Question 1: Define the term 'neutralisation' in the context of acids and alkalis.

Answer: Neutralisation is a chemical reaction where an acid reacts with an alkali to produce a salt and water, resulting in a solution with a pH close to 7.

Question 2: Explain the mechanism of how acids and alkalis neutralise each other at the molecular level.

Answer: Acids release hydrogen ions (H⁺) in solution, while alkalis release hydroxide ions (OH⁻). When combined, these ions form water (H₂O), and the remaining ions form a salt.

Question 1: Calculate the volume of 0.1 mol/dm³ hydrochloric acid (HCl) required to neutralise 25 cm³ of a 0.2 mol/dm³ sodium hydroxide (NaOH).

Answer: V₁ = (C₂ × V₂) / C₁ = (0.2 mol/dm³ × 25 cm³) / 0.1 mol/dm³ = 50 cm³

Question 2: A student titrates 25.0 cm³ of hydrochloric acid with 0.1 mol/dm³ NaOH. If 30.0 cm³ of NaOH is required to neutralise the acid, what is the concentration of the acid?

Answer: C₁ = (C₂ × V₂) / V₁ = (0.1 mol/dm³ × 30.0 cm³) / 25.0 cm³ = 0.12 mol/dm³

Question 1: Describe the key safety precautions that should be taken when carrying out a neutralisation experiment in the laboratory.

Answer: Wear safety goggles and gloves, work in a well-ventilated area, handle acids and alkalis carefully to avoid spills, and dispose of chemicals properly.

Question 2: Outline a basic procedure for performing a titration to determine the concentration of an unknown acid.

Answer: Fill a burette with the standard alkali solution, record its initial volume. Add the acid to a conical flask with a few drops of indicator. Slowly titrate the acid with the alkali until a colour change indicates neutralisation. Record the final volume. Repeat to obtain consistent results and calculate the concentration of the acid using the titres.

Question 1: A solution of sulphuric acid (H₂SO₄) is neutralised by 40 cm³ of 0.5 mol/dm³ potassium hydroxide (KOH). Calculate the amount of sulphuric acid in moles present in the solution.

Answer: Moles of KOH = C × V = 0.5 mol/dm³ × 40 cm³ = 0.5 mol/dm³ × 0.040 dm³ = 0.020 mol Since 1 mol of H₂SO₄ reacts with 2 mol of KOH, moles of H₂SO₄ = 0.020 mol / 2 = 0.010 mol.

Question 2: In an industrial process, neutralisation is used to treat wastewater containing acidic pollutants. Explain why neutralisation is a suitable method and what products are formed.

Answer: Neutralisation is suitable because it safely raises the pH of the wastewater by reacting acids with alkalis, producing harmless salts and water, thus reducing environmental harm.

Question 1: Explain how neutralisation reactions are used in the treatment of indigestion using antacids.

Answer: Antacids contain bases like magnesium hydroxide that react with excess stomach acid (hydrochloric acid), forming water and salts, thereby relieving indigestion.

Question 2: Describe a modern technological application that relies on neutralisation reactions and explain its significance.

Answer: Neutralisation reactions are used in the manufacture of salts for pharmaceuticals and agriculture, such as producing ammonium nitrate fertiliser. This process ensures the salts are safe and suitable for their intended use.