Department of Pharmacy
SOUTHEASTUNIVERSITY
Pharm 1015: Physical Pharmacy I Sessional
Experiment No: 1
Name of the experiment:
Preparation of standard 0.1 N 100 ml Na2CO3 solution
Principle: For analysis, standard solution is necessary. There are some substances in nature which called primary standard. 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. It must be easy to obtain, to purify, to dry and to preserve in pure state. These substances should be unaltered during weighing, relatively high molecular weight and readily soluble under the condition in which it is applied. Sodium carbonate, sodium tetraborate, potassium hydrogenphthalate, potassium dichromate, potassium iodate, sodium oxalate, oxalic acid etc are the example of primary standard substance.
The number of gram equivalent weight of a substance present in 1 liter of a solution represents its normality.
Normality = Gram equivalent weight of solute/Volume of solution in liter
The number of gram molecular weight of a substance dissolved in water to form a solution occurring a volume of 1000 ml at a specified temperature, is called molarity.
Molarity = Number of moles of solute/Volume of solution in liter.
Chemicals/Reagents and apparatus:
A. 1. PureNa2CO3
2. Distilled water
B. 4. 100 ml volumetric flask
5. Measuring cylinder
6. Analytical balance
7. Funnel
8. Spatula
Procedures:
For 0.1 N 100 mlNa2CO3solution:
i) Calculate the amount of Na2CO3 required to prepare 0.1 N 100 mlNa2CO3 solution.
ii) Required (530 mg) amount of Na2CO3 is weighted accurately.
iii) Then it is taken in a 100 ml volumetric flask.
iv) About 30-40 ml distilled water is added and shake gently until make a solution.
v) Then level the volumetricflaskto the markby adding distilled water.
Calculation:
For 0.1 N 100 ml Na2CO3 solution:
Molecular weight of Na2CO3 = 106
So, equivalent wt of Na2CO3 = 106/2
=53
1000 ml 1 N solution containing = 53 gm of Na2CO3
1 ml1 N,, ,, =53/100 ““
100 ml0.1 N ,, ,, =53× 100 ×0.1/1000
= 530 mg
Results andComments:
By the above way the standard solution of various normality and Molarity of the compound Na2CO3 and oxalic acid is prepared. We know that Na2CO3 is a primary standard substance. That is why the strength if Na2CO3 is not changes with time.
Experiment No: 2
Name of the experiment:
Preparation of standard 0.1 N 100 ml Oxalic acid solution
Principle: For analysis, standard solution is necessary. There are some substances in nature which called primary standard. 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. It must be easy to obtain, to purify, to dry and to preserve in pure state. These substances should be unaltered during weighing, relatively high molecular weight and readily soluble under the condition in which it is applied. Sodium carbonate, sodium tetraborate, potassium hydrogenphthalate, potassium dichromate, potassium iodate, sodium oxalate, oxalic acid etc are the example of primary standard substance.
The number of gram equivalent weight of a substance present in 1 liter of a solution represents its normality.
Normality = Gram equivalent weight of solute/Volume of solution in liter
The number of gram molecular weight of a substance dissolved in water to form a solution occurring a volume of 1000 ml at a specified temperature, is called molarity.
Molarity = Number of moles of solute/Volume of solution in liter.
Chemicals/Reagents and apparatus:
A. 1. PureOxalic Acid
2. Distilled water
B. 4. 100 ml volumetric flask
5. Measuring cylinder
6. Analytical balance
7. Funnel
8. Spatula
Procedures:
Calculation:
Results andComments:
By the above way the standard solution of various normality and Molarity of the compound oxalic acid is prepared.
Experiment No: 3
Name of the experiment:
Preparation of indicator(Methyl orange and Phenolphthaeline) solution.
Principle:
Indicators, as the name implies, are organic dyes which indicate, generally by their color change, the equivalence or end point of the titration.
Phenolphthaeline and methyl orange are two common example of acid base indicator. Phenoloptheline is pink in basic solution and colorless in acidic solution. Methyl orange is red in acidic solution and yellow in basic solution.
Chemicals/Reagent and Apparatus:
A. 1. Methyl orange indicator powder
2. Phenolphthalein indicator powder
3. Alcohol
4. Distilled water
B. 5. Measuring cylinder
6. Glass rod
7. Reagent bottle
8. Analytical balance
9. Funnel
10. Filter paper
Procedure:
A)For Methyl Orange indicator:
i) In a 100 ml volumetric flask 50 mg of methyl orange powder is taken. Then about 50 ml water is added to the volumetric flask.
ii) Shake the volumetric flask until dissolve the powder.
iii) Dissolve 0.5 gm of Na salt (usually NaCl).
iv) Add 7.5 ml of 0.1 M HCl solution.
v) Then level the volumetricflaskto the mark by adding distilled water If precipitation appears then filter.
B)For Phenoloptheline indicator:
i) In a 100 ml volumetric flask 1 gm of Phenoloptheline powder is taken.
ii) Then 50 ml ethanol is added and more 10 to 20 ml water is added.
iii) Shake the volumetric flask until dissolve in the water.
vi) Then level the volumetricflaskto the mark by adding distilled water.
v) Filter the solution if precipitation appears.
Comments: Theindicators prepared by above way are used for acid-base titration.
Experiment No: 4
Name of the experiment:
Preparation and standardization of approximately 0.1 N 100 ml Sodium hydroxide (NaOH) solution by standard 0.1 N Oxalic acid solutions.
Principle:There are some substances in nature by which standard solution can not be prepared. They loss purity when they are exposed to environment. So, by weighing, standard solution of these substances can not be prepared. These substances are called secondary standard. Sodium hydroxide is one of them. For analysis, standardization of these solutions by other standard solution is necessary.
The process of addition, the standard solution to standardized solution until the reaction is just complete, is known as titration. By this process, the concentration of an unknown solution is determined against a known concentration solution with the help of indicator. The acid base titration is done according to the following reaction
2NaOH + C2H2O4 === C2Na2O4
Chemicals/Reagents and Apparatus:
A. 1. Oxalic acid
2. Sodium Hydroxide (NaOH)
3. Phenoloptheline (Indicator)
B. 1. Burette
2. Pipette
3. Conical flask
4. Volumetric flask
5. Measuring cylinder
6. Weighing bottle.
Procedures:
- Preparation of approximate 0.1 N 100 ml of NaOH solution:
Calculate the amount of NaOH required to prepare 0.1 N 100 ml of NaOH solution. Requited (400 mg) amount of NaOHis weighted accurately then it is taken in a 100 ml volumetric flask. About 30-40 ml distilled water is added and shake gently until make a solution. Level the volumetricflaskto the mark by adding distilled water.
2. Preparation of standard 0.1 N 100 ml of Oxalic acid solution:
Calculate the amount of Oxalic acid required to prepare 0.1 N 100 ml of Oxalic acid solution. Requited (1260 mg) amount of Oxalic acidis weighted accurately then it is taken in a 100 ml volumetric flask. About 30-40 ml distilled water is added and shake gently until make a solution. Level the volumetricflaskto the mark by adding distilled water.
3. Titration of NaOH solution with standard oxalic acidsolution:
Fill the burette with prepared NaOH Solution. Take 10 ml of the standard Oxalic acid solution in a conical flask and dilute with about 10 ml of distilled water. Then add 2-3 drops of Phenolphthaeline indicator to it and titrate with the acid solution until the color of the indicator changes from colorless to pink. This indicates the end point of titration. At least other two titrationsare to be performed.
Calculation:
Preparation of NaOH solution:
Molecular weight of NaOH = 40
Equivalent weight of NaOH = 40
Gm equivalent weight of NaOH = 40/1 = 40.0 gm
So, the amount of NaOH required to prepare 0.1 N 100 ml of NaOH solution
= 0.4 gm
= 400 mg
Preparation of Oxalic acid solution:
Molecular weight of oxalic acid(C2H2O4.2H2O) = 126
Equivalent weight of Oxalic acid = 126/2 = 63.0
Gm equivalent weight of oxalic acid = 126/2 = 63.0 gm
So, the amount of Oxalic acid required to prepare 0.1 N 100 ml of oxalic acid solution = 0.63 gm
= 630 mg
So, the amount of Oxalic acid has to weighted = 630 mg
Titration data table:
No of observations / Volume of Oxalic Acid (V1) in ml / Initial burette reading / Final burette reading / NaOH required in ml / Average NaOH required in ml (V2)01 / 10
02 / 10
03 / 10
According to acid base titration, we know S1V1 = S2V2
Where, S1 = Strength of oxalic acid solution
V1= Volume of oxalic acid solution
S2 = Strength of NaOH solution
V2 = Volume of NaOH solution
We have V2 = …….. ml
S2 = V1 S1/ V2
= …………. N
= …….. N
Result:
The exact strength of the prepared Sodium hydroxide is ……. N.
Experiment no.05
Name of the experiment:
Preparation and standardization of approximately 0.1 N Hydrochloric acid (HCl) Solution by standard 0.1 N Na2CO3 solution.
Principle: There are some substances in nature by which standard solution can not be prepared. They loss purity when they are exposed to environment. So, by weighing, standard solution of these substances can not be prepared. These substances are called secondary standard. Hydrochloric acid is one of them. For analysis, standardization of these solutions by other standard solution is necessary.
The process of addition, the standard solution to standardized solution until the reaction is just complete, is known as titration. By this process, the concentration of an unknown solution is determined against a known concentration solution with the help of indicator. The acid base titration is done according to the following reaction
HCl + Na2CO3 === NaCl + H2CO3
Chemicals/Reagents and Apparatus:
A. 1. Hydrochloric acid
2. Anhydrous sodium carbonate (Na2CO3)
3. Methyl Orange (Indicator)
B.1. Burrete
2. Pipette
3. Conical flask
4. Volumatric flask
5. Measuring cylinder
6. Weighing bottle.
7. Analytical balance
Procedure:
1. Preparation of approximate 0.1 N 100 ml of HCl solution:
Calculate the amount of HCl required to prepare 0.1 N 100 ml of HCl solution. The requited (1.2 ml of 32 % solution) amount of HClis taken accurately and transfer it in 100 ml volumetric flask. About 30-40 ml distilled water is added and shake gently then level the volumetricflaskto the mark by adding distilled water.This will give a solution that is approximately 0.1 N HCl solution.
2. Preparation of standard 0.1 N 100 ml of Na2CO3 solution:
Calculate the amount of Na2CO3require to prepare 0.1 N 100 ml of Na2CO3solution. Required (530 mg) Na2CO3 is weighted accurately and taken in a 100 ml volumetric flask. About 30-40 ml distilled water is added and shake gently until make a solution. Level the volumetricflaskto the mark by adding
Titration of HCl with Na2CO3 :
Fill the burette with prepared HCl Solution. Take 10 ml of the standard sodium carbonate (Na2CO3) solution in a conical flask and dilute with about 10 ml of distilled water. Then add 2-3 drops of methyl orange indicator to it and titrate with the acid solution until the color of the indicator changes from yellow to orange. This indicates the end point of titration. At least other two titrations are to be performed.
Calculation:
Preparation of 0.1 N 100 ml of HCl Solution:
Equivalent weight of HCl = 36.5
The amount of HCl required to prepare 0.1 N 100 ml of HCl solution = 0.365 gm HCl
The strength of supplied HCl solution is 32% w/v
So required volume of HCl to prepare the solution = 100x 0.365/ 32 ml
= 1.15 ml HCl
Preparation of 0.1 N 100 ml of Na2CO3 solution:
Equivalent weight of Na2CO3 = 106/2 = 53
So, the amount of Na2CO3 require to prepare 0.1 N 100 ml of Na2CO3 solution = 0.53 gm
=530 mg
So, the amount of Na2CO3has to weight = 530 mg
Titration data table:
No. of observations / Volume of Na2CO3 (V1) used in ml / Initial burette reading / Final burette reading / Volume of HCl required in ml / Average volume of HCl used (V2) in ml01 / 10
02 / 10
03 / 10
According to acid base titration, we know S1V1 = S2V2
Where, S1 = Strength Na2CO3 of solution
V1= Volume of Na2CO3 solution
S2 = Strength of HCl solution
V2 = Volume of HCl solution
We have, V2 = ……….. ml
S2 = V1 S1/ V2
= …………. N
= ………N
Result:The exact strength of the prepared hydrochloric acid is ……. N
Experiment No: 5
Name of the experiment:
Standardization of potassium permanganate solution by standard oxalic acid solution.
Principle: There are some substances in nature by which standard solution can not be prepared. They loss purity when they are exposed to environment. So, by weighing, standard solution of these substances can not be prepared. These substances are called secondary standard. Potassium permanganateis one of them. For analysis, standardization of these solutions by other standard solution is necessary.The concentration of an unknown solution(KMnO4) is determined against a known concentration solution of oxalic acid solution .Here KMnO4act as a oxidizing agent i.e. accept electron and oxalic acid act as a reducing agent . Beside theseKMnO4 act as a self indicator.
Chemicals/ Reagents & Apparatus:
A. 1. Potassium permanganate
2. Oxalic acid
3. Sulfuric acid
B. 1. Burette
2. Pipette
3. Conical flask
4. Volumetric flask
5. Measuring cylinder
6. Weighing bottle.
7. Analytical balance
Procedure:
1`. Preparation of 0.01M KMnO4 solution.:
100 mL0.01M was prepared by dissolving 0.158 g of Potassium permanganate in 100mL of distil water in a100mL volumetric flask . The solution was heated for about 10-15 minutes and then cooled to the laboratory temperature. The solution is then filtrates and was stored in the dark.
2. Preparation of standard 0.01M 100 ml of Oxalic acid solution:
Calculate the amount of Oxalic acid required to prepare 0.01 M 100 ml of Oxalic acid solution. Required 0.126g amount of Oxalic acidis weighed accurately then it is taken in a 100 ml volumetric flask. About 30-40 ml distilled water is added and shake gently until make a solution. Level the volumetricflaskto the mark by adding distilled water.
Standardization of potassium permanganate solution by standard oxalic acid solution.:
10 ml of0.01 M 100 ml of Oxalic acid solution was taken in the volumetric flask and acidified with 1mL of 1M Sulfuric acid solution and heated about 70-800C .Potassium permanganate solution was added drop wise from the burette and the titration was continued constant stirring until the first faint pink color was obtained. The temperature of the solution near the end point was not allowed to fall 500C . The experiment was repeated two more times.
Reaction:
2KMnO4 +5 C2H2O4.2H2O + 6H+ = 2Mn+ + 10CO2 + 8H2O
Calculation:
1`. Preparation of 0.01MKMnO4 solution.:
Molecular weight ofKMnO4 = 158
1000mL 1M solution containing = 158 g KMnO4
100 ml 0.01M “ “ = 158* 100* 0.01/1000
= 0.158 g.
2.Preparation of Oxalic acid solution:
Molecular weight of oxalic acid (C2H2O4.2H2O) = 126
1000mL 1M solution containing = 126 g C2H2O4.2H2O
100 ml 0.01M “ “ = 126* 100* 0.01/1000
= 0.126g.
Titration data table for the standardization of KMnO4 solution.
No of observations / Volume of Oxalic Acid (V1) in ml / Initial burette reading in mL / Final burette reading in mL / Required KMnO4in ml / Average KMnO4 required in ml (V2)
01 / 10
02 / 10
03 / 10
we know
2 S1V1 =5 S2V2
Where, S1 = Strength of oxalic acid solution
V1= Volume of oxalic acid solution
S2 = Strength of KMnO4 solution
V2 = Volume of KMnO4 solution
Experiment No: 6
Name of the experiment: Preparation of an acidic buffer solution of desired pH.
Introduction: it is often necessary to maintain a certain pH of a solution in laboratory and industrial processes. This is achieved with the help of buffer solutions, buffer systems or simply buffers.
A buffer solution is one which maintains its pH fairly constant even upon the addition of small amount of acid or base. In other words, a buffer solution resists a change in its pH. That is, we can add a small amount of an acid or base to a buffer solution and the pH will change very little. Two common types of buffer solution are: a) a weak acid together with a salt of the same acid with a strong base. These are called Acid buffers e.g. CH3COOH + CH3COONa, b) a weak base and its salt with a strong base. These are called Basic buffers. e.g. NH4OH + NH4Cl
To prepare a desired buffer we use Henderson-Hasselbalch equation. For acidic buffer the equation is
pH = pKa + log [salt]/[acid]
or, pH = -logKa +log [salt]/[acid]
Reagent/Chemicals and Apparatus:
A. 1.Acetic acid
2. Sodium acetate
3. Distilled water
- 1. Volumetric flask (100ml)
2. Funnel
3. Analytical Balance
4. pH paper
5. Pipette
Procedure:
1. Calculation of the amount of acetic acid and Sodium acetate to prepare 100 ml buffer solution of 6.5 pH by using Henderson-Hasselbalch equation as we know,
pH = -logKa + log [salt]/[acid]
or, 6.5 = -log (1.8x 10-5) + log [salt]/[acid] ( Ka for acetic acid is 1.8x 10-5)
or, 6.5 = 4.7447 + log [salt]/[acid]
or, 6.5 – 4.7447 = log [salt]/[acid]
or, 1.755 = log [salt]/[acid]
or, 5.69 = [salt]/[acid] , taking anti log on both side.
If we take 0.1 M of salt then the require amount of acid is 5.69x 0.1 = 0.569 M.
The molecular weight of sodium acetate is 82.
So we need 82x 0.1 = 8.2 gm of sodium acetate.
The molecular weight of acetic acid is 60.
So we need 60x 0.569 = 34.14 gm of acetic acid
The weight of 1ml of acetic acid (supplied acetic acid in the laboratory) is 1.05 gm
So the required acetic acid for the buffer = 34.14/1.05 ml
= 32.5 ml
2. Pipette out 32.5 ml of acetic acid and transfer it in the 100ml volumetric flask.
3. By the analytical balance, weigh 8.2 gm of sodium acetate and transfer it to the same volumetric flask and sake gently to prepare the solution.
4. Finalize the volume up to the mark of the volumetric flask by adding distilled water.
Result: analyze the pH of the solution by pH paper and pH meter.
Experiment No: 7
Name of the experiment: Preparation of a basic buffer solution of desired pH.
Introduction: it is often necessary to maintain a certain pH of a solution in laboratory and industrial processes. This is achieved with the help of buffer solutions, buffer systems or simply buffers.
A buffer solution is one which maintains its pH fairly constant even upon the addition of small amount of acid or base. In other words, a buffer solution resists a change in its pH. That is, we can add a small amount of an acid or base to a buffer solution and the pH will change very little. Two common types of buffer solution are: a) a weak acid together with a salt of the same acid with a strong base. These are called Acid buffers e.g. CH3COOH + CH3COONa, b) a weak base and its salt with a strong base. These are called Basic buffers. e.g. NH4OH + NH4Cl
To prepare a desired buffer we use Henderson-Hasselbalch equation. For acidic buffer the equation is
pOH = pKb + log [salt]/[base]
or, pOH = -logKb + log [salt]/[base]
Reagent/Chemicals and Apparatus:
A. 1. Ammonium hydroxide solution
2. Ammonium chloride
3. Distilled water
- 1. Volumetric flask (100ml)
2. Funnel
3. Analytical Balance
4. pH paper
5. Pipette
Procedure:
1. Calculate the amount of ammonium hydroxide and ammonium chloride need to prepare 100 ml buffer solution of 7.5 pH by using Henderson-Hasselbalch equation as we know,
pH + pOH = 14
or, pOH = 14 - 7.5
pOH = 6.5
and pOH = -logKb + log [salt]/[base]
6.5 = -log (1.8x 10-5) + log [salt]/[base] ( Kb for NH4OH is 1.8x 10-5)
or, 6.5 = 4.7447 + log [salt]/[base]
or, 6.5 – 4.7447 = log [salt]/[base]
or, 1.755 = log [salt]/[base]
or, 5.69 = [salt]/[base] , taking anti log on both side.
If we take 0.1 M of salt then the require amount of acid is 5.69x 0.1 = 0.569 M.
The molecular weight of ammonium chloride is 53.5.
So we need 53.5x 0.1 = 5.35gm of ammonium chloride.
The molecular weight of ammonium hydroxide is 35.
So we need 35x 0.569 = 19.915 gm of ammonium hydroxide
The weight of 1ml of ammonium hydroxide (supplied acetic acid in the laboratory) is 0.91 gm
So the required ammonium hydroxide for the buffer = 19.915/0.91ml
= 21.88 ml
2. Pipette out 21.88 ml of ammonium hydroxide solution and transfer it in the 100ml volumetric flask.
3. By the analytical balance, weigh 5.35 gm of ammonium chloride and transfer it to the same volumetric flask and sake gently to prepare the solution.
4. Finalize the volume up to the mark of the volumetric flask by adding distilled water.
Result: analyze the pH of the solution by pH paper and pH meter.
Experiment No: 8
Name of the experiment: Determination of partition coefficient of benzoic acid in water and carbon tetrachloride solution.
Introduction: when a solution is shaken up with two liquids which are immiscible with each other but in both of which the solute is soluble, then the solute is found to distribute itself between two solvents in such a way that the ratio of the concentration of the solute in the two liquids, is a constant at a constant temperature. This state meet is known as “ Nernst’s distribution law”.