Chemistry Practicals For HSC-Part 1 (XI -Science Group) - Experiment No.2 : Titration (KOH - H2SO4)

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TITRATION

By D. J. Sindh Govt. Science College
OBJECT:
(2) You are given decinormal (1/ 10 or 0.1 N) solution of KOH. Find out Normality and amount of given H₂SO₄ per 100 cm³ solution.

APPARATUS:

• Burette
• Pipette
• Conical Flask
• Burette Stand with clamp
• Dropper

THEORY:
ACID-BASE TITRATION:
Titration is a process in which the fixed volume of a solution taken by means of pipette into conical flask is compared with the volume of solution added from burette upto end point which is obtained by colour change. The purpose of titration is to determine the unknown strength of a solution.
Base is a substance which gives OH^- ion in aqueous solution. When an acid solution reacts completely with base solution or vice versa, a salt is formed.
Neutralization: In acid-base reaction H⁺ ion from an acid and OH^- ion from a base form H₂O, salt is also formed process is called neutralization. This is indicated by end point and is observe by suitable indicator those having different colors in different medium.
Strong acid and strong base ionize completely in aqueous solution while weak acid and weak base ionize only partially in aqueous solution, these form acidic and basic salts which hydrolyse to change the pH.

Na₂CO₃ + H₂O ⟶ NaOH + H₂CO₃

(Basic salt) + (water) (Strong base) (Weak acid)

where Basic (pH > 7)

EQUATIONS

2KOH + H₂SO₄ ⟶ K₂SO₄ + 2H₂O

IONIC EQUATION

2K⁺ + 20H^- + 2H⁺ + SO₄^2- ⟶ 2K⁺ + SO₄^2- + 2H₂0

PROCEDURE:
Strong Base Vs Stron Acid

1. Wash burette, pipette and conical flask with water.
2. Rinse burette with base (KOH) and pipette with acid (H₂SO₄).
3. Do not rinse the conical flask.
4. Fill the burette with base (KOH) upto Zero mark.
5. Pipette out 10 cm³ of acid (H₂SO₄) into the conical flask.
6. Add one or two drops of phenolphthalein as indicator.
7. Titrate the base in the burette with the acid in conical flask. The burette solution is added drop by drop into the conical flask.
8. Shake the flask after each addition, The end point is indicated by colour change from colourless to light pink.
9. Note the end point.
10. Wash the conical flask with water and repeats the same process for at least three times to obtain two similar readings.

OBSERVATION:
Volume Of NaOH = V₁ = ___ cm³

S.NO.	Initial reading	Final reading	Difference	Concordant reading
1.	cm3	cm3	cm3	___ cm3
2.	cm3	cm3	cm3
3.	cm3	cm3	cm3

• Solution in burette = KOH
• Solution in conical flask = H₂SO₄
• Indicator used = Phenophthalein
• Normality of standard solution Of KOH = N₁ = 0.1 N
• Volume of KOH solution = V₁ = _____ ml (Hint: Burette - concordant reading)
• Normality of H₂SO₄ solution = N₂ = ___ N
• Volume of standard solution Of H₂SO₄ = V₂ = 10 ml
• Amount of H₂SO₄ = _____ gms/cm³

CALCULATION:
Normality Of H₂SO₄

N_{1 (base)} x V_1(base) = N_{2 (acid)} x V_{2 (acid)}
OR

N₂ = (N₁ x V₁) / V₂ = ____ N

Equivalent mass Of H₂SO₄ = Molecular mass / Basicityty of H₂SO₄
Molecular mass of H₂SO₄ = 2(1) + 1(32) + 4(16) = 2 + 32 + 64 = 98 a.m.u
Equivalent mass Of H₂SO₄ = 98 / 2 = 49

Amount Of H₂SO₄ In gms / 100 cm³

RESULT:
Normality of given H₂SO₄ solution = ______ N
Amount of H₂SO₄ in the given solution = ____ gms / 100 cm³

OR

By Mubarka Naz
EXPERIMENT NO.2
OBJECT: You are provided N/18 (0.56 N) Solution of H₂SO₄. Determine the normality and amount of KOH in gms per 600 cm³.

APPARATUS:

• Burette
• Pipette
• Conical Flask
• Burette Stand
• Clamp with holder
• Dropper

THEORY:
When an acid reacts completely with an equivalent amount of a base neutralization takes place. Neutralization is the process in which H⁺ ion of an acid and OH^- ions of a base combine to form water. A salt is also formed during the neutralization.

Acid + Base ⟶ Water + Salt

EQUATION:

2KOH + H₂SO₄ ⟶ K₂SO₄ + 2H₂O

IONIC EQUATION:

2K⁺ + 20H^- + 2H⁺ + SO₄^2- ⟶ 2K⁺ + SO₄^2- + 2H₂0

PROCEDURE:
1) Wash the burette pipette and conical flask with water.
2) Rinse the burette with KOH solution and pipette with H₂SO₄ solution.
3) Fill the burette with KOH solution leaving no air bubble in the jet and clamp it vertically in to the burette stand.
4) Pipette out 10 cm³(ml) of H₂SO₄ solution in to the conical flask and add 1 - 2 drops of phenolphthalein indicator.
5) Start adding the solution of NaOH drop by drop from the burette into the conical flask, shake the flask after each addition.
6) Continue the addition of KOH solution till a permanent pinkish tinge appears with a single drop of KOH solution, record the reading of burette at this stage.
7) Wash the conical flask with water and repeat the process of titration till concordant reading is obtained.

OBSERVATIONS:
i) Solution in burette: KOH
ii) Solution in conical flask: H₂SO₄
iii) Normality of Acid: 0.56 N
iv) Normality of base: ? N
v) Indicator: Phenolphthalein
vi) Change of colour: colourless to pink

OBS Nos.	Initial Reading Of Burette (X cm3)	End Point Reading Of Burette (Y cm3)	Volume consume (Y - X cm3)
1.	0	8.0	8.0
2.	8.0	16.0	8.0
3.	16.0	24.0	8.0
4.

Concordant Reading = (8.0 + 8.0 + 8.0) / 3 = 8.0 cm³

CALCULATIONS:

RESULT:
Normality of the given KOH solution = ___ N or gram.equivalent/ltr/dm³.
Amount of KOH in the given solution = _____ gms / 600 cm³.

Chemistry Practicals For HSC-Part 1 (XI -Science Group) - Experiment No.1 : Titration (NaOH - HCl)

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TITRATION

By D. J. Sindh Govt. Science College
OBJECT:
(1) You are given 0.2 N solution of HCl. Find out Normality and amount of given NaOH per 1200 cm³ solution.

APPARATUS:

• Burette
• Pipette
• Conical Flask
• Burette Stand with clamp
• Dropper

THEORY:
ACID-BASE TITRATION:
Titration is a process in which the fixed volume of a solution taken by means of pipette into conical flask is compared with the volume of solution added from burette upto end point which is obtained by colour change. The purpose of titration is to determine the unknown strength of a solution.
Base is a substance which gives OH^- ion in aqueous solution. When an acid solution reacts completely with base solution or vice versa, a salt is formed.
Neutralization: In acid-base reaction H⁺ ion from an acid and OH^- ion from a base form H₂O, salt is also formed process is called neutralization. This is indicated by end point and is observe by suitable indicator those having different colors in different medium.
Strong acid and strong base ionize completely in aqueous solution while weak acid and weak base ionize only partially in aqueous solution, these form acidic and basic salts which hydrolyse to change the pH.

Na₂CO₃ + H₂O ⟶ NaOH + H₂CO₃

(Basic salt) + (water) (Strong base) (Weak acid)

where Basic (pH > 7)

EQUATIONS

NaOH + HCl ⟶ NaCl + H₂O

IONIC EQUATION

Na⁺ + OH^- + H⁺ + Cl^- ⟶ Na⁺ + Cl^- + H₂O

PROCEDURE:
Strong Base Vs Stron Acid

1. Wash burette, pipette and conical flask with water.
2. Rinse burette with base (NaOH) and pipette with acid (HCl).
3. Do not rinse the conical flask.
4. Fill the burette with base (NaOH) upto Zero mark.
5. Pipette out 10 cm³ of acid (HCl) into the conical flask.
6. Add one or two drops of phenolphthalein as indicator.
7. Titrate the base in the burette with the acid in conical flask. The burette solution is added drop by drop into the conical flask.
8. Shake the flask after each addition, The end point is indicated by colour change from colourless to light pink.
9. Note the end point.
10. Wash the conical flask with water and repeats the same process for at least three times to obtain two similar readings.

OBSERVATION:
Volume Of NaOH = V₁ = ___ cm³

S.NO.	Initial reading	Final reading	Difference	Concordant reading
1.	cm3	cm3	cm3	___ cm3
2.	cm3	cm3	cm3
3.	cm3	cm3	cm3

• Solution in burette = NaOH
• Solution in conical flask = HCl
• Indicator used = Phenophthalein
• Normality of standard solution Of NaOH = N₁ = ____ N
• Volume of NaOH solution = V₁ = _____ ml (Hint: Burette - concordant reading)
• Normality of HCl solution = N₂ = 0.2N
• Volume of standard solution Of HCl= V₂ = 10 ml
• Amount of NaOH = _____ gms/cm³

CALCULATION:
Normality Of NaOH

N₁ x V₁ = N₂ x V₂
OR

N₁ = (N₂ x V₂) / V₁ = ____ N

Equivalent mass Of NaOH = Molecular mass / Acidity of NaOH
Equivalent mass Of NaOH = 40 / 1 = 40

Amount Of NaOH In gms / 1200 cm³

RESULT:
Normality of given NaOH solution = ______ N
Amount of NaOH in the given solution = ____ gms / 1200 cm³

OR

By Mubarka Naz
EXPERIMENT NO.1
OBJECT: Given 0.1 N Solution of NaOH. Determine the normality and amount of HCl in gms per 800 cm³.

APPARATUS:

• Burette
• Pipette
• Conical Flask
• Burette Stand
• Clamp with holder
• Dropper

THEORY:
When an acid reacts completely with an equivalent amount of a base neutralization takes place. Neutralization is the process in which H⁺ ion of an acid and OH^- ions of a base combine to form water. A salt is also formed during the neutralization.

Acid + Base ⟶ Water + Salt

EQUATION:

HCl + NaOH ⟶ H₂O + NaCl

IONIC EQUATION:

H⁺ + OH^- + Na⁺ + OH^- ⟶ H₂O + Na⁺ + Cl^-

PROCEDURE:
1) Wash the burette pipette and conical flask with water.
2) Rinse the burette with NaOH solution and pipette with HCl solution.
3) Fill the burette with NaOH solution leaving no air bubble in the jet and clamp it vertically in to the burette stand.
4) Pipette out 10 cm³(ml) of HCl solution in to the conical flask and add 1 - 2 drops of phenolphthalein indicator.
5) Start adding the solution of NaOH drop by drop from the burette into the conical flask, shake the flask after each addition.
6) Continue the addition of NaOH solution till a permanent pinkish tinge appears with a single drop of NaOH solution, record the reading of burette at this stage.
7) Wash the conical flask with water and repeat the process of titration till concordant reading is obtained.

OBSERVATIONS:
i) Solution in burette: NaOH
ii) Solution in conical flask: HCl
iii) Normality of Acid: ?
iv) Normality of base: 0.1 N
v) Indicator: Phenolphthalein
vi) Change of colour: colourless to pink

OBS Nos.	Initial Reading Of Burette (X cm3)	End Point Reading Of Burette (Y cm3)	Volume consume (Y - X cm3)
1.	0	6.1	6.1
2.	6.1	12.2	6.1
3.	12.2	18.2	6.0
4.	18.2	24.3	6.1

Concordant Reading = (6.1 + 6.1 + 6.0 + 6.1) / 4 = 6.075 cm³

CALCULATIONS:

RESULT:
Normality of the given HCl solution = ___ N or gram.equivalent/Itr/dm³.
Amount of HCl in the given solution = _____ gms / 800 cm³.

Chemistry Practicals For HSC-Part 1 (XI -Science Group) - Titration - General Descripttion - DJ Sindh Government College

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TITRATION

EQUATIONS:
NaOH + HCl ⟶ NaCl + H₂O
2KOH + H₂SO₄ ⟶ K₂SO₄ + 2H₂O
KOH + HNO₃ ⟶ KNO₃ + H₂O
2NaOH + H₂C₂0₄.2H₂0 ⟶ Na₂C₂0₄ + 4H₂0
2KOH + H₂C₂0₄.2H₂0 ⟶ ₂C₂0₄ + 4H₂0
2HCI + Na₂CO₃ ⟶ 2NaCI + CO₂ + H₂O
H₂SO₄ + K₂CO₃ ⟶ K₂SO₄ + CO₂ + 2H₂0
2HNO₃ + K₂CO₃ ⟶ 2KNO₃ + CO₂ + H₂O
H₂SO₄ + Na₂CO₃ ⟶ Na₂SO₄ + CO₂ + H₂O

2KMn0₄ + 3H₂SO₄ ⟶ K₂SO₄ + 3H₂0 + 2MnSO₄ + 5[O]
10FeSO₄ + 5H₂SO₄ + 5[0] ⟶ 5Fe₂(SO₄)₃ + 5H₂0

---

2KMn0₄ + 10FeSO₄ + 8H₂SO₄ ⟶ K₂SO₄ + 2MnSO₄ + 5Fe₂(SO₄)₃ + 8H₂0

---

2KMn0₄ + 3H₂SO₄ ⟶ K₂SO₄ + 3H₂0 + 2MnSO₄ + 5[O]
5H₂C₂O₄ + 5[0] ⟶ 10CO₂ + 5H₂0

---

2KMn0₄ + 5H₂C₂O₄ + 3H₂SO₄ ⟶ K₂SO₄ + 2MnSO₄ + 10CO₂ + 8H₂0

---

IONIC EQUATION FOR ACID — BASE
Na⁺ + OH^- + H⁺ + Cl^- ⟶ Na⁺ + Cl^- + H₂O

2K⁺ + 20H^- + 2H⁺ + SO₄^2- ⟶ 2K⁺ + SO₄^2- + 2H₂0

K⁺ + OH^- + H⁺ + NO₃^- ⟶ K⁺ + NO₃^- + H₂0

2Na⁺ + 20H^- + 2H⁺ + C₂0₄^2- ⟶ 2Na⁺ + C₂0₄^2- + 2H₂0

2K⁺ + OH^- + 2H⁺ + C₂0₄^2- ⟶ 2K⁺ + C₂0₄^2- + 2H₂0

2H⁺ + 2Cl^- + 2Na⁺ + CO₃^2- ⟶ 2Na⁺ + 2Cl^- + CO₂ + H₂O

2H⁺ + SO₄^2- + 2K⁺ + CO₃^2- ⟶ 2K⁺ + SO₄^2- + CO₂ + H₂O

2H⁺ + 2NO₃^- + 2K⁺ + CO₃^2- ⟶ 2K⁺ + 2NO₃^- + CO₂ + H₂0

2H⁺ + SO₄^2- + 2Na⁺ + CO₃^2- ⟶ 2Na⁺ + SO₄^2- + CO₂ + H₂O

THEORY
FOR ACID-BASE TITRATION:
Titration is a process in which the fixed volume of a solution taken by means of pipette into conical flask is compared with the volume of solution added from burette upto end point which is obtained by colour change. The purpose of titration is to determine the unknown strength of a solution.
Base is a substance which gives OH^- ion in aqueous solution. When an acid solution reacts completely with base solution or vice versa, a salt is formed.
Neutralization: In acid-base reaction H⁺ ion from an acid and OH^- ion from a base form H₂O, salt is also formed process is called neutralization. This is indicated by end point and is observe by suitable indicator those having different colors in different medium.
Strong acid and strong base ionize completely in aqueous solution while weak acid and weak base ionize only partially in aqueous solution, these form acidic and basic salts which hydrolyse to change the pH.

Na₂CO₃ + H₂O ⟶ NaOH + H₂CO₃

(Basic salt) + (water) (Strong base) (Weak acid)

where Basic (pH > 7)

FOR REDOX TITRATION:
Titration is a process in which the fixed volume of a solution taken by means of pipette Into conical flask is compared with the volume of solution added from burette upto end point which is obtained by colors change. The purpose of titration is to determine the unknown strength of a solution using standard solution.
Redox Titration: The type of titration in which one reactant in oxidized and other reduced are know as redox titration.
Oxidation is a process in which reducing Agent loses electron and absorbs oxygen, so the oxidation number increases.

Fe²⁺ ⟶ Fe³⁺ + e^-

2FeSO₄ + H₂SO₄ + [0] ⟶ Fe₂(SO₄)₃ + H₂O

Reduction is a process in which oxidizing agent gain electron and loses oxygen, so oxidation number decreases

Mn⁷⁺ + 5e^- ⟶ Mn²⁺

2KMnO₄ + H₂SO₄ ⟶ K₂SO₄ 2MnSO₄ + 3H₂0 + 5[O]

Oxidizing strength of KMnO₄ increase in the presence of H₂SO₄, so H₂SO₄ is required in redox titration. KMnO₄ is coloured substance so it acts as indicator, therefore no external indicator is required.
Reaction between KMnO₄ and oxalic acid is very slow because oxalic acid is a covalent organic compound. So the Titration flask is heated to about 60 — 70°C before adding KMnO₄ from the burette.

PROCEDURE:
Strong Base Vs Stron Acid OR Strong Base Vs Weak Acid

1. Wash burette, pipette and conical flask with water.
2. Rinse burette with base and pipette with acid.
3. Do not rinse the conical flask.
4. Fill the burette with base upto Zero mark.
5. Pipette out 10 cm³ of acid into the conical flask.
6. Add one or two drops of phenolphthalein as indicator.
7. Titrate the base in the burette with the acid in conical flask. The burette solution is added drop by drop into the conical flask.
8. Shake the flask after each addition, The end point is indicated by colour change from colourless to light pink.
9. Note the end point.
10. wash the conical flask with water and repeats the same process for at least three times to obtain two similar readings.


Strong Acid Vs Weak Base
1. Wash burette, pipette and conical flask with water.
2. Rinse burette with acid and pipette with base solution.
3. Do not rinse the conical flask.
4. Fill the burette with acid upto Zero mark.
5. Pipette out 10 cm³ of base into the conical flask.
6. Add one or two drops of methyl orange as indicator.
7. Titrate the acid in the burette with the base in conical flask. The burette solution is added drop by drop into the conical flask.
8. Shake the flask after each addition, The end point is indicated by colour change from yellow to light pink.
9. Note the end point.
10. Wash the conical flask with water and repeat the same process for at least three times to obtain two similar readings.


Redox Titration
1. Wash all the apparatus with water.
2. Rinse burette with KMnO₄ and pipette with reducing agent.
3. Do not rinse the conical flask.
4. Fill the burette with base upto Zero mark.
5. Pipette out 10 cm³ of given reducing agent into the conical flask.
6. Add one test tube of dilute H₂SO₄ (In case of oxalic acid as a reducing agent titration flask is heated to about 60 - 70°C before titration).
7. Titrate KMnO₄ in the burette with the reducing agent in conical flask. The burette solution is added drop by drop into the conical flask.
8. Shake the flask after each addition, the end point is indicated by colour change from colourless to light pink.
9. Note the end point.
10. Wash the conical flask with water repeat the same process for at least three times to obtain two similar readings.

OBSERVATION:

S.NO.	Initial reading	Final reading	Difference	Concordant reading
1.	cm3	cm3	cm3	cm3
2.	cm3	cm3	cm3
3.	cm3	cm3	cm3

• Solution in burette = ______ (Strong Acid Or Strong Base)
• Solution in conical flask = ________ (Weak Acid Or Weak Base)
• Indicator used =  _______ (e.g Phenophthalein for Strong Base)
• Normality of standard solution = N₁ = ___ N
• Volume of standard solution = V₁ = 10 ml
• Normality of solution of unknown strength = N₂ = ?
• Volume of solution of unknown strength = V₂ = ?

CALCULATION:
Standard solution Vs Solution of unknown strength

N₁V₁ = N₂V₂

Equivalent mass = Molecular mass / Acidity or Basicity or change in oxidation state
Amount = Normality (N₂) x equivalent mass x volume of solution in cm³ / 1000

RESULT:

• Normality of given ________ solution = ______ N
• Amount of ______ in the given solution = gms / ___ dm³ or ____ cm³

Acidity Of Base:

Base	Acidity	Equivalent mass
NaOH	1	40
KOH	1	56
Na2CO3	2	53
K2CO3	2	69

Basicity Of Acid

Acid	Basicity	Equivalent mass
HCl	1	36.5
H2SO4	2	49
HNO3	1	63
C2H204.2H20	2	63

Oxidizing Agent

Oxidizing agent	No. of electron gained	Equivalent mass
KMnO4	5	31.6

Reducing Agent

Reducing Agent	No. of electron lost	Equivalent mass
FeSO4.7H20	1	278
FeSO4.(NH4)2SO4.6H20	1	392
H2C204.2H20	2	63