In its latest sample paper for 2026 boards examniantions, Cbse has made drastic changes in the scheme of science question paper.
Now you will a get 4 marker questions. And to solve them you should have a sound practice of simillar questions. Acids Bases And Salts 4 Marks Questions Answers is created for the same. Read NCERT diligently and prepare our short notes to score great marks in the examination.
Neutralization & Indicators
Q1. Sara took 2 mL of dilute NaOH solution in a test tube and added two drops of phenolphthalein solution to it. The solution turned pink in colour.
She added dilute HCl to the above solution drop by drop until the solution in the test tube became colourless. 30 drops of dilute HCl were used for the change in colour from pink to colourless.
When Sara added a drop of NaOH to the solution, the colour changed back to pink again.
Sara now tried the activity with different volumes of NaOH and recorded her observation in the table given below:
| S. No. | Volume of dil. NaOH taken (mL) | Drops of dil. HCl used |
|---|---|---|
| 1 | 2 | 30 |
| 2 | 3 | 45 |
| 3 | 4 | 60 |
Answer the following questions based on the above information:
A. If Sara used concentrated HCl in place of dilute HCl, how many drops will be required for the change in colour to be observed?
(a) 30 (b) < 30 (c) > 30
Justify your answer.
B. Sara measured 30 drops of dil. HCl and found its volume to be 1.5 mL. If Sara observed a colour change by using 4 mL of HCl, how many mL of NaOH did she add initially?
OR
Sara takes 15 drops of dilute HCl in a test tube and adds two drops of phenolphthalein. Then she adds NaOH dropwise. She observes a colour change after adding 30 drops of NaOH. What change in colour would she observe and why?
C. Write a balanced chemical equation for the reaction taking place. Which of the following is true and why?
The reaction is a:
(a) neutralisation and double displacement reaction
(b) neutralisation and precipitation reaction
(c) precipitation and double displacement reaction
(d) neutralisation, double displacement as well as precipitation reaction
Show Answer
A. (b) < 30 drops will be required because concentrated HCl has more H⁺ ions per drop, so fewer drops are needed to neutralise the same amount of base.
B. Volume of HCl used = 4 mL → Drops = (30 drops / 1.5 mL) × 4 mL = 80 drops
From the table, 30 drops neutralise 2 mL NaOH ⇒ 80 drops neutralise (2/30)×80 ≈ 5.33 mL NaOH
OR
The solution will change from colourless to pink because phenolphthalein is colourless in acidic medium and turns pink in basic medium when excess NaOH is added.
C. NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)
Correct option: (a) because H⁺ and OH⁻ combine to form water (neutralisation) and Na⁺ and Cl⁻ exchange partners (double displacement). No precipitate is formed, so it is not a precipitation reaction.
Q2. Ravi took 3 mL of dilute H₂SO₄ in a test tube and added two drops of methyl orange.
The solution turned red. He added aqueous NaOH drop by drop until the colour changed to yellow. 40 drops of NaOH were used. He then added one more drop of H₂SO₄, and the colour turned red again.
He repeated the experiment with different volumes of acid and recorded:
| Vol. of H₂SO₄ (mL) | Drops of NaOH used |
|---|---|
| 2 | 26 |
| 3 | 40 |
| 4 | 53 |
A. If Ravi used dilute CH₃COOH instead of H₂SO₄ (same volume and concentration in mol/L), would more or fewer drops of NaOH be needed? Justify.
B. If 40 drops = 2 mL NaOH, how much NaOH (in mL) is needed to neutralise 6 mL of H₂SO₄?
OR
If Ravi adds methyl orange to NaOH first (yellow), then adds CH₃COOH dropwise, what colour change occurs and why?
C. Write the balanced equation. Is this reaction exothermic? Justify.
Show Answer
A. More drops of NaOH will be needed because CH₃COOH is a weak acid and partially ionises, producing fewer H⁺ ions, requiring more base for neutralisation.
B. From data: 3 mL acid → 2 mL NaOH ⇒ 6 mL acid → 4 mL NaOH
OR
Colour changes from yellow to red because methyl orange is yellow in base and red in acid due to increase in H⁺ ion concentration.
C. H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)
Yes, it is exothermic because neutralisation reactions release heat.
Q3. Priya tested dry HCl gas on dry and wet blue litmus paper. No change with dry, but wet litmus turned red.
A. Why no change with dry litmus?
B. What ion causes acidity in water?
OR
Will dry NaOH turn red litmus blue? Why?
C. Write ionic equation for HCl in water. What is this process called?
Show Answer
A. H⁺ ions are not formed without water, so no acidic behaviour.
B. H₃O⁺ (hydronium ion)
OR
No, because OH⁻ ions are not released without water.
C. HCl + H₂O → H₃O⁺ + Cl⁻ — called ionisation.
Q4. A student added CuO to dilute H₂SO₄. Black solid dissolved, solution turned blue.
A. What salt is formed?
B. Why is CuO called a basic oxide?
OR
Will CO₂ behave similarly? Why?
C. Write equation. What type of reaction?
Show Answer
A. Copper(II) sulphate
B. Because it reacts with acid to form salt and water, like a base.
OR
No, CO₂ is an acidic oxide; it reacts with bases, not acids.
C. CuO + H₂SO₄ → CuSO₄ + H₂O — neutralisation-like reaction.
Q5. A cloth stained with turmeric turns reddish-brown when soap is applied and yellow after washing.
A. What does this indicate about soap?
B. Why does colour return after washing?
OR
Can vanilla be used similarly? How?
C. What type of indicator is turmeric? Give one more example.
Show Answer
A. Soap is basic in nature.
B. Washing removes the base, restoring the original yellow colour of turmeric.
OR
Yes, vanilla odour disappears in base but remains in acid — it’s an olfactory indicator.
C. Natural indicator; another example: litmus.
Q6. A student electrolysed distilled water — bulb didn’t glow. After adding NaOH, gases evolved at electrodes.
A. Why didn’t bulb glow initially?
B. Which gases are formed at cathode and anode?
OR
Why is acid/base added for electrolysis?
C. Write equation. What type of reaction?
Show Answer
A. Distilled water has no ions, so it does not conduct electricity.
B. H₂ at cathode, O₂ at anode
OR
To provide ions for conduction.
C. 2H₂O(l) → 2H₂(g) + O₂(g) — electrolytic decomposition.
Q7. Bee sting causes pain. Relief is obtained by applying baking soda paste.
A. What acid is injected?
B. Why does baking soda help?
OR
What if vinegar is applied?
C. Write word equation. What type of reaction?
Show Answer
A. Methanoic acid
B. Baking soda (a base) neutralises the acid, reducing pain.
OR
Pain increases because vinegar is acidic.
C. Methanoic acid + Sodium hydrogencarbonate → Salt + Water + CO₂ — neutralisation.
Q8. Toothpaste prevents decay.
A. Why does decay start below pH 5.5?
B. How does toothpaste help?
OR
What is enamel made of?
C. Why should we not eat sweets before bed?
Show Answer
A. Enamel (calcium hydroxyapatite) corrodes below pH 5.5.
B. Toothpaste is basic, so it neutralises acids formed by bacteria.
OR
Calcium hydroxyapatite
C. Bacteria degrade sugar overnight, producing acids that lower pH and cause decay.
pH experiments with universal indicator (soil, saliva, rainwater, antacids)
Q11. Neha tested the pH of her saliva before and after having a meal using universal indicator paper. She observed that the colour of the pH paper changed from green (before meal) to orange (after meal).
She repeated the test on three different days and recorded consistent results.
A. What does the change in colour from green to orange indicate about the nature of saliva after a meal?
B. Why does the pH of saliva decrease after eating food?
OR
If Neha had tested lemon juice instead of saliva, what colour would the pH paper show and why?
C. Write one way by which the acidic effect in the mouth can be neutralised. Which type of substance is used for this purpose?
Show Answer
A. The change from green to orange indicates that saliva becomes acidic after a meal.
B. Bacteria present in the mouth degrade sugar and food particles, producing acids, which lower the pH of saliva.
OR
The pH paper would turn red or pink because lemon juice contains citric acid, making it strongly acidic.
C. The acidic effect can be neutralised by cleaning the mouth with basic toothpaste. A mild base like baking soda is used for this purpose.
Q12. A student collected rainwater from his rooftop after a heavy shower and tested its pH using universal indicator paper. The paper turned yellow, indicating a pH of about 6. After a week of industrial activity in the area, he collected another sample of rainwater and found the pH paper turned orange (pH ≈ 4.5).
A. What is rainwater with pH less than 5.6 called?
B. Why is such rainwater harmful to aquatic life in rivers?
OR
Why is normal rainwater slightly acidic even without pollution?
C. Name two acidic gases released by industries that cause this type of rain.
Show Answer
A. Rainwater with pH less than 5.6 is called acid rain.
B. Acid rain lowers the pH of river water, making it difficult for aquatic organisms to survive, as they are sensitive to pH changes.
OR
Normal rainwater is slightly acidic because carbon dioxide from air dissolves in it, forming carbonic acid.
C. Sulphur dioxide (SO₂) and nitrogen dioxide (NO₂) are two acidic gases that cause acid rain.
Q13. A farmer tested the pH of soil from his field using universal indicator paper. The paper turned red, indicating a pH of about 4.
He also observed that crops were not growing well. He decided to add slaked lime to the soil.
A. What does the red colour of the pH paper indicate about the soil?
B. Why did the farmer choose slaked lime?
OR
What would happen if the farmer added ammonium sulphate instead?
C. Name one other substance that can be used to reduce acidity of soil.
Show Answer
A. The red colour indicates that the soil is highly acidic.
B. Slaked lime [Ca(OH)₂] is a base, which neutralises the acidity of the soil and makes it suitable for crop growth.
OR
Ammonium sulphate is acidic in nature, so it would increase soil acidity, worsening the problem.
C. Quicklime (CaO) or chalk (CaCO₃) can also be used to reduce soil acidity.
Q14. Ravi suffers from acidity after eating spicy food. His mother gives him a teaspoon of baking soda dissolved in water, and he gets immediate relief. He tests the pH of the baking soda solution using universal indicator paper and observes a greenish-blue colour.
A. What does the greenish-blue colour indicate about baking soda solution?
B. How does baking soda provide relief in acidity?
OR
Why is milk of magnesia also used for the same purpose?
C. Write the common name and chemical name of baking soda.
Show Answer
A. The greenish-blue colour indicates that baking soda solution is basic in nature.
B. Baking soda neutralises the excess hydrochloric acid in the stomach, providing relief from pain and irritation.
OR
Milk of magnesia contains magnesium hydroxide, a mild base, which also neutralises excess stomach acid.
C. Common name: baking soda; Chemical name: sodium hydrogencarbonate.
Q15. Priya tested different household substances with universal indicator paper and recorded the following observations:
| Substance | Color of pH Paper | Approx. pH |
| Tap water | Green | 7 |
| Coffee | Orange | 5 |
| 1M NaOH | Dark blue | 14 |
| Carrot juice | Yellow-green | 6 |
A. Which substance is neutral? Justify.
B. Why is coffee acidic in nature?
OR
If Priya tested vinegar, what colour would she observe and why?
C. What does a pH of 14 indicate about 1M NaOH?
Show Answer
A. Tap water is neutral because its pH is 7, and the pH paper shows green colour, which corresponds to neutral substances.
B. Coffee contains organic acids, which make it mildly acidic.
OR
She would observe red or pink colour because vinegar contains acetic acid, making it acidic.
C. A pH of 14 indicates that 1M NaOH is a strongly basic solution.
Reactions of metals with acids/bases (Zn, Al, Cu, Fe)
Q16. A student added zinc granules to dilute hydrochloric acid in a test tube. Brisk effervescence was observed.
The gas evolved through a soap solution and a burning candle was brought near the gas-filled bubble—it burnt with a pop sound. The student repeated the experiment with dilute acetic acid and observed that effervescence was much slower.
A. Why is the effervescence slower with dilute acetic acid compared to dilute hydrochloric acid?
B. If the student uses copper pieces instead of zinc granules with dilute HCl, will any gas be evolved? Justify your answer.
OR
What would be observed if the student adds zinc granules to sodium hydroxide solution and warms it?
C. Write a balanced chemical equation for the reaction between zinc and dilute hydrochloric acid. Which of the following is true and why? The reaction is a
(a) displacement and redox reaction
(b) neutralisation and double displacement reaction
(c) decomposition and redox reaction
(d) combination and displacement reaction
Show Answer
A. Acetic acid is a weak acid and ionises less in water, producing fewer H⁺ ions, so the reaction with zinc is slower than with hydrochloric acid, which is a strong acid.
B. No gas will be evolved because copper is less reactive than hydrogen and cannot displace hydrogen from acids.
OR
Hydrogen gas will be evolved, as zinc reacts with sodium hydroxide on heating to form sodium zincate and hydrogen gas.
C. Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
Correct option: (a) because zinc displaces hydrogen from HCl (displacement) and zinc is oxidised while hydrogen is reduced (redox reaction).
Q17. A teacher demonstrated the reaction of aluminium foil with dilute sodium hydroxide solution. On warming, gas bubbles were observed. The gas burnt with a pop sound when tested with a burning splinter.
The same aluminium foil was added to dilute sulphuric acid, and again gas was evolved that gave a pop sound.
A. Why does aluminium react with both acids and bases to produce hydrogen gas?
B. Will iron filings react similarly with sodium hydroxide solution? Why or why not?
OR
Why is aluminium used in cooking utensils despite being a reactive metal?
C. Write a balanced chemical equation for the reaction of aluminium with sodium hydroxide. Which of the following best describes this reaction?
(a) only a displacement reaction
(b) a reaction of an amphoteric metal with a base
(c) a neutralisation reaction
(d) a precipitation reaction
Show Answer
A. Because aluminium is an amphoteric metal, it reacts with both acids and bases to produce hydrogen gas.
B. No, because iron is not amphoteric and does not react with bases like sodium hydroxide.
OR
Aluminium forms a protective oxide layer that prevents further reaction, making it suitable for utensils.
C. 2Al(s) + 2NaOH(aq) + 6H₂O(l) → 2NaAl(OH)₄(aq) + 3H₂(g)
(Note: NCERT uses simplified form for zinc: 2NaOH + Zn → Na₂ZnO₂ + H₂; for Al, the concept is analogous.)
Correct option: (b) because aluminium is amphoteric and this is not a neutralisation or precipitation reaction.
Q18. In a laboratory activity, a student placed iron nails in two test tubes:
Test tube A: dilute HCl
Test tube B: dilute NaOH
He observed effervescence only in test tube A. The gas from test tube A burnt with a pop sound. No change was seen in test tube B even after warming.
A. Why is hydrogen gas evolved in test tube A but not in test tube B?
B. What salt is formed in test tube A? What is its colour in solution?
OR
If the student uses zinc instead of iron in test tube B, what would he observe?
C. Write the balanced chemical equation for the reaction in test tube A. Which of the following is correct? The reaction shows that
(a) iron is below hydrogen in the reactivity series
(b) iron is amphoteric
(c) iron is above hydrogen in the reactivity series
(d) iron oxide is basic
Show Answer
A. Iron reacts with acids to produce hydrogen gas, but does not react with bases because it is not an amphoteric metal.
B. Iron(II) chloride is formed, which gives a pale green solution.
OR
Hydrogen gas would be evolved, as zinc is amphoteric and reacts with NaOH on heating.
C. Fe(s) + 2HCl(aq) → FeCl₂(aq) + H₂(g)
Correct option: (c) because iron displaces hydrogen from acid, proving it is above hydrogen in the reactivity series.
Q19. A student tested four metals—zinc, copper, iron, and aluminium—with dilute hydrochloric acid. He recorded his observations as follows:
| Metal | Observation |
| Aluminium | Moderate effervescence |
| Iron | Slow effervescence |
| Copper | No effervescence |
A. Why does copper not produce any effervescence with dilute HCl?
B. Why is the effervescence with aluminium slower than expected despite its high reactivity?
OR
Arrange the four metals in decreasing order of reactivity with acids.
C. Write a balanced equation for the reaction of iron with dilute HCl. Which of the following statements is true?
(a) All metals react with acids to produce hydrogen
(b) Only metals above hydrogen in reactivity series react with acids
(c) Copper reacts with acids but very slowly
(d) Hydrogen gas is not produced in metal-acid reactions
Show Answer
A. Because copper is below hydrogen in the reactivity series and cannot displace hydrogen from acids.
B. Because aluminium forms a protective oxide layer that initially prevents the reaction; once the layer is removed, reaction proceeds.
OR
Reactivity order: Aluminium > Zinc > Iron > Copper
C. Fe(s) + 2HCl(aq) → FeCl₂(aq) + H₂(g)
Correct option: (b) because only metals more reactive than hydrogen can displace it from acids.
Q20. A student added granulated zinc to dilute sulphuric acid in a test tube fitted with a delivery tube. The gas evolved was passed through soap water, forming bubbles.
A burning candle brought near a bubble caused it to burn with a pop sound. The student then added the same zinc to aqueous sodium hydroxide and warmed the mixture—again, a gas was produced that gave a pop sound.
A. What does the pop sound confirm about the gas in both cases?
B. Why does zinc react with both an acid and a base?
OR
Will magnesium react with sodium hydroxide to produce hydrogen gas? Justify.
C. Write balanced chemical equations for both reactions. Which of the following is correct? Zinc is
(a) a basic metal
(b) an amphoteric metal
(c) a noble metal
(d) a non-metal
Show Answer
A. The pop sound confirms the gas is hydrogen in both cases.
B. Because zinc is an amphoteric metal, it reacts with both acids and bases to produce hydrogen gas.
OR
No, because magnesium is not amphoteric and does not react with bases.
C.
With acid: Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)
With base: Zn(s) + 2NaOH(aq) → Na₂ZnO₂(aq) + H₂(g)
Correct option: (b) because zinc reacts with both acids and bases, a property of amphoteric metals.
Neutralisation titrations (H₂SO₄/NaOH, HNO₃/KOH)
Q26. Sara took 2 mL of dilute NaOH solution in a test tube and added two drops of phenolphthalein solution to it. The solution turned pink in colour.
She added dilute H₂SO₄ drop by drop until the solution became colourless. 40 drops of dilute H₂SO₄ were used. When she added one more drop of NaOH, the pink colour reappeared.
She repeated the experiment with different volumes of NaOH and recorded:
| SL No. | Volume of dil. NaOH | Drop of dil. H₂SO₄ used |
| 1 | 2 ml | 40 |
| 2 | 3 ml | 60 |
| 3 | 4 ml | 80 |
A. If Sara used concentrated H₂SO₄ instead of dilute H₂SO₄, how many drops would be required to neutralise 2 mL of NaOH?
(a) 40 (b) < 40 (c) > 40
Justify your answer.
B. Sara found that 40 drops of dil. H₂SO₄ = 2 mL. If she used 3 mL of H₂SO₄ to neutralise a NaOH solution, what volume of NaOH was used?
OR
Sara adds 20 drops of dil. H₂SO₄ to a test tube, adds phenolphthalein, then adds NaOH dropwise. After 40 drops of NaOH, the colour changes. What colour change is observed and why?
C. Write a balanced chemical equation for the reaction. Which of the following is true and why? The reaction is a
(a) neutralisation and double displacement reaction
(b) neutralisation and precipitation reaction
(c) displacement and redox reaction
(d) decomposition and neutralisation reaction
Show Answer
A. (b) < 40 drops, because concentrated H₂SO₄ has more H⁺ ions per drop, so fewer drops are needed to neutralise the same amount of base.
B. 2 mL H₂SO₄ neutralises 2 mL NaOH ⇒ 3 mL H₂SO₄ neutralises 3 mL NaOH
OR
The solution changes from colourless to pink because phenolphthalein is colourless in acid and turns pink in base when excess NaOH is added.
C. 2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)
Correct option: (a) because H⁺ and OH⁻ combine to form water (neutralisation) and ions exchange partners (double displacement). No precipitate is formed, so it is not a precipitation reaction.
Q27. Ravi performed a neutralisation experiment using dilute KOH and dilute HNO₃ with phenolphthalein as indicator. He recorded:
| Volume of KOH | Drops of HNO₃ used |
| 2 ml | 30 |
| 3 ml | 45 |
| 5 ml | 75 |
A. Why does the pink colour disappear when HNO₃ is added to KOH containing phenolphthalein?
B. If 30 drops = 1.5 mL, how much KOH is neutralised by 4.5 mL of HNO₃?
OR
What would happen if Ravi used methyl orange instead of phenolphthalein? Describe the colour change.
C. Write the balanced equation. Is this a neutralisation reaction? Justify.
Show Answer
A. The pink colour disappears because phenolphthalein is pink in basic medium and becomes colourless when the solution is neutralised or becomes acidic due to addition of HNO₃.
B. 1.5 mL HNO₃ → 2 mL KOH ⇒ 4.5 mL HNO₃ → 6 mL KOH
OR
With methyl orange, the colour would change from yellow (in base) to red (in acid) at the endpoint.
C. KOH(aq) + HNO₃(aq) → KNO₃(aq) + H₂O(l)
Yes, it is a neutralisation reaction because an acid and a base react to form salt and water.
Q28. A student added dilute NaOH to dilute HCl containing phenolphthalein. Initially colourless, the solution turned pink after adding excess NaOH.
She reversed the experiment:
added HCl to NaOH with phenolphthalein—pink turned colourless at endpoint. Her data:
| NaOH | Drops of HCl |
| 2 ml | 24 |
| 4 ml | 48 |
| 6 ml | 72 |
A. Why is phenolphthalein not suitable for titrating a weak base with a strong acid?
B. If 24 drops = 1.2 mL HCl, how much HCl is needed to neutralise 5 mL NaOH?
OR
Can this experiment be done with turmeric as indicator? Why or why not?
C. Write the balanced equation. What type of salt is formed?
Show Answer
A. Because phenolphthalein changes colour around pH 8–10, but the equivalence point of a weak base–strong acid titration is below pH 7, so no sharp colour change occurs.
B. 1.2 mL HCl → 2 mL NaOH ⇒ for 5 mL NaOH, HCl required = (1.2/2) × 5 = 3 mL
OR
Yes, turmeric is a natural indicator that turns reddish-brown in base and yellow in acid, so it can indicate neutralisation.
C. NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)
Sodium chloride, a neutral salt formed from strong acid and strong base.
Q29. Ananya used KOH and H₂SO₄ in a neutralisation activity with phenolphthalein. She observed that 3 mL KOH required 45 drops of H₂SO₄.
She repeated with acetic acid instead of H₂SO₄—same volume of KOH required more drops.
A. Why are more drops of acetic acid needed compared to H₂SO₄ for the same volume of KOH?
B. If 45 drops of H₂SO₄ = 3 mL, how many mL of H₂SO₄ neutralise 7 mL KOH?
OR
What safety precaution must be followed while diluting H₂SO₄ before use?
C. Write balanced equations for both reactions. Classify each reaction.
Show Answer
A. Because acetic acid is a weak acid and ionises less, producing fewer H⁺ ions, so more volume is needed to neutralise the same amount of base.
B. 3 mL H₂SO₄ → 3 mL KOH ⇒ 7 mL KOH → 7 mL H₂SO₄
OR
Acid must be added slowly to water with constant stirring, because dilution is highly exothermic and adding water to acid may cause splashing and burns.
C.
With H₂SO₄: 2KOH + H₂SO₄ → K₂SO₄ + 2H₂O
With CH₃COOH: KOH + CH₃COOH → CH₃COOK + H₂O
Both are neutralisation and double displacement reactions.
Q30. In a lab, students neutralised NaOH with HNO₃ using phenolphthalein. Data:
| NaOH | Drops of HNO₃ |
| 2 ml | 20 |
| 3 ml | 30 |
| 5 ml | 50 |
A. What does the reappearance of pink colour on adding one drop of NaOH after endpoint indicate?
B. If 10 drops = 0.5 mL, how much NaOH is neutralised by 2.5 mL HNO₃?
OR
Why is the salt formed in this reaction neutral?
C. Write the equation. Which ions combine to form water?
Show Answer
A. It indicates that the solution has become basic again due to excess OH⁻ ions from added NaOH, confirming the reversibility of the indicator colour change.
B. 0.5 mL HNO₃ → 2 mL NaOH ⇒ 2.5 mL HNO₃ → 10 mL NaOH
OR
Because it is formed from a strong acid (HNO₃) and a strong base (NaOH), so the salt does not hydrolyse and pH = 7.
C. NaOH(aq) + HNO₃(aq) → NaNO₃(aq) + H₂O(l)
H⁺(aq) and OH⁻(aq) ions combine to form water.
Q31. A student tested aqueous solutions of four salts using red and blue litmus papers and recorded the following observations:
| Salt solution | Effect On Red Litmus | Effect On Blue Litmus |
| Sodium chloride (NaCl) | No change | No change |
| Aluminium chloride (AlCl₃) | No change | Turns red |
| Sodium carbonate (Na₂CO₃) | Turns blue | No change |
| Ammonium chloride (NH₄Cl) | No change | Turns red |
A. Why does sodium chloride solution not change the colour of either litmus paper?
B. Why does sodium carbonate solution turn red litmus blue?
OR
Why does aluminium chloride solution turn blue litmus red?
C. Which of the following statements is correct and why?
(a) Salts of strong acid and strong base are neutral
(b) Salts of weak acid and strong base are acidic
(c) Salts of strong acid and weak base are basic
(d) All salts are neutral
Show Answer
A. Because sodium chloride is formed from a strong acid (HCl) and a strong base (NaOH), so its solution is neutral (pH = 7) and does not change litmus colour.
B. Because sodium carbonate is formed from a strong base (NaOH) and a weak acid (carbonic acid), so its solution is basic, turning red litmus blue.
OR
Because aluminium chloride is formed from a strong acid (HCl) and a weak base (Al(OH)₃), so its solution is acidic, turning blue litmus red.
C. Correct option: (a) because salts of strong acid and strong base do not hydrolyse and give neutral solutions, as confirmed by NaCl.
Q32. Priya prepared aqueous solutions of NaCl, NH₄Cl, Na₂CO₃ and AlCl₃ and tested their pH using universal indicator paper. She observed:
| Salt | Color of pH paper | Approx. pH |
| NaCl | Green | 7 |
| NH₄Cl | Orange | 4 |
| Na₂CO₃ | Blue | 10 |
| AlCl₃ | Red | 3 |
A. Why is the pH of NH₄Cl solution less than 7?
B. What acid and base were used to form Na₂CO₃?
OR
If Priya tested KNO₃ solution, what pH would she expect? Why?
C. Write the general rule for predicting the nature (acidic/basic/neutral) of a salt solution based on the strength of the parent acid and base.
Show Answer
A. Because NH₄Cl is formed from a strong acid (HCl) and a weak base (NH₄OH), so it hydrolyses to give an acidic solution (pH < 7).
B. Carbonic acid (weak) and sodium hydroxide (strong).
OR
She would expect pH = 7 (green) because KNO₃ is formed from strong acid (HNO₃) and strong base (KOH), so it is neutral.
C.
Strong acid + Strong base → Neutral salt (pH = 7)
Strong acid + Weak base → Acidic salt (pH < 7)
Weak acid + Strong base → Basic salt (pH > 7)
Q33. In a lab activity, students added phenolphthalein to solutions of NaCl, Na₂CO₃, and NH₄Cl. Only the Na₂CO₃ solution turned pink.
A. Why did only Na₂CO₃ turn phenolphthalein pink?
B. Will blue litmus turn red in NH₄Cl solution? Justify.
OR
What would be observed if methyl orange is added to AlCl₃ solution?
C. Which of the following salts will give a basic solution and why?
(a) CuSO₄ (b) Na₂CO₃ (c) KCl (d) NH₄NO₃
Show Answer
A. Because Na₂CO₃ solution is basic, and phenolphthalein turns pink in basic medium. The other two salts give neutral or acidic solutions.
B. Yes, because NH₄Cl solution is acidic, so it turns blue litmus red.
OR
Methyl orange will turn red, as AlCl₃ solution is acidic.
C. Correct option: (b) Na₂CO₃, because it is a salt of strong base and weak acid, so its solution is basic.
Q34. A teacher asked students to classify salts based on their pH. They tested NaCl, CH₃COONa, AlCl₃ and KNO₃.
| Salt | pH |
| NaCl | 7 |
| CH₃COONa | 9 |
| AlCl₃ | 3 |
| KNO₃ | 7 |
A. Why is CH₃COONa solution basic even though it contains no OH⁻ ions?
B. Why do both NaCl and KNO₃ give neutral solutions?
OR
What is common between AlCl₃ and NH₄Cl in terms of their parent acid and base?
C. Write word equations for the formation of CH₃COONa and AlCl₃. Classify each salt as acidic, basic or neutral.
Show Answer
A. Because CH₃COONa is formed from strong base (NaOH) and weak acid (CH₃COOH), so it hydrolyses to produce OH⁻ ions, making the solution basic.
B. Because both are formed from strong acids and strong bases (NaCl: HCl + NaOH; KNO₃: HNO₃ + KOH), so they are neutral.
OR
Both are formed from a strong acid and a weak base, so their solutions are acidic.
C.
Acetic acid + Sodium hydroxide → Sodium acetate + Water → Basic salt
Hydrochloric acid + Aluminium hydroxide → Aluminium chloride + Water → Acidic salt
Q35. A student was given four salt solutions labelled P, Q, R and S. He tested them with litmus and found:
- P : no change in colour
- Q : turns red litmus blue
- R : turns blue litmus red
- S : no change
He was told the salts were NaCl, Na₂CO₃, NH₄Cl and KNO₃.
A. Identify P, Q, R and S.
B. Why does Q give a basic solution?
OR
Why does R give an acidic solution?
C. Which of the following is true and why?
(a) All salts are neutral
(b) Salts can be acidic, basic or neutral depending on the parent acid and base
(c) Only salts of metals are neutral
(d) Salts of non-metals are always acidic
Show Answer
A.
| Sample | Nature of Solution | Example Salt |
|---|---|---|
| P | Neutral | NaCl or KNO₃ |
| Q | Basic | Na₂CO₃ |
| R | Acidic | NH₄Cl |
| S | Neutral | KNO₃ or NaCl |
B. Because Na₂CO₃ is formed from a strong base (NaOH) and a weak acid (H₂CO₃), so its solution is basic.
OR
Because NH₄Cl is formed from a strong acid (HCl) and a weak base (NH₄OH), so its solution is acidic.
C. Correct option: (b) because the nature of a salt solution depends on the strength of the acid and base from which it is formed, as seen in the examples.
Industrial chemicals from NaCl (bleaching powder, baking soda, washing soda, POP)
Q36. A student visited a chemical factory where bleaching powder is manufactured. He observed that chlorine gas is passed over dry slaked lime to produce bleaching powder. The factory stores the product in airtight containers.
A. Why is dry slaked lime used instead of wet slaked lime in this process?
B. Why must bleaching powder be stored in airtight containers?
OR
What happens if bleaching powder is exposed to atmospheric CO₂?
C. Write a balanced chemical equation for the preparation of bleaching powder. Which of the following is correct and why? Bleaching powder is
(a) used only for bleaching clothes
(b) an oxidising agent and a disinfectant
(c) a reducing agent
(d) a neutral salt
Show Answer
A. Because water reacts with chlorine to form hydrochloric and hypochlorous acids, which reduces the yield of bleaching powder; dry conditions ensure proper reaction with Ca(OH)₂.
B. Because bleaching powder reacts with moisture and CO₂ from air, losing its chlorine content and becoming ineffective.
OR
It decomposes to release chlorine and forms calcium carbonate, reducing its bleaching power.
C. Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O
Correct option: (b) because it is used to bleach cotton, disinfect water, and as an oxidising agent in industries, as stated in the NCERT text.
Q37. In a school lab, students prepared baking soda by reacting sodium chloride with ammonia, carbon dioxide and water. They observed that the product was sparingly soluble and precipitated out. On heating this solid, they observed gas evolution that turned lime water milky.
A. Why is baking soda sparingly soluble in water?
B. What gas is evolved on heating baking soda? How is this property useful in baking?
OR
Why is baking soda used in antacids?
C. Write a balanced chemical equation for the thermal decomposition of baking soda. Which of the following is true and why? Baking soda is
(a) a strong base
(b) a neutral salt
(c) a mild non-corrosive basic salt
(d) an acidic salt
Show Answer
A. Baking soda (sodium hydrogencarbonate) has low solubility in water, so it precipitates out during preparation.
B. Carbon dioxide gas is evolved. This gas makes bread or cake soft and spongy by creating bubbles in the dough.
OR
Because it is a mild base that neutralises excess stomach acid, providing relief from acidity.
C. 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂
Correct option: (c) because it is basic in nature, non-corrosive, and used in food and medicine, as per the NCERT text.
Q38. A student heated gypsum crystals in a boiling tube and observed that the crystals turned into a white powder. When he added water to this powder, it hardened into a solid mass within minutes.
A. What is the white powder formed after heating gypsum?
B. Why does the powder harden when water is added?
OR
Why is this powder called “Plaster of Paris”?
C. Write a balanced chemical equation for the reaction when water is added to the white powder. Which of the following is correct and why? The white powder is
(a) calcium sulphate dihydrate
(b) calcium sulphate hemihydrate
(c) anhydrous calcium sulphate
(d) calcium carbonate
Show Answer
A. The white powder is Plaster of Paris (calcium sulphate hemihydrate).
B. Because it reacts with water to reform gypsum, which sets into a hard solid mass.
OR
Because large deposits of gypsum were found in Paris, and it was historically used there for making statues and plaster.
C. CaSO₄·½H₂O + 1½H₂O → CaSO₄·2H₂O
Correct option: (b) because heating gypsum at 373 K removes 1½ molecules of water, forming calcium sulphate hemihydrate, as stated in the NCERT text.
Q39. A factory produces washing soda by heating baking soda and then recrystallising the product. The final product appears as white crystals. A student tested its solution with red litmus and observed it turned blue.
A. What does the colour change of litmus indicate about washing soda solution?
B. What does “10H₂O” in Na₂CO₃·10H₂O signify?
OR
Why is washing soda used to remove permanent hardness of water?
C. Write the balanced equation for the formation of washing soda from baking soda. Which of the following is true and why? Washing soda is
(a) an acidic salt
(b) a neutral salt
(c) a basic salt
(d) a dehydrating agent
Show Answer
A. The colour change indicates that washing soda solution is basic in nature.
B. It signifies water of crystallisation—10 water molecules are present in one formula unit of sodium carbonate.
OR
Because it reacts with calcium and magnesium salts in hard water to form insoluble carbonates, thus removing permanent hardness.
C. Na₂CO₃ + 10H₂O → Na₂CO₃·10H₂O
Correct option: (c) because it is formed from a strong base (NaOH) and weak acid (H₂CO₃), making it basic, as per NCERT.
Q40. In the chlor-alkali process, an aqueous solution of common salt is electrolysed. Three useful products are obtained: a gas at the anode, a gas at the cathode, and a solution near the cathode.
A. Name the three products formed in this process.
B. Why is this process called “chlor-alkali”?
OR
Give one use of each product.
C. Write the balanced chemical equation for the chlor-alkali process. Which of the following is correct and why?
(a) Only chlorine is useful
(b) All three products have industrial uses
(c) Hydrogen is the main product
(d) NaOH is formed at the anode
Show Answer
A. Chlorine gas (anode), hydrogen gas (cathode), and sodium hydroxide solution (near cathode).
B. Because “chlor” refers to chlorine and “alkali” refers to sodium hydroxide, the two main products.
OR
Chlorine: for bleaching powder; Hydrogen: as fuel; NaOH: in soap and paper industries.
C. 2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + Cl₂(g) + H₂(g)
Correct option: (b) because all three products are used in industries, as shown in Figure 2.8 of the NCERT text.
