Chapter 7.2 - Preparation and Properties of Hydrogen chloride

In the previous section, we discussed about sulphuric acid. In this section we will discuss about Hydrogen chloride.

Hydrogen chloride

Hydrogen chloride is an important compound of hydrogen and chlorine. The preparation of hydrogen chloride in the laboratory is shown in fig.7.3 below:

Fig.7.3

The reactants used are sodium chloride (NaCl) and concentrated sulphuric acid. First, NaCl is taken in a flask. Then the concentrated sulphuric acid is added through the thistle funnel. The mixture is heated. Equation of the reaction is:
NaCl + H₂SO₄ → NaHSO₄ + HCl. This is a balanced equation
• The HCl gas formed is passed through concentrated sulphuric acid. Why is this?
We know that concentrated sulphuric acid is a drying agent. We have seen the details here. So we will get dry HCl in the gas jar.
• HCl gas is denser than air. So it will be collected in the gas jar by the upward displacement of air.
• HCl gas is colourless. So how do we know whether the gas jar is full or not? The method is as follows:
We know that:
    ♦ Acids turn blue litmus paper red
    ♦ Bases turn red litmus paper blue
HCl is acidic. So show a wet blue litmus paper at the mouth of the gas jar. If the gas jar is full, the HCl gas will come in contact with the litmus paper, and it's colour will become red.

Solubility of Hydrogen chloride gas in water

• We have seen the fountain experiment for ammonia gas. Details here. It proved that the ammonia gas is soluble in water. 
• The same experiment can be used to prove the solubility of HCl gas in water. The apparatus is shown in the fig.7.4 below:

Fig.7.4

• The only difference is that, blue litmus is added to the water in the trough. This blue colour will change to red colour when it becomes the fountain. We can explain this colour change as follows:
When the water enters the flask, the HCl gas will dissolve in it. The water will thus become acidic. But blue litmus is already added to the water. So this blue colour will change to red. Thus we get a red coloured fountain.

Properties of Hydrogen chloride gas

• It is a colourless gas • It has a pungent smell • It is denser than air • It is soluble in water • It is acidic in nature

Identification of hydrogen chloride gas

Introduce a glass rod dipped in ammonia solution to the HCl gas. Thick white fumes of ammonium chloride shows the presence of HCl gas. The fumes are due to the formation of ammonium chloride (NH₄Cl). The equation of the reaction is:
NH₃ + HCl → NH₄Cl
This is a balanced equation
■ Note that the same reaction was used earlier in the identification of ammonia. Details here. 
• There we dipped the glass rod in ammonia and introduced it to HCl
• Here we dip the glass rod in HCl and introduce it to ammonia

Hydrochloric acid

• The aqueous solution of hydrogen chloride gas is hydrochloric acid. Molecular formula of hydrochloric acid is HCl. 
• It is a volatile acid. That is., the acid will easily change to gaseous state. If the bottle containing concentrated HCl is kept open, white fumes will be formed. This is due to the dissolution of HCl gas in the water vapour of the air. A video can be seen here.

Some reactions of hydrochloric acid:
HCl reacts with almost all metals and metallic compounds. Let us see some examples:
■ Reaction with the metal zinc:
• We have seen the above reaction in the preparation of hydrogen. Details here.
• The balanced equation is: Zn + 2HCl → ZnCl₂ + H₂ .
• So, when HCl reacts with a metal, two things happen:
    (i) The chloride of the metal is formed
    (ii) Hydrogen is formed
■ Reaction with sodium hydroxide:
We have seen the above reaction when we learned neutralisation. Details here.
■ Reaction with calcium carbonate:
• This reaction gives calcium chloride. The balanced equation is:
2HCl + CaCO₃ → CaCl₂ + H₂O + CO₂.

Identification of chloride salts

The following procedure can be used:
1. Make an aqueous solution of the given salt.
2. Add a little silver nitrate (AgNO₃) solution to this
3. If a curdy white precipitate is formed, the given salt can be a chloride salt. The balanced equation is:
NaCl + AgNO₃ → AgCl ↓ +  NaNO₃.
4. But we need to confirm. For that, add ammonium hydroxide solution to the white curdy precipitate

5. If the white precipitate dissolves, we can confirm that, the given salt is a chloride salt.
A video showing the formation of white curdy precipitate can be seen here.

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In the next section, we will see Nitric acid. 

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Chapter 7.1 - Sulphuric acid

In the previous section, we discussed about ammonia. In this section we will a few more non-metallic compounds.

Sulphuric acid

Sulphuric acid is a chemical of utmost importance in industry. It's various industrial uses are listed below:

•Manufacture of fertilisers •Manufacture of explosives •Manufacture of paints •Manufacture of fibres •Manufacture of other chemicals •Used for dehydration

■ Sulphuric acid is called the King of chemicals.

Industrial preparation of Sulphuric acid

Sulphuric acid is industrially prepared by Contact process. Let us see the various stages in the process:

Stage 1: Sulphur is burnt in air to produce sulphurdioxide (SO₂). The equation is:
S + O₂ → SO₂. This is a balanced equation.

Stage 2: This SO₂ is allowed to combine with more oxygen. High temperature is required for this process. Also vanadium pentoxide (V₂O₅) is used as a catalyst. The product is sulphur trioxide (SO₃).

Let us write the equation:

Reactants:

    ♦ Sulphur dioxide. One molecule is SO₂

    ♦ Oxygen. One molecule is O₂

Products:

    ♦ Sulphur trioxide. One molecule is SO₃

• So skeletal equation is:

SO₂ + O₂ → SO₃. This is not a balanced equation. The steps for writing the balanced equation are shown below:
Step 1: SO₂ + O₂ → SO₃
Step 2: 2SO₂ + O₂ → 2SO₃

	Reactants			Products
	S	O		S	O
Step 1	1	4		1	3
Step 2	2	6		2	6

So the balanced equation is: 2SO₂ + O₂ → 2SO₃

Stage 3: SO₃ is dissolved in concentrated sulphuric acid. The product is H₂S₂O₇. It is called oleum. The equation is: SO₃ + H₂SO₄ → H₂S₂O₇

This is a balanced equation.
Stage 4: This Oleum is mixed with water to produce sulphuric acid of the required concentration. Let us write the equation:

Reactants:

    ♦ Oleum. One molecule is H₂S₂O₇. 

    ♦ Water. One molecule is H₂O

Products:

    ♦ Sulphuric acid. One molecule is H₂SO₄.

• So skeletal equation is:

H₂S₂O₇ + H₂O → H₂SO₄. This is not a balanced equation. The steps for writing the balanced equation are shown below:
Step 1: H₂S₂O₇ + H₂O → H₂SO₄
Step 2: H₂S₂O₇ + H₂O → 2H₂SO₄

	Reactants				Products
	H	S	O		H	S	O
Step 1	4	2	8		2	1	4
Step 2	4	2	8		4	2	8

So the balanced equation is: H₂S₂O₇ + H₂O → 2H₂SO₄.

• Theoretically, there is no need for stages 3 and 4. Let us see why this is so:

(i) At the end of stage 2, we get sulphur trioxide (SO₃). When this is dissolved in water, we should get sulphuric acid. The equation is: SO₃ + H₂O → H₂SO₄. This is a balanced equation.

(ii) But there are practical difficulties in carrying out this reaction. The dissolution of SO₃ in water is an exothermic process. 
(iii) The H₂SO₄ which is initially formed, becomes fine smog like particles. These particles hinder further dissolution. So the complete dissolution according to the equation does not take place. 
(iv) So, in the industrial production, we go through all the four stages. 

The stages can be presented in the form of a flow chart given below:

Physical properties of sulphuric acid

The important physical properties of H₂SO₄ are: •Colourless •viscous in nature •Highly corrosive •Denser than water •Water soluble

Chemical properties of sulphuric acid

Affinity towards water: 
• Take 5 ml of water in a test tube and slowly add concentrated sulphuric acid to it. 
• If we touch the bottom of the test tube, we can feel the heat. 
• This is because, the reaction between water and H₂SO₄ is highly exothermic.

■ While diluting sulphuric acid, the acid should be added to water in very small quantities, while stirring it. We should not do the other way round. That is., we should not add water to the acid. If we do it, there will be spurting and may cause burns to our body.

Concentrated H₂SO₄ has great affinity towards water. It has the ability to absorb moisture from the substances that come in contact with it. It will absorb water even if the water is chemically combined with the substance. 
Let us see an example:

1. Take a little sugar in a watch glass. Add a few drops of concentrated sulphuric acid to it. A reaction takes place and a black residue is left in the watch glass. Let us write the equation: 
C₁₂H₂₂O₁₁  → 12C + 11H₂O. This is a balanced equation. 
2. The 12 carbon atoms present in the left side as 'C₁₂' is present as '12C' on the right side. 
3. Now look at the other elements: Oxygen and hydrogen: 
    ♦ In the left side, there are 22 Hydrogen atoms and 11 Oxygen atoms. 
    ♦ So we can say: In a sugar molecule, the number of H atoms and number of O atoms are in the ratio 22:11 which is same as 2:1. 
    ♦ This is the same 'ratio of hydrogen and oxygen' in water. 
   ♦ On the right side, we see that 11 molecules of water are formed. Those 11 water molecules contain the same 22 H and 11 O on the left  side 
4. So, when sulphuric acid reacts with a substance, the H and O atoms get separated from that substance. Those separated atoms combine together to form water.
5. The residue that remains, after the removal of water is carbon. 

■ After the separation, we say that the substance is dehydrated. Thus, H₂SO₄ has the ability to dehydrate substances.

• Similar to sugar, glucose (C₆H₁₂O₆) and fructose (C₆H₁₂O₆) also reacts with H₂SO₄ and get dehydrated. [Note that, glucose and fructose have the same chemical formula. But the arrangement of atoms is different]
• Number of H atoms : Number of O atoms ratio is the same 2:1 in glucose and fructose also.

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The examples of sugar, glucose and fructose that we saw above are those in which, water is present in a chemically combined form. 
Another case is there, in which water is not chemically combined. But it is present in crystalline form. That is., molecules of water are attached to the molecules of the substance by physical bonding. 
• Copper sulphate is an example. 
    ♦ It contains water in crystalline form. The chemical formula is: CuSO₄.5H₂O.
    ♦ It is called hydrous copper sulphate. It is blue in colour 
    ♦ The H₂SO₄ reacts with CuSO₄.5H₂O and causes dehydration.
    ♦ After dehydration, it is called 'anhydrous copper sulphate' (CuSO₄). It is white in colour.

In the previous section, we saw the application of quick lime (CaO) as a drying agent in the preparation of ammonia. Now we see that H₂SO₄ also can absorb water. So it can also be used as a drying agent. In fact, it is used as a drying agent in the manufacture of chlorine (Cl₂), Sulphur dioxide (SO₂) and Hydrogen chloride (HCl). We saw the case of Cl₂ in a previous section (see fig.5.4). 

But it cannot be used as a drying agent in the manufacture of ammonia because of the following reason: H₂SO₄ is an acid, while ammonia is alkaline. The two will react together.

Reaction of sulphuric acid with salts

• We have seen that salts are formed when acids and alkalies react together. 
• Now, if we take such a salt, and allow it to react with H₂SO₄, we will get back the acid. Let us see an example:

• We know that when the acid HCl react with the alkali NaOH, we get the salt NaCl. 
• Now, this NaCl is allowed to react with H₂SO₄. We will get back the acid HCl. The reaction is as shown below:
NaCl + H₂SO₄ → NaHSO₄ + HCl. This is a balanced equation.

• Another example:
Sulphuric acid reacts with potassium nitrate (KNO₃). Nitric acid is obtained as a product. The equation is: 
KNO₃ + H₂SO₄ → KHSO₄ + HNO₃. This is a balanced equation.

■So we can say: Concentrated sulphuric acid can displace acids from their salts. This method is employed in the manufacture of hydrochloric acid, nitric acid etc., 

Oxidising nature of sulphuric acid

We have seen oxidation and reduction in a previous chapter. Details here. Based on that discussion, we can write the 'oxidation number' of any element in a compound.

Consider the following reaction:

C + H₂SO₄ → CO₂ + 2H₂O + 2SO₂.

This is the reaction between carbon and sulphuric acid. The reaction is carried out as follows:

• Concentrated sulphuric acid is added to a small quantity of carbon in a test tube and heated. 

From the equation, we can see that, the products are CO₂, H₂O and SO₂. Let us write the oxidation number of each element on both sides of the equation:

C⁰ + H₂⁺¹S⁺⁶O₄^-2 → C⁺⁴O₂^-2 + 2H₂⁺¹O^-2 + 2S⁺⁴O₂^-2.

From this we can see that, the oxidation state of carbon increased from 0 to +4. That means, the carbon is oxidised. The sulphuric acid acts as the oxidising agent.

Another example:

Cu + 2H₂SO₄ → CuSO₄ + 2H₂O + SO₂.

Let us write the oxidation number of each element on both sides of the equation:

Cu⁰ + 2H₂⁺¹S⁺⁶O₄^-2 → Cu⁺²S⁺⁶O₄^-2 + 2H₂⁺¹O^-2 + S⁺⁴O₂^-2.

From this we can see that, the oxidation state of copper increased from 0 to +2. That means, the copper is oxidised. The sulphuric acid acts as the oxidising agent.

In the above examples, sulphuric acid oxidised carbon (a non-metal) and copper (a metal). So we get a common property of sulphuric acid:

■ Sulphuric acid oxidises some metals and non-metals

Identification of sulphate salts

• We know that, when hydrochloric acid reacts with alkalies, we get chlorides.Example:

HCl + NaOH → NaCl + H₂O.

• HCl also reacts with metals to form chlorides. Example:

Zn + 2HCl  → ZnCl₂ + H₂.

■ We can write many examples where hydrochloric acid takes part in reactions to give chlorides. 

■ In the same way, sulphuric acid also takes part in chemical reactions to give sulphates. 

Let us see some examples:

• Sulphuric acid reacts with the alkali sodium hydroxide (NaOH) to give sodium sulphate. The equation is:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O.

• Sulphuric acid reacts with the metal zinc to give zinc sulphate. The equation is:

Zn + H₂SO₄ → ZnSO₄ + H₂O.

• Sulphuric acid reacts with calcium carbonate to give calcium sulphate. The equation is:

CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂.

So we saw some methods by which sulphates are formed. Let us now see how these sulphate salts can be identified:

1. Prepare an aqueous solution of the given salt. We want to check whether this salt is sulphate salt or not.

2. Add some barium chloride solution to it.

3. If it is a sulphate salt, white precipitate of barium sulphate will be formed. The equation is:

BaCl₂ + Na₂SO₄ → BaSO₄↓ +  2NaCl.

4. Can we confirm that it is a sulphate salt? That is., will any other salt give a white precipitate when barium chloride is added?

In fact, like sulphate salts, carbonate salts also give white precipitate. The equation is:
BaCl₂ + Na₂CO₃ → BaCO₃↓ +  2NaCl.

5. So, when we test a given salt by adding barium chloride, and if we get a white precipitate, the given salt can either be sulphate or carbonate. We want to know which one.
6. For that, add concentrated HCl to the precipitate.
7. If the precipitate dissolves in HCl and bubbles of CO₂ are formed, it is a carbonate.
8. If the precipitate does not dissolve, it is a sulphate.

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In the next section, we will see Hydrogen chloride. 

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Chapter 7 - Compounds of Non-metals - Ammonia

In the previous section, we completed the discussion on acids, alkalies and salts. In this section we will see Compounds of Non-metals.

Ammonia
We know that, nitrogen is an important element required for the growth of plants. Details here. We have also seen that plants do not get the required quantities of nitrogen by natural means alone. We have to supply nitrogen through fertilisers. So we have to produce large quantities of such fertilisers. In other words, we have to produce such fertilisers industrially. For the industrial production of nitrogenous fertilisers, we have to produce ammonia first.

Preparation of ammonia in the laboratory

Fig.7.1 shows the diagram for the preparation of ammonia in the laboratory.

Fig.7.1

1. Ammonium chloride (NH₄Cl) and Calcium hydroxide (Ca(OH)₂) are mixed well and heated.
Let us write the equation:

Reactants:

    ♦ Ammonium chloride. One molecule is NH₄Cl. 

    ♦ Calcium hydroxide. One molecule is Ca(OH)₂.

Products:

    ♦ Calcium chloride. One molecule is CaCl₂.

    ♦ Water. One molecule is H₂O
    ♦ Ammonia. One molecule is NH₃

• So skeletal equation is:

NH₄Cl + Ca(OH)₂ → CaCl₂ + H₂O + NH₃. This is not a balanced equation. The steps for writing the balanced equation are shown below:
Step 1: NH₄Cl + Ca(OH)₂ → CaCl₂ + H₂O + NH₃
Step 2: 2NH₄Cl + Ca(OH)₂ → CaCl₂ + H₂O + NH₃
Step 3: 2NH₄Cl + Ca(OH)₂ → CaCl₂ + H₂O + 2NH₃
Step 4: 2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2H₂O + NH₃ 

	Reactants						Products
	N	H	Cl	Ca	O		N	H	Cl	Ca	O
Step 1	1	6	1	1	2		1	5	2	1	1
Step 2	2	10	2	1	2		1	5	2	1	1
Step 3	2	10	2	1	2		2	8	2	1	1
Step 4	2	10	2	1	2		2	10	2	1	2

So the balanced equation is: 2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2H₂O + NH₃ 

2. From the equation, we can see that water is also formed as a product. So we have to prevent the newly formed ammonia (NH₃) from dissolving in the newly formed water. 
3. For that, the test tube is kept in a slanting position. So that, the newly formed water will not collect at the bottom of the test tube. 
4. Still, there will be water in the vapour form. This will move out through the delivery tube, along with the NH₃. That means, the gas which comes out of the delivery tube will be a mixture of water vapour and ammonia. 
5. We have to remove the water vapour. For that, the delivery tube enters a drying tower. The upper chamber of the drying tower contains quick lime (CaO). This CaO, which is a drying agent, will absorb the water vapour. So the delivery tube that comes out of the drying tower will contain NH₃ only. 
6. This delivery tube enters a gas jar. Thus the ammonia gas is collected in the gas jar. We can see that the gas jar is kept in an inverted position. This is because, ammonia gas is lighter than air. That is., the density of ammonia is lesser than the density of air. So it will rise up. This rising ammonia will displace the air present in the jar, and will occupy it’s top position in the inverted jar.

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Drying agent

Drying agents are substances capable of absorbing moisture from substances. After the absorbtion, ths substances will become ‘dry’. In the above experiment, CaO is used as the drying agent. CaO is alkaline. The ammonia is also alkaline. The two will not react with each other, and so, the ammonia comes out of the drying tower.

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Ammonia is highly soluble in water. This can be proved using the 'fountain experiment'. The arrangement is shown in the fig.7.2 below: 

Fig.7.2

• The flask which is placed in inverted position is filled with ammonia gas. A jet tube enters the flask from the bottom. The bottom end of the jet tube is dipped in water contained in a beaker. To this water, some phenolphthalein is already added.

• A few drops of water is added into the flask using the syringe. We can see a fountain of pink water in the flask. How is this fountain formed? Why is it pink coloured? Let us analyse:

1. When a few drops of water is added into the flask using the syringe, Some of the ammonia gas dissolves in that water. This will create some vacuum in the flask. So the out side atmospheric pressure will be greater than the pressure inside the flask. This atmospheric pressure will push down on the water in the beaker. So the water rushes up through the jet tube. 
2. When more water enters the flask in this way, more ammonia gas will dissolve in that water. Thus new vacuum is created. Because of this additional vacuum, the atmospheric pressure will again push down on the water in the beaker. 
3. This continues as a cyclic process. Thus a fountain will be formed. The cycle will continue until all the ammonia is dissolved. When all the ammonia is dissolved, there will not be any formation of 'new vacuum'.

■ Now we will see the reason for the pink colour: The ammonia gas present in the flask dissolves in the water which rushes into the flask. So the water becomes a solution. A ‘solution of ammonia in water’. This solution is alkaline in nature. So the phenolphthalein that is already present, will turn pink. Thus we get a pink fountain.
A video showing the demonstration can be seen here.

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This experiment shows that ammonia gas is readily soluble in water. The equation is:

NH₃ + H₂O → NH₄OH. This is a balanced equation.

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A highly concentrated solution of ammonia in water is called liquor ammonia.

Ammonia gas can be easily liquefied by applying pressure. Liquefied ammonia is called liquid ammonia.

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We know that, ammonia is alkaline in nature. So let us see it’s reaction with an acid:

■ Introduce the tip of a glass rod which is dipped in concentrated hydrochloric acid, into a glass jar filled with ammonia. We can see the formation of thick white fumes. This is due to the formation of ammonium chloride (NH₄Cl). Let us write the equation:
NH₃ + HCl → NH₄Cl
This is a balanced equation
• The white fumes are caused by small white particles of NH₄Cl

■ In this way, ammonia reacts with acids yielding ammonium salts, which are used as chemical fertilisers

Properties of ammonia

• It has no colour

• It has a pungent smell

• It is alkaline in nature

• It is highly soluble in water

• It is lighter than air

Identification of ammonia gas

We can use the following tests:

1. The basic test is by the smell. Ammonia has a characteristic pungent smell.

2. Ammonia turns wet red litmus paper into blue colour, showing it’s alkaline property

3. When the tip of a glass rod which is dipped in concentrated hydrochloric acid is introduced into a glass jar filled with ammonia, thick white fumes are formed.

Identification of ammonium salts

1. Prepare a solution of the given ammonium salt. Take 5 ml of nesslers reagent in a test tube. 
2. Add a few drops of the prepared salt solution into this. 
3. If a brownish orange precipitate is formed, then the given salt is a salt of ammonia.

Industrial preparation of Ammonia

1. Nitrogen and hydrogen are taken in the ratio 1:3. This ratio is obvious because, in the final product which is ammonia (NH₃), There are 3 atoms of hydrogen for every one atom of nitrogen.
2. They are made to react with each other. Very high temperature and pressure is required for the reaction to take place. This is because, nitrogen does not take part in reactions very easily, due to it's triple bond.
3. Spongy iron is used as a catalyst.
4. This is known as the Haber process. The equation of the reaction is:
N₂ + 3H₂ → 2NH₃. This is a balanced equation.

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In the next section, we will see sulphuric acid. 

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