Experiment 1 standardized a 500 ml HCl solution against 1 N NaOH solution. The HCl solution was titrated with 1 N NaOH using a burette and phenolphthalein indicator. Calculations using the titration data found the HCl solution to be 1.015 N. Experiment 2 determined the % purity of a KOH sample. A KOH solution was prepared from a 2 gm sample and titrated against the 1.015 N HCl solution from Experiment 1 using methyl orange indicator. Calculations based on the titration data found the % purity of the KOH sample to be 84.65%, close to the expected 84% purity.

1. EXPERIMENTS
STANDARDIZATION OF ACID (HCl) &
% PURITY OF BASE (KOH)

2. RECAP
1
2
3
4
5
6
UNITS FOR
PHYSICAL
MEASUREMENT
MASS, VOLUME,
DENSITY &
TEMPERATURE
ATOMIC MASS
MOLECULAR
MASS
MOLE CONCEPT
Mole, Avogadro
Number, amu & gram
MOLAR MASS

3. RECAP
7
8
9
10
11
12
MASS % OF AN
ELEMENT
PERCENTAGE BY
WEIGHT
PERCENTAGE BY
VOLUME
MOLE FRACTION
OF COMPONENT
MOLARITY
MOLALITY

4. RECAP
13
14
15
16
17
18
NORMALITY
EQUIVALENT
WEIGHT
ACIDITY, BASICITY &
RELATED CONCEPTS
CALCULATION OF
EQUIVALENT
WEIGHT
NORMALITY
Vs.
MOLARITY
NORMALITY &
MOLARITY
RELATIONSHIP

5. RECAP
19
20
21
22
23
24
STANDARDIZATION
DEFINITION & NEED
PRIMARY &
SECONDARY
STANDARD
VOLUMETRIC
ANALYSIS/ TITRATION
STANDARDISATION
OF NaOH
PRACTICAL
DEMONSTRATION
NORMALITY
CALCULATION

6. RECAP
25
26
27
28
29
30
% PURITY
DEFINITION
PERCENT PURITY
OF HCl
COMMON FORMULA &
TITRATION FORMULA
EXAMPLES OF PURITY
IMPORTANCE OF
PURITY
PURITY OF
CALCULATION

7. Experiment -1
Standardization of 500 ml. HCl
With 1 N NaOH

8. Transfer To
500 ml.
Vol. Flask
& Make Up
Volume Up
To 500 ml.
Mark
Standardization Of 500 ml. HCl With 1 N NaOH
1. Preparation Of 500 ml., 1 N NaOH Solution For
Standardization of HCl Solution.
+ 

20 gm NaOH 2 gm NaOH
As Compensation
For 96% Purity
NaOH
Dissolve In
Approx. 250 ml.
Water

9. Standardization Of 500 ml. HCl With 1 N NaOH
2. Preparation Of 1 N 500 ml. HCl Solution For
Standardization Against 1N NaOH.
NaOH + HCl  NaCl + H2O
 Eq. Wt. of HCL = 36.5 gm/Eq., Purity of HCl = 37% Density of HCl = 1.13 g /ml
Density =
𝑀𝑎𝑠𝑠
𝑉𝑜𝑙𝑢𝑚𝑒
 Volume =
𝑀𝑎𝑠𝑠
𝐷𝑒𝑛𝑠𝑖𝑡𝑦
=
36.5 𝑔
1.13 𝑔/𝑚𝑙
= 32.3 ml
 32.3 ml HCl required for 36.5 gm of HCl
 If Purity of HCl is 100% then we required 32.3 ml HCl for 1 N, 1000 ml Solution.
 But here Purity of HCl = 37%
So, more amount of HCl required.
% purity of HCl HCl in ml req. for 1 N
100 % 32.3 ml
37% (Actual Purity of HCL) ?
=
100  32.3
37
= 87.3 ml  88 ml HCl required for 1 N 1000 ml
% Purity & Qty Of HCL To Be
Used Are Inversely Related
Hence, We Perform Straight
Multiplication.

10. Standardization Of 500 ml. HCl With 1 N NaOH
2. Preparation Of 1 N 500 ml. HCl Solution For
Standardization Against 1N NaOH.
1. Above calculation (% and ml.) is inversely proportional as one amount
increases, another amount decreases (Inverse Variation).
2. So, we perform straight multiplication
3. We take 88 ml of HCl approximately for 1 N 1000 ml HCl solution.
4. We take 44 ml of HCl approximately for 1 N 500 ml HCl solution.
5. Take 100 ml water approximately in 500 ml beaker then add 44 ml HCl very
slowly in 500 ml beaker by using measuring cylinder, Stirring well with glass
rode .
6. Take clean and dry standard measuring flask then transfer solution into SMF
using a funnel & greasing a stopper of SMF.

11. Standardization Of 500 ml. HCl With 1 N NaOH
2. Preparation Of 1 N 500 ml. HCl Solution For
Standardization Against 1N NaOH.
A. Wash Funnel Several Times With Distilled Water By Using A Wash Bottle To
Transfer The Sticking Particles If Any Into The Standard Measuring Flask.
B. Add Distilled Water And Swirl The Standard Measuring Flask Till HCl Is
Completely Soluble.
C. Make 500 ml Solution.
D. NOTE:
When You Mix Acid With Water, It's Extremely Important To Add The Acid To The
Water. If You Add Water To Acid, They React In A Vigorous Exothermic Reaction
& Form An Extremely Concentrated Solution Of Acid Initially. So Much Heat Is
Released That The Solution May Boil Very Violently, Releasing Heat & Splashing
Concentrated Acid Out Of The Container.

12. Standardization Of 500 ml. HCl With 1 N NaOH
3. Titration For Standardization.
Burette:-Sodium Hydroxide Solution ( 1N NaOH Solution)
Iodine Flask :- ……. N HCl (25 ml).
Indicator :- Phenolphthalein
Color Change :- Colorless To Light Pink.
Procedure :-
 Clean The Burette And Funnel With Distilled Water .
 Rinse With NaOH And Fill It With Standardized 1 N NaOH Solution In Burette Up To 0.0 ml Mark
Level.
 Take Iodine Flask Which Contain 25 ml Of HCl Solution From SMF.
 Add 2 To 3 Drops Of Phenolphthalein Indicator And Shake Well.
 Titrate Against 1N NaOH Solution Drop Wise From The Burette Slowly With Constant Shaking.
 The End Point Is Marked By Appearance Of Permanent Light Pink Color.
 Repeat The Titration Till 3 Concordant Reading.

13. CALCULATION
B.R. = 23.7 ml of 1N NaOH solution is required for complete neutralization of 25 ml HCl
solution.
N1V1 = N2V2
N1 =Normality of NaOH solution = 1.0665 N
V1 = Volume of NaOH solution = B.R. = 23.7 ml
N2 = Normality of HCl solution = ……?…. N
V2 = Volume of HCl solution = 25 ml
N1V1 = N2V2
1.0665 x 23.7 = N2 x 25  N2 =
1.0665  23.7
25
= 1.015 N
Result :- Normality of HCl solution is 1.015 N

14. Experiment -2
Determine % Purity of KOH

15. HOW TO CALCULATE PURITY ?
Determine %
Purity of KOH
1 N HCl Solution
(Available From
Experiment - 1)
Preparation
Of KOH
Solution

16. Preparation Of KOH Solution (≡ 2 gm KOH Approximately)
Why Only 2 gm KOH?
We Have To Plan The Experiment So That Burette Reading Should Be Between
20 ml. to 30 ml. If The Burette Reading Is Less Than 10 ml., It Can Cause Major Errors,
Hence To Avoid This Issue, (Assuming 1 N HCl) Quantity Of 2 gm. KOH Should Be Used.
HCl Volume KOH
1 mole neutralize with 1 mole
36.5 gm neutralize with 56 gm
1000 ml (36.5gm) neutralize with 56 gm
20 ml neutralize with ? =
20  56
1000
= 1.12 gm KOH
Above calculation (ml and gm) is directly proportional as one amount increases, another amount increases
 If Purity of KOH is 100% then we required 1.12 gm KOH for 1 N, 20 ml HCl Solution.
 But here Purity of KOH = 84 %
 So, more amount of KOH required.

17. Standardization Of 1 N 500 ml. HCl With 1 N NaOH
% purity of KOH Wt. of KOH
100 % 1.12 gm
84% (Actual Purity of KOH) ? =
𝟏𝟎𝟎  𝟏.𝟏𝟐
𝟖𝟒
= 1.33  2 gm KOH
Above calculation (% and gm) is inversely proportional as one amount increases, another amount decreases
So we take 100/84 instead of 84/100
Here we take 2 gm KOH, so B.R. will get above 30 ml (increase)
We take Wt. of KOH = 2 gm approximately
2 gm KOH approx. + 80 ml Water = KOH Solution
Instead of 80 ml water, we can take 100 ml but it should not exceed volume by 150 ml.
Then titrate it against 1.015 N HCl Solution.
Indicator : Methyl Orange
Colour Change : Yellow to Light Orange

18. CALCULATION
B.R. HCl = 33.4 ml
Normality of HCl = 1.015 N
Weight of KOH = 2.2425 gm.
Eq. Wt. of KOH = 56 gm. /eq
% purity =
𝑩𝒖𝒓𝒆𝒕𝒕𝒆 𝑹𝒆𝒂𝒅𝒊𝒏𝒈 𝒕𝒊𝒕𝒓𝒂𝒏𝒕 × 𝑵𝒐𝒓𝒎𝒂𝒍𝒊𝒕𝒚 𝒕𝒊𝒕𝒓𝒂𝒏𝒕 × 𝑬𝒒.𝑾𝒕. (𝒂𝒏𝒂𝒍𝒚𝒕𝒆)
𝟏𝟎 × 𝑾𝒆𝒊𝒈𝒉𝒕 𝒐𝒓 𝑽𝒐𝒍𝒖𝒎𝒆 (𝒂𝒏𝒂𝒍𝒚𝒕𝒆)
% purity of KOH =
𝟑𝟑.𝟒 × 𝟏.𝟎𝟏𝟓 × 𝟓𝟔
𝟏𝟎 × 𝟐.𝟐𝟒𝟐𝟓
= 84.65 % (Actual purity of KOH = 84 %)

19. SUMMARY
Experiment – 1:
Prepare 500 ml
HCl Solution
Standardize The
HCl Solution With
1 N NaOH
EXAMPLES OF
Practical Video HCl
 1 N NaOH
CALCULATION:
N1V1 = N2V2
Experiment – 2:
Prepare KOH
Solution
CALCULATION
Practical Video KOH
 1 N HCL

20. THANK YOU VERY MUCH