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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Chapter 6.2 - The pH scale

In the previous section, we saw that hydronium ions are responsible for the acidic properties. We also saw mono, di and tri basic acids. In this section we will learn about alkalies. Later in this section, we will learn about the pH scale.

Common factor in Alkalies

Let us write the chemical formula of some commonly known alkalies:

Name of alkali	Chemical formula
Sodium hydroxide	NaOH
Calcium hydroxide	Ca(OH)2
Ammonium hydroxide	NH4OH
Magnesium hydroxide	Mg(OH)2
Potassium hydroxide	KOH

• In the above list, we find that, a particular 'group' of two elements is present in all alkalies. The group we see is: 'OH', consisting of oxygen and hydrogen. This is called the hydroxyl group.

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The 'hydroxyl group' is the common factor present in all alkalies, and is responsible for the common properties of alkalies

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When an alkali is kept in a bottle, it has no effect. But when it comes into contact with other substances, it will show it's true nature. For example, when an alkali is dissolved in water, it will ionise. This ionisation process will liberate (OH)^- ions. These ions are called hydroxide ions, and are responsible for the alkaline nature. Some examples of this type of ionisation are given below:

NaOH → Na⁺ + (OH)^-

Ca(OH)₂ → Ca²⁺ + 2(OH)^-

KOH → K⁺ + (OH)^-

NH₄OH → NH₄⁺ + (OH)^-

Substances that increase the concentration of (OH)^- ions in it's water solution are alkalies.

Neutralisation reaction

We have seen the preliminary details about acids and alkalies. We will now see what happens when these two react with each other.

We know that acids release H⁺ ions and alkalies release (OH)^- ions. When acids and alkalies react together, the H⁺ of acids and the (OH)^- of alkalies combine together to form water.

H⁺ + (OH)^-  → H₂O

This water is only one of the two products. The other product is a salt. Let us see an example:

Here the salt formed is sodium chloride (NaCl). Both the products are neutral:

• NaCl does not have any acidic or alkaline character. It is neutral

• H₂O does not have any acidic or alkaline character. It is neutral

So we see that, acid and alkali react together to form neutral products. Such a reaction is called Neutralisation reaction.

But a natural question arises:

■ Can we guarantee that acidic and alkaline characters are completely neutralised?

In other words: Is the resulting solution, completely neutral?

• The answer is that, for complete neutralisation, 'appropriate quantities' of acid and alkali should be taken. In simple terms we can say this:

    ♦ The number of H⁺ ions available must be sufficient to neutralise all the (OH)^- ions

    ♦ The number of (OH)^- ions available must be sufficient to neutralise all the H⁺ ions

If any of these ions is in greater quantity, complete neutralisation will not take place.

Let us do an experiment to demonstrate this:
1. Take 200 ml of dilute NaOH solution in a conical flask. Add two drops of phenolphthalein into this. Phenolphthalein is an indicator. It is pink in alkali. So when we add it, the solution becomes pink in colour. 

2. Take dilute HCl in a burette. Hold the conical flask below the burette and add HCl gradually. Mix the solution well by shaking the conical flask continuously. The arrangement is shown in fig.6.2 below:

Fig.6.2

3. Observe the colour change taking place in the NaOH solution. We can see that the pink colour is gradually decreasing. Continue adding HCl gradually. The pink colour also continues to decrease.

4. Then a stage will be reached in which the solution becomes nearly colourless. At that stage, reduce the addition of HCl. From now on, HCl should be added drop by drop. Continue shaking the conical flask. 

5. Stop adding the HCl, when the colour disappears completely with just one drop. We must note down the volume of HCl required to make the colour completely disappear. We can calculate this volume from the markings on the side of the burette.

What do we infer from the experiment?

• Phenolphthalein is an indicator. It is pink in alkalies. It indicates whether a solution is alkaline or not. 

• We added two drops of it to the NaOH solution. The solution turned pink. From that colour change, we know that the solution taken in the conical flask is alkaline in nature. 

• But when we add HCl, the H+ ions released from the HCl, reacts with the (OH)^- ions present in the conical flask. The number of free (OH)^- ions decreases. This decreases the alkaline nature of the solution. So the pink colour decreases. 

• When we add more HCl, more (OH)^- ions gets neutralised. This causes further decrease in the pink colour. 

• Finally, we reach a stage where the number of ions is so less that, they are neutralised by a single drop of HCl. So with the last added drop of HCl, we have a neutral solution in the conical flask.

Our experiment is complete. But we can do a few more trials:

■ To the neutral solution obtained in the conical flask, add some NaOH solution. We can see that the pink colour reappears. What is the reason?

Ans: In the neutral solution there is no H⁺ or (OH)^-. To this, we have added some NaOH. So new (OH)^- ions will get released from the newly added NaOH. Because of this new (OH)^- ions, the solution will become alkaline again. So the phenolphthalein will show pink colour.

■ Let us continue this trial: Add dil.HCl again, drop by drop. We can see that the pink colour disappears again. What is the reason?

Ans: The newly added HCl neutralise the effect of newly added NaOH.
■ A video demonstrating a similar experiment can be seen here. We will learn more details in higher classes.

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Now we will see a different situation in the experiment. Start the experiment from the beginning. That is., we are going to repeat the steps from (1):

1. So we have 200 ml of dilute NaOH solution in a conical flask. We are going to neutralise it using HCl from the burette. 

2. The only difference is that, this time, the HCl in the burette should not be dilute. It should be a mixture of dilute HCl and concentrated HCl. 

3. Complete all the steps as before. That is., stop the addition of HCl when the pink colour completely disappears with a single drop.

4. This time we can notice that, the volume of HCl required to neutralise 200 ml of NaOH is reduced.
5. This is because, a smaller volume of concentrated HCl is able to supply the required number of H⁺ ions to neutralise all the (OH)^- ions in the conical flask. That means, concentrated HCl has greater number of H⁺ ions than dilute HCl.

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pH Scale

So we have seen the details about the neutralisation reaction. In many scientific and engineering fields, we will encounter situations in which the substance given to us may be acidic or alkaline. In such situations, we must be able to do two things:

■ Determine whether the given substance is acidic or alkaline.

■ If it is acidic, determine how much acidity is present. If it is alkaline, determine how much alkalinity is present

• So it is clear that, being able to classify a substance as 'acidic' or 'alkaline' is not good enough. We must be able to determine 'how much' acidity or alkalinity is present also. 

• This is because, if there is a large number of H⁺ ions in a given sample, it will be more acidic. Similarly, if there is a large number of (OH)^- ions in a given sample, it will be more alkaline. So 'how much' is very important. Let us see how it is done:

• To determine 'how much', we need a scale. Just like:

    ♦ A scale of cm or m is used to measure 'how much' distance

    ♦ A scale of gram or kilogram is used to measure 'how much' weight

■ For our present case, we use the pH scale. To learn about this scale, we need to understand two terms: (i) mol (ii) logarithm. First we will take mol.

• 'Mol' is used to denote a number. It is similar to 'dozen'.

    ♦ 1 dozen apples = 12 apples
    ♦ 1 mol apples = 6.022 × 10²³ apples

That is a very large number of apples. But when we use 'mol' for counting particles like atoms, molecules, ions etc., it is not a very large quantity. For example, in just one gram of carbon-12, there will be 6.022 × 10²³ molecules of carbon-12. So mol is an appropriate unit for counting atoms, molecules, ions etc.,

In our present discussion, we are considering the number of H⁺ ions in a solution. Let us see how many ions will be present in a ‘standard solution’. A standard solution has to be specified so that, people all around the world will use that solution as a ‘standard’ or a ‘benchmark’. The bench mark that we use is pure distilled water. This water, at room temperature (25^o c), does not have any acidic or alkaline properties. But still, even at room temperature, there will be a small number of ions. At room temperature, 1 litre of distilled water will contain 10^-7 mols of H⁺ ions. Let us calculate how many is that in ordinary numbers:

• We know that 1 mol ions = 6.022 × 10²³ ions

• So 10^-7 mols = 10^-7 × 6.022 × 10²³ = 6.022 × 10¹⁶ ions

This 6.022 × 10¹⁶ is the actual number of ions. But we do not need this actual number. What we need is the number 'in terms of mols'. 10^-7 mols mentioned above is sufficient. So the information should be written as:

■ At room temperature, 1 litre of distilled water will contain 10^-7 mols of ions

• Note that, the number is for 1 litre. That is., per litre. So it is a ‘concentration’. We can say:

• The concentration of H⁺ ions in water is 10^-7 mols/litre

• Symbolically, this is represented as: [H⁺] = 10^-7 mols/litre

• We know that 10^-7 = 0.0000001. There are six zeros after the decimal point. It is not very convenient to write this every time. Neither is it convenient to write 10^-7. It would be better if we can use ordinary numbers. Thus comes the use of logarithm.

• Logarithm is a maths topic. For our present discussion, we need to know only some of it’s basic details:

• Consider the number 16. It can be written as 2⁴. That is., 16 = 2⁴

• In logarithm, we write this as: log ₂16 = 4

• This is read as: Logarithm of 16 to the base 2 is 4

• Another example:

243 = 3⁵ ⇒ log ₃243 = 5 

This is read as: Logarithm of 243 to the base 3 is 5

• Another example:

1000 = 10³ ⇒ log ₁₀1000 = 3 

This is read as: Logarithm of 1000 to the base 10 is 3

• Another example:

0.00001 = 10^-5 ⇒ log ₁₀(0.00001) = -5 

This is read as: Logarithm of 0.00001 to the base 10 is -5
■ So logarithm is another way of writing exponents

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• Our number is 0.0000001. This is the [H⁺]. It is equal to 10^-7

• Let us take logarithm:

0.0000001 = 10^-7 ⇒ log ₁₀(0.0000001) = -7

There we have it. A simple number: -7

• We can say this: Base 10 logarithm of the H⁺ ion concentration (represented as [H⁺] ) is -7

• But still there is a '-' sign. why not take 'negative of the logarithm'?

• Then we can say this: Negative of the base 10 logarithm of [H⁺] is 7. (∵ -(-7) = 7) 

■ So '7' represents a neutral solution. Because it corresponds to distilled water.

■ What if the solution that we take is a little acidic?

• If it is acidic, the concentration of H⁺ ions will be greater than 10^-7 

• That is., [H⁺] will be greater than 10^-7. It will take values like 10^-6, 10^-5, 10^-4 etc.,

• We can write:

[H⁺] = 10^-7 → Neutral

[H⁺] = 10^-6 → A little acidic

[H⁺] = 10^-5 → More acidic

[H⁺] = 10^-4 → Very acidic

. . . so on

Note that, when the acidity increases, the 'numeric value' in the exponent decreases. This is due to the '-' sign in the exponent. For example, from math classes, we know that, 10-15 is less than 10-11

• Let us write the above in logarithmic form:

Negative logarithm of [H⁺] = 7 → Neutral

Negative logarithm of [H⁺] = 6 → A little acidic

Negative logarithm of [H⁺] = 5 → More acidic

Negative logarithm of [H⁺] = 4 → Very acidic

. . . so on

Note that, when logarithm is taken to base 10, the 'base 10' part need not be written. For example, 'log101000 = 3' is same as 'log 1000 = 3'

■ So, when the negative logarithm value decreases from 7, the acidity increases. The logarithm value can decrease further and reach zero. A zero value would be highly acidic.

• So we moved downwards from '7', and reached up to zero. Now we will look at the other side of 7.

7 is neutral. From 7 to zero, it is acidic. So the other side will be alkaline.This can be written as:

Negative logarithm of [H⁺] = 8 → A little alkaline

Negative logarithm of [H⁺] = 9 → More alkaline

Negative logarithm of [H⁺] = 10 → Very alkaline

■ So, when the negative logarithm value increases from 7, the alkalinity increases. The logarithm value can increase further and reach a maximum of 14. A 14 value would be highly alkaline. It is like a number line as shown below:

• An important difference from the number line is that, there are no arrows at the ends. This is because, unlike the number line, here we do not have values up to infinity. The values are only from zero to 14. 
■ This is the basis of pH scale. it was developed by the Danish scientist Sorensen. The letter 'p' stands for 'power'. And 'H' stands for hydrogen. So it is the 'power of hydrogen'. We will now see how this scale can be put to a practical use.

Consider the following scenario:

We have a solution in hand. We want to know whether it is acidic or alkaline. Also we want to know where the solution will fall in the pH scale. That is., what number from 0 to 14, will be appropriate for the solution that we have.
To find the number, we need the number of H⁺ ions in 1 litre of the solution. But scientists have developed an easier method. We do not have to find the actual number of H⁺ ions.

The following procedure is usually adopted:

• Place a drop of the solution on a pH paper with the help of a fine dropper.

• Observe the colour produced in the pH paper. Match it with the standard colour pH chart.

• Note down the number of the colour in the chart, that matches most closely with the colour produced in the pH paper. This is the number that we require.
A pH chart is shown in the fig.6.3 below. It is obtained from wikimedia commons.

Fig.6.3

Example: If the colour obtained in the pH paper matches with the orange colour (just below the top most red), the number is 1. We can say: [H⁺] = 10^-1 mols/litre

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In the next section, we will see a few more practical situations where acidity or alkalinity will have to be considered. We will also learn more about salts. 

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