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Laboratory Manual Course Code: PHARM 1206 Prepared By Md. Imran Nur Manik Lecturer Department of Pharmacy Northern University Bangladesh

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Page 1: Physical Pharmacy-I Lab MANIK

Laboratory Manual

Course Code: PHARM 1206

Prepared ByMd. Imran Nur Manik

Lecturer

Department of Pharmacy

Northern University Bangladesh

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Manual: Physical Pharmacy-I Lab

Prepared By: Md. Imran Nur Manik Page 0

Lecturer; Department of Pharmacy; Northern University Bangladesh.

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Sl. No. Date Name of the experiment Page No.

01. Preparation of solutions of different concentrations. 01 – 02

02. Standardization of NaOH solution by Potassium Hydrogen

Phthalate (KHP). 03 – 05

03. Standardization of HCl Solution by NaOH Solution. 06 – 07

04. Determination of pKa value of weak acid. 08 – 10

05. Preparation of constant pH buffer. 11 – 12

Bibliography and Some commonly used Lab Equipment. 12

Md. Imran Nur Manik

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Manual: Physical Pharmacy-I Lab

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Experiment No. 01 Date:

Name of the experiment: Preparation of solutions of different concentrations.

Principle:

The concentration of a solution is defined as: the amount of solute present in a given amount of solution.

The concentration of a solution can be expressed as molarity, molality normality, mole fraction etc.

Molarity (M): Molarity is the number of moles of solute dissolved per liter of solution. It can be mathematically

expressed as, solutionofLitre

soluteofMoles)MMolarity(

Example: Calculate the molar concentration of sodium sulphate (Na2SO4; MW=142 gm).That contains 2.8 gm of

Na2SO4 in 1500mL solution.

SOLUTION:

)i.........(..........litre)V(in

nMor,

solutionofLitre

soluteofMoles)MMolarity(That,Know We

Here, Molecular mass of Na2SO4 = (232 + 32 + 164) gm=142 gm

Calculation of moles (n) of Na2SO4: 142 gm Na2SO4= 1mol

2.8 gm Na2SO4= 142

8.2mol= 0.01972 mol

Calculation of volume (V) in litres:

1000 millilitre =1litre

1500 millilitres = 1000

1500

litre =1.5 litre

Calculation of Molarity:

mole/L 0.013151.5

0.01972M

V

nMget, we) i (equation thein valuesthesePutting

Thus the solution is 0.01315 M

Problem: Calculate the amount required to prepare 250 mL, 0.25M HCl. Purity of acid is 32%.

Normality (N): Normality is the number of gram equivalent (eq.) weights of solute (solu.) per liter of solution.

Equivalent mass (Eq. M.)/Gram equivalent Wight: Equivalent mass of a substance can be calculated

by dividing its molar mass per the number of active units in one molecule of this substance (k) thus:

The active unit in the acid-base reactions is the number of hydrogen ions liberated by a single molecule of an acid

or reacted with a single molecule of a base.

For e.g. 1 mole of NaOH (m.wt. = 40) can combine with 1 mole of hydrogen ion, therefore the equivalent weight

of NaOH is 40÷1=40 gm. Thus the concentration of 1N NaOH is same as 1M NaOH.

On the other hand H2SO4 has two ionisable hydrogen atoms; its equivalent mass will be, g492

98Eq.M.

Thus, the concentration of 1M H2SO4 is same as 2N H2SO4.

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Relationship between Normality and Molality

Relationship between molarity and normality for the same solute in the same solution is expressed by the

following equation

N=nM

Where, N=normality M=Molarity and n=Valence

Problem: Calculate the amount required to prepare 0.1N H2SO4.Having density of 1.19 g/mL and 98%

purity.

Concentration can also be calculated by dilution method when the concentration and volume of one of the

two reactants is known.

Dilution: It is the addition of solvent to decrease the concentration of solute. The solution volume changes, but

the amount of solute remains constant.

M1V1 = M2V2

Initial values Final values

Problem: How to made 100 mL 0.1M NaCl from 5M NaCl solution?

Calculation: All calculations are to be done as per the solutions to be made, needed for the experiment.

Apparatus: 1. Electrical balance

2. Pipette and pipette filler

3. Funnel

4. Volumetric flask

5. Beaker

6. Measuring cylinder

7. Spatula

8. Stirrer

9. Permanent marker

Chemicals/Reagents: Write all the reagents required to prepare the solutions (as per the solutions to be made for the experiment)

Experimental procedures: (Specimen)

Preparation of 1M 50 mL NaOH solution:

1. Weigh out accurately W gm (To be calculated) of NaOH.

2. Transfer it to 50 mL volumetric flask.

3. At first completely dissolve it with small volume of distilled water (DW).

4. Finally adjust the volume up to the 50 mL mark with DW Q.S.

Preparation of 1N 100 mL Na2CO3 solution:

1. Weigh out accurately W gm

(To be calculated) of Na2CO3.

2. Transfer it to 100 mL volumetric flask.

3. At first completely dissolve it with small

volume of distilled water (DW).

4. Finally adjust the volume up to the

mark with DW Q.S.

Fig: Preparation of Solution.

Precautions:

Comments:

Md. Imran Nur Manik

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Manual: Physical Pharmacy-I Lab

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Experiment No. 02 Date:

Name of the experiment: Standardization of NaOH solution by Potassium Hydrogen Phthalate (KHP).

Terminologies Used in Titrimetric Analysis

The term 'titrimetric analysis' refers to quantitative chemical analysis. It is carried out by determining the

volume of a solution of accurately known concentration, required to react quantitatively with a measured volume

of a solution of the substance to be determined. The solution of accurately known strength is called the standard

solution.

In titrimetric analysis the reagent of known concentration is called the titrant and the substance being

titrated is termed the titrand or analyte.

The process of adding the standard solution until the reaction is just complete is termed a titration, and

the substance to be determined is titrated. The point at which this occurs is called the equivalence point or the

theoretical (or stoichiometric) end point.

The completion of the titration is detected more usually, by the addition of an auxiliary reagent, known

as an indicator. After the reaction between the substance and the standard solution is practically complete, the

indicator should give a clear visual change (either a colour change or the formation of turbidity) in the liquid

being titrated. The point at which this occurs is called the end point of the titration.

Principle In titrimetry, chemicals used as reference solutions known as primary standards or secondary standards.

Primary Standards A primary standard is a compound of sufficient purity from which a standard solution can be prepared by

direct weighing of a quantity of it, followed by dilution to give a defined volume of solution. The solution

produced is called a primary standard solution.

A primary standard should satisfy the following requirements.

1. It must be easy to obtain, purify, dry (preferably at 110-120°C), and to preserve in a pure state.

(This requirement is not usually met by hydrated substances, since it is difficult to remove surface moisture completely without effecting partial decomposition.)

2. It should be unaltered in air during weighing; that means, it should not be hygroscopic, oxidised by air, or

affected by carbon dioxide.

3. It should maintain an unchanged composition during storage.

4. It should be capable of being tested for impurities by qualitative and other tests of known sensitivity.

(The total amount of impurities should not, in general, exceed 0.01-0.02%) 5. It should be readily soluble under the conditions in which it is employed.

6. The reaction with the standard solution should be stoichiometric and practically instantaneous.

7. The titration error should be negligible, or easy to determine accurately by experiment.

In practice, an ideal primary standard is difficult to obtain, and a compromise between the above ideal

requirements is usually necessary.

The substances commonly employed as primary standards are indicated below:

(a) Acid-base reactions: Sodium carbonate (Na2CO3); sodium tetraborate (Na2B4O7),

potassium hydrogen phthalate (C8H5O4K), constant boiling point hydrochloric acid, potassium hydrogeniodate

KH(IO3)2, benzoic acid (C6H5COOH) etc.

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(b) Complex formation reactions: Silver, Silver nitrate (AgNO3), Sodium chloride (NaCl), various metals

(e.g. Spectroscopically pure Zinc, Magnesium, Copper, and Manganese) and salts, depending upon the reaction

used.

(c) Precipitation reactions: Silver, Silver nitrate (AgNO3), Sodium chloride(NaCl), Potassium chloride (KCl),

and Potassium bromide(KBr) (prepared from potassium bromate).

(d) Oxidation-reduction reactions: Potassium dichromate (K2Cr2O7), Potassium bromate (KBrO3),

Potassium iodate (KIO3), Potassium hydrogeniodate KH(IO3)2, Sodium oxalate (Na2C2O4),

Arsenic(III) oxide (As2O3), and pure Iron.

Secondary Standards Solutions that are prepared by standardisation against a primary standard are referred to as secondary standards.

It follows that, a secondary standard solution is a solution in which the concentration of dissolved solute has not

been determined from the weight of the compound dissolved, but by the reaction (titration) of a volume of the

solution, against a measured volume of a primary standard solution.

Examples: Sodium tetraborate Na2B4O7.10H2O , Copper sulphate Cu2SO4.5H2O etc.

In the present experiment, NaOH solution is to be used secondary standard chemical and needs to be

standardized by the primary standard chemicals. Here the primary standard chemical is KHP.

This is a 1:1 titration therefore, one mole of base, will titrate one mole of acid.

The end point of the solution would be determined by an indicator, phenolphthalein.

Reaction:

+Na+, K+

Phenolphthalein: Phenolphthalein exists in two tautomeric forms: (i) the benzenoid form which is yellow and

present in basic solution; and (ii) the quinonoid form which is pink and present in acid solution.

Fig. Two forms of Phenolphthalein.

Apparatus:

1. Electrical balance

2. Spatula

3. Volumetric flask

4. Measuring cylinder

5. Burette

6. Pipette and pipette filler

7. Conical flask

8. Beaker

9. Funnel

Chemicals / Reagents:

1. 0.1 N Sodium hydroxide (NaOH) solution

2. 0.1 N Potassium hydrogen phthalate (C8H5O4K)

3. 0.5% Phenolphthalein indicator

4. Distilled Water

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Calculation (Sample): Full calculations required to make the reagents, for the experiment must be written down

Gram Equivalent weight of NaOH = (40÷ 1)= 40 gm

250 mL 0.1N NaOH = {(40250 0.1) ÷1000} gm = 1gm (for 100% pure NaOH)

Gram Equivalent weight of KC8H5O4 = ( 204.22÷ 1) gm = 204.22 gm

50 mL 0.1N KHC8H4O4 = {(204.22500.1) ÷1000} gm = 1.021 gm (for 100% pure KHC8H4O4)

Reagents and their preparations:

1. 0.1 N 50 mL Potassium Hydrogen Phthalate solution: Take 1.021gm Potassium hydrogen phthalate

(KHP) in a 50 mL volumetric flask; now dissolve it completely with small portion of DW. Finally add distilled

water Q.S. to make 50 mL solution.

2. 0.1 N 250 mL NaOH solution: Take 1 gm NaOH in a 250 mL volumetric flask. Completely dissolve

it with small portion of DW (e.g.125 mL). Finally adjusted the volume up to the 250 mL mark by Q.S. of distilled

water.

3. 0.5% 100 mL Phenolphthalein solution: 50 mL Ethanol+ 0.5 gm Phenolphthalein+ 50 mL DW.

Procedures:

1. Standardization of NaOH solution by Potassium Hydrogen Phthalate (KHP) solution:

i. Fill the burette with the prepared 0.1 N NaOH solution.

ii. Take 10 mL standard 0.1N KHP (C8H5O4K) solution in a conical flask.

iii. Add 1-2 drops of Phenolphthalein indicator in the acid solution and titrate it with the NaOH solution,

until the colour of the acid solution changes from colourless to faint pink.

iv. Perform another two titrations and calculate the result.

Table-1: Data for the standardization of NaOH solution:

No. of

observations

Volume of KHP

(V1 mL)

Volume of NaOH solution

(mL)

Difference

(FBR-IBR)

(mL)

Mean

volume

(V2 mL) IBR FBR

1 10

2 10

3 10

2. Calculation of strength of NaOH solution:

We know that, V1S1 = V2S2 Here, Volume of KHP, V1 = 10 mL

S2 =

2

11

V

SV Strength of KHP,S1 = 0.1 N

Volume of NaOH, V2 = mL (Mean)

= Strength of NaOH, S2= ?

= N

Result: The strength of NaOH= N.

Precautions:

Comments:

Md. Imran Nur Manik

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Experiment No. 03 Date:

Name of the experiment: Standardization of HCl Solution by NaOH Solution.

Principle An acid-base titration involves the addition of a titrant solution to an analyte solution. In titrimetry, chemicals used

as reference solutions known as primary standards or secondary standards.

A primary standard is a reagent that is extremely pure, stable, has no water of hydration, and has a high

molecular weight. A secondary standard solution refers to a solution that has its concentration measured by

titration with a primary standard solution.

The secondary standard chemicals are not absolute pure and may contain some impurities that come during their

synthesis. Therefore the secondary standard chemical should standardize by using the primary pure chemicals.

In the present experiment, NaOH solution is to be used secondary standard chemical and needs to be

standardized by the primary standard chemicals. Here the primary standard chemical is Oxalic acid. After being

standardized by the Oxalic acid, NaOH is to be used for the standardization of HCl.

The end point of the solution will be determined by an indicator, phenolphthalein.

Reaction:

1. H2C2O4 (aq) + 2NaOH (aq) ⇌ Na2C2O4 (aq) + 2H2O (l)

2. HCl (aq) + NaOH (aq) ⇌H2O (l) + NaCl (aq)

Apparatus:

1. Electrical balance

2. Spatula

3. Volumetric flask

4. Measuring cylinder

5. Burette

6. Pipette and pipette filler

7. Conical flask

8. Beaker

9. Funnel

Chemicals / Reagents:

1. 0.1 N Sodium hydroxide (NaOH) solution

2. 0.1 N Oxalic acid (dihydrate) [H2C2O4.2H2O]

3. 0.1 N Hydrochloric acid (HCl) solution

4. 0.5% Phenolphthalein indicator

5. Distilled Water

Calculation (Sample): Full calculations required to make the reagents, for the experiment must be written down. Gram equivalent weight of H2C2O4 = (126.08 ÷ 2) gm = 63.040 gm

100 mL 0.1N H2C2O4 = {(63.0401000.1)÷ ÷1000 }gm = Y gm (for 100% pure H2C2O4)

Reagents and their preparations:

1. 0.1 N 100 mL Oxalic acid solution: Completely dissolve 0.636 gm of Oxalic acid (99%) in a 100 mL

volumetric flask with a small volume of DW. Finally adjust the volume up to 100 mL mark by adding

distilled water Q.S.

2. 0.1 N 1000 mL NaOH solution: Completely dissolve 4.12 gm of NaOH (97%) in a1000 mL volumetric

flask with a small volume of DW. Finally adjust the volume up to 1000 mL mark by adding distilled water

Q.S.

3. 0.1N 100 mL HCl solution: Take 1.14 mL of HCl (32%) and completely dissolve it in 100 mL

volumetric flask in 100 mL DW.

4. 0.5% 100 mL Phenolphthalein solution: 50 mL Ethanol+ 0.5 g Phenolphthalein+ 50 mL DW.

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Procedures: 1. Standardization of NaOH solution by Oxalic acid Solution:

a) Fill the burette with the prepared 0.1 N NaOH solution.

b) Take 10 mL standard 0.1N Oxalic acid solution in a conical flask.

c) Add 1-2 drops of Phenolphthalein indicator in the acid solution and titrate it with the NaOH solution,

until the colour of acid solution changes from colourless to faint pink. d) Perform another two titrations and calculate the result.

Table-1: Data for standardization of NaOH solution:

No. of

observations

Volume of Oxalic acid

(V1 mL)

Volume of NaOH solution

(mL)

Difference

(FBR-IBR)

(mL)

Mean

volume

(V2 mL) IBR FBR

1 10

2 10

3 10

2. Calculation of strength of NaOH solution:

We know that, V1S1 = V2S2 Here, Volume of Oxalic acid, V1 = 10 mL

S2 =

2

11

V

SV Strength of Oxalic Acid, S1= 0.1 N

Volume of NaOH, V2 = mL (Mean)

= Strength of NaOH, S2 = ?

= N

The strength of NaOH= N.

3. Standardization of HCl solution by NaOH solution:

a) Fill the burette with the standardized NaOH solution.

b) Take 10 mL prepared 0.1N HCl solution in a conical flask.

c) Add 1-2 drops of Phenolphthalein indicator in the acid solution and titrate it with the NaOH solution,

until the colour of acid solution changes from colourless to faint pink. d) Perform another two titrations and calculate the result.

Table-2: Data for standardization of HCl solution:

No. of

observations

Volume of HCl

(V1 mL)

Volume of NaOH solution

(mL)

Difference

(FBR-IBR)

(mL)

Mean volume

(V2 mL)

IBR FBR

1 10

2 10

3 10

4. Calculation of strength of HCl solution:

We know that, V1S1 = V2S2 Here, Volume of HCl, V1 = 10 mL

S1 =1

22

V

SV Strength of HCl, S1 = ?

Volume of NaOH, V2 = mL (Mean)

= Strength of NaOH, S2 = N (Known Earlier)

= N

Result: The strength of HCl= N.

Precautions:

Comments:

Md. Imran Nur Manik

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Experiment No. 04 Date:

Name of the experiment: Determination of pKa value of weak acid.

Principle A buffer solution is one which resists (or buffers) a change in its pH. If acid is added then, within reason, the pH

does not fall; if base is added, the pH does not rise. Buffers are usually composed of a mixture of weak acids or

weak bases and their salts and function best at a pH equal to the pKa of the acid or base involved in the buffer.

The equation that predicts the behaviour of buffers is known as the Henderson–Hasselbalch equation

(named after chemists Lawrence Joseph Henderson and Karl Albert Hasselbalch). It is derived as follows, by

considering a weak acid that ionises in solution.

[Acid]

[Salt]logpHpKa or,

[Acid]

[Salt]logpKapH

Where, pKa=–log Ka; and Ka=Decomposition constant of acid.

An example of a buffer is a mixture of acetic acid and sodium acetate, which ionises and acts as follows:

CH3COOH⇌CH3COO– +H+

CH3COO–Na+⇌CH3COO– +Na+

Apparatus:

1. Electrical balance

2. Spatula

3. Volumetric flask

4. Measuring cylinder

5. Burette

6. Pipette and pipette filler Fig. Mechanism of Buffer action of an acid buffer.

7. Conical flask

8. Beaker

9. Funnel

Chemicals / Reagents:

1. 0.1 N Sodium hydroxide (NaOH) solution

2. 1% Acetic acid (CH3COOH) solution

3. 0.1 N Oxalic acid (dihydrate)[ H2C2O4.2H2O]

4. 0.5% Phenolphthalein indicator

5. Distilled Water

Calculation: Full calculations required to make the reagents, for the experiment must be written down.

Preparation of Reagents:

A. 1% 100 mL Acetic Acid Solution: Take 1 mL glacial acetic acid (100%) in a 100 mL volumetric flask, and

then adjust the volume up to 100 mL mark by adding distilled water.

B.0.1 N 1000 mL NaOH Solution: Completely dissolve 4.12 gm NaOH (97%) pellets in a 1000 mL volumetric

flask at first with a small portion of DW, and finally adjust the volume up to the mark by adding DW Q.S.

C.0.1 N 100 mL Oxalic Acid Solution: Completely dissolve 0.636 gm oxalic acid (99%) in a 100 mL volumetric

flask at first with a small portion of DW, and finally make the volume up to mark by adding distilled water Q.S.

D.0.5% Phenolphthalein: 50 mL Ethanol + 0.5 gm phenolphthalein + 50 mL distilled water.

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Procedures:

1. Standardization of NaOH solution by Oxalic acid Solution:

a) Fill the burette with the prepared 0.1 N NaOH solution.

b) Take 10 mL standard 0.1N Oxalic acid solution in a conical flask.

c) Add 1-2 drops of Phenolphthalein indicator in the acid solution and titrate it with the NaOH solution,

until the colour of the acid solution changes from colourless to faint pink. d) Perform another two titrations and calculate the result.

Table-1: Data for standardization of NaOH solution:

No. of

observations

Volume of Oxalic acid

(V1 mL)

Volume of NaOH solution

(mL)

Difference

(FBR-IBR)

(mL)

Mean

volume

(V2 mL) IBR FBR

1 10

2 10

3 10

2. Calculation of strength of NaOH solution:

We know that, V1S1 = V2S2 Here, Volume of Oxalic acid,V1 = 10 mL

S2 =

2

11

V

SV Strength of Oxalic acid, S1 = 0.1 N

Volume of NaOH, V2 = mL (Mean)

= Strength of NaOH, S2= ?

= N

The strength of NaOH = N.

3. Determination of the strength of CH3COOH solution by NaOH solution:

a) Fill the burette with the standardized NaOH solution.

b) Take 10 mL prepared Acetic acid (CH3COOH) solution in a conical flask.

c) Add 1-2 drops of Phenolphthalein indicator in the acid solution and titrate it with the NaOH solution,

until the colour of the acid solution changes from colourless to faint pink. d) Perform another two titrations and calculate the result.

Table-2: Data for standardization of CH3COOH solution:

No. of

observations

Volume of CH3COOH

(V1 mL)

Volume of NaOH solution

(mL)

Difference

(FBR-IBR)

(mL)

Mean

volume

(V2 mL) IBR FBR

1 10

2 10

3 10

4. Calculation of strength of Acetic acid solution:

We know that, V1S1 = V2S2 Here, Volume of CH3COOH, V1 = 10 mL

S1 =1

22

V

SV Strength of CH3COOH, S1 = ?

Volume of NaOH, V2 = mL (Mean)

= Strength of NaOH, S2 = N (Known Earlier)

= N

The strength of CH3COOH = N.

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5. Determination of pH of Buffer solution:

1) Take 40 mL 1% Acetic acid solution and 20 mL standardized NaOH solution in a 100 mL beaker and

mix them very well.

2) Then measure the pH of the buffer solution using a pH meter.

The pH of the solution=

6. Determination of pKa of the solution:

Now calculate pKa value is by the following equation,

volumeTotal

COOHCH ofStrength COOHCH Remaining of VolumeVolume Total

NaOH ofStrength NaOH of Volume

logpHpKa

getw)equation(1thefromThus

VolumeTotal

COOHCHofStrengthCOOHCH Remaining of Volume[Acid]Acid,ofionconcentratexperimentpresenttheIn

VolumeTotal

COOHCHofStrengthCOOHCHofVolume[Acid]Acid,ofionconcentrattheAnd

VolumeTotal

NaOHofStrengthNaOHofVolume[Salt]Salt,ofionconcentrattheHere,

)1..(..............................................................................................................[Acid]

[Salt]logpHpKa

33

33

33

e

Result: The pKa value of the solution =

Precautions:

Comments:

Md. Imran Nur Manik

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Experiment No. 05 Date:

Name of the experiment: Preparation of constant pH buffer.

Principle:

A solution which resists changes in pH when small quantities of an acid or an alkali are added to it, is called

buffer solution. Most of the buffer solutions usually consist of a mixture of weak acid and one of its salts or a

weak base and one of its salts. The resistance to a change in pH is known as buffer action. A buffer solution

contains large and equal concentrations of an acid and its conjugate base. The pH of this solution is

approximately equal to the pKa of the acid. Addition of small amounts of acid or base results in the mopping up

or the release of protons by the conjugate base or the acid as necessary, which keeps the solution pH constant.

TYPES OF BUFFERS

Buffers are classified into following types

1. Simple Buffers: Simple buffers are categorized into three different ways.

A) Salts of weak acid and a weak base

Example: Ammonium acetate-CH3COONH4, Ammonium cyanide-NH4CN.

B) Proteins and amino acids.

C) A mixture of an acid salt and a normal salt formed from polybasic acid

Example: Na2HPO4 and Na3PO4

2. Mixed Buffers Acidic Buffer: A solution of weak acid and its salt with strong base.

Example: CH3COOH +CH3COONa, H2CO3 +Na2CO3

Basic Buffer: A solution of weak base and its salt with strong acid.

Example: NH4OH+NH4Cl, NH4OH+NH4NO3

3. Natural Buffers A solution is said to be naturally buffered if it contains buffering compounds

as it exists in nature. Blood is an example of a naturally buffered solution.

Fig.Citric acid monohydrate

Fig. pH meter Apparatus:

1. Electrical balance

2. Spatula

3. Volumetric flask

4. Conical flask

5. Beaker

6. Measuring cylinder

7. pH meter

8. Stirrer

Chemicals/Reagents:

1. 0.1 M Na-Acetate (CH3COONa) solution

2. Citric acid (monohydrate)

3. Distilled Water

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Calculation: Full calculations required to make the reagents, for the experiment must be written down.

Preparation of Reagents:

0.1 M 100 mL CH3COONa Solution: Completely dissolve 1.374 gm CH3COONa (99%) in a 100 mL volumetric

flask with a small amount of DW. Later on adjust the volume up to the mark by adding distilled water Q.S.

Procedures:

1. Take 100 mL 0.1 M CH3COONa solution in a 250 mL beaker.

2. Carefully submerge the detector of pH meter into the CH3COONa solution.

3. Take the reading of pH meter.

4. Now add crystalline citric acid (C6H8O7.H2O) to the CH3COONa solution with continuous stirring by

the stirrer.

5. Carefully submerge the detector of pH meter into the solution.

6. Take the reading of pH meter simultaneously while adding citric acid.

7. Continue the step 4 until the pH is adjusted to 6 or above.

Table-1: Data for pH adjustment:

No. of observation pH of CH3COONa Solution pH of final solution

1

Result: The pH value of the solution=

Precautions:

Comments:

Bibliography:

1. G H Jeffery, J Bassett, J Mendham, and R C Denney: Vogel's textbook of

quantitative chemical analysis, 5th edn, ( 1989). Longman Group UK Limited, Longman Scientific & Technical.

2. Arun Bahl, B.S Bahl, and G.D Tuli: Essentials of Physical Chemistry, Multi colour edn. (2011). New Delhi, S.Chand and Company Limited. 3. Donald Cairns: Essentials of

Pharmaceutical Chemistry, 3rd edn. (2008). Pharmaceutical Press, UK.

Md. Imran Nur Manik