Unit 18: Salts — Short Questions

Ans. The strength of electrostatic forces between ions depends upon:
i. The magnitude of the charges: If magnitude of charge of ion is high, the electrostatic forces will be stronger.
ii. The size of ions: If size of ions will be small, the electrostatic forces will be stronger.

Ans. Lead (II) chloride is slightly soluble in water because the high lattice energy of its crystals is not fully overcome by the hydration energy released when water interact with lead (II) chloride. Solubility of lead chloride is 1 g at 25°C. Solubility of lead chloride may be increased by increasing the temperature of solution. At 100°C solubility of PbCl₂ is 3.3 – 3.5 g.

Ans. Following lead salts are insoluble in water:
Lead (II) Sulphate PbSO₄ Insoluble
Lead (II) Carbonate PbCO₃ Insoluble
Lead (II) Phosphate Pb₃(PO₄)₂ Insoluble
Lead (II) Sulphide PbS Insoluble
Lead (II) Hydroxide Pb(OH)₂ Insoluble
Lead (II) Chromate PbCrO₄ Insoluble

So we can say lead (II) salts are generally insoluble except nitrates, acetate and chlorates.

Ans. Insoluble carbonates The carbonates of most of the metals are insoluble except carbonates of sodium, potassium and ammonium.

Two insoluble carbonates are calcium carbonate CaCO₃ and Lead (II) carbonate PbCO₃.

Ans. Crystal lattice A crystal lattice is a regular repeating three dimensional arrangement of particles (atom, ions or molecules) in a solid.

It is the ordered pattern that makes up a crystal.

Example In sodium chloride (NaCl)

Sodium ions (Na⁺) and chloride ions (Cl⁻) are arranged in a repeating face – centered cube like pattern.

Ans. It is a reaction in which acids neutralize bases or alkalis, salt and water are formed.

Example Sodium hydroxide, an alkali, neutralizes hydrochloric acid to form sodium chloride and water. This reaction is called neutralization reaction.

Ans. Salt An ionic compound which is formed by the neutralization reaction of an acid and a base, along with water, is called a salt.

Example Sodium chloride, potassium iodide, and barium chloride etc.

Ans. The crystal lattice of sodium chloride having face centred cube is given:

Ans. The melting points of ionic compounds depend upon the charges present on the ions and their sizes. Higher charges and smaller ion sizes lead to stronger attractions and hence the higher melting points.

Ans. Ionic compounds do not conduct electricity in solid state due to absence of free ion. In the solid state ions are held together by strong forces of attractions, they are not free to move and, therefore, cannot conduct electricity.

Ans. When ionic compounds are heated to their melting points, their ions become free to move around and hence start conducting electricity. Similarly, when ionic compounds are dissolved in water, the forces of attraction between ions break down and the ions are again free to move independently. So, they start conducting electricity.

Ans. All chloride of metallic ions except silver and lead (II) chlorides are soluble in water.

Examples NaCl, KCl, MgCl, etc.

Ans. Carbonates of sodium, potassium and ammonium are soluble in water and others are insoluble.

Ans. The soluble salts are:

• All sodium, potassium, and ammonium salts are soluble in water e.g. NaCl, Na₂SO₄, K₂CO₃.
• All nitrate salts are generally soluble in water e.g. potassium nitrate KNO₃, calcium nitrate Ca(NO₃)₂, etc.

Ans. The reactions are given:
(i) 2NaOH(aq) + H₂SO₄(aq) ⟶ Na₂SO₄(aq) + 2H₂O(l)
(ii) Zn(s) + H₂SO₄(aq) ⟶ ZnSO₄(aq) + H₂(g)

Ans.
Formation of copper (II) chloride
Cu(OH)₂(s) + 2HCl(aq) ⟶ CuCl₂(aq) + 2H₂O(l)

Formation of Magnesium sulphate
Mg(s) + H₂SO₄(aq) ⟶ MgSO₄(aq) + H₂(g)

Formation of Calcium chloride
CaCO₃(s) + 2HCl(aq) ⟶ CaCl₂(aq) + CO₂(g) + H₂O(l)

Ans. A soluble salt can be prepared using an excess of an insoluble base by its reaction with an acid like sulphuric acid.

Example Copper (II) sulphate is prepared by adding insoluble CuO to sulphuric acid.

CuO(s) + H₂SO₄(aq) ⟶ CuSO₄(aq) + H₂O(l)

Ans. Water molecules, being polar, surround and separate the ions in the salt, reducing the electrostatic attraction between them. This allows the ions to move freely get dissolved and participate in conduction of electricity.

Ans. Ionic compounds are hard because the strong ionic bonds form a stiff lattice structure. Each ion is tightly surrounded by oppositely charged ions resisting deformation and giving the material its hardness.

Ans. Soluble salts can be prepared by reacting an acid with an excess amount of metal. The metal reacts with the acid to produce a salt and hydrogen gas. The excess metals ensures complete reaction.

Example Zn(s) + H₂SO₄(aq) ⟶ ZnSO₄(aq) + H₂(g)

Ans. The purpose of titration:

• We can measure the exact volume of acid and base used to neutralize the given volume of other reacting substance.
• Titration also helps us to determine the concentration of the solution.

Ans. Phenolphthalein is used because it changes color form colorless to help pink at the end point of the reaction. It indicates that the strong acid and strong alkali have fully reacted by changing its colour

Ans. Excess zinc ensures that all the acid has reacted completely. When no more bubbles are observed, it indicates the acid is fully consumed.

Ans. Carbon dioxide gas (CO₂) is evolved during the reaction between calcium carbonate CaCO₃ and hydrochloric acid observed as effervescence (bubbling).

CaCO₃(s) + 2HCl(aq) ⟶ CaCl₂(aq) + H₂O(l) + CO₂(g)

Ans. When CaCO₃ and HCl are reacted, after the effervescence stops, the mixture is filtered to remove excess of CaCO₃. The filtrate contains calcium chloride CaCl, and is thus evaporated and cooled to obtain calcium chloride crystals.

CaCO₃(s) + 2HCl(aq) ⟶ CaCl₂(aq) + CO₂(g) + H₂O(l)

Ans. The melting points of ionic salts are very high because of the strong electrostatic forces of attraction between oppositely charged ions.

Explanation

• Ionic salts like (NaCl, MgO) consists of positive and negative ions.
• These ions are arranged in a regular and repeating pattern to give three dimensional network structures called crystal lattice.
• Each ion is surrounded by number of oppositely charged ions.
• To melt a salt, a large amount of energy is needed to overcome these strong forces.

Example Melting point of sodium chloride is 801°C and that of MgO is 2852°C.

Ans. Physical state of Ionic salts: Ionic salts exist in the solid state at room temperature because of the strong electrostatic forces of attraction between their oppositely charged ion.

Explanation

• Ionic salts are made of positive (cations) and negative (anions)
• Oppositely charges attract each other strongly and arrange themselves into a giant crystal lattice. The forces holding the ions together are very strong in all directions.

So strong ionic bonds and giant lattice structure are responsible for the ionic salts to be solids at room temperature.

Sodium chloride has higher melting point than magnesium chloride.

Although Mg²⁺ has a higher charge (+2) than Na⁺¹. But MgCl₂ shows some covalent character because.

• Mg²⁺ is small and highly charged.
• It strongly polarizes the Cl⁻ ion.
• It reduces purely ionic bonding strength

As a result MgCl₂ melts at a slightly lower temperature than NaCl.

Ans. Preparation of calcium sulphate (Insoluble salt): We can prepare calcium sulphate in the laboratory by the reaction of calcium carbonate with dilute sulphuric acid.

CaCO₃(s) + H₂SO₄(aq) ⟶ CaSO₄(s) + CO₂(g) + H₂O(l)

Procedure

• Take calcium carbonate (marble clips) in a beaker or a flask.
• Add dilute sulphuric acid slowly.
• Effervescence occurs due to evolution of CO₂ gas.
• Calcium sulphate forms as a white solid Calcium sulphate is insoluble in water so it precipitates out.

Ans. Comparison

Melting point of NaCl is 801°C.

Melting point of MgCl₂ is 714°C.

Ans. Preparation of pure crystals of copper sulfate

Pure crystals of copper sulphate can be prepared from sulphuric acid (H₂SO₄) and copper (II) oxide CuO.

Reaction
CuO(s) + H₂SO₄(aq) ⟶ CuSO₄(aq) + H₂O(l)

Method

• Measure 25cm³ dilute sulphuric acid into a beaker
• Add excess copper (II) oxide to dil H₂SO₄ in a beaker.
• The reaction produces copper (II) sulphate and water
• Continue adding CuO until all the acid has reacted.
• Separation: Filter the mixture to remove unreacted CuO.
• Crystallization: Evaporate the filtrate to obtain a hot saturated solution. Allow the solution to cool and crystallize, then filter to collect crystals. Dry the crystals in desiccator or oven.

Result Pure crystals of copper sulphate (CuSO₄) salts are prepared.

Ans. See Q.1 from theory.

Ans. See Q.5 from theory.

Ans. (i) Preparation of K₂SO₄

Reaction
2KOH(aq) + H₂SO₄(aq) ⟶ K₂SO₄(aq) + 2H₂O(l)

Method

• Take 50 cm³ of KOH solution in a beaker or conical flask with the help of pipette.
• Add a few drops of indicator (phenolphthalein) to get pink colour
• Add dilute sulphuric acid slowly with constant stirring until the solution becomes colourless (neutral).
• Note the volume of sulphuric acid to neutralize potassium hydroxide.
• Repeat the experiment taking the same volumes of sodium hydroxide and sulphuric acid without using indicator.
• Heat the neutral solution in an evaporating dish gently until one third volume is left.
• Cool down the solution for crystallization to occur completely.
• Filter and dry the crystals of potassium sulphate.

Result Pure crystals of potassium sulphate (K₂SO₄) are prepared.

(ii) Preparation of copper (II) chloride (CuCl₂)

This method involves neutralization reaction between the soluble acid and an insoluble base like copper hydroxide and dilute sulphuric acid.

Reaction
Cu(OH)₂(s) + 2HCl(aq) ⟶ CuCl₂(s) + 2H₂O(l)

Method

• Take a small amount of copper (II) hydroxide in a beaker
• Add dilute hydrochloric acid slowly while stirring.
• The blue solution gradually diminishes forming a green solution of copper (II) chloride.
• Continue adding acid until the base is fully utilized forming salt.
• Filter if any impurity remains.
• Evaporate the filtrate to get a saturated solution.
• Allow it to cool to obtain green crystals of copper (II) chloride CuCl₂.
• Filter and dry the crystals carefully.

Result Copper (II) chloride is prepared in the form of green hydrated crystals.

(iii) Preparation of calcium chloride (CaCl₂)

Calcium chloride (CaCl₂) is prepared from hydrochloric acid (HCl) and calcium carbonate (CaCO₃)

Reaction
CaCO₃(s) + 2HCl(aq) ⟶ CaCl₂(aq) + CO₂(g) + H₂O(l)

Method

• Measure 25cm³ dilute hydrochloric acid into a beaker.
• Add excess calcium carbonate to dil HCl in the beaker.
• The reaction produces calcium chloride, carbon dioxide and water
• Carbon dioxide evolved with effervescence locating all the acid has reacted.
• Filter the mixture to remove unreacted CaCO₃.
• Evaporate the filtrate to obtain a hot saturated solution. Allow the solution to cool and crystallize.

Result Pure crystals of calcium chloride (CaCl₂) are prepared.

Ans. See Q.4 from theory.