There are two ways to make a solution of known molarity.
The first and most convenient way to make the solution,
by dissolving exactly 1
mole of solute in a
litre (dm^{3}) of solution. The second way is to
make up a solution quickly, using an estimated amount of
solute and an estimated amount of solution, and then
determine the solution's exact molarity by titration.

*
Titration is the chemical process by
which we can determine the concentration of unknown
solution, that reacts with a standard solution, whose
concentration is known.*

Titrations play a major role in determining, amounts of
solutes present in a solution. Titration process is an
important tool of the analytical chemist.

**
9.6.1 Molarity (M) and Molar Solution:**

The most generally useful means of expressing a
concentration of solution is molarity.

*
It is defined as the number of moles of solute dissolved
per 1 litre (1dm*^{3}) of a solution; it is
denoted by (M).

Thus 1 mole of H_{2}SO_{4} (i.e. its
gram formula mass) 98g dissolved in one litre (1dm^{3})
of solution is said to be 1 molar (1M) solution, if only
one-half the mole i.e. 49g of H_{2}SO_{4}
are dissolved in one litre of solution, the solution is
said to be one-half molar (i.e. 0.5 M)

The molarity of any solution is found by dividing the
number of moles of solute by the number of moles/litres
of the solution.

Molarity (M) =
Moles of solute/Litres of solution

We known that,

(i) Number of moles = Mass of solute/
Gram formula mass

(ii) Litres of solution = Volume of
solution in cm^{3} (ml)/1000

Molarity (M) = Mass of solute/ Gram formula mass x 1000/
Volume of solution in cm^{3} (ml)

**
For example:**

(i) Calculate the molarity of a solution, containing
1.5g of NaOH in 250 cm^{3} of solution?

**
Solution**

Molarity (M) = Mass of solute/ Gram formula mass x 1000/
Volume of solution in cm^{3} (ml)

**
Data **

(i) Molarity
= M
= ?

(ii) Mass of solute (NaOH)
= l.5g

(iii) Gram formula mass of solute (23+16+1) =
40g

(iv) Volume of solution = 250 cm^{3}

M = 1.5g x 10000/40 x 250 = 0.15

Result: Molarity of NaOH solution is = M = 0.15M

**
Example. 2**

i) What mass of (NaOH) must be dissolved in 500cm^{3}
of solution to make 1.5 M-solution?

**
Solution**

M = Mass of solute/ Gram formula mass x
1000/cm^{3} of solution

By cross-multiplication, we get,

Mass of solute =
M x G.F. mass x cm^{3} of solution/1000

(i) Mass of solute (NaOH) =
?

(ii) Molarity
= M = 1.5 M

(iii) G.F mass of solute (23+16+1) = 40
g

(vi) Cm^{3} of solution
= 500 cm^{3}

Mass of solute = 1.5gx 40 x 500cm^{3} 1000 cm^{3}/1000cm^{3}

Result mass of solute i.e. NaoH = 30g

**
9.6.2 Preparation of Solution of Known Molarity:**

For example, how to prepare 1 M-solution of NaOH

To prepare 1M solution of NaOH you first weigh put one
mole of NaOH i.e. 23 +16 + 1 - 40g of NaOH, place 40g of
NaOH in a 1 litre (1 dm^{3}) volumetric flask,
and add some water to bring the level of solution to
calibration mark as shown in figure given below, finally
the solution is shaken until uniform

**
9.6.3 Standard Solution:**

A solution whose molarity or strength is known is called
standard solution. For example 1M.KOH solution, contains
1mole of (KOH) i.e. formula mass expressed in grams
=39+16+1 = 56g of KOH are dissolved in 1 litre of
solution is said to be 1 molar (1M). If we use only
one-half mole i.e. 28g of KOH in 1-litre of solution our
solution would be one-half molar i.e. 0.5 M.

**
9.6.4 Acid-Base Titration:**

In acid-base titration, a solution of known
concentration (say base) is added gradually to a
solution of unknown concentration (say an acid) so as to
determine the concentration of unknown solution. The
point at which the reaction is completed is called the
end point.

**
Steps for Carrying out Titration:**

1.
Fill the burette with the given solution of NaOH; read
and record the initial reading of the burette (Remember
to read the lower meniscus) if the burette is not full
refill it with NaOH solution. Then let the burette drain
into beaker down to the zero mark before using it. The
solution in the burette is called "**Titre**".

2.
Pipette out 10 cm of HCl in a conical flask, (titration
flask) and add one or two drops phenolphthalein
indicator. The solution in titration flask is called **
"Titrant**". A suitable acid-base indicator, such a
methyl orange, litmus or phenolphthalein is used to
detect the end point.

S.No. |
Indicator |
Colour in acids |
pH-range |
Colour in bases |

l. |
Methyl orange |
Red |
3 - 5 |
Yellow |

2. |
Litmus |
Red |
6-8 |
Blue |

3. |
Phenolphthalein |
Colour less |
8- 10 |
Pink (red) |

3.
Add slowly the NaOH solution, from the burette into the
conical flask (titration flask) with-constant shaking.
Stop adding the NaOH solution, when the mixture in the
titration flask becomes light pink. Record the final
burette reading. If you have been adding the reactant
rapidly and think you have over run the end point,
repeat the entire titrations slowing down when you think
you are nearing the end point. Rinse out the flask with
water before the next sample is titrated.

4.
Record the initial and final burette readings for each
experiment.

To see how titration works, let us consider that we have
(HCL) solution acid) whose concentration we want to find
by allowing it to react with a BI> lard base such as
NaOH.

We begin- the titration by measuring out a known volume
of HCL in a conical flask (titration flask) and adding
few drops of an indicator, such as phenolphthalein (colourless
liquid in acidic solutions but turns pink colour in
basic solutions).

Next, we fill a burette with the(NaOH) standard solution
(of known concentration) and we slowly add the NaOH to
the HCL, until the phenolphthalein just begins to turn
pink, indicating that all the HCL has been reacted and
that the solution is starting to become basic. Then by
recording the reading from the burette, to find the
volume of the NaOH (standard solution) that has reacted
with a known volume of HCL, we can calculate the
concentration (molarity) of the HCL.

Let us assume for example that we take 10 mls of
HCl-solution and find that we have to add 30 mls of 0.1
M NaOH from the burette to obtain complete reaction.
Using the following formula we can calculate the
molarity of unknown solution of HCl?

HCl (aq) + NaOH (aq)
------à
NaCl (aq) + H_{2}O (l)

The balanced equation shows that:

1 mole of HCL reacts with 1 mole of NaOH

1-HCl
------à
1-NaOH

Burette = NaOH solution molarity =
M = 0.1 M.

Pipette = HCl solution molarity = M
= ?

Indicator = Phenolphthalein.

Colour change = Colourless to pink.

V_{1} - Volume of HCL taken =
10 mls

M_{1} - Molarity of HCL taken
= ?

V_{2} - Volume of NaOH solution =
30 mls

M_{2} - Molarity of NaOH solution =
0.1M

n_{1} -
Number of moles of HCL = 1

n_{2} - Number of moles of NaOH
=
1

V_{1}XM_{1}/n_{1} = V_{2}XM_{2}/n_{2}

10x?/1 = 30x0.1/1 = 30x0.1/10 =3/10

M_{1} = .3M

Molarity
of HCL = 0.3M

We can also find out the amount of HCL dissolved per
1-litre = (1dm^{3}) of a solution by the
following formula.

Amount of HCl per dm^{3} = molarity x F. mass in
grams

Amount of HCl per dm^{3} = 0.3 x 36.5 = 10.95g

(As F. mass of HCl = 1+35.5 = 36.5g)

Result: Molarity of HCL = 0.3M |

Amount of HCl/dm^{3} = 10.95g |

**
Example.**

A flask contain 30 ml of NaOH solution, it require 50ml
of 0.15M-H_{2}SO_{4}, to complete the
reaction. Calculate the molarity of NaOH and how many
grams of NaOH were in the flask?

**
Solution.**

The balance equation for the reaction is:

2NaOH (aq) + H_{2}SO_{4} (aq)
------à
Na_{2}SO_{4} (aq) + 2H_{2}O (l)

2 mole 1 mole

Since 2 mole of NaOH ^{_____}^{ }
1mole of H_{2}SO_{4}we need twice the
amount of NaOH to react completely with H_{2}SO_{4}
solution as needed for the same concentration of HCl
solution.

M_{1}xV_{1/}n_{1} = M_{2}xV_{2}/n_{2}

Where:

V_{1} = Volume of H_{2}SO_{4} =
50 ml

M_{1} = Molarity of H_{2}SO_{4}
= 0.15M

n_{1} = Number of moles of H_{2}SO_{4}
= 1

V_{2} = Volume of NaOH = 30 ml

M_{2} = Molarity of NaOH = ?

n_{2} = Number of moles of NaOH = 2

M_{1}xV_{1/}n_{1} = M_{2}xV_{2}/n_{2}
0.15x50/1 =M_{2}x30/2

M_{2 }= 0.15x50x2/30 = 0.5M

Amount of NaOH per 30 ml = MxG.F.mass
x
volume of solution/1000

0.5x40x30/1000 = 0.6g

Result: Molarity of NaOH = 0.5M |

Amount of NaOH in the flask = 0.6g |