Class 9 Science Chapter 4: Acids, Bases and Salts – 18 Important Questions with Complete Answers

The restructured CBSE Class 9 Science curriculum prioritises experimental skills and daily-life applications. Chapter 4 carries 8–10% weightage in board exams, split across MCQs (1 mark each), short answers (2 marks), descriptive questions (3 marks), and long-form problem-solving (5 marks). Examiners test three core competencies: (1) Identification and classification—recognising acids, bases, and salts by their properties; (2) Experimental reasoning—predicting indicator colour changes in litmus, turmeric, and china rose solutions; (3) Application—linking neutralisation to acid rain, soil remediation, and industrial chemistry. Questions on natural indicators are high-frequency because they integrate lab skills with theoretical knowledge. Acid rain causation and pH calculations appear in 3–5 mark questions, demanding detailed explanations and balanced chemical equations. Practising these 18 questions ensures you master all question types and gain confidence for both term exams and board assessments.

**Q1. Which of the following indicators turns red in acidic solutions?** A) Turmeric B) Litmus (blue) C) China rose D) Methyl orange **Answer: B** – Blue litmus paper turns red in acidic conditions (pH < 7) due to H⁺ ion concentration. Turmeric remains yellow in acids but turns brown in bases. China rose petals shift from pink to colourless in acids. **Q2. What is the pH of a neutral solution at 25°C?** A) 0 B) 7 C) 10 D) 14 **Answer: B** – pH 7 represents neutrality where [H⁺] = [OH⁻] = 10⁻⁷ mol/L. Below 7 is acidic, above 7 is basic. **Q3. Which salt is produced when dilute HCl reacts with NaOH?** A) NaCl and HCl B) NaCl and H₂O C) NaOH and Cl₂ D) NaCl only **Answer: B** – The neutralisation reaction is: HCl + NaOH → NaCl + H₂O. Only one salt and water form; no acid or base remains. **Q4. Acid rain is primarily caused by:** A) HCl vapours B) SO₂ and NO₂ emissions C) Water vapour D) O₂ reactions **Answer: B** – Burning fossil fuels releases SO₂ and NO₂, which dissolve in atmospheric water to form dilute H₂SO₄ and HNO₃, lowering rainwater pH to ~5.6 or lower. **Q5. Which indicator shows different colours in acids, bases, and neutral solutions?** A) Litmus B) China rose C) Turmeric D) Methyl orange **Answer: B** – China rose petals: pink (neutral/slightly acidic), colourless (acidic), green (basic). Litmus and turmeric show two-colour changes, not three.

**Q1. Why is blue litmus used to test for acids and red litmus for bases? Explain with colour changes.** **Answer:** Blue litmus detects acids because H⁺ ions denature the litmus dye, turning it red. This colour change is visible and immediate. Red litmus detects bases because OH⁻ ions restore the litmus dye structure, turning it blue. Using opposite litmus colours ensures sensitivity: if we used red litmus for acids, no visible change occurs because red remains red. This principle extends to all indicators—the indicator must have a dye that changes colour in the expected pH range. **Q2. Write the balanced equation for neutralisation between H₂SO₄ and KOH. Identify the salt formed.** **Answer:** H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O. The salt is potassium sulphate (K₂SO₄). Sulphuric acid is dibasic (two replaceable H⁺ ions), so one mole reacts with two moles of the monoacidic base KOH to produce the normal salt K₂SO₄. **Q3. State three everyday applications of neutralisation reactions in daily life.** **Answer:** (1) Antacid tablets contain bases like Mg(OH)₂ or CaCO₃ that neutralise excess HCl in the stomach, relieving acidity. (2) Soil remediation—acidic soils are treated with lime (CaO or Ca(OH)₂) to raise pH and improve crop yield. (3) Wastewater treatment—industrial acidic effluents are neutralised with bases before discharge to meet environmental standards and prevent water pollution. **Q4. What causes acid rain? How does pH of acid rain differ from normal rain?** **Answer:** Acid rain forms when SO₂ and NO₂ from vehicle emissions and power plants dissolve in atmospheric moisture: SO₂ + H₂O → H₂SO₃ → H₂SO₄ (in presence of O₂). Normal rainwater is slightly acidic (pH ≈ 5.6) because of dissolved CO₂ forming carbonic acid (H₂CO₃). Acid rain has pH < 5.6, sometimes as low as 3.0–4.5, causing damage to aquatic ecosystems, buildings, and forests. **Q5. Turmeric is yellow in acids and brown in bases. Explain this colour change at the molecular level.** **Answer:** Turmeric contains curcumin, a natural dye molecule. In acidic conditions (high H⁺), the curcumin structure remains unchanged, keeping the solution yellow. In basic conditions (high OH⁻), OH⁻ ions interact with curcumin's chemical bonds, altering its electron configuration. This change in the dye's conjugated system shifts the absorbed light wavelength, producing the brown colour. The change is reversible if acid or base is neutralised.

**Q1. An experiment involves testing three unknown solutions (A, B, C) with litmus paper, turmeric solution, and china rose solution. Solution A turns red litmus blue and turmeric brown. What is A? Predict its effect on china rose petals and explain.** **Answer:** Solution A is a base (alkali). Red litmus turns blue in bases due to OH⁻ ions neutralising H⁺ impurities on the litmus dye. Turmeric turns brown in bases because curcumin's molecular structure changes under basic conditions. China rose petals will turn green in solution A because basic pH (OH⁻) causes colour shift from pink to green. This occurs because bases alter the pH-dependent electron transitions in the china rose anthocyanin dye molecules, shifting light absorption toward longer wavelengths (green region). **Q2. A farmer's field has soil pH = 4.2 (acidic). He adds calcium oxide (CaO) as a neutralising agent. Write the equation and explain why this improves crop yield.** **Answer:** CaO + H₂O → Ca(OH)₂ (calcium hydroxide forms in moist soil). Ca(OH)₂ + 2H⁺ → Ca²⁺ + 2H₂O. The base Ca(OH)₂ neutralises excess H⁺ ions, raising soil pH toward 6.5–7.0 (optimal range). This pH increase improves nutrient availability (phosphorus, potassium, magnesium become more soluble), enhances microbial activity for nutrient cycling, and reduces aluminium toxicity. Plants absorb nutrients more efficiently, resulting in better growth and higher crop yield. **Q3. In a school laboratory, a student accidentally mixed 50 mL of HCl (pH = 1) with 50 mL of NaOH (pH = 13). Calculate the final pH of the resulting solution. Show your reasoning.** **Answer:** HCl (pH = 1) has [H⁺] = 10⁻¹ = 0.1 mol/L; moles of H⁺ = 0.1 × 0.05 = 0.005 mol. NaOH (pH = 13) has pOH = 1, so [OH⁻] = 10⁻¹ = 0.1 mol/L; moles of OH⁻ = 0.1 × 0.05 = 0.005 mol. Neutralisation: H⁺ + OH⁻ → H₂O. Moles of H⁺ = moles of OH⁻ = 0.005, so they completely neutralise. Total volume = 100 mL; remaining [H⁺] ≈ 0, final pH ≈ 7 (neutral, assuming negligible water autoionisation at this level). **Q4. Acid rain with pH = 4 falls on a limestone (CaCO₃) statue. Write the equation for corrosion and explain the damage mechanism.** **Answer:** CaCO₃ + 2H⁺ → Ca²⁺ + H₂O + CO₂↑. The H⁺ ions from acid rain (HNO₃ and H₂SO₄) react with carbonate ions in limestone, dissolving the stone into soluble calcium ions and releasing CO₂ gas. The statue's surface gradually erodes and becomes porous. Additionally, the released CO₂ can recombine with water to form carbonic acid, prolonging the corrosion. This irreversible chemical weathering degrades the statue's fine details and structural integrity over years, destroying cultural heritage.

**Q1. Explain the concept of pH and the pH scale. How do natural indicators help classify substances as acids, bases, or salts? Support your answer with at least two indicator examples and their colour changes in acids, bases, and neutral solutions.** **Full Solution:** The pH scale measures the concentration of hydrogen ions (H⁺) in a solution. Defined as pH = −log₁₀[H⁺], where [H⁺] is in mol/L, the pH scale ranges from 0 to 14 at 25°C: - pH < 7: Acidic (e.g., lemon juice pH ≈ 2.0, vinegar pH ≈ 3.0) - pH = 7: Neutral (e.g., pure water at 25°C) - pH > 7: Basic/Alkaline (e.g., baking soda pH ≈ 8.3, ammonia solution pH ≈ 11.0) Natural indicators are organic compounds whose colour depends on pH. They classify substances by exhibiting distinct colour changes: **Litmus (natural dye from lichen):** - Acidic (pH < 7): Blue litmus turns red (H⁺ ions denature dye structure) - Neutral (pH = 7): Litmus remains purple/blue - Basic (pH > 7): Red litmus turns blue (OH⁻ ions restore dye structure) **Turmeric (curcumin compound):** - Acidic (pH < 7): Yellow (unchanged dye) - Neutral (pH = 7): Yellow - Basic (pH > 7): Brown (electron redistribution in curcumin alters light absorption) **China Rose (anthocyanin pigment):** - Acidic (pH < 7): Colourless (protonation of dye, conjugation broken) - Neutral (pH ≈ 7): Pink (neutral form of anthocyanin) - Basic (pH > 7): Green (deprotonation extends conjugation, shifts absorption) To classify an unknown substance: Add a few drops to indicator solutions. If blue litmus turns red and turmeric turns brown, it is a base. If red litmus turns blue, it is an acid. If no colour change occurs, it is likely neutral or a salt (most salts are pH-neutral). This systematic approach combines observation with chemical theory. --- **Q2. A school laboratory receives three unmarked bottles of solutions. Test them using natural indicators and classify them. Then, design a neutralisation experiment using one acid and one base from your classification. Write balanced equations and explain how to identify the point of complete neutralisation.** **Full Solution:** **Step 1: Classification using natural indicators** Take small samples of solutions P, Q, R. Add blue litmus, red litmus, turmeric, and china rose: - Solution P: Blue litmus → red, turmeric → brown → Classification: **Base** - Solution Q: Red litmus → blue, turmeric → yellow, china rose → colourless → Classification: **Acid** - Solution R: No colour change with any indicator → Classification: **Neutral/Salt solution** **Step 2: Neutralisation experiment (Acid Q + Base P)** Apparatus: Burette, conical flask, white tile, dropper, measuring cylinder. Procedure: 1. Measure 20 mL of base P in a conical flask. 2. Add 3–4 drops of china rose indicator (turns solution green because P is basic). 3. Slowly add acid Q drop by drop, swirling continuously. 4. Observe colour change: green → pink (endpoint of neutralisation). 5. Record the volume of acid Q used (say, 16 mL). **Chemical equations (assuming HCl in Q and NaOH in P):** HCl + NaOH → NaCl + H₂O OR (if H₂SO₄ in Q and Ca(OH)₂ in P): H₂SO₄ + Ca(OH)₂ → CaSO₄ + 2H₂O **Step 3: Identifying the neutralisation point** The endpoint occurs when: - Indicator colour shift is permanent (pink remains stable for 30 seconds) - pH ≈ 7 (complete consumption of H⁺ and OH⁻ ions) - H⁺ + OH⁻ → H₂O (all ions neutralised into water) Verification: Test the final solution with both blue and red litmus—both remain unchanged, confirming pH = 7. --- **Q3. Acid rain (pH = 3.8) falls on a mixed ecosystem containing limestone caves, forest soil, and a marble statue. Explain the effects on each component with relevant chemical equations. Propose two methods to mitigate acid rain damage.** **Full Solution:** **Part A: Effects of acid rain (pH = 3.8, [H⁺] = 10⁻³·⁸ ≈ 0.0000158 mol/L)** **1. Limestone caves (CaCO₃):** Reaction: CaCO₃ + 2H⁺ → Ca²⁺ + H₂O + CO₂↑ Effect: H⁺ from HNO₃ and H₂SO₄ in rain dissolves carbonate rock, enlarging caves and forming sinkholes. The calcium ions leach into groundwater, altering cave morphology irreversibly. **2. Forest soil:** Reaction: 2H⁺ + MgCO₃ → Mg²⁺ + H₂O + CO₂↑ (for soil carbonates) Effect: Acid rain lowers soil pH (< 5.5), reducing availability of essential nutrients (Ca²⁺, Mg²⁺, K⁺). Aluminium ions (Al³⁺) become soluble at low pH, damaging plant root cells and inhibiting water uptake. Forest growth slows; leaf yellowing (chlorosis) occurs due to Mg deficiency. **3. Marble statue (CaCO₃):** Reaction: CaCO₃ (marble) + 2H⁺ → Ca²⁺ + H₂O + CO₂↑ Effect: Surface erosion removes fine sculptural details. The statue becomes porous, allowing water penetration, which triggers freeze-thaw cycles and further structural damage. Heritage value is permanently lost. **Part B: Mitigation methods** **Method 1: Emission reduction (source control)** - Implement stricter emission standards for SO₂ and NO₂ from vehicles and power plants. - Use catalytic converters in cars: 2NO₂ → N₂ + 2O₂ (breaking down nitrogen oxides). - Switch to renewable energy (solar, wind) to replace coal-based power. - Promote fuel efficiency standards. **Method 2: Neutralisation (site-specific remediation)** - Apply lime (CaO) or limestone powder to acidic soils: CaO + H₂O → Ca(OH)₂ followed by Ca(OH)₂ + 2H⁺ → Ca²⁺ + 2H₂O (pH raised from 4.5 to 6.5). - Coat marble statues with protective resins to prevent acid contact. - Install rainwater harvesting systems with alkaline filters to pre-treat runoff in vulnerable areas. Long-term goal: Reduce atmospheric SO₂ and NO₂ by 80% to prevent acid rain formation entirely (pH > 5.6).

**Case Study: A food manufacturing factory discharges acidic wastewater (pH = 2.0) into a nearby river. Local farmers irrigate their fields with this river water, and fish populations have declined by 60% in three months. Environmental officers measure river pH (now 5.2) downstream. Investigate the problem and design a solution.** **Part A: Analysis** **Step 1: Calculate H⁺ ion concentration change** Factory discharge: pH = 2.0 → [H⁺] = 10⁻² = 0.01 mol/L (highly acidic) Downstream river: pH = 5.2 → [H⁺] = 10⁻⁵·² ≈ 0.00000631 mol/L The river's buffering capacity (natural carbonates) has partially neutralised factory acid, but pH remains suboptimal. **Step 2: Identify causes of fish decline** - Low pH damages fish gill tissue and disrupts osmoregulation (ion balance). - Acidic conditions dissolve sediment-bound heavy metals (Pb²⁺, Cd²⁺, Zn²⁺), making them bioavailable to fish, causing toxicity. - Algae growth inhibited at pH < 5.6 → reduced dissolved oxygen and food chain disruption. - Acid precipitation on soil leaches more Al³⁺ into river, further lowering pH and poisoning aquatic organisms. **Step 3: Farmer field impacts** - Crops struggle to absorb micronutrients (Fe, Zn, Mn) at pH < 5.0. - Soil acidification reduces populations of beneficial bacteria (nitrogen-fixers), lowering soil fertility. - Aluminium toxicity damages crop roots, stunting growth. **Part B: Solution Design** **Option 1: Wastewater pre-treatment at the factory** Add slaked lime Ca(OH)₂ to acidic wastewater before discharge: 2H⁺ + Ca(OH)₂ → Ca²⁺ + 2H₂O Target: Raise pH from 2.0 to 6.5–7.0 using 0.05 kg Ca(OH)₂ per litre of waste (calculate exact dose by titration: HCl + NaOH → NaCl + H₂O analogy). Cost: ₹50–100 per tonne of waste. Result: Factory discharge becomes neutral; river pH recovers to 6.5–7.0 within 2 weeks. **Option 2: Constructed wetland filter downstream** Build a wetland with limestone beds and aquatic plants (water hyacinth, cattails) between factory outlet and river: - Limestone dissolves slowly, releasing HCO₃⁻ to neutralise acid: CaCO₃ + H⁺ → Ca²⁺ + HCO₃⁻ - Plants absorb excess metals and produce oxygen. - Cost: ₹200,000 initial; low ongoing maintenance. - Timeline: 3–4 weeks for pH stabilisation; river ecosystem recovery in 2–3 months. **Expected outcomes:** - River pH stabilises at 6.5–7.2. - Fish populations recover within 3 months as dissolved oxygen increases. - Farmer fields return to normal productivity within 1 growing season. - Long-term: Factory adopts ISO 14001 compliance for wastewater management.

Chapter 4 requires mastery of both theoretical concepts and practical reasoning. The 18 questions above span all board exam formats—MCQs test instant recall of indicator colours, short answers demand balanced equations and definitions, while 3–5 mark questions require synthesis of multiple concepts and real-world applications like acid rain, soil remediation, and wastewater treatment. Many students memorise indicator colour changes but fail to explain the *molecular reason* why litmus turns red in acid (H⁺ denatures the dye structure) or why china rose turns green in base (OH⁻ alters anthocyanin conjugation). This conceptual gap costs marks on descriptive questions. At cbsetutor.ai, our AI tutor drills exactly these patterns—students attempt a 1-mark MCQ on litmus, receive instant feedback, then immediately see the 3-mark follow-up question asking them to *explain why* blue litmus turns red. Over 10–15 practice sessions, muscle memory forms. Students build speed (solve 10 MCQs in 4 minutes) while deepening conceptual clarity. The AI tracks which topics (acid rain, neutralisation, pH calculations) need reinforcement and auto-generates similar questions at increasing difficulty. Daily 15-minute drills on cbsetutor.ai ensure you score 28–30/30 on Chapter 4 assessments. Start a 3-day free trial at cbsetutor.ai to access all 18 questions with step-by-step solutions, interactive indicator colour simulators, and topic-wise performance analytics.