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.
• 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.
The 'hydroxyl group' is the
common factor present in all alkalies, and is responsible for the
common properties of alkalies
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.
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 |
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.
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:
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.
 |
| Fig.6.2 |
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.
■ A video demonstrating a similar experiment can be seen here. We will learn more details in higher classes.
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.
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.
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
♦ 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 (25o 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
• Our number is 0.0000001. This is the [H+]. It is equal to 10-7
■ So logarithm is another way of writing exponents
• 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
• Let us write the above in logarithmic form:
| 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 |
|---|
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
■ 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.
| 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 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.
■ 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.
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:
• 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.
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
A pH chart is shown in the fig.6.3 below. It is obtained from wikimedia commons.
 |
| Fig.6.3 |
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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