# Preparation of Standard Solution From Liquid Substance ## Free Guide to the Preparation of Standard Solution From Liquid Substance

Preparation of Standard Solution From Liquid Substance: When dealing with liquid substances, the following information about the stock liquid is very important and must be noted:
1. Specific density of the liquid in g/cm3
2. Specific gravity of the stock liquid
3. %purity by mass of the stock liquid
4. Molar mass of the stock liquid
The above information is obtained from the label of the stock liquid usually provided by the manufacturer.
When the above information is available, the preparation can be done in either of the two ways:
1. Determining the molarity of the stock liquid substance (acid) first, followed by applying the dilution principle
2. calculating the volume of stock (acid) require for the preparation of a required volume of standard solution.

Below is a table showing the specification of some common acids.

1. DETERMINE THE MOLARITY OF THE STOCK FOLLOW BY
DILUTION PRINCIPLE
The general formula for calculating the molarity of a stock acid is given as;

#### Cs = 10pd/m.m

Where Cs = molarity of stock liquid
P = %by mass of the liquid
d = density in g/cm3 or specific gravity
m.m = molar mass of the liquid
While dilution principle formula is given as:
C1V1= C2V2
Where C1 = conc. of the stock solution
V1 = volume of stock solution
C2 = desired conc. to be prepared
V2 = desired volume of the standard solution
Example: How can you prepared 500ml, 0.031M of HCl whose %by mass = 36,
specify gravity = 1.18 and m.mass = 36.5g/mole?
Solution
Firstly determine the molarity of the stock solution by using the formula

#### Cs = 10pd/m.m

Cs =?, P=35, d=1.18 and m.mass = 36.5g/mole
Cs = 10 x 35 x 1.18/36.5
= 11.64mol/dm3
So, Molarity of the stock acid = 11.64mol/dm3

The molarity of the stock acid is now known to be 11.64mol/dm3.
Therefore, the volume of the stock acid required to prepare 500cm3 of 0.031M must be calculated using the dilution formula.
C1V1 = C2V2
Where C1 = 11.64M, V1=? C2 = desire conc. = 0.031M
V2 = desire volume = 500cm3
11.64M x V1 = 0.031M x 500cm3

#### V1 = 0.03M x 500cm3 / 11.64M

= 1.336cm3
:. V1 ≈ 1.3cm3

Preparation of Standard Solution From Liquid Substance

Now

Prepare 500 cm3 of 0.031M HCl, transfer 1.3cm3 of the stock acid to 30 cm3 portion of distilled water in a beaker. Transfer the resulting solution into 500cm3 volumetric flasks and fill to mark with distilled water, cork, and shake very well. You now have 500cm3 of 0.031M HCl solution.
Example
How can you prepare 200cm3 of 1.2M HNO3 acid? Given that, the specific gravity
= 1.426, % by mass = 70 and m.mass = 63g/mole

#### Cs = 10pd/m.m

Where C =?, P = 70, d = 1.41 and m.mass = 63g/mole

Cs = 10 x 70 x 1.42 / 63
= 15.8M
:. Cs = 15.8M

Dilution principle
C1V1 = C2V2
C1 = 15.8M, V1 = ?, C2 = 1.2M, V2 = 200cm^3
15.8M x V = 1.2M x 200cm^3

#### V1 = 1.2Mx200cm / 15.8M

V1 = 15.2cm^3
Therefore, 15.2cm3 of conc. HNO3 should be measured and transfer into 30cm^3 of distilled water. Transfer the resulting solution into 200cm3 volumetric flasks and fill to mark with distilled water.

### 1. CALCULATING THE VOLUME OF STOCK REQUIRED FORTHE PREPARATION OF REQUIRED VOLUME OF STA”DARDSOLUTION.

Calculating the volume of the stock acid required can be done using the formula below.

#### Cs = CrVr m.mass / 10pd

Where Cs = volume of stock solution
Cr = required conc.
Vr = required volume of standard solution
P = % by mass of the acid
d = specific gravity of the acid
m.mass = molar mass of acid
Example
How can you prepare 500cm^3 of 0.23M H2SO4? When the acid has the following specifications; m.mass 98g/mole, %by mass = 98%, and specific gravity ≈ 1.80

#### Solution

The volume of stock acid required to prepare 500cm3 of 0.23M H2SO4 can be calculated using the formula.

#### Vs = 0.23 x 500 x 98 / 96

Vs = 0.0064dm^3
1dm3 ≈ 1000cm^3
0.0064dm3 = xcm^3

x = 0.0064dm^3 1000cm^3 / 1dm^3

Vs= 6.4cm^3

Therefore, measure 6.4cm3 of stock (acid) into a beaker containing 30cm3 of distilled water. Transfer the resulting solution into 500cm3 volumetric flasks and fill to mark with distilled water. In some cases, you will be asked to prepare a standard solution in which its concentration is expressed in g/dm3 from a liquid substance. In such cases, the given conc. in g/dm3 must first be converted to mol/dm3, followed by calculation of the volume of stock solution required for the preparation using normal formula.
Example
How can you prepare 300cm3 18.2g/dm3 of HNO3?

(m.mass =63g/mole)

#### Solution

It is not convenient to weigh liquid substances; therefore, conversion of a unit of
expression from g/dm3 to mol/dm3 is necessary.

#### Conc. in mol/dm^3 = (18.2g/dm^3) / (63g/mole)

= 0.2889mol/dm3≈ 0.29mol/dm3

:. 300cm3 of 18.2g/dm3 of HNO3 is equivalent to300cm3 0.29mol/dm3of HNO3.
Now, how to prepare 300cm3 of 0.29mol/dm3.

#### Vs = CrVr m.m / 10pd

where Vr = 300cm^3 = 0.3dm^3

#### Vs = (0.29Mx0.3dm x63g/mole) / (10x70x1.41)

Vs = 0.0056dm3 ≈5.6cm3

Therefore, measure 5.6cm3 of conc. HNO3 into 30cm3 of distilled water in a beaker. Transfer the resulting solution into 300cm3 volumetric flasks and fill to mark with distilled water, now you have 300cm3 18.2g/dm3 of HNO3 solution.