How to prepare solutions for liquid samples

How to prepare solutions for liquid samples

1. solutions
Section 1

2. Content
1.Preparation of molar solution
2.Preparation of normal solution
3.Preparation of buffer solution
4.Determination of PH of solution

3. 1) Preparation of molar solution
Molar Solution:
A molar solution implies concentration in terms of Moles/
liter.
One molar solution (1M) means one mole of substance per
liter of solution.
What is MOLE
Mole means molecular weight of substance in gram.
For preparing Molar solution, You must know Molecular
weight
You can calculate M.W (Sum of atomic weights)
(At. Wt. Na = 23, O = 16, H = 1)
•One Na atom = 23
•One O atom = 16
•One H atom = 1
M.W. of NaOH = 23 + 16 + 1 = 40

4. Prepare 1M NaOH solution (1000ml)
Molarity =
����� �� ������
������ �� �����??????�� (�)

M =
�??????��
���??????� �??????��
×
�
������ (�)

•Molar mass of NaOH = 40g/mol
•Volume = 1 L
•Molarity = 1 M
•Mass = ?
Mass = Molarity × Molar mass × volume (L)
= 1 × 40 × 1 = 40 g
Dissolve 40g NaOH in solvent , then make up the final
volume up to 1000ml (1L)

5. Prepare 0.1M solution of NaOH
dissolved in 750 ml of solution
Molarity =
����� �� ������
������ �� �����??????�� (�)

M =
�??????��
���??????� �??????��
×
�
������ (�)

•Molar mass of NaOH = 40g/mol
•Volume = 750 ml
•Molarity = 0.1 M
•Mass = ?
Mass =
Molarity × Molar mass × volume
����

=
0.1 × 40 × 750
����
= 3 g
Dissolve 3g NaOH in solvent , then make up the final
volume up to 750ml

6. 0.1Molar solution of NaOH(1000ml)
Molarity =
����� �� ������
������ �� �����??????�� (�)

M =
�??????��
���??????� �??????��
×
�
������ (�)

•Molar mass of NaOH = 40g/mol
•Volume = 1000 ml
•Molarity = 0.1 M
•Mass = ?
Mass = Molarity × Molar mass × volume
= 0.1 × 40 × 1 =4 g
So, Dissolve 4g NaOH in solvent , then make up the final
volume up to 1000ml

7. Required Apparatus and Chemicals

8. Let's Start Preparation
Distilled water

9. Prepare 0.1M HCI Solution
% age purity of HCI in our labs is 37% ‚So On the basis
of that molarity calculation has been done
Dilute 370 ml HCI in 1000 ml of water (37%)
•Density of HCI = 1.19 g/ml
•Molecular weight of HCI = 36.5 Therefore,
•Molarity =
% age of Hcl× Denisty of HCL
M.W of HCL
×10 =
37×1.19
36.5
×10 = 12 M
M
1V
1= M
2V
2
0.1 × 1000 = 12 × V
2
V
2= 8.33 ml
Therefore 8.33ml of 37% HCI in 1000 ml of water is
required pare 0.1M HCI solution

10. Required Apparatus and Chemicals
Hydrochloric
acid
Measuring flask
(1000ml)
Distilled water
pipette

11. Let's Start Preparation
Distilled water
8.33 ml
Distilled water
1000 ml

12. 2) Preparation of normal solution
Normal Solution
Normal solution implies the concentration in terms of
equivalent weight of solute per liter of solvent.
Equivalent weight means Molecular weight of substance
divided by number of reacting OH or H.
Equivalent weight =
Molecular weight
Number of reacting OH or H

For preparing Normal solution, You must know Equivalent
weight
For NaOH E.W. can be calculated as
E.W. =
M.W
Number of replaceable H+ ions
=
40
1
= 40gm
E.W of NaOH = 40 gm

13. Prepare 0.1N NaOH solution (1000ml)
Normality =
�� �� ��??????� ���??????�??????���� �� ������
������ �� �����??????�� (�)

N =
�??????��
���??????�??????���� ��
×
�
������ (�)

•????????????�??????�????????????????????????� �� of NaOH = 40g
•Volume = 1 L
•Normality = 0.1 N
•Mass = ?
Mass = Normality × ????????????�??????�????????????????????????� �� × volume (L)
= 0.1 × 40 × 1 = 4 g
So, Dissolve 4g NaOH in solvent , then make up the final
volume up to 1000ml (1L)

14. Prepare 0.1N HCL solution (500ml)
% age purity of HCI in our labs is 37% ‚So On the basis of that
molarity calculation has been done
Dilute 370 ml HCI in 1000 ml of water (37%)
•Density of HCI = 1.19 g/ml
•Molecular weight of HCI = 36.5 Therefore,
•Molarity =
% age of Hcl× Denisty of HCL
Molecular weight of HCL
×10 =
37×1.19
36.5
×10 = 12 M
•N = Molarity × acidity/basicity (HCL=1)
= 12 × 1 = 12
N
1V
1= N
2V
2
12 × V
1 = 0.1 × 500
V
1= 4.16 ml
Therefore 4.16ml of 37% HCI in 1000 ml of water is required pare
0.1N HCI solution

15. Required Apparatus and Chemicals
Hydrochloric
acid
Measuring flask
(500ml)
Distilled water
pipette

16. Let's Start Preparation
Distilled water
Distilled water

17. 3) Preparation of buffer solution
Buffer Solutions
Buffer solutions are solutions which resist the change in the
pH of solution upon addition of small amount of strong acid
or strong base.
Types of Buffer solutions:
1-Acidic buffer
2- Basic buffer
1-Acidic buffer
Consists of weak acid and its salt of strong base. e.g. Acetic
acid and sodium acetate (CH
3COOH / CH
3COONa)
Upon addition of small amount of acid (H
+
), sodium acetate
reacts with it giving weakly ionized acetic acid.
H
+
+ CH
3COONa CH
3COOH + Na
+

Upon addition of small amount of a base (OH
−
), acetic acid
reacts with it and non ionized water is formed.
OH
−
+ CH
3COOH CH
3COO
−
+ H
2O

18. 2- Basic buffer
Consists of weak base and its salt of strong
acid , e.g. Ammonium hydroxide and
ammonium chloride (NH
4OH / NH
4Cl)
Upon addition of a small amount of acid:
H
+
+ NH
4OH NH
4
−
+ H
2O
Upon addition of a small amount of base:
OH
−
+ NH
4Cl NH
4OH + Cl
−

Henderson Equation for calculation of
the pH of Buffer solutions:
pH of acidic buffer:
pH = pKa + log Cs / Ca
pH of basic buffer:
pOH = pKb + log Cs / Cb

19. How to make a buffer
1- choose the acid conjugate base pair that
pKa is closest to the pH that you want
2- Use henderson hasselbalch to calculation
the ratio of acid + conjugate base to get
the ph you want
3- calculate both the mass and volume of
the acid and conjugate base pair that you
need to get the ratio
4- combine the acid + base & you have
your puffer

20. Want 500ml of a PH 5 buffer that have
0.1M ????????????
�??????
�??????
� + Na??????
�??????
�??????
� solid (M.W = 82.04
gm/mol) (PKA is 4.74)
pH = pKa + log
CB
A
5 = 4.74 + log
CB
A

0.26 = log
CB
A
10
0.26
=
CB
A

1.82
1
=
CB
A

(0.1M) of acid=
no of mole (?)
volume litre(0.5)

no of mole of acid= 0.1×0.5= 0.05mol/L
1.82
1
=
x
0.05 mol

X( no of mole of base )= 1.82 × 0.05 = 0.091mol/L
Mass of base = no of mole × molar mass
= 0.091 × 82.04 = 7.47 gm

21. How many gram of NaCN must be added to 1L
of 0.5M HCN solution to make a buffer of PH=
8.51? (PKA is 9.21)
pH = pKa + log
CB
A
8.51 = 9.21 + log
CN
−
HCN

-0.7 = log
????????????
−
0.5
10
−0.7
=
????????????
−
0.5
0.20=
????????????
−
0.5

????????????
−
= 0.5 ×0.2 = 0.10 M
(0.1M) of ????????????
−
=
no of mole (?)
volume litre(1L)

no of mole of ????????????
−
= 0.1×1= 0.1mol/L
1 mole of NaCN have 1 mol of ????????????
−

Mass of NaCN = no of mole × molar mass
= 0.1 × 49 = 4.9 gm NaCN

22. 4) Determination of PH of
solution
PH value can be measured by using following methods:
ph stripes
ph indicators
calorimetric
Electrometric (ph meter)

23. PH stripes
Aim: a strip of litmus paper with which
you can measure the pH value of a liquid.
Principle: The substance in the paper causes the paper to
show a different colour at different acidities.
Requirement: acetic acid, pH paper strip, standard colour
pH chart, dropper and a white tile
Procedure:
1.Take a pH paper strip and place it on a white tile.
2.Take a small quantity of acetic acid using a droper
3.Pour one drop of acetic acid on the pH paper.
4.Observe the colour produced on the pH paper.
5.The colour of the pH paper changes to red.
6.Compare the colour produced on the pH paper with the
different colour shades of a standard colour pH chart and
note down the pH from the colour chart.
7.The pH of acetic acid is 3.
Result : colour of the pH paper changes according to PH of
the solution

25. pH indicators
Aim: measure the pH value of a liquid.
Principle: Indicators are the organic
compounds of natural or synthetic origin whose color is dependent
on pH.
Requirement: acetic acid, universal indicator, a standard colour
pH chart, dropper and a test tube.
Procedure:
1.Take a small quantity of acetic acid using a dropper and pour it
into a test tube.
2.Take a small quantity of universal indicator solution
3.Pour a few drops of universal indicator into the test tube
containing acetic acid.
4.Shake the test tube well.
5.The colour of the solution changes to red.
6.Compare the colour produced in the test tube with the standard
colour pH chart and note down the pH from the colour chart.
7.The pH of acetic acid is 3.
Result : colour of the solution changes according to PH of the
solution

27. Calorimetric
Principle: Calorimetric method is based
on the property of acid - base indicator
dyes ,which produces colour depending
on pH of the sample. The color change
can be measured as an
absorbance change
spectrophotometrically

28. Electometric (pH meter)
Aim: most accurate of the methods employed for the
determination of Hydrogen Ion Concentration.(pH measurements
accurate to 0.1 to 0.001 pH)
Principle: determination of the activity of the hydrogen ion by
potentiometric measurement using a standard hydrogen electrode
and a reference electrode.
Requirement: PH meter, distilled water, solution in a peaker
Procedure:
1.Before use, remove electrode from storage solution, rinse, and
blot, dry with a soft tissue paper.
2.Calibrate the instrument with standard buffer solution. (Ex:
KCI solution of pH 7.0) or wash with distilled water
3.Once the instrument is calibrated remove the electrode from
standard solution; rinse, blot and dry.
4.Dip the electrode in the sample whose pH has to be measured.
Stir the sample to ensure homogeneity
5.Note down the reading (pH) from the pH meter.