Study of Compounds — Nitric Acid

(a) The common anion is the nitrate ion.

(b) The cation in X is either sodium or potassium

(c) The metal ion present in Y is the lead ion.

(d) Z contains the ammonium ion

(e) The balanced equations are:

1. Reaction between Potassium nitrate and concentrated sulphuric acid (below 200° C):

$\underset{[\text{conc.}]}{\text{KNO}_3} + \underset{[\text{conc.}]}{\text{H}_2\text{SO}_4} \xrightarrow{\lt 200 \degree\text{C}} \underset{[\text{acid salt}]}{\text{KHSO}_4} + \text{HNO}_3$

2. 2Pb(NO₃)₂ $\xrightarrow{\Delta}$ 2PbO + 4NO₂ + O₂

3. Cu + 4HNO₃ (conc.) ⟶ Cu(NO₃)₂ + 2H₂O + 2NO₂.

Explanation

(a) On adding copper to acidified nitrates and heating, dense reddish brown fumes of nitrogen dioxide [NO₂] are evolved.

(b) The alkali metal nitrates such as sodium or potassium nitrates when heated melts into colourless liquids which decompose on heating to give oxygen gas.

$2\text{KNO}_3 \xrightarrow{\Delta} 2\text{KNO}_2 + \text{O}_2$

$2\text{NaNO}_3 \xrightarrow{\Delta} 2\text{NaNO}_2 + \text{O}_2$

(c) Lead nitrate decomposes on heating to give nitrogen dioxide, oxygen gas and yellow residue of lead oxide which fuses with the glass of the test tube.

2Pb(NO₃)₂ $\xrightarrow{\Delta}$ 2PbO + 4NO₂ + O₂

(d) Ammonium nitrate decomposes explosively on heating leaving behind no residue.

NH₄NO₃ $\xrightarrow{\Delta}$ N₂O + 2H₂O

Ammonium nitrate reacts with sodium hydroxide to give sodium nitrate, ammonia, and water. Ammonia gas turns moist red litmus paper blue.

NH₄NO₃ + NaOH ⟶ NaNO₃ + NH₃ + H₂O

(a) In the catalytic oxidation of ammonia by Ostwald's process, the temperature should be 800° C and catalyst used is platinum.

$4\text{NH}_3 + 5\text{O}_2 \xrightarrow[800 \degree\text{C}]{\text{Pt}} 4\text{NO} + 6\text{H}_2\text{O} + \text{Heat}$

(b) 4NO₂ + 2H₂O + O₂ ⟶ 4HNO₃

(a)

1. To etch designs on copper and brassware.
   Property : Acts as solvent for a large number of metals except noble metals
2. To purify gold.
   Property : Gold may contain Cu, Ag, Zn, Pb etc. as impurities which dissolve in nitric acid.
3. To prepare aqua regia.
   Property : It dissolves noble metals

(b) Brown ring test for nitric acid

To the aq. solution of a nitrate or nitric acid :

1. Add freshly prepared saturated solution of iron [II] sulphate.
2. Now add conc. sulphuric acid carefully from the sides of the test tube, so that it should not fall drop-wise in the test tube.
3. Cool the test tube in water.
4. A brown ring appears at the junction of the two liquids.

Equations for brown ring test:

6FeSO₄ + 3H₂SO₄ + 2HNO₃ (dil. ) ⟶ 3Fe₂(SO₄)₃ + 4H₂O +2NO

FeSO₄ + NO ⟶ FeSO₄.NO
[Nitroso Ferrous sulphate, a brown compound]

(c) A freshly prepared ferrous sulphate solution is used, because on exposure to the atmosphere, it is oxidised to ferric sulphate, which will not give the brown ring test.

(a) X is copper nitrate [Cu(NO₃)₂],
     Y is nitrogen dioxide [NO₂] and
     Z is hydrogen sulphide [H₂S].

(b) 2Cu(NO₃)₂ $\xrightarrow{\Delta}$ 2CuO + 4NO₂ + O₂

(c) When a H₂S gas, which has a rotten egg smell, is passed through a solution of Cu(NO₃)₂, a black ppt. of CuS is formed.
Cu(NO₃)₂ + H₂S ⟶ CuS + 2HNO₃

(a) Dilute nitric acid is considered a typical acid except for it's reaction with metals since it does not liberate hydrogen. It is a powerful oxidising agent and the nascent oxygen formed oxidises the hydrogen to water.

3Cu + 8HNO₃ ⟶ 3Cu(NO₃)₂ + 4H₂O + 2NO

3Zn + 8HNO₃ ⟶ 3Zn(NO₃)₂ + 4H₂O + 2NO

3Fe + 8HNO₃ ⟶ 3Fe(NO3)2 + 4 H₂O + 2NO

(b) Reaction with dil. HNO₃:

3Cu + 8HNO₃ ⟶ 3Cu(NO₃)₂ + 4H₂O + 2NO

Reaction with conc. HNO₃:

Cu + 4HNO₃ (conc.) ⟶ Cu(NO₃)₂ + 2H₂O + 2NO₂.

(a) A — concentrated sulphuric acid,

B — sodium nitrate or potassium nitrate and

C — is nitric acid.

(b) Nitric acid decomposes to liberate nitrogen dioxide along with oxygen and form water.

4HNO₃ $\xrightarrow{\Delta}$ 4NO₂ + 2H₂O + O₂

(c) Copper metal is oxidised by concentrated nitric acid to copper nitrate.

Cu + 4HNO₃ $\xrightarrow{\Delta}$ Cu(NO₃)₂ + 2H₂O + 2NO₂

(a) Freshly prepared FeSO₄ is used in the brown ring test, because upon exposure to the atmosphere, it is oxidised to ferric sulphate which will not give brown ring.

(b) The brown ring of nitroso ferrous sulphate decomposes on disturbing the test tube. The heat evolved decomposes the unstable brown ring.

(c) Lead nitrate (Pb(NO₃)₂) contains Pb²⁺ ions, which can form an insoluble precipitate of PbSO₄ when H₂SO₄ (sulphuric acid) is added.

This precipitate interferes with the formation of the necessary layers and prevents proper diffusion and interaction with Fe²⁺, thus suppressing the brown ring formation.

(d) Nitroso ferrous sulphate - FeSO₄.NO

KNO₃

Reason — The alkali metal nitrates such as sodium or potassium nitrates when heated melts into colourless liquids which decompose on heating to give oxygen gas.

$2\text{KNO}_3 \xrightarrow{\Delta} 2\text{KNO}_2 + \text{O}_2$ ↑

FeSO₄

Reason — In the brown ring test the conc. sulphuric acid being heavier settles down and the iron [II] sulphate layer remains above it resulting in formation of the brown ring at the junction.

6FeSO₄ + 3H₂SO₄ + 2HNO₃ (dil. ) ⟶ 3Fe₂(SO₄)₃ + 4H₂O +2NO

FeSO₄ + NO ⟶ FeSO₄.NO

[Nitroso Ferrous sulphate, a brown compound]

NO₂

Reason — Lead nitrate decomposes on heating to give lead oxide, nitrogen dioxide and oxygen.

2Pb(NO₃)₂ $\xrightarrow{\Delta}$ 2PbO + 4NO₂ + O₂

conc. H₂SO₄

Reason — In the laboratory preparation of Nitric acid by distillation process, potassium nitrate (nitre) is heated with conc. sulphuric acid (H₂SO₄).

$\underset{[\text{conc.}]}{\text{KNO}_3} + \underset{[\text{conc.}]}{\text{H}_2\text{SO}_4} \xrightarrow{\lt 200 \degree\text{C}} \underset{[\text{acid salt}]}{\text{KHSO}_4} + \text{HNO}_3$

Ostwald process

Reason — Nitric acid is manufactured by Ostwald process developed by a german chemist Ostwald, in 1914.

Pt

Reason — The catalyst used in the manufacture of HNO₃ by Ostwald's process is Platinum.

NO

Reason — Cold and dilute HNO₃ reacts with Copper and oxidises it to give Nitric oxide (NO)

3Cu + 8HNO₃ ⟶ 3Cu(NO₃)₂ + 4H₂O + 2NO

NO₃^-

Reason — The brown ring test is a qualitative test used to detect the presence of nitrate ions (NO₃^-) in a solution. A brown ring forms at the junction of the two liquids if nitrate ions are present.

H₂SO₄

Reason — Conc. Nitric acid (HNO₃) reacts with non-metals like sulphur to give sulphuric acid and nitrogen dioxide.

S + 6HNO₃ ⟶ H₂SO₄ + 2H₂O + 6NO₂

Freshly prepared FeSO₄

Reason — The solvent for NO is freshly prepared FeSO₄, because upon exposure to the atmosphere, it is oxidised to ferric sulphate which will not give brown ring.

HNO₃

Reason — Nitric acid (HNO₃) is used to etch designs on brassware because it acts as solvent for a large number of metals except noble metals.

decompose in the presence of light

Reason — Nitric acid (HNO₃) decomposes in presence of sunlight even at room temperature and turn into yellow colour solution. To avoid the decomposition, nitric acid is normally stored in reagent bottles.

NH₄NO₃

Reason — Ammonium nitrate (NH₄NO₃) decomposes explosively leaving behind no residue.

Only R

Reason — On heating zinc nitrate, the products formed are zinc oxide, nitrogen dioxide and oxygen. Zinc oxide is yellow when hot, white when cold.

2Zn(NO₃)₂ $\xrightarrow{\Delta}$ 2ZnO + 4NO₂ + O₂

Both A and R are true and R is the correct explanation of A.

Explanation— In the laboratory preparation of Nitric acid, nitric acid is obtained by distilling the reactants, conc. sulphuric acid (H₂SO₄) and potassium nitrate (KNO₃). Hence, the assertion (A) is true.
Sulphuric acid is a non-volatile acid and produces volatile nitric acid on reacting with potassium nitrate or sodium nitrate. Hence, the reason (R) is true.
Reason (R) explains why conc. sulphuric acid (H₂SO₄) and potassium nitrate (KNO₃) are used as reactants for the preparation of nitric acid. Hence, reason (R) is a correct explanation of assertion (A).

A is true but R is false.

Explanation— Pure nitric acid is colourless but the acid obtained in the laboratory is slightly yellow. The yellow colour is due to dissolution of reddish brown coloured nitrogen dioxide gas in the acid. This gas is produced due to the thermal decomposition of a portion of nitric acid. Hence, the assertion (A) is true.

Brown gas is not used in nitric acid preparation but brown coloured nitrogen dioxide gas is produced due to the thermal decomposition of portion of nitric acid. Hence, the reason (R) is false.

A is true but R is false.

Explanation— Catalytic oxidation of Ammonia by Ostwald's process does not require any external heat because the reaction itself is exothermic and the heat evolved maintains the temperature of the catalytic chamber. Hence, the assertion (A) is true.

Catalytic oxidation of ammonia is an exothermic process. Hence, the reason (R) is false.

Both A and R are true and R is the correct explanation of A.

Explanation — The yellow colour of nitric acid is removed by bubbling carbon dioxide gas or dry air into it. Hence, the assertion (A) is true.
The yellow colour will disappear because carbon dioxide gas or dry air drive out NO₂ gas from the warm nitric acid. Hence, the reason (R) is true.

Reason (R) explains how the use of carbon dioxide and dry air turn yellow colour nitric acid into a colourless compound. Hence, reason (R) is a correct explanation of assertion (A).

Both A and R are true and R is the correct explanation of A.

Explanation — Nitric acid reacts with alkalies like NaOH/KOH and neutralises it to form salt and water. Hence, the assertion (R) is true.

NaOH + HNO₃ ⟶ NaNO₃ + 2H₂O

Alkalies react with acids to form salt and water. Such type of reactions are called as neutralisation reaction. Hence, the reason (R) is true.

Since, reason (R) depicts the neutralisation reaction that takes place between Nitric acid and NaOH/KOH, it is the correct explanation of assertion (A).

Both A and R are true but R is not the correct explanation of A.

Explanation — Cold and dilute nitric acid oxidises metals to their nitrates and nitric oxide. Hence, the assertion (A) is true.

3Cu + 8HNO₃ ⟶ 3Cu(NO₃)₂ + 4H₂O + 2NO

Hot dilute nitric acid or concentrated nitric acid liberates nitrogen dioxide. Hence, the reason (R) is true.

Cu + 4HNO₃ ⟶ Cu(NO₃)₂ + 2H₂O + 2NO₂

However, reason (R) doesn't explain the action of Cold and dilute nitric acid on metals. So, it is not the correct explanation of assertion (A).

(a) All glass apparatus is used in the laboratory preparation of nitric acid since the vapours of nitric acid being highly corrosive attack rubber, cork, etc.

(b) When nitric acid is kept in a reagent bottle for a long time, it turns dark yellowish brown in colour.
This is because nitric acid is unstable hence, it decomposes slightly even at ordinary temperatures and in the presence of sunlight. When it is kept for a long time, the decomposition is complete resulting in the formation of reddish brown nitrogen dioxide [NO₂] that dissolves in the acid giving it a darker yellowish brown colour.

4HNO₃ ⟶ O₂ + 2H₂O + 2NO₂

(c) Iron or aluminium are rendered passive on reaction with fuming HNO₃ due to formation of a thin oxide coating on the surface of the metal which prevents further reaction

(d) Dilute nitric acid is generally considered a typical acid except for it's reaction with metals since it generally does not liberate hydrogen on reaction with metals. Nitric oxide on decomposition forms nascent oxygen which oxidizes the hydrogen to water.

(e) Pure nitric acid [HNO₃] is colourless and unstable and decomposes slightly even at ordinary temperatures and in the presence of sunlight.
The decomposition results in formation of reddish brown nitrogen dioxide [NO₂] which remains dissolved in the acid thus imparting a slight yellowish brown colour.

HNO₃ (Nitric acid)

3Cu + 8HNO₃ ⟶ 3Cu(NO₃)₂ + 4H₂O + 2NO

(a) When concentrated nitric acid is reacted with sulphur, dense brown fumes of nitrogen dioxide are observed.

S + 6HNO₃ [conc.] ⟶ H₂SO₄ + 2H₂O + 6NO₂

(b) Lead nitrate decomposes on heating to give nitrogen dioxide, oxygen gas and yellow residue of lead oxide which fuses with the glass of the test tube.

2Pb(NO₃)₂ $\xrightarrow{\Delta}$ 2PbO + 4NO₂ + O₂

(c) Reddish brown nitrogen dioxide gas is evolved on heating zinc nitrate crystals.

$2\text{Zn(NO}_3)_2 \xrightarrow{\Delta} 2\text{ZnO} + \text{O}_2 + 4\text{NO}_2$

(d) Reddish brown nitrogen dioxide gas is evolved.

Black coloured residue of CuO is obtained in the test tube.

$2\text{Cu(NO}_3)_2 \xrightarrow{\Delta} 2\text{CuO} + \text{O}_2 + 4\text{NO}_2$

When nitric acid undergoes decomposition it yields nascent oxygen. This nascent oxygen oxidises copper allowing it to react with nitric acid. Hence, the oxidising nature of nitric acid helps it to react with copper.

(a) CuCO₃ + 2HNO₃ [dil.] ⟶ Cu(NO₃)₂ + H₂O + CO₂

(b) C + 4HNO₃ ⟶ CO₂ + 2H₂O + 4NO₂

(c) Balanced equation for laboratory preparation of nitric acid:

$\underset{[\text{conc.}]}{\text{KNO}_3} + \underset{[\text{conc.}]}{\text{H}_2\text{SO}_4} \xrightarrow{\lt 200 \degree\text{C}} \underset{[\text{acid salt}]}{\text{KHSO}_4} + \text{HNO}_3$

(d) Balanced equation for action of heat on a mixture of copper and concentrated nitric acid:

$\text{Cu} + 4\underset{[\text{conc.}]}{\text{HNO}_3} \longrightarrow \text{Cu(NO}_3 \text{)}_2 + 2\text{H}_2 \text{O} + 2\text{NO}_2$

(a) Sodium nitrate
2NaNO₃ $\xrightarrow{\Delta}$ 2NaNO₂ + O₂↑

(b) Ammonium nitrate
NH₄NO₃ $\xrightarrow{\Delta}$ N₂O + 2H₂O

(c) Calcium nitrate
2Ca(NO₃)₂ $\xrightarrow{\Delta}$ 2CaO + 4NO₂ + O₂

(d) Silver nitrate
2AgNO₃ $\xrightarrow{\Delta}$ 2Ag + 2NO₂ + O₂

(e) Freshly prepared ferrous sulphate
6FeSO₄ + 3H₂SO₄ + 2HNO₃ (dil.) ⟶ 3Fe₂(SO₄)₃ + 4H₂O +2NO

FeSO₄ + NO ⟶ FeSO₄.NO

(f) Nitric oxide
2NO + O₂ ⟶ 2NO₂ (brown gas)

Nitric oxide can be prepared by catalytic oxidation of ammonia as shown below:

$4\text{NH}_3 + 5\text{O}<em>2 \xrightarrow[700-800 \degree\text{C}]{\text{Pt}} \text{4NO} + \text{6H}</em>\text{2}\text{O} + 21.5 \text{K cals}$

(g) Nitrogen dioxide Cu + 4HNO₃ ⟶ Cu(NO₃)₂ + 2H₂O + 2NO₂

(a) Nitrogen dioxide gas

S + 6HNO₃ ⟶ H₂SO₄ + 2H₂O + 6NO₂

(b) Oxygen gas

$2\text{KNO}_3 \xrightarrow{\Delta} 2\text{KNO}_2 + \text{O}_2$

(a) Nitric acid

S + 6HNO₃ ⟶ H₂SO₄ + 2H₂O + 6NO₂

(b) Nitric acid

$4\text{NH}_3 + 5\text{O}_2 \xrightarrow[700-800 \degree\text{C}]{\text{Pt}} 4\text{NO} + 6\text{H}_2\text{O} + 21.5 \text{K cals}$

$2\text{NO} + \text{O}_2 \xrightarrow{50 \degree\text{C}} \text{2NO}_\text{2}$

4NO₂ + 2H₂O + O₂ ⟶ 4HNO₃

(a) Potassium nitrate

(b) Ammonium nitrate

(c) Calcium nitrate

(d) Lead nitrate

Complete table is as follows:

Cold, dilute nitric acid reacts with copper to form nitric oxide.

Nitric acid (HNO₃) was formerly known aqua fortis, meaning strong water. It is so called because it reacts with nearly all metals. It can even dissolve silver which does not dissolve in other acids.

Conc. nitric acid (1 part by volume) when mixed with conc. hydrochloric acid (3 parts by volume) gives the mixture called Aqua Regia (meaning royal water).

The conversion of free atmospheric nitrogen (N₂) into useful nitrogenous compounds in the soil is known as fixation of nitrogen.

During thunderstorms, the nitrogen present in the atmosphere reacts with oxygen to form nitric oxide.

N₂ + O₂ ⟶ 2NO

Nitric oxide is further oxidized to nitrogen dioxide.

2NO + O₂ ⟶ 2NO₂

The nitrogen dioxide dissolves in atmospheric moisture or rain water in the presence of oxygen of the air and forms nitric acid which is washed down during rain and combines with the salt present on the surface of the earth.

4NO₂ + 2H₂O + O₂ ⟶ 4HNO₃

(a) Dry air (free from carbon dioxide and dust particles) and dry ammonia produced in Haber’s process.

(b) Platinum gauze

(c) Oxygen

(d) 1:10 i.e., 1 volume of ammonia and 10 volume of air.

(e) Quartz is acid resistant and when packed in layers, it slows down the movement of the gaseous NO₂ so that it can dissolve in water uniformly.

Below is the equation for the lab. preparation of HNO₃ from potassium nitrate and conc. H₂SO₄:

$\underset{[conc.]}{\text{KNO}_3} + \underset{[conc.]}{\text{H}_2\text{SO}_4} \xrightarrow{\lt 200 \degree\text{C}} \underset{[acid \space salt]}{\text{KHSO}_4} + \text{HNO}_3$

Conc. HCl is not used in place of conc. Sulphuric Acid (H₂SO₄) because HCl is volatile and hence nitric acid vapous will carry HCl vapours.

Concentrated nitric acid prepared in the laboratory is yellow in color due to dissolution of reddish brown nitrogen dioxide gas in the acid. This gas is produced due to the thermal decomposition of a portion of nitric acid.

4HNO₃ ⟶ 2H₂O + 4NO₂ + O₂

The yellow color of the acid is removed :

1. If dry air or CO₂ is bubbled through the yellow acid, the latter turns colourless because it drives out NO₂ gas from warm acid, which is further oxidised to nitric acid.
2. By adding excess water, nitrogen dioxide gas dissolves in water and thus yellow color of acid is removed.

The temperature of the reaction should not increase above 200°C, because sodium sulphate formed at higher temperature, forms a hard crust which sticks to the walls of the retort and is difficult to remove, although the yield of nitric acid is higher.

2NaNO₃ + H₂SO₄ $\xrightarrow{\gt 200 \degree\text{C}}$ Na₂SO₄ + 2HNO₃

The higher temperature

• may damage the glass apparatus.
• decomposition of nitric acid can also occur.
• wastage of fuel.

(a) An aqueous solution of nitric acid (68% concentration) forms a constant boiling mixture at 121°C. A constant boiling mixture is one which boils without change in composition.

Hence on boiling further, the mixture evolves out the vapours of both acid and water in the same proportion as in the liquid.

Therefore, dilute HNO₃ cannot be concentrated beyond 68% by boiling.

(b) Iron becomes passive or inert when treated with pure concentrated nitric acid due to the formation of the extremely thin layer of insoluble iron oxide (passivity) which stops the reaction.

Fe + 6HNO₃ ⟶ Fe(NO₃)₃ + 3H₂O + 3NO₂

This inert iron is known as passive iron.

Passivity can be removed by rubbing the surface layer with sandpaper or by treating with strong reducing agent.

(a) The products formed when carbon and conc. nitric acid is heated are carbon dioxide, nitrogen dioxide and water

C + 4HNO₃ ⟶ CO₂ + 2H₂O + 4NO₂

(b) The products formed when dilute HNO₃ is added to copper are copper nitrate, nitric oxide and water

3Cu + 8HNO₃ ⟶ 3Cu(NO₃)₂ + 4H₂O + 2NO

(a) Action on non-metals:

C + 4HNO₃ ⟶ CO₂ + 2H₂O + 4NO₂

S + 6HNO₃ ⟶ H₂SO₄ + 2H₂O + 6NO₂

(b) Reaction with alkalis:

K₂O + 2HNO₃ ⟶ 2KNO₃ + H₂O

CuO + 2HNO₃ ⟶ Cu(NO₃)₂ + H₂O

(c) Nitric acid acting as oxidising agent:

C + 4HNO₃ ⟶ CO₂ + 2H₂O + 4NO₂

3Cu + 8HNO₃ [cold and dil.] ⟶ 3Cu(NO₃)₂ + 4H₂O + 2NO

(a) Sodium hydrogen carbonate when treated with nitric acid forms sodium nitrate, carbon dioxide and water.

NaHCO₃ + HNO₃ ⟶ NaNO₃ + H₂O + CO₂ ↑

(b) Cupric oxide reacts with dilute nitric acid, it forms copper nitrate and water.

CuO + 2HNO₃ ⟶ Cu(NO₃)₂ + H₂O

(c) Zinc reacts with dil. nitric acid to form zinc nitrate, nitric oxide and water

3Zn + 8HNO₃ ⟶ 3Zn(NO₃)₂ + 4H₂O + 2NO

(d) When concentrated nitric acid is heated it forms nitrogen dioxide gas, water and oxygen gas

4HNO₃ $\xrightarrow{\Delta}$ 4NO₂ + 2H₂O + O₂

(a) NaOH + HNO₃ [dil.] ⟶ NaNO₃ + H₂O

(b) CuO + 2HNO₃ [dil.] ⟶ Cu(NO₃)₂ + H₂O

(c) Pb + 4HNO₃ [conc.] ⟶ Pb(NO₃)₂ + 2H₂O + 2NO₂

(d) Mg(OH)₂ + 2HNO₃ [dil.] ⟶ Mg(NO₃)₂ + 2H₂O

(e) Fe + 6HNO₃ [conc.] ⟶ Fe(NO₃)₃ + 3H₂O + 3NO₂

(f) Aqua regia is a freshly prepared mixture of concentrated hydrochloric acid and concentrated nitric acid mixed in the ratio of 3:1 by volume.

3HCl + HNO₃ ⟶ NOCl + 2H₂O + 2[Cl]

A : 3Cu + 8HNO₃ [cold and dil.] ⟶ 3Cu(NO₃)₂ + 4H₂O + 2NO

B : 2Cu(NO₃)₂ $\xrightarrow{\Delta}$ 2CuO + 4NO₂ + O₂

C : 2Cu + O₂ ⟶ 2CuO

D : CuO + H₂ $\xrightarrow{\Delta}$ Cu + H₂O

(a) HNO₃ is a strong oxidizing agent

(b) NaNO₃ gives NaNO₂ and O₂ on heating.

(c) Constant boiling nitric acid contains 68% nitric acid by weight

(d) Nitric acid turns yellow when exposed to light.

(e) Magnesium reacts with very dilute (about 1%) nitric acid to liberate hydrogen gas.