and water only. This type of reaction is known as ……………………..
Answer:
Positively charged
Negatively charged
Salt
Neutralization reaction
In simple words: Acids give positive ions, bases give negative ions, and when they meet they form salt and water in a neutralization reaction.

📝 Teacher's Note: Emphasize that H₃O⁺ is positively charged and OH⁻ is negatively charged. This helps students understand ionic nature of acids and bases.

🎯 Exam Tip: Remember the pattern: acid + base → salt + water, and this reaction is always called neutralization.

Question. 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.
Answer:
(a) hydronium
(b) hydroxide
(c) salt
(d) water
(e) Hydrogen
In simple words: Solution X is acidic (has hydronium ions), Y is basic (has hydroxide ions), they neutralize to form salt and water, and acids react with magnesium to give hydrogen gas.

📝 Teacher's Note: Connect litmus color change to ion presence - blue to red means H₃O⁺ ions present, red to blue means OH⁻ ions present.

🎯 Exam Tip: Link litmus tests to chemical behavior - acidic solutions turn blue litmus red and react with metals to give hydrogen gas.

Question. Match the following:

Answer:
(a) E - Sodium hydrogen carbonate
(b) A - Sodium potassium carbonate
(c) D - Sodium zincate
(d) B - Alum
(e) C - Sodium carbonate
In simple words: Different types of salts have different structures - some have hydrogen still attached, some have multiple metals, and some have complex arrangements.

📝 Teacher's Note: Use the formulas to show students how to identify salt types - acid salts have H, mixed salts have two metals, double salts crystallize together.

🎯 Exam Tip: Learn to identify salt types from their formulas - look for hydrogen (acid salt), multiple metals (mixed salt), or complex arrangements.

Question. 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
Answer:
(a) \( \text{Pb(NO}_3\text{)}_2 + \text{H}_2\text{SO}_4 \longrightarrow \text{PbSO}_4 + 2\text{HNO}_3 \)
(b) \( \text{Cu} + \text{H}_2\text{SO}_4 \longrightarrow \text{CuSO}_4 + \text{H}_2 \)
(c) \( \text{Pb(NO}_3\text{)}_2 + 2\text{NaCl} \longrightarrow \text{PbCl}_2 + 2\text{NaNO}_3 \)
(d) \( 2\text{NH}_3 + \text{H}_2\text{SO}_4 \longrightarrow \text{(NH}_4\text{)}_2\text{SO}_4 \)
(e) \( \text{Na}_2\text{CO}_3 + 2\text{HCl} \longrightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \)
In simple words: These are chemical equations showing how different salts are formed by mixing acids with metals, carbonates, or other compounds. Each reaction follows the law of conservation of mass with equal atoms on both sides.

📝 Teacher's Note: Start by identifying the products first, then balance the equations step by step. Emphasize that lead compounds are toxic and should only be discussed theoretically. Use colored markers for reactants and products to help visual learners.

🎯 Exam Tip: Always check that the number of each type of atom is equal on both sides - this is the most common mistake in balancing equations.

Question. 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.
Answer:
(i) Complex salts
(ii) Alkali
In simple words: Complex salts have metal ions that are attached to other groups like a metal core with partners around it. Alkali is just a fancy name for bases that dissolve in water, like washing soda.

📝 Teacher's Note: Use simple analogies - complex salts are like a king (metal ion) surrounded by courtiers (ligands). For alkali, relate to common household items like soap and baking soda that feel slippery when wet.

🎯 Exam Tip: Remember that all alkalis are bases, but not all bases are alkalis - only the water-soluble ones get the special "alkali" name.

Question. 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).
Answer:
(i) \( \text{Zn} + \text{H}_2\text{SO}_4 \longrightarrow \text{ZnSO}_4 + \text{H}_2 \)
(ii) \( \text{Na}_2\text{SO}_3 \longrightarrow \text{Na}_2\text{O} + \text{SO}_2 \)
(iii) \( \text{Na}_2\text{CO}_3 + \text{H}_2\text{SO}_4 \longrightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} + \text{CO}_2 \)
(iv) First step: \( \text{Zn} + \text{H}_2\text{SO}_4 \longrightarrow \text{ZnSO}_4 + \text{H}_2 \)
Second step: \( \text{ZnSO}_4 + \text{Na}_2\text{CO}_3 \longrightarrow \text{Na}_2\text{SO}_4 + \text{ZnCO}_3 \)
In simple words: We can get different gases by choosing the right chemical reactions - acids with metals give hydrogen, heating sulphites gives sulphur dioxide, and acids with carbonates give carbon dioxide. For zinc carbonate, we first make zinc sulphate, then swap it with sodium carbonate.

📝 Teacher's Note: Demonstrate the hydrogen test with a burning splint - it makes a 'pop' sound. For the two-step zinc carbonate formation, explain it as a "chemical swap" where ions exchange partners like in a dance.

🎯 Exam Tip: When questions ask for "two steps required," always show both equations clearly labeled as Step 1 and Step 2, and ensure the product of the first reaction is used in the second reaction.