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Quiz Chapter 6: Acid, Base and Salt

10 questions ยท Form 4 Chemistry Bab 4: Acid, Base and Salt

Question 1 of 10Score: 0

What gas is evolved when ammonium chloride is heated with sodium hydroxide solution?

Full Question List & Answer Key

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1. What gas is evolved when ammonium chloride is heated with sodium hydroxide solution?

  1. Ammonia gas
  2. Chlorine gas
  3. Nitrogen dioxide gas
  4. Hydrogen gas
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Answer: A

Alkalis react with ammonium salts upon heating to produce salt, water, and ammonia gas (NH3).

2. What is the pH value of a 0.01 mol dm-3 hydrochloric acid (HCl) solution?

  1. 1.0
  2. 2.0
  3. 12.0
  4. 13.0
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Answer: B

pH = -log[H+]. For 0.01 mol dm-3 HCl (10-2 mol dm-3), pH = -log(0.01) = 2.0.

3. A solid salt X is heated strongly. It yields a brown gas that turns moist blue litmus paper red, and leaves a residue that is YELLOW when hot and WHITE when cold. What is salt X?

  1. Copper(II) nitrate
  2. Lead(II) nitrate
  3. Zinc nitrate
  4. Zinc carbonate
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Answer: C

Brown gas is NO2 (from nitrate). Oxide residue that is yellow when hot and white when cold is Zinc oxide (ZnO). Thus, salt X is Zinc nitrate.

4. What observation confirms the formation of a brown ring in the confirmatory test for nitrate ions (NO3-)?

  1. A brown precipitate settling at the bottom of the test tube
  2. A brown gas escaping from the mouth of the test tube
  3. A brown ring forming at the boundary between two liquid layers
  4. The entire solution turning dark brown immediately
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Answer: C

The nitrate brown ring test forms a distinct brown ring of [Fe(H2O)5(NO)]2+ at the junction of the concentrated sulfuric acid and aqueous layer.

5. When sodium hydroxide (NaOH) solution is added until excess to an unknown cation solution, a white precipitate forms which DISSOLVES in excess NaOH. Adding aqueous ammonia (NH3) until excess also produces a white precipitate which DISSOLVES in excess NH3. Which cation is present?

  1. Al3+
  2. Pb2+
  3. Zn2+
  4. Mg2+
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Answer: C

Zinc ion (Zn2+) is the only cation whose white hydroxide precipitate dissolves in EXCESS of both NaOH and NH3 solutions.

6. Which reagent is used to confirm the presence of chloride ions (Cl-) in an aqueous solution?

  1. Barium chloride solution acidified with hydrochloric acid
  2. Silver nitrate solution acidified with dilute nitric acid
  3. Iron(II) sulfate solution and concentrated sulfuric acid
  4. Potassium iodide solution
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Answer: B

Acidifying with dilute HNO3 and adding AgNO3 yields a white precipitate of AgCl if Cl- ions are present.

7. Which reagent can be used to distinguish between Aluminium ion (Al3+) and Lead(II) ion (Pb2+) solutions?

  1. Sodium hydroxide solution
  2. Potassium iodide solution
  3. Dilute nitric acid
  4. Ammonia solution
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Answer: B

Potassium iodide (KI) reacts with Pb2+ to form a bright yellow precipitate (PbI2), whereas no precipitate forms with Al3+.

8. What reaction technique should be used to prepare a pure sample of Barium Sulfate (BaSO4)?

  1. Neutralisation titration between an acid and alkali
  2. Reaction of acid with excess insoluble metal oxide
  3. Double decomposition (precipitation) by mixing two soluble salt solutions
  4. Direct combination of barium metal and sulfur powder
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Answer: C

Barium sulfate is an insoluble salt, so it must be prepared via double decomposition (precipitation).

9. In a titration experiment, 25.0 cm3 of 0.1 mol dm-3 NaOH solution requires 20.0 cm3 of H2SO4 for complete neutralisation. What is the molarity of H2SO4?

  1. 0.03125 mol dm-3
  2. 0.0625 mol dm-3
  3. 0.125 mol dm-3
  4. 0.250 mol dm-3
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Answer: B

Balanced eq: H2SO4 + 2NaOH -> Na2SO4 + 2H2O (a=1, b=2). Using MaVa/MbVb = ab => Ma * 200.1 * 25 = 12 => Ma = 0.1 * 2540 = 0.0625 mol dm-3.

10. What is the ionic equation for the neutralisation reaction between nitric acid and potassium hydroxide?

  1. K+(aq) + NO3-(aq) -> KNO3(s)
  2. H+(aq) + OH-(aq) -> H2O(l)
  3. HNO3(aq) + KOH(aq) -> KNO3(aq) + H2O(l)
  4. H+(aq) + KOH(aq) -> K+(aq) + H2O(l)
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Answer: B

The ionic equation for any strong acid-strong base neutralisation is H+(aq) + OH-(aq) -> H2O(l).

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