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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Chapter 6.1 - Basic details about Acids

In the previous section, we saw the relation between non-metallic oxides and acids. In this section we will see metallic oxides.
Let us do an experiment:
For this experiment, we need 'quick lime'. Quick lime is the common name for calcium oxide (CaO). It is obtained by heating naturally occuring substances like limestone or sea shells. Limestone and sea shells are calcium carbonates (CaCO₃). When it is heated, carbon dioxide escapes, and CaO is left behind. Let us write the equation:

Reactants:
    ♦ Calcium carbonate. One molecule is CaCO₃

Products: 
    ♦ Calcium oxide. One molecule is CaO

    ♦ Carbon dioxide. One molecule is CO₂

• So the skeletal equation is: CaCO₃ → CaO + CO₂ . This is a balanced equation.

1. So we have an oxide CaO. An oxide of a metal calcium. So it is a metallic oxide. We are going to do the experiment with it. 

2. Take water in a beaker. Add some quick lime to it and stir well. A reaction takes place. It is between CaO and water. If we touch the sides of the beaker, we can feel heat. This is because the reaction releases heat. In other words, it is an exothermic reaction.  Let us write the equation:

Reactants:
    ♦ Calcium oxide. One molecule is CaO

    ♦ Water. One molecule is H₂O

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

• So the skeletal equation is: CaO + H₂O → Ca(OH)₂. This is a balanced equation.

3. How are we going to use this calcium hydroxide?

• We added quick lime to water. The quick lime was converted into Ca(OH)₂ (calcium hydroxide).
• Some water will be used up for the reaction. But there will still be some excess water. Because all the water will not be used up. 
• So the product, which is Ca(OH)₂, will dissolve in the excess water. And we get a solution of Ca(OH)₂ in water. 
• But all the Ca(OH)₂ will not dissolve in water. Some of it will precipitate to the bottom of the beaker. The clear portion above the precipitate is called 'supernate'.

■ We see a separation as a precipitate and a supernate because, there is not enough water for all the Ca(OH)₂ to dissolve. That portion of Ca(OH)₂, which is unable to dissolve, will form the precipitate. That means, the supernate is a solution of Ca(OH)₂ in water. 

Take some of this supernate in a test tube. Add a drop of red litmus to it. We can see that, a blue colour appears in the solution. So we can infer that Ca(OH)₂ is alkaline in nature. Let us analyse what we have done:

• We took quick lime. It is an oxide of calcium, which is a metal. So quick lime is a metallic oxide.
• This metallic oxide, when added to water, gave an alkali. We can say this: Metallic oxides are generally alkaline in nature.

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■ So, if we are given a list of oxides, we can classify them as acidic or alkaline. Given below is a list:

SO₃, NO₂, CaO, K₂O, P₂O₅, Na₂O, CO₂, MgO

Let us classify them into acidic and alkaline:

Acidic oxides	Alkaline oxides
SO3	CaO
P2O5	K2O
CO2	Na2O
NO2	MgO

The above classification is based on the following two facts:
■ Non-metallic oxides are acidic in nature
■ Metallic oxides are alkaline in nature

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Following are some of the properties of acids and alkalies:

■ Acids:

• Have sour taste

• Changes blue litmus to red

• Produces carbon dioxide gas on reacting with carbonates

• Produces hydrogen gas on reacting with metals like Fe, Mg etc.,

■ Alkalies:

• Have bitter taste

• Changes red litmus to blue

• Slippery to touch

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Let us explore some more about acids:

Acids can be classified as Mineral acids and Organic acids

■ Mineral acids are also known as inorganic acids. They are prepared from one or more inorganic compounds, and are stronger than organic acids. Most common mineral acids are Sulphuric acid, Nitric acid and Hydrochloric acid. Mineral acids should not be tasted.
■ Organic acids are those acids found in naturally occuring organic compounds. They are generally weak acids. The sour taste of some of the food items is due to the presence of organic acids. Some examples are given below:

• Lime – Citric acid

• Tamarind – Tartaric acid

• Tomato – Oxalic acid

• Sour milk – Lactic acid

• Vinegar – Acetic acid

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The common factor in acids

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

Name of acid	Chemical formula
Hydrochloric acid	HCl
Nitric acid	HNO3
Acetic acid	CH3COOH
Carbonic acid	H2CO3
Sulphuric acid	H2SO4

• From the above list, we find that hydrogen is present in all the acids. So hydrogen is a common factor in acids.

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Hydrogen is the component present in all acids and is responsible for the common properties of acids. 

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Let us see a reaction in which an acid is a reactant:

Hydrochloric acid reacts with iron to give ferric chloride and hydrogen. Let us write the equation:

Reactants:

    ♦ Iron. One molecule is Fe

    ♦ Dilute Hydrochloric acid. One molecule is HCl.

Products:

    ♦ Ferric chloride. One molecule is FeCl₂.

    ♦ Carbon dioxide. One molecule is CO₂.

• So skeletal equation is:

Fe + HCl → FeCl₂ + H₂. This is not a balanced equation. The steps for writing the balanced equation are shown below:
Step 1: Fe + HCl → FeCl₂ + H₂
Step 2: Fe + 2HCl → FeCl₂ + H₂

	Reactants			Products
	Fe	H	Cl	Fe	H	Cl
Step 1	1	1	1	1	2	2
Step 2	1	2	2	1	2	2

■ So the balanced equation is: Fe + 2HCl → FeCl₂ + H₂

We can say this:

The acid HCl, reacts with the metal iron to liberate hydrogen. In fact, it is a common property of all acids:

■ All acids react with metals to liberate hydrogen

We saw another example in the preparation of hydrogen in the lab. Details here.

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We will now discuss about acids in detail:
We know that in HCl:

• There is a covalent bonding between H and Cl.
• Cl is more electronegative, and so attracts the shared pair of electrons towards itself
• So the hydrogen has a partial positive charge, and chlorine has a partial negative charge. 
• We have seen the above details here. 
• When HCl is mixed with water, the bond between H and Cl breaks. The hydrogen will have to leave with out taking it's electron.
• This is because, Cl is more electronegative, and so, will keep both the shared electrons
• The H atom thus becomes H⁺ ion and Cl becomes Cl^-
• The H⁺ ions will go towards the water molecules. They will attach with the water molecules. 
• The water molecules will then become ions because of the positive charge brought by the H⁺ ions. 
• The new ion is written as H₃O⁺. It is called the hydronium ion. 
• These hydronium ions are responsible for the characteristics of acids.

Consider the following scenario:

We have a situation in which we want the effects of an acid. We have a bottle of HCl. If we keep the HCl in the bottle itself, or pour it into another suitable container, there is no effect. To see the effects of acid, the acid must come in contact with other materials.

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Note that it is very important to handle acids carefully. All safety measures should be taken. Some of the compulsory safety measures are:

• Wearing lab coat or lab apron

• Wearing safety goggles for protection of eyes

• Using acid resistant gloves

Experiments should be performed only under the guidance of authorised professionals

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• We have seen the effect when HCl comes into contact with water: The hydronium ions (H₃O⁺)are formed, and that gives the acidic effect.

• If more H+ ions are liberated, more hydronium ions will be formed, and there will be greater acidic effect

■ If one molecule of an acid gives one H⁺ ion, It is a monobasic acid

Eg: HCl → H⁺ + Cl^-.

HNO₃ → H⁺ + NO₃^-

■ If one molecule of an acid gives two H⁺ ions, It is a dibasic acid

Eg: H₂SO₄ → 2H⁺ + SO₄^2-

It may be noted that, the two H⁺ ions are obtained as a result of two steps as shown below:

Step 1: H₂SO₄ dissociates into H⁺ and HSO₄^-

Step 2: HSO₄^- dissociates into H⁺ and SO₄^2-

Step 1 gives one H⁺ and one HSO₄^-. But after step 2, the HSO₄^- is no longer present. It has dissociated into H⁺ and SO₄^2-

So we get a total of two H⁺ ions

Another example: 

H₂CO₃ → 2H⁺ + CO₃^2-.

Here also, the two H⁺ ions are obtained as a result of two steps as shown below:

Step 1: H₂CO₃ dissociates into H⁺ and HCO₃^-

Step 2: HCO₃^- dissociates into H⁺ and CO₃^2-.

So we get a total of two H⁺ ions

■ If one molecule of an acid gives three H⁺ ions, It is a tribasic acid

Eg: H₃PO₄ → 3H⁺ + PO₄^3-

Here, the three H⁺ ions are obtained as a result of three steps as shown below:

Step 1: H₃PO₄ dissociates into H⁺ and H₂PO₄^-

Step 2: H₂PO₄^- dissociates into H⁺ and  HPO₄^2-.

Step 3: HPO₄^2- dissociates into H⁺ and  PO₄^3-.

So we get a total of three H⁺ ions

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We have seen some of the important details about acids. In the next section, we will discuss about alkalies. 

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Chapter 6 - Acids, Alkalies and Salts

In the previous section, we completed the discussion on Hydrogen and Chlorine. In this chapter we will see Acids and Alkalies.

We know that, when metals or non-metals react with oxygen, various oxides are formed.

• When metals react with oxygen, we get metallic oxides.

• When non-metals react with oxygen, we get non-metallic oxides.

We will first consider non-metallic oxides.

We know that carbon is a non-metal. Nitrogen and sulphur are also non-metals. When these non-metals react (the reaction can be combustion or ordinary reaction) with oxygen, their oxides are produced. And these oxides are ‘non-metallic oxides’.

Here we have to learn about 3 non-metallic oxides. CO₂, NO₂ and SO₂. We will first consider CO₂. 

Let us do an experiment:

1. Take some calcium carbonate (CaCO₃) in a boiling tube. Add 5 ml dilute hydrochloric acid (dil.HCl) to it using a thistle funnel. See fig.6.1 below:

Fig.6.1

2. The calcium carbonate reacts with HCl. We get calcium chloride (CaCl₂) and carbon dioxide (CO₂) gas as products. Let us write the equation:

Reactants:

    ♦ Calcium carbonate. One molecule is CaCO₃. 

    ♦ Dilute Hydrochloric acid. One molecule is HCl.

Products:

    ♦ Calcium chloride. One molecule is CaCl₂.

    ♦ Carbon dioxide. One molecule is CO₂.
    ♦ Water. One molecule is H₂O

• So skeletal equation is:

CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O. This is not a balanced equation. The steps for writing the balanced equation are shown below:
Step 1: CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O 
Step 2: CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

	REACTANTS								PRODUCTS
	Ca	C	O	H	Cl				Ca	C	O	H	Cl
Step 1	1	1	3	1	1				1	1	3	2	2
Step 2	1	1	3	2	2				1	1	3	2	2

■ So the balanced equation is: CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

Thus we successfully produced a non-metallic oxide CO₂.  There is more to do:

3. Pass this CO₂ through water in a test tube. When we do that, we get a solution. The solution formed by dissolving CO₂ in water. 
4. Now dip a blue litmus paper in the solution. The blue litmus turns into red. 
5. So we can conclude that, the solution is acidic in nature.

■ That means, an acid is formed when CO₂ dissolves in water. 
• In fact, when the CO₂ dissolves in water, a reaction takes place. The product formed is carbonic acid (H₂CO₃). Let us write the equation:

Reactants:
    ♦ Carbon dioxide. One molecule is CO₂

    ♦ Water. One molecule is H₂O

Product: 
    ♦ Carbonic acid. One molecule is H₂CO₃

• So the skeletal equation is: H₂O + CO₂ → H₂CO₃. This is a balanced equation.

■ From the above experiment, we get the following information:

The non-metallic oxide CO₂ dissolves in water to give an acid

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Generally, CO₂ will not dissolve in water easily. If we want to dissolve large quantities of CO₂ in water, we will have to apply pressure. Soda water is produced in this way.

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Now we consider another non-metallic oxide. That of nitrogen:

Nitrogen does not take part in reactions easily. This is because of the triple bond in the nitrogen molecule. But under the action of thunder and lightning, nitrogen in the atmosphere combines with oxygen. Thus oxides of nitrogen are produced. We saw the details in the previous chapter when we saw Nitrogen cycle. We will write the steps again:
1. First, nitrogen combines with oxygen to form nitric oxide. The balanced equation is:
N₂ + O₂ → 2NO

2. The NO thus formed will combine with more oxygen to become Nitrogen dioxide NO₂. The balanced equation is:
2NO + O₂ → 2NO₂
3. Thus we reach the stage of a non-metallic oxide NO₂. We want to know the further actions of this NO₂.

4. The NO₂ dissolves in rain water. It may be noted that, oxygen is required for this dissolution to take place. The balanced equation is: 
4NO₂ + 2H₂O + O₂ → 4HNO₃.

5. So, as a result of the 'dissolving of nitric oxide in water', we get HNO₃, which is nitric acid. 
■ Thus so far, we have seen two cases:

• The non-metallic oxide CO₂ dissolves in water to give carbonic acid

• The non-metallic oxide NO₂ dissolves in water to give nitric acid.

Now we will see yet another non-metallic oxide. The oxide of sulphur. It is called sulphur dioxide (SO₂) We have seen it's formation when we studied about the reactions of oxygen with non-metals. Details here. The balanced equation for the formation of SO₂ is:

S + O₂ → SO₂.

• This oxide also dissolves in water to form an acid. The acid formed is sulphurous acid (H₂SO₃). Let us write the equation:

Reactants:

    ♦ Sulphur dioxide. One molecule is SO₂. 

    ♦ Water. One molecule is H₂O.

Products:

    ♦ Sulphurous acid. One molecule is H₂SO₃.

• So skeletal equation is:

SO₂ + H₂O → H₂SO₃. This is a balanced equation.

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So we have one more case to add to our list. The modified list is given below:

■ Three cases seen so far:

• The non-metallic oxide CO₂ dissolves in water to give carbonic acid

• The non-metallic oxide NO₂ dissolves in water to give nitric acid.

• The non-metallic oxide SO₂ dissolves in water to give Sulphurous acid.

We can say this: Non-metallic oxides generally give acids when they dissolve in water

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Let us now see the ill-effects of such acid formation:

■ Factories, thermal power plants, automobiles etc., send out large quantities of non-metallic oxides like NO₂ and SO₂. These acids dissolve in rain water, and form acids. When large quantities of acids come down with rain, we call it an acid rain. Following are the ill-effects of acid rains:

• Acid rains destroy the leaves of plants. The plants then will no longer be able to produce food by photosynthesis.

• Acid rains destroys the greenery of a region. This will increase air pollution

• Acid rains can cause deaths of fish and other marine organisms

• Acid rains cause destruction of substances containing calcium carbonate (CaCO₃). Because, CaCO₃ reacts with acids

    ♦Corals contain CaCO₃. When acids reach them, reaction takes place, and the corals will be destroyed. The effects can be seen here.

     ♦ Marble contain CaCO₃. When acids reach them, reaction takes place, and the beauty of marble is lost.
     ♦ Limestone is an important naturally occuring veriety of stone. It is essential for many industries. They are obtained from limestone quarries. The quarries as a whole will be destroyed if the acid rains reach them. The destruction takes place due to the reaction of CaCO₃ with the acids. The effects can be seen here.

We have to take steps for the prevention of acid rains. Following are some of the effective steps:

• Reduce the use of fossil fuels like coal and petroleum

• Use those fossil fuels from which sulphur compounds have been removed. In this way, oxides of sulphur will not be formed during the combustion of the fuel.

• Regularly check that, the gases emitted from automobiles does not cause pollution above the accepted level.

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We have completed our present discussion on the formation of acids from non-metallic oxides. In the next section, we will discuss about metallic oxides. 

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