---
title: "Acids, Bases and Salts: CBSE Class 10 Science previous year questions"
url: https://www.swavid.com/cbse/class-10/science/pyq/acids-bases-and-salts
---

# Acids, Bases and Salts: CBSE Class 10 Science previous year questions

16 questions from CBSE Class 10 board papers, newest first, each with full working.

## CBSE Class 10 Science Question Paper 2026 (Set 31/1/1) with Solutions

### Question 18

*1 mark · MCQ*

Which one of the following can be used as an acid-base indicator by a visually impaired (blind) student ?

- Turmeric
- Vanilla essence
- Methyl orange
- Litmus

**Solution**

1. Olfactory indicators are substances whose odour changes in acidic or basic media.
2. Vanilla essence is an olfactory indicator and can be used by visually impaired students, whereas turmeric, methyl orange, and litmus are visual indicators.

**Answer:** (b) Vanilla essence

> Common mistake: Choosing colour-based indicators like litmus or methyl orange.

### Question 22

*1 mark · MCQ*

The natural sources of oxalic acid, lactic acid and methanoic acid respectively are :

- tomato, curd, ant-sting
- tomato, orange, nettle-sting
- orange, milk, ant-sting
- orange, sour milk, nettle-sting

**Solution**

1. Oxalic acid is naturally present in tomato.
2. Lactic acid is present in curd (sour milk), and methanoic (formic) acid is present in ant-sting.

**Answer:** (a) tomato, curd, ant-sting

> Common mistake: Confusing lactic acid source with orange or citrus fruits.

### Question 27

*3 marks · Short answer*

Explain chlor-alkali process with chemical equation. Name the products formed at anode and cathode.

**Solution**

1. When electricity is passed through an aqueous solution of sodium chloride (brine), it decomposes to form sodium hydroxide, chlorine gas, and hydrogen gas.
2. Chemical equation: $2\text{NaCl}(aq) + 2\text{H}_2\text{O}(l) \longrightarrow 2\text{NaOH}(aq) + \text{Cl}_2(g) + \text{H}_2(g)$.
3. Product formed at anode: Chlorine gas ($\text{Cl}_2$).
4. Product formed at cathode: Hydrogen gas ($\text{H}_2$).

**Answer:** Chlorine at anode, hydrogen at cathode, and sodium hydroxide near cathode.

> Common mistake: Interchanging the products formed at anode and cathode.

### Question 27 (OR)

*3 marks · Short answer*

Write the preparation of following compounds with balanced chemical equation :
(i) Baking soda
(ii) Bleaching powder
(iii) Plaster of Paris

**Part (i)**

1. Baking soda (Sodium hydrogen carbonate) is prepared by reacting cold and concentrated sodium chloride solution with ammonia and carbon dioxide.
2. Equation: $\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 + \text{NH}_3 \longrightarrow \text{NH}_4\text{Cl} + \text{NaHCO}_3$.

Answer (i): $\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 + \text{NH}_3 \longrightarrow \text{NH}_4\text{Cl} + \text{NaHCO}_3$

**Part (ii)**

1. Bleaching powder is prepared by the action of chlorine on dry slaked lime.
2. Equation: $\text{Ca}(\text{OH})_2 + \text{Cl}_2 \longrightarrow \text{CaOCl}_2 + \text{H}_2\text{O}$.

Answer (ii): $\text{Ca}(\text{OH})_2 + \text{Cl}_2 \longrightarrow \text{CaOCl}_2 + \text{H}_2\text{O}$

**Part (iii)**

1. Plaster of Paris is prepared by heating gypsum at $373\text{ K}$.
2. Equation: $\text{CaSO}_4 \cdot 2\text{H}_2\text{O} \xrightarrow{373\text{ K}} \text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O} + 1\frac{1}{2}\text{H}_2\text{O}$.

Answer (iii): $\text{CaSO}_4 \cdot 2\text{H}_2\text{O} \xrightarrow{373\text{ K}} \text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O} + 1\frac{1}{2}\text{H}_2\text{O}$

**Answer:** Prepared using NaCl, slaked lime, and gypsum respectively.

> Common mistake: Writing incorrect water of crystallisation coefficients for gypsum and Plaster of Paris.

## CBSE Class 10 Science Question Paper 2025 (Set 31/1/1) with Solutions

### Question 3

*1 mark · MCQ*

The following table shows the pH values of four solutions A, B, C and D on a pH scale:
The solutions A, B, C and D respectively are of a

- Strong acid, weak acid, neutral, strong base
- Weak acid, neutral, weak base, strong base
- Weak acid, neutral, strong base, weak base
- Weak acid, neutral, strong base, strong acid

**Solution**

1. A pH value less than 7 indicates an acidic solution (lower pH means stronger acid).
2. A pH value of 7 is neutral, and values greater than 7 indicate basic solutions (higher pH means stronger base).

**Answer:** (b) Weak acid, neutral, weak base, strong base

> Common mistake: Confusing the strength of acids and bases with respect to the numerical value of pH.

### Question 4

*1 mark · MCQ*

Consider the following reactions :
(i) Dilute hydrochloric acid reacts with sodium hydroxide.
(ii) Magnesium oxide reacts with dilute hydrochloric acid.
(iii) Carbon dioxide reacts with sodium hydroxide.
It is found that in each case :

- Salt and water is formed.
- Neutral salts are formed.
- Hydrogen gas is formed.
- Acidic salts are formed.

**Solution**

1. Acid-base neutralisation reactions (such as HCl + NaOH) produce salt and water.
2. Metal oxides reacting with acids (such as MgO + HCl) form salt and water.
3. Non-metal oxides reacting with bases (such as $CO_2$ + NaOH) form salt and water.

**Answer:** (a) Salt and water is formed.

> Common mistake: Assuming non-metal oxides with bases do not form water.

### Question 37

*4 marks · Case-based*

Seawater contains many salts dissolved in it. Common salt is separated from these salts. Deposits of solid salt are also found in several parts of the world. These large crystals are often brown due to impurities. This is called rock salt and is mined like coal. The common salt is an important raw material for chemicals of daily use.
(a) Write balanced chemical equations to show the products formed during electrolysis of brine.
(b) List two uses of any one product obtained during electrolysis of brine.
(c) (i) A mild non-corrosive basic salt 'A', used for faster cooking, is strongly heated to produce a compound 'B', that is used for removing permanent hardness of water. Identify A and B and also write the equation for the reaction that occurs when A is heated.

**Part (a)**

1. 1. Write the balanced chemical equation for the chlor-alkali process where electricity is passed through an aqueous solution of sodium chloride.
2. 2. $2\text{NaCl (aq)} + 2\text{H}_2\text{O (l)} \rightarrow 2\text{NaOH (aq)} + \text{Cl}_2\text{ (g)} + \text{H}_2\text{ (g)}$

Answer (a): $2\text{NaCl (aq)} + 2\text{H}_2\text{O (l)} \rightarrow 2\text{NaOH (aq)} + \text{Cl}_2\text{ (g)} + \text{H}_2\text{ (g)}$

**Part (b)**

1. 1. Choose any product such as chlorine gas.
2. 2. Two uses of chlorine gas are water treatment (swimming pools) and production of polyvinyl chloride (PVC) or bleaching powder.

Answer (b): Water treatment and manufacturing of PVC

**Part (c) (i)**

1. 1. Identify the mild non-corrosive basic salt A used for faster cooking as sodium hydrogen carbonate (baking soda, $\text{NaHCO}_3$).
2. 2. Identify compound B used for removing permanent hardness of water as sodium carbonate ($\text{Na}_2\text{CO}_3$).
3. 3. Write the thermal decomposition equation: $2\text{NaHCO}_3 \xrightarrow{\text{Heat}} \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2$

Answer (c) (i): A is $\text{NaHCO}_3$, B is $\text{Na}_2\text{CO}_3$, and the reaction is $2\text{NaHCO}_3 \xrightarrow{\text{Heat}} \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2$

**Answer:** Electrolysis of brine produces hydrogen, chlorine, and sodium hydroxide; A is sodium hydrogen carbonate and B is sodium carbonate.

> Common mistake: Confusing baking soda ($\text{NaHCO}_3$) with washing soda ($\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$).

### Question 37 (OR)

*2 marks · Case-based*

(c) (ii) Define water of crystallisation. Give two examples of salts that have water of crystallisation.

**Part (i)**

1. 1. Define water of crystallisation as the fixed number of water molecules present in one formula unit of a salt.
2. 2. Give two examples: Copper sulphate pentahydrate ($\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$) and Sodium carbonate decahydrate ($\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$).

Answer (i): Water of crystallisation is the fixed number of water molecules in one formula unit of a salt; examples are $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$ and $\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$.

**Answer:** Water of crystallisation is the fixed number of water molecules present in one formula unit of a salt.

> Common mistake: Forgetting to mention that the number of water molecules is fixed for a particular salt.

## CBSE Class 10 Science Question Paper 2024 (Set 31/1/1) with Solutions

### Question 5

*1 mark · MCQ*

Consider the following compounds :
$\text{FeSO}_4$; $\text{CuSO}_4$; $\text{CaSO}_4$; $\text{Na}_2\text{CO}_3$
The compound having maximum number of water of crystallisation in its crystalline form in one molecule is :

- $\text{FeSO}_4$
- $\text{CuSO}_4$
- $\text{CaSO}_4$
- $\text{Na}_2\text{CO}_3$

**Solution**

1. $\text{FeSO}_4$ has $7\text{H}_2\text{O}$ (ferrous sulphate heptahydrate).
2. $\text{CuSO}_4$ has $5\text{H}_2\text{O}$, $\text{CaSO}_4$ has $2\text{H}_2\text{O}$, and $\text{Na}_2\text{CO}_3$ has $10\text{H}_2\text{O}$ (sodium carbonate decahydrate), which is the maximum.

**Answer:** (d) $\text{Na}_2\text{CO}_3$

> Common mistake: Confusing with copper sulphate pentahydrate which has only 5 molecules of water.

### Question 28

*3 marks · Short answer*

(i) The $\text{pH}$ of a sample of tomato juice is $4.6$. How is this juice likely to be in taste ? Give reason to justify your answer.
(ii) How do we differentiate between a strong acid and a weak base in terms of ion-formation in aqueous solutions ?
(iii) The acid rain can make the survival of aquatic animals difficult. How ?

**Part (i)**

1. The pH of tomato juice is $4.6$, which is less than $7$.
2. Therefore, it is acidic in nature and will taste sour.

Answer (i): Sour in taste because a pH of $4.6$ indicates an acidic nature.

**Part (ii)**

1. A strong acid dissociates completely in an aqueous solution to yield a high concentration of hydronium ions ($\text{H}_3\text{O}^+$).
2. A weak base dissociates only partially in an aqueous solution to yield a low concentration of hydroxide ions ($\text{OH}^-$).

Answer (ii): Strong acids dissociate completely into ions, whereas weak bases dissociate only partially in aqueous solutions.

**Part (iii)**

1. When acid rain flows into rivers, it lowers the pH of the river water.
2. This acidic water makes the survival of aquatic animals difficult.

Answer (iii): Acid rain lowers the pH of aquatic bodies, making it difficult for aquatic life to survive.

**Answer:** Tomato juice is sour in taste due to its acidic pH, strong acids dissociate completely while weak bases partially dissociate in water, and acid rain lowers water pH making aquatic survival difficult.

> Common mistake: Confusing the ion types released by acids and bases.

## CBSE Class 10 Science Question Paper 2023 (Set 31/1/1) with Solutions

### Question 1

*1 mark · MCQ*

In the experimental setup given below, it is observed that on passing the gas produced in the reaction in the solution 'X' the solution 'X' first turns milky and then colourless.

[Diagram showing dilute hydrochloric acid reacting with sodium carbonate, passing gas into test tube 'X']

The option that justifies the above stated observation is that 'X' is aqueous calcium hydroxide and

- it turns milky due to carbon dioxide gas liberated in the reaction and after sometime it becomes colourless due to formation of calcium carbonate.
- it turns milky due to formation of calcium carbonate and on passing excess of carbon dioxide it becomes colourless due to formation of calcium hydrogen carbonate which is soluble in water.
- it turns milky due to passing of carbon dioxide through it. It turns colourless as on further passing carbon dioxide, sodium hydrogen carbonate is formed which is soluble in water.
- the carbon dioxide liberated during the reaction turns lime water milky due to formation of calcium hydrogen carbonate and after some time it turns colourless due to formation of calcium carbonate which is soluble in water.

**Solution**

1. Dilute hydrochloric acid reacts with sodium carbonate to evolve carbon dioxide gas, which turns lime water (aqueous calcium hydroxide, X) milky due to the formation of insoluble calcium carbonate.
2. On passing excess carbon dioxide, the milky solution becomes colourless due to the formation of soluble calcium hydrogen carbonate.

**Answer:** (b) it turns milky due to formation of calcium carbonate and on passing excess of carbon dioxide it becomes colourless due to formation of calcium hydrogen carbonate which is soluble in water.

> Common mistake: Confusing calcium carbonate with calcium hydrogen carbonate regarding solubility in water.

### Question 4

*1 mark · MCQ*

The table below has information regarding $\text{pH}$ and the nature (acidic/basic) of four different solutions. Which one of the options in the table is correct ?

- Lemon juice | Orange | 3 | Basic
- Milk of magnesia | Blue | 10 | Basic
- Gastric juice | Red | 6 | Acidic
- Pure water | Yellow | 7 | Neutral

**Solution**

1. Pure water is neutral with a pH of 7.
2. Milk of magnesia is basic with a pH of around 10, making option (b) the correct pairing from standard NCERT data.

**Answer:** (b) Milk of magnesia | Blue | 10 | Basic

> Common mistake: Memorizing incorrect pH values for common laboratory solutions.

### Question 6

*1 mark · MCQ*

Select washing soda from the following :

- $\text{NaHCO}_3$
- $\text{Na}_2\text{CO}_3 \cdot 5\text{H}_2\text{O}$
- $\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$
- $\text{NaOH}$

**Solution**

1. Washing soda is sodium carbonate decahydrate, which has the chemical formula $\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$.

**Answer:** (c) $\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$

> Common mistake: Confusing washing soda formula with baking soda ($\text{NaHCO}_3$) or other hydrates.

### Question 21

*2 marks · Very short answer*

(a) (i) A compound 'X' which is prepared from gypsum has the property of hardening when mixed with proper quantity of water. Identify 'X' and write its chemical formula.
(ii) State the difference in chemical composition between baking soda and baking powder.

**Part (i)**

1. Compound X is Plaster of Paris.
2. Its chemical formula is $\text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O}$.

Answer (i): Plaster of Paris, $\text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O}$

**Part (ii)**

1. Baking soda is chemically sodium hydrogen carbonate ($\text{NaHCO}_3$).
2. Baking powder is a mixture of baking soda and a mild edible acid such as tartaric acid.

Answer (ii): Baking soda is pure $\text{NaHCO}_3$, while baking powder is $\text{NaHCO}_3$ mixed with a mild edible acid.

**Answer:** Compound X is Plaster of Paris with formula $\text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O}$. Baking soda is sodium hydrogen carbonate, whereas baking powder is a mixture of baking soda and a mild edible acid like tartaric acid.

> Common mistake: Confusing Plaster of Paris formula with gypsum, or stating baking powder is just baking soda.

### Question 21 (OR)

*2 marks · Very short answer*

(b) Write balanced chemical equation for the reaction that occurs when :
(i) blue coloured copper sulphate crystals are heated and
(ii) Sodium hydrogen carbonate is heated during cooking.

**Part (i)**

1. Blue copper sulphate crystals contain water of crystallisation ($\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$).
2. On heating, they lose water to become white anhydrous copper sulphate.

Answer (i): $$\text{CuSO}_4 \cdot 5\text{H}_2\text{O}_{(s)} \xrightarrow{\text{Heat}} \text{CuSO}_4_{(s)} + 5\text{H}_2\text{O}_{(g)}$$

**Part (ii)**

1. Sodium hydrogen carbonate on heating decomposes to give sodium carbonate, water and carbon dioxide.

Answer (ii): $$2\text{NaHCO}_3_{(s)} \xrightarrow{\text{Heat}} \text{Na}_2\text{CO}_3_{(s)} + \text{H}_2\text{O}_{(l)} + \text{CO}_2_{(g)}$$

**Answer:** Balanced equations: (i) $\text{CuSO}_4 \cdot 5\text{H}_2\text{O} \xrightarrow{\text{Heat}} \text{CuSO}_4 + 5\text{H}_2\text{O}$, (ii) $2\text{NaHCO}_3 \xrightarrow{\text{Heat}} \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2$.

> Common mistake: Forgetting to balance the equation for the thermal decomposition of sodium hydrogen carbonate.

### Question 28

*3 marks · Short answer*

(a) Suggest one remedial measure each to counteract the change in $\text{pH}$ in human beings in following cases :
(i) Production of too much acid in stomach during indigestion
(ii) Stung by a honey bee / nettle leaves
(b) Fresh milk has a $\text{pH}$ of 6. When it changes into curd will its $\text{pH}$ increase or decrease? Why?

**Part (a)(i)**

1. Use antacids such as milk of magnesia which are basic in nature to neutralize excess stomach acid.

Answer (a)(i): Antacid (like Milk of Magnesia)

**Part (a)(ii)**

1. Apply a mild base like baking soda on the affected area to neutralize the acid injected by the bee or nettle sting.

Answer (a)(ii): Baking soda solution

**Part (b)**

1. When fresh milk changes into curd, lactic acid is produced.
2. Due to the presence of acid, the pH of the solution decreases.

Answer (b): pH decreases because lactic acid is produced during curd formation.

**Answer:** (a) Antacids and mild bases, (b) pH decreases due to formation of lactic acid.

> Common mistake: Writing that pH increases when acid is formed.

## Related pages

- [Acids, Bases and Salts: NCERT solutions](https://www.swavid.com/science/class/10/chapter/acids-bases-and-salts/ncert-solutions)
- [All CBSE Class 10 Science papers](https://www.swavid.com/cbse/class-10/science/previous-year-papers)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student