Acids, Bases and Salts – Case-based Questions with Answers
Case-Based Questions with Answers
Each case presents a short scenario followed by 2–3 questions. Answers are given immediately after each question to aid self-assessment. Chemical formulas use <sub> and <sup> tags for accuracy.
Case 1 — Antacid at home
Ravi has heartburn and takes an antacid containing magnesium hydroxide, Mg(OH)2. After taking it, his symptoms reduce quickly.
Q1.1: Write the chemical reaction between Mg(OH)2 and hydrochloric acid (HCl) in the stomach.
Answer: Mg(OH)2 + 2HCl → MgCl2 + 2H2O.
Q1.2: State the type of reaction and explain why symptoms reduce.
Answer: This is a neutralisation reaction (acid + base → salt + water). HCl is neutralised, lowering acidity in the stomach, which relieves irritation and heartburn.
Case 2 — School titration
A student titrates 25.0 mL of unknown HCl with 0.10 M NaOH and finds the end point at 18.5 mL of NaOH (1:1 reaction).
Q2.1: Calculate the molarity of the HCl solution.
Answer: M1V1=M2V2; MHCl×25.0 mL = 0.10 M × 18.5 mL → MHCl = (0.10×18.5)/25.0 = 0.074 M.
Q2.2: Which indicator is suitable for this titration and why?
Answer: Phenolphthalein is suitable because it changes colour in the basic region near the equivalence point for strong acid–strong base titrations and gives a clear end point.
Case 3 — Carbonate test
In a lab, dilute HCl is added to an unknown white powder and brisk effervescence is observed and the gas turns limewater milky.
Q3.1: Identify the likely compound.
Answer: The compound is likely a carbonate, e.g., CaCO3, because reaction with acid produces CO2 which turns limewater milky.
Q3.2: Write the balanced chemical equation for reaction with HCl.
Answer: CaCO3 + 2HCl → CaCl2 + CO2 + H2O.
Case 4 — Soil treatment
A farmer finds his soil too acidic for crops. He is advised to add quicklime (CaO) to improve fertility.
Q4.1: Write the reaction of CaO with water and subsequent reaction that neutralises soil acidity.
Answer: CaO + H2O → Ca(OH)2 (slaked lime). Then Ca(OH)2 neutralises acidity: Ca(OH)2 + 2HCl → CaCl2 + 2H2O (example with HCl as acid).
Q4.2: Explain how this improves crop growth.
Answer: Neutralising soil acidity raises pH towards neutral, making nutrients more available and reducing toxic metal solubility, thereby improving conditions for root growth and microbial activity.
Case 5 — Bleach production
A factory produces bleaching powder by reacting chlorine with slaked lime. The product is used for water purification.
Q5.1: Give the chemical equation for producing bleaching powder.
Answer: Ca(OH)2 + Cl2 → CaOCl2 + H2O (simplified representation for bleaching powder formation).
Q5.2: Why is bleaching powder useful in water treatment?
Answer: Bleaching powder releases hypochlorite ions, which disinfect water by oxidising and killing bacteria and pathogens, making water safe for drinking when used correctly.
Case 6 — Acid spill in lab
A student accidentally spills a small amount of concentrated HCl on the bench. The teacher suggests neutralising the spill with sodium bicarbonate (NaHCO3).
Q6.1: Write the neutralisation reaction and identify the gas produced, if any.
Answer: HCl + NaHCO3 → NaCl + H2O + CO2. Carbon dioxide gas (CO2) evolves.
Q6.2: Mention two safety precautions while neutralising the spill.
Answer: Wear gloves and goggles, add NaHCO3 gradually to avoid vigorous bubbling, ventilate the area, and clean the residue after neutralisation.
Case 7 — Titration indicator choice
A student must titrate a weak acid with a strong base. He wonders whether to use methyl orange or phenolphthalein.
Q7.1: Which indicator should be used and why?
Answer: Phenolphthalein is preferred because the equivalence point for weak acid–strong base titration is in the basic pH range where phenolphthalein changes colour (around pH 8.2–10).
Q7.2: What would happen if methyl orange were used?
Answer: Methyl orange changes at lower pH (~3.1–4.4) and may give a premature end point, leading to an overestimation of acid concentration.
Case 8 — Salt from sea water
A coastal community harvests salt by flooding shallow pans with sea water and allowing evaporation.
Q8.1: Explain why NaCl crystallises out on evaporation.
Answer: As water evaporates, the concentration of dissolved ions increases; when the solubility product of NaCl is exceeded, NaCl nucleates and crystallises out of solution.
Q8.2: Name one impurity and a common purification step.
Answer: Impurity: gypsum (CaSO4) or other salts. Purification: dissolve crude salt in water, filter to remove insolubles and recrystallise NaCl by controlled evaporation.
Case 9 — Acidic rain effect
A lake near an industrial area shows decline in fish population. Analysis shows lowered pH due to industrial emissions.
Q9.1: Name two gaseous pollutants that cause acid rain and the acids they form.
Answer: SO2 forms H2SO4 (sulfuric acid) and NOx (NO, NO2) forms HNO3 (nitric acid) in the atmosphere.
Q9.2: Suggest one mitigation measure to reduce lake acidity.
Answer: Liming—adding CaCO3 or Ca(OH)2 to neutralise acidity, restoring pH suitable for aquatic life; also control emissions at source.
Case 10 — Production of CuSO4
A student dissolves copper in dilute sulfuric acid with heating and obtains a blue solution.
Q10.1: Write the chemical equation for the reaction when copper reacts with hot, concentrated H2SO4.
Answer: Cu + 2H2SO4 (conc.) → CuSO4 + SO2 + 2H2O.
Q10.2: Why is SO2 produced instead of H2 when copper reacts with sulfuric acid?
Answer: Concentrated H2SO4 is an oxidising acid; it oxidises copper metal to Cu2+ while itself gets reduced to SO2, so hydrogen gas is not liberated.
Case 11 — Salt hydrolysis observation
Students test solutions of NaCl, NH4Cl and Na2CO3 with indicators and obtain different colours.
Q11.1: Predict the pH nature (acidic/neutral/basic) of these three solutions and briefly explain.
Answer: NaCl — neutral (salt of strong acid & strong base); NH4Cl — acidic due to NH4+ hydrolysis producing H+; Na2CO3 — basic due to CO32− producing OH− on hydrolysis.
Q11.2: Give one chemical equation showing hydrolysis of NH4+.
Answer: NH4+ + H2O ⇌ NH3 + H+.
Case 12 — Choosing an indicator for titration
A chemist is titrating a strong base against a weak acid; the equivalence point is expected to be basic.
Q12.1: Which indicator is suitable and why?
Answer: Phenolphthalein is suitable because it changes colour in the alkaline pH range and the equivalence point of weak acid–strong base titration lies in the basic region.
Q12.2: State a practical tip to obtain an accurate end point during titration.
Answer: Add titrant dropwise near the end point and swirl continuously; repeat titration to obtain concordant readings and take the average for accuracy.
Case 13 — Disposal of acidic waste
A small industry needs to neutralise acidic wastewater before release. They plan to add lime (CaO) to the effluent.
Q13.1: Explain the chemical process of neutralisation and why lime is suitable.
Answer: Lime reacts with water to form Ca(OH)2, which neutralises acids: CaO + H2O → Ca(OH)2; Ca(OH)2 + 2HCl → CaCl2 + 2H2O. Lime is cheap, effective and raises pH, precipitating some impurities as insoluble hydroxides.
Q13.2: Mention an environmental precaution when neutralising large volumes.
Answer: Neutralise gradually while monitoring pH to avoid over-alkalinisation; control exothermic heating and treat precipitates before discharge to prevent secondary pollution.
Case 14 — Identification of unknown liquid
A sample turns blue litmus red and reacts with Zn to produce bubbles that extinguish a glowing splint.
Q14.1: What is the likely nature of the sample and which gas is evolved?
Answer: The sample is acidic (turns blue litmus red); reaction with Zn produces hydrogen gas (H2), which extinguishes a glowing splint.
Q14.2: Write a general equation for metal reaction with acid producing hydrogen.
Answer: Metal + Acid → Salt + H2 (e.g., Zn + 2HCl → ZnCl2 + H2).
Case 15 — Preparing a sample of salt
A student is asked to prepare dry sodium chloride crystals starting from HCl and NaOH solutions.
Q15.1: Outline the steps including the equation to obtain dry NaCl crystals.
Answer: Neutralise measured HCl with NaOH: HCl + NaOH → NaCl + H2O. Evaporate the resulting solution to concentrate, then allow to cool so NaCl crystallises. Filter, wash and dry crystals to obtain pure NaCl.
Q15.2: State one observation during neutralisation and one precaution.
Answer: Observation: slight warming (exothermic) and neutral pH at end point. Precaution: add NaOH slowly with stirring and monitor pH/indicator to avoid overshooting.
Concluding note:
These Case-Based Questions and Answers are prepared strictly as per the NCERT syllabus and provide application-oriented practice suitable for CBSE Class 10 Board exam preparation.