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.
• 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’).
In the next section, we will see a few more solved examples.
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.
• The positive ions are called
cations, and the negative ions are called anions
The following table shows some cations and 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+
|
• 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 Cx+
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 Cx+
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 Ay-.
• 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 Ay-.
5. Now assemble the cation and anion together. The cation should be written first. So we get:
Cx+ Ay-.
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: Cmx+ Any- .
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:
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+ + SO42-.
3. Assemble the cation and anion with cation first: Mg2+ SO42- .
4. Do the interchanging:
• Number of Mg2+ ions = Number of charges in SO42- ion = 2
• Number of SO42- 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:
Cx+ Ay-.
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: Cmx+ Any- .
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
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.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 | |
Solved example 6.2
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+ + SO42-.
3. Assemble the cation and anion with cation first: Mg2+ SO42- .
4. Do the interchanging:
• Number of Mg2+ ions = Number of charges in SO42- ion = 2
• Number of SO42- 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
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
Step 2: Mg(OH)2 + H2SO4 → MgSO4 + 2H2O
In the next section, we will see a few more solved examples.
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