Chapter 6.4 - Chemical formulae of Salts

In the previous section, we saw how to predict the name of the salt that will be formed from a neutralisation reaction. We have also seen how to write it's chemical formula. In this section we will see a few more solved examples.

Solved example 6.3
Some cations and anions are given below:
Cations: • Ca²⁺ (Calcium ion)    • NH₄⁺ (Ammonium ion)
Anions: • Cl^- (Chloride ion)    • SO₄^2- (Sulphate ion)    • PO₄^3- (Phosphate ion)
Write the chemical formulae of all the salts possible by combining them
Solution:
In this problem, we do not have to split the acid and alkali into cations and anions. They are already given separately
Case 1: Combination between Ca²⁺ and Cl^-.
1. Assemble the cation and anion with cation first: Ca²⁺  Cl^- .
2. Do the interchanging:
• Number of  Ca²⁺ ions = Number of charges in Cl^- ion = 1 
• Number of Cl^- ions = Number of charges in Ca²⁺ ion = 2
3. So chemical formula of the final salt is: CaCl₂.
Case 2: Combination between Ca²⁺ and SO₄^2-.
1. Assemble the cation and anion with cation first: Ca²⁺  SO₄^2- .
2. Do the interchanging:
• Number of  Ca²⁺ ions = Number of charges in SO₄^2- ion = 2 
• Number of SO₄^2- ions = Number of charges in Ca²⁺ ion = 2
3. So chemical formula of the final salt is: Ca₂(SO₄)₂.
4. Here, the subscripts (2 and 2) have a common factor. The common factor is '2'. In such cases we must divide the subscripts by the common factor. We get 2 ÷ 2 = 1
5. We can write: Ca₁(SO₄)₁. When the subscript is '1', it is not usually written.
6. So final chemical formula is CaSO₄.
Case 3: Combination between Ca²⁺ and PO₄^3-.
1. Assemble the cation and anion with cation first: Ca²⁺  PO₄^3- .
2. Do the interchanging:
• Number of  Ca²⁺ ions = Number of charges in PO₄^3- ion = 3 
• Number of PO₄^3- ions = Number of charges in Ca²⁺ ion = 2
3. So chemical formula of the final salt is: Ca₃(PO₄)₂.
Case 4: Combination between NH₄⁺ and Cl^-.
1. Assemble the cation and anion with cation first: NH₄⁺  Cl^- .
2. Do the interchanging:
• Number of  NH₄⁺ ions = Number of charges in Cl^- ion = 1
• Number of Cl^- ions = Number of charges in NH₄⁺ ion = 1
3. So chemical formula of the final salt is: (NH₄)₁(Cl)₁.
4. When the subscript is '1', it is not usually written.
5. So final chemical formula is NH₄Cl.
Case 5: Combination between NH₄⁺ and SO₄^2-.
1. Assemble the cation and anion with cation first: NH₄⁺  SO₄^2- .
2. Do the interchanging:
• Number of  NH₄⁺ ions = Number of charges in SO₄^2- ion = 2
• Number of SO₄^2- ions = Number of charges in NH₄⁺ ion = 1
3. So chemical formula of the final salt is: (NH₄)₂(SO₄)₁.
4. When the subscript is '1', it is not usually written.
5. So final chemical formula is (NH₄)₂SO₄.
Case 6: Combination between NH₄⁺ and PO₄^3-.
1. Assemble the cation and anion with cation first: NH₄⁺  PO₄^3- .
2. Do the interchanging:
• Number of  NH₄⁺ ions = Number of charges in PO₄^3- ion = 3
• Number of PO₄^3- ions = Number of charges in NH₄⁺ ion = 1
3. So chemical formula of the final salt is: (NH₄)₃(PO₄)₁.
4. When the subscript is '1', it is not usually written.
5. So final chemical formula is (NH₄)₃PO₄

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Uses of salts

We come across salts on many occassions in our day to day life. Some of them which occur naturally are listed below:

■ Sodium chloride (NaCl). It's common name is table salt.
• It is mainly used for cooking purposes. 
• Another important use is in the making of freezing mixtures 

■ Potassium chloride (KCl). It's common name is Sylvite/Muriate of potash. It occurs naturally at some places. 
• Industrial production of KCl is done using naturally occuring sylvite. 
• It is used as a fertilizer. It is also used in the laboratory for various experiments.
■ Potassium nitrate (KNO₃). It is one of the several nitrogen containing compounds. These nitrogen containing compounds are collectively called as saltpetre. 
• It's main use is in the making of fertilizers. 
■ Sodium nitrate (NaNO₃). It's common name is Chile salt peter or Peru saltpetre. These names are derived because of the vast deposits of NaNO₃ compounds in the atacama desert in Chile and Peru. 
• It is mainly used in the making of fertilizers
■ Calcium sulphate (CaSO₄). It is a hygroscopic substance. That means, it has the ability to attract and hold water molecules from the surroundings. 
• So it is used as a desiccant. A desiccant is a substance that is used in certain containers to keep it's contents dry
• When it reacts attracts and reacts with water, it forms a hydrate (CaSO₄.2H₂O). This is known as gypsum. It occurs naturally at some places.
• Another hydrate is commonly known as plaster of paris. It is used for moulding various objects
■ Calcium carbonate (CaCO₃). It occurs naturally as limestones and in the shells or marine organisms. It's main use in the manufacture of cement.

Now we will see some salts which are made artificially:
■ Sodium carbonate (Na₂CO₃.10H₂O). It's common name is washing soda. It's main uses are in the manufacture of soaps, detergents, glass etc.,
■ Sodium bicarbonate (NaHCO₃). It's common name is baking soda. It's main uses are in the making of antacids, fire extinguishers etc., It is also used for baking purposes.
■ Copper sulphate (CuSO₄.5H₂O). It's common name is blue vitriol. It's main use is in the preparation of fungicides. It is also used in the laboratory for various experiments.

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We have completed the discussion on acids, alkalies and salts. Now we will see some solved examples in general from this chapter

Solved example 6.4 

Complete the chemical equations for the following ionisation reactions:

KCl → K⁺ + Cl^-.

HNO₃ → H⁺ + NO₃^-

Mg(OH)₂ → Mg²⁺ + 2(OH)^-

H₂SO₄ → 2H⁺ + SO₄^2-

NH₄Cl → NH₄⁺ + Cl^-

CaSO₄ → Ca²⁺ + SO₄^2-.
Solved example 6.5
Identify the symbols of ions given below, and write their names
SO₃^2-, NO₃^-, HCO₃^-, OH^-, CO₃^2-, HSO₄^-
Carbonate - CO₃^2-
Bisulphate - HSO₄^-
Sulphite - SO₃^2-
Nitrate - NO₃^-
Hydroxide - OH^-
Bicarbonate - HCO₃^-

Solved example 6.6
A little distilled water is taken in a beaker
(a) What is the pH value of distilled water?
(b) What happens to the pH value when the following substances are added to the water in the beaker? Justify your answer.
(i) Caustic soda
(ii) Vinegar
Solution:
(a) Distilled water is neutral. It is neither acidic nor alkaline. So it's pH value is 7
(b.i) Caustic soda is sodium hydroxide. It is an alkali. When it is added to water concentration of (OH)^- ions will increase. That means, the solution will become alkaline. So the pH value will increase.
(b.ii) Vinegar is acetic acid. When it is added to water concentration of H⁺ ions will increase. That means, the solution will become acidic. So the pH value will decrease.

Solved example 6.7
Some salts are given in column A. Their chemical formulae and uses are given in column B and column C irregularly. Match the columns by identifying the correct chemical formulae and uses of the salts.

A	B	C
Salt	Chemical formula	Use
Washing soda	CuSO4.5H2O	Fire extinguisher
Gypsum	NaHCO3	Fungicide
Blue vitriol	Na2CO3.10H2O	Cement manufacture
Baking soda	CaSO4.2H2O	Glass manufacture

Solution:
The corrected table is given below:

A	B	C
Salt	Chemical formula	Use
Washing soda	Na2CO3.10H2O	Glass manufacture
Gypsum	CaSO4.2H2O	Cement manufacture
Blue vitriol	CuSO4.5H2O	Fungicide
Baking soda	NaHCO3	Fire extinguisher

Solved example 6.8
The pH values of some substances are given in the table. Analyse the table and answer the questions that follow.

Substance	pH value
Vinegar	4.2
Lime water	10.5
Milk	6.4
Water	7
Tooth paste	8.7
Blood	7.36

(a) Is blood acidic or alkaline in nature
(b) The pH value of pure milk is 6.4. Does the pH value increase or decrease when milk changes to curd?. Justify your answer
(c) Among the substances given in the table,
(i) Which one is strongly alkaline?
(ii) Which one has weak acidic nature?
Solution:
(a) The pH value of blood is given as 7.36. This is greater than 7. So blood is alkaline in nature
(b) The pH value of pure milk is given as 6.4. This is less than 7. So pure milk is acidic. When it changes to curd, the pH value will be come less than 6.4. This is because curd is acidic
(c) i. Substances which have pH greater than 7 are all alkaline. So lime water, tooth paste, and blood are alkaline. Among them, lime water has the greatest difference from 7. So it is the most alkaline in the given table
ii. Substances which have pH less than 7 are all acidic. So vinegar and milk are acidic. Among them, milk has the smallest difference from 7. So it is the weakest acidic nature in the given table 
More precisely it can be written as follows:
• Vinegar has pH value 4.2. So [H⁺] = 10^-4.2. That means there are 10^-4.2 mols of H⁺ ions in 1 litre of vinegar
• Milk has pH value 6.4. So [H⁺] = 10^-6.4. That means there are 10^-6.4 mols of H⁺ ions in 1 litre of milk
• 10^-6.4 < 10^-4.2. So the number of H⁺ ions is less in Milk. So it is a weaker acid than vinegar.

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In the next chapter, we will see compounds of non-metals. 

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Chapter 6.3 - Salts formed from Neutralisation reaction

In the previous section, we saw the details about acidity and alkalinity. We also saw the pH scale. In this section we will see some practical situations where acidity or alkalinity will have to be considered. We will also learn more about salts.

In agriculture, soils with acidic nature are suitable for some crops. While soils with alkaline nature are suitable for some other crops. So it is important to test the soils before beginning the cultivation. Sample of soil is taken in a special manner, prescribed by the agricultural officer. This sample is mixed with distilled water. The mixture thus obtained is kept undisturbed for some time. The soil particles will settle down. The sample for testing is taken from the clear portion at the top. The pH value of this sample is determined. From the pH value, the acidity/alkalinity of the soil can be calculated. Based on this result, the officer will prescribe the suitable crop that can be planted in that soil. He can also determine the ‘quantity of acidity or alkalinity’. So he can prescribe whether any treatments have to be done to the soil, to make it suitable for cultivation.

Some times farmers spread powdered slaked lime. Slaked lime is Ca(OH)₂. We have seen it’s preparation when we studied the basics about alkalies at the beginning of a previous section here. It is used in soils which are highly acidic in nature. When the Ca(OH)₂ comes in contact with the rain water, (OH)^- ions will be released. These (OH)^- ions will neutralise H⁺ ions. Thus the acidity of the soil will be reduced.

Similarly, when the alkalinity of the soil is high, non-metallic oxides like SO₂ is added. These oxides produce acids when they come in contact with rain water. That means, H⁺ ions will be produced. They will neutralise the excess (OH)^- ions present in the soil.

When the acid level in the stomach increases, we feel acidity. Medicines used for reducing the acid level in stomach are called antacids. They are alkaline substances. They neutralise the excess acids.

Salts

• We have seen that, when an acid and an alkali react together, we get a salt and water. We have seen the example of the reaction between HCl and NaOH. The products are NaCl and H₂O. The equation is: NaOH + HCl → NaCl + H₂O

• We can write the above equation in terms of ions: Na⁺OH^- + H⁺Cl^- → Na⁺Cl^- + H⁺(OH)^-

We can note the following points:

1. The positive ion Na⁺ in the resulting salt, comes from the alkali.

2. The negative ion Cl^- in the resulting salt comes from the acid.

■ In fact, this is a common property in all neutralisation reactions. We can write it as:

• The positive ion in the resulting salt, comes from the alkali.

• The negative ion in the resulting salt comes from the acid.

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• Salts are usually ionic compounds. They dissociate into positive and negative ions when dissolved in water or on fusion (fusion is another term for ‘melting’).

• The positive ions are called cations, and the negative ions are called anions
The following table shows some cations and anions:

Name of Cation	Symbol		Name of Anion	Symbol
Potassium ion	K+		Hydroxide ion	OH-
Zinc ion	Zn2+		Carbonate ion	CO32-
Ferrous ion	Fe2+		Bicarbonate	HCO3-
Ferric ion	Fe3+		Nitrate ion	NO3-
Cuprus ion	Cu+		Sulphate ion	SO42-
Cupric ion	Cu2+		Bisulphate ion	HSO4-
Ammonium ion	NH4+		Phosphate ion	PO43-
Manganus ion	Mn2+		Dihydrogen phosphate ion	H2PO4-
Magnesium ion	Mg2+

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• If we know the reactants in a neutralisation reaction:

    ♦ We will be able to predict the name of the salt which will be formed

    ♦ We will also be able to write the chemical formula of the salt which will be formed

For doing the above two things, we must first do a careful analysis:
1. Our aim is to obtain the chemical formula of the salt
2. The data that we have is: Names of the reactants (acid and alkali)
3. We have seen that the cation in the final salt, comes from the alkali. So the first step is to split the alkali into cation and anion.
• From that, take out the cation
• Let the cation be represented by the letter 'C'. 
• A cation will be having positive charge.
• Let the number of positive charges be 'x'. 
• So our required cation is C^x+

4. We have seen that the anion in the final salt, comes from the acid. So the second step is to split the acid into cation and anion.
• From that, take out the anion
• Let the anion be represented by the letter 'A'. 
• An anion will be having negative charge. 
• Let the number of negative charges be 'y'. 
• So our required anion is A^y-.

5. Now assemble the cation and anion together. The cation should be written first. So we get:
C^x+   A^y-. 
6. We know that, the final salt is electrically neutral. That means, the net charge is zero. So x must be equal to y. 
7. This may not be always possible. 
For example, when we assemble Fe3+ and SO₄2-, x= 3 and y = 2. So the charges will not neutralise completely.
8. In such cases, we must assemble 'suitable numbers' of cations and anions.
Let 'm' be the number of cations required
Let 'n' be the number of anions required
Then (5) will become: C_m^x+   A_n^y- .   
9. Now total number of positive charges = mx, and total number of negative charges = ny
10. These must be equal. So we get mx = ny
11. To satisfy this equation, 
• m must be equal to y
• n must be equal to x
• Then we will get yx = xy, and thus, (10) will be satisfied
12. Thus we find that, there is a sort of 'interchanging'. 
• The number of cations required (m) is the number of charges in anion (y)
• The number of anions required (n) is the number of charges in cation (x)
■ When we give the above required number of ions, we will get a neutral salt. The solved example given below will demonstrate the procedure

Solved example 6.1
In the neutralisation reaction between the alkali Magnesium hydroxide (Mg(OH)₂) and the Hydrochloric acid (HCl), write the chemical formula of the salt formed. Also write the balanced equation of the neutralisation reaction.
Solution:
1. The cation in the final salt, comes from the alkali. So the first step is to split the alkali into cation and anion: Mg(OH)₂ → Mg²⁺ + 2(OH)^- .
2. The anion in the final salt, comes from the acid. So the second step is to split the acid into cation and anion: HCl → H⁺ + Cl^-.
3. Assemble the cation and anion with cation first: Mg²⁺  Cl^- .
4. Do the interchanging:
• Number of  Mg²⁺ ions = Number of charges in Cl^- ion = 1 
• Number of Cl^- ions = Number of charges in Mg²⁺ ion = 2
5. So chemical formula of the final salt is: MgCl₂.

Balanced equation for the neutralisation reaction:

Reactants:

    ♦ Magnesium hydroxide. One molecule is Mg(OH)₂. 

    ♦ Hydrochloric acid. One molecule is HCl.

Products:

    ♦ Magnesium chloride. One molecule is MgCl₂.

    ♦ Water. One molecule is H₂O

• So skeletal equation is:

Mg(OH)₂ + HCl → MgCl₂ + H₂O. This is not a balanced equation. The steps for writing the balanced equation are shown below:
Step 1: Mg(OH)₂ + HCl → MgCl₂ + H₂O
Step 2: Mg(OH)₂ + 2HCl → MgCl₂ + H₂O
Step 3: Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O

	Reactants					Products
	Mg	O	H	Cl		Mg	O	H	Cl
Step 1	1	2	3	1		1	1	2	2
Step 2	1	2	4	2		1	1	2	2
Step 3	1	2	4	2		1	2	4	2

So the balanced equation is: Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O

Solved example 6.2

In the neutralisation reaction between the alkali Magnesium hydroxide (Mg(OH)₂) and the Sulphuric acid (H₂SO₄), write the chemical formula of the salt formed. Also write the balanced equation of the neutralisation reaction.
Solution:
1. The cation in the final salt, comes from the alkali. So the first step is to split the alkali into cation and anion: Mg(OH)₂ → Mg²⁺ + 2(OH)^- .
2. The anion in the final salt, comes from the acid. So the second step is to split the acid into cation and anion: H₂SO₄ → 2H⁺ + SO₄^2-.
3. Assemble the cation and anion with cation first: Mg²⁺  SO₄^2- .
4. Do the interchanging:
• Number of Mg²⁺ ions = Number of charges in SO₄^2- ion = 2 
• Number of SO₄^2- ions = Number of charges in Mg²⁺ ion = 2
5. So chemical formula of the final salt is: Mg₂(SO₄)₂.
6. Here, the subscripts (2 and 2) have a common factor. The common factor is '2'. In such cases we must divide the subscript by the common factor. We get 2 ÷ 2 = 1
7. We can write: Mg₁(SO₄)₁. When the subscript is '1', it is not usually written.
8. So final chemical formula is MgSO₄.

Balanced equation for the neutralisation reaction:

Reactants:

    ♦ Magnesium hydroxide. One molecule is Mg(OH)₂. 

    ♦ Sulphuric acid. One molecule is H₂SO₄.

Products:

    ♦ Magnesium sulphate. One molecule is MgSO₄.

    ♦ Water. One molecule is H₂O

• So skeletal equation is:

Mg(OH)₂ + H₂SO₄ → MgSO₄ + H₂O. This is not a balanced equation. The steps for writing the balanced equation are shown below:
Step 1: Mg(OH)₂ + H₂SO₄ → MgSO₄ + H₂O
Step 2: Mg(OH)₂ + H₂SO₄ → MgSO₄ + 2H₂O

	Reactants					Products
	Mg	O	H	S		Mg	O	H	S
Step 1	1	6	4	1		1	5	2	1
Step 2	1	6	4	1		1	6	4	1

So the balanced equation is:
Step 2: Mg(OH)₂ + H₂SO₄ → MgSO₄ + 2H₂O

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In the next section, we will see a few more solved examples. 

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