Lab .1

Concentration expression

The concentration of a solution may be expressed either in terms of the

quantity of solute in a definite volume of solution or as the quantity of

solute in definite mass of solvent or solution.

The expressions used are Morality, Normality, Mole fraction, Mole

percent, percent by weight, percent by volume, percent weight in volume,

milligram percent

Morality, Normality and percent expressions are the most commonly

used in analytical work.

Experimental work:

Prepare the following solution using volumetric flasks & pipettes:

A. 50 ml of 0.5 M NaCL.

B . 50 ml of 2N NaCL.

C. 50 ml of 0.1N Na2CO3.

D. 50 ml of 0.1 M Na2CO3.

E . 50 gm of 2% w/w NaCL solution.

F. 50 ml of 10% w/v NaOH or NaCL.

G. 50 ml of 10% v/v alcohol.

Note: Morality is gram molecular weight per one liter solution.

Normality: is gram equivalent weight per one liter solution.

Dilution

-From stock solution 1% NaCL prepares 50 ml 0.5% solution using equation:

V1 * C1= V2* C2

Materials and equipment

-NaCL, Na2CO3, NaOH, Alcohol, H20.

-Volumetric flasks (50cc), pipettes.

Lab.2

TWO COMPONENT SYSTEM CONTAINING LIQUID PHASES

We may define a phase as a homogenous, physically distinct portion system, which is separated from other portions of system by bounding surfaces e.g. a system containing water and its vapor is a two phase system. An equilibrium mixture of ice, water and water vapor is a three phase system .

We know ethyl alcohol and water are miscible in all proportions, while water and mercury are completely immiscible.

Between these two extremes lie whole ranges of system which exhibit partial miscibility. Such a system is phenol and water.

Their miscibility is affected by 2 factors Conc. And temp.

Bimodal curve: is the curve that separate two phase area from one phase area.

Tie line: is the line drawn across the region containing two phases, is

All system prepared along tie line at equilibrium separated into two phases of constant composition. These phases are termed conjugate phases.

Mass ratio: is a relative amount by weight of conjugate phases.

It depends on the position of the original system along the tie line.

Procedure:

1. Prepare the following percent W/W phenol/water(10 gm total) 2%,7%,9%,11% ,24%,40%,55 %,63%,70%,75%.

2. Put test tube in afixed temperature in water bath (25 C0) or (left test tube at room temp.) and keep it for 10 minutes at that temp.

3. Take the test tubes out and before their temp has changed record which one has 2 phases and which has one phase.

4. Repeat the work at higher temp using the following temp.40C0, 50C0, 70C0.

5. Draw a curve temp verses concentrations showing your 2 phases area and one phase area in the curve.

6. Draw tie line for each temp.7. Take 40% W/W for example to find the mass ratio and the

composition of each phase at different temp.8. Mention the upper consulate temp.

Material: phenol and water.

Instruments and equipment: water bath, test tubes, glass stirrer

Three component systemLab.3

The rules that relate to use of triangular co-ordinates graph paper

1 -Each of the three corners or apexes of triangular represent 100% by weight of one component and zero% of other two components

2 -The three lines joining the corner points represent two component mixtures &each line is divided into one hundred equal units .In going along a line e bounding the triangles so as to represent concentrations in a two component system , it does not matter whether we proceed in a clock wise or counter clock wise direction around the triangle , provided we are consistent.

3 -The area within the triangle represents all possible combinations of three components to give three components systems.

4 -If a line is drawn through any apex to appoint on opposite side, then all system represented by points on such aline have a constant ratio of two components.

5-Any line drawn parallel to one side of tiangle, represent ternary systems in which the proportion (or percent by weight) of one component is constant.

Procedure:

1. Prepare 10 gm of the following combination of HAc & CHCl3:5%, 10%, 20%, 30%, 40%, 50%, 60%, 70%, 80%, and 90% w/w HAc: HCCl3 in a small clean &dry flask which form one single phase.

2. To these mixtures slowly add water from a burette until a turbidity just appears. Check the weight of water (which is equal to its volume).

3. Obtain a miscibility curve by calculating the percent w/w of each component in the turbid mixture and plot this triangular diagram.

Note:

To prepare samples in step no.1, the required amount of HAc &CHCl3 from burettes by converting the weight in to volume according to the law:

Specific gravity (sp.gr) = weight/volume

Sp.gr of HAc =1.009 and for CHCl3 =1.3

Materials and equipment:

-Acetic acid (HAc), CHCl3 distilled water.

-Conical flasks, burette.

Lab.4

Tie Line for three component system

In the three component system , the direction of tie lines are related to the shape of the bimodal which, in turn ,depend on the relative solubility of the third component in the other two components.

Only when added component equally on the other two components to bring them into solution will the bimodal perfectly symmetrical and the tie lines run parallel to the base line.

Procedure:

1. In a small reparatory funnel prepare 50 gm of a mixture having composition giving rise to a two phase system (e.g. 4gm HAc +16 gm CHCl3 +30gm H2O).

2. Separate each layer in two conical flasks.3. Weigh 10 gm for each layer.4. Titrate each layer with standard 1N NaOH solution using

phenolphthalein as indicator. The end point from colorless to pink.

5. Obtain tie line, calculate the percent W/W of HAc in each layer and locate the values on the miscibility curve. The straight line joining these points should pass through compositions of the two phase system.

Calculation:

HAc + NaOH→ NaAc + H2O

1 M.Wt. of HAc = 1 M.Wt. of NaOH

1 eq.wt of HAc = 1eq.wt of NaOH

60 = 1000 ml 1N NaOH

60/1000 = 1 ml 1N NaOH

So, each 1 ml of 1N NaOH is equivalent to 0.06 gm, this is the chemical factor (it is the no. of gms of substance which is equivalent to 1 ml of standard solution).

E.P 1 x o.o6 =gm HAc in 10 gm aqueous layer (upper layer).

E.P 2 x o.o6 =gm HAc in 10 gm CHCl3 layer (lower layer).

Change these values to percent.

Note:

- For the upper layer (100% HAc) represent mostly water with little chloroform.

This layer represents aqueous layer.

- For the lower layer (100% HAc) represent mostly chloroform with little water.

- This layer represents chloroformic layer.

Materials & equipments:

1. H2O, HAc, CHCl3, 1N NaOH solution, phenolphthalein indicator.2. Burettes, separatory funnel, conical flasks, balance.

Lab.5 Partition coefficient

Distribution of solute between immiscible solvent:

If an excess of liquid or solid is added to mixture of two immiscible liquids, it will distribute itself between the phases so that each becomes saturated.

If the substance is added to the immiscible solvent in an amount insufficient to saturate the solutions, it will still become distributed between the two layers in a definite conc. ratio.

If C1&C2 are the equilibrium concentrations of the substance in solvent (1) and solvent (2) ,the equilibrium expression becomes:

C1/C2 =K

The equilibrium constant K is known as distribution ratio, distribution coefficient, or partition coefficient, which is defined as solubility or concentration ratio at constant temp.

Experimental work :

Determination the partition coefficient of iodine between water and chloroform.

Procedure:

1. in dry Stoppered conical flask (iodine flask) put 20 ml of 1% iodine in chloroform (use burette).

2. Add 50 ml D.W. to it.3. The flask is the thoroughly shaken from time to time half hour

after equilibrium is established ,allow to stand for complete phase separation, this need another half an hour.

4. 10 ml of the sample are taken from the upper aqueous layer, care is taken to avoid touching the chloroforming layer. Then titrate against 0.02 N sodium thiosulphate.the end point is the disappearance of light brownish color.

5. 5 ml are taken from the organic layer (lower layer) .the inside wall of the pipette must be kept dry as it passes through aqueous phase by placing the finger tightly over the upper end of the pipette. Then titrate against 0.1 N sodium thiosulphate. Before titration, add 5 ml of 10% pot. Iodide to facillate extraction of I2 from the organic layer and it's titration with aqueous sodium thiosulphate. The end point is the disappearance of the brownish color.

Calculation:

Iodine distributed between the aqueous phase and chloroformic phase.

Aqueous phase:-the no. of ml of sodium thiosulphate (0.02N) consumed in the titration is equivalent to the amount of Iodine present.

(Na2S2O3) V1 X C1 = V2 X C2 (iodine)

E.P X 0.02 N =10 X N2

N2 =conc. Of iodine in water

Chloroformic phase:-the no. of ml of sodium thiosulphate

(0. 1N) consumed in the titration is equivalent to the amount of Iodine present.

(Na2S2O3) V1 X C1 = V2 X C2 (iodine)

E.P2 X 0.1N = 5 X N2

N2 =conc. Of iodine in chloroform

Conc. Of iodine in CHCl3

Partition coefficient = --------------------------------------- Conc. Of iodine in water

Material &equipment : Iodine, potassium iodide, chloroform, water, sodium

thiosulphate. Solutions: 10%w/v KI , 0.02N&0.1N sodium thiosulphate ,

1% I2 / CHCL3

Conical flask (iodine flask) , pipette , burette .

Home work :-1. What idea will the p.c value give you about the solubility?2. What is the importance of p.c to the pharmacist?

Lab .6

Solubility Methods of solubility

Solubility: is defined in quantitative term as the concentration of solute in a saturated solution at certain temperature.

In a quantitative way it may be stated as spontaneous interaction of two or more substances to form a homogenous molecular dispersion.

The solubility of a compound depend upon the physical and chemical properties of solute and solvent , as well as upon such factors as temperature , pressure , pH of solution and to lesser extent the state of

subdivision of solute .

Methods of solubility:

1-Solvent combinations:

The solubility of the solute is qualitatively related to dielectric constant of the solvent system. For example given solute will have qualitatively similar solubility profile with respect to dielectric constant for various co- solvent combinations.

The objective of this experiment is to increase solubility of salicylic acid (weak organic acid, slightly soluble in water) by solvent combination, changing the dielectric constant of the solvent system used.

Procedure:

1 -Put 0.1 gm salicylic acid (S.A)in conical flask

2-Add 10 ml distilled water and shake the flask to see its solubility.

3-Add from burette drop by drop absolute alcohol with continuous shaking until all crystals of S.A dissolve.

4 -Measure the amount of alcohol used.

Calculate the percent of alcohol in the final mixture.

Express the solubility, as1part of S.A is soluble in x parts of y% of alcohol .

2 - Salt formation (pH control) :

Most weak electrolytes can be retained in solution by adjusting pH so as to keep drug in ionized form.

The aim of this experiment is to increase the solubility of S.A by salt formation using Na2CO3.

Procedure:

1 -Put 0.1 gm salicylic acid (S.A) in conical flask.

2-Add 10 ml distilled water and shake the flask to see its solubility.

3-add 0.1 gm Na2CO3 to the flask with shaking and observe the result.

4-add 5 ml dilute HCL (10%) slowly.

Observe the result and develop an equation to account for the observation in step 3 and 4 .

3 - solubilization by complexation :

It has been found that insoluble drug can form a soluble complex with some compound.

In organic and organic materials which do not themselves ionize may be rendered soluble in polar solvent by complexation with electrolytes.

The aim of this experiment is to solubilize I2 in water with KI.

Procedure

1-put 0.1 gm I2 in conical flask .

2-add 10 ml water, shake and observe.

3-add 0.2 gm KI.

Observe the result and write an equation for it.

Materials &equipments:

Salicylic acid ,water , alcohol , iodide , potassium iodide , sodium carbonate , 10% HCL solution , 2conical flask , pipette , burette.

Home work

1 -How solubility express?

2-what is the saturated solution?

3-what is the difference between saturated and unsaturated solution?

4-what is relation between the dielectric constant and polarity of solvent?

5-what is the importance of solubility to the pharmacist?

Lab.7

Buffer solution

Buffer solutions are solutions that tend to resist changes in pH when acids or bases are added.

Buffer solutions usually contain a salt of weak acid or weak base and corresponding acid or base.

Buffer equation: the pH of Buffer solution and the change in pH upon the addition of an acid or base may be calculated by use of Buffer equation .this expression is developed by considering the effect of a salt on the ionization of a weak acid when the salt and acid have an ion in common.

The buffer equation or Henderson- Hassel balks equation, for a weak acid and its salt:

[Salt]

PH= pka +log ------------- [Acid]

For weak base and its salt:

[Base]

PH= pkw- pkb+ log ------------- [Salt]

Buffer capacity :- is the ability of a buffer solution to resist pH change .

The smaller PH change caused by the addition of a given amount of acid or alkaline, the greater the buffer capacity of a solution.

The buffer has its greatest capacity before any base is added where [salt]/ [acid] =1 (when equimolar amount of acid and salt are used) therefore, according to buffer equation, pH = pKa.

The buffer capacity is also influenced by an increase in the total concentration of the buffer constituents since obviously a greater concentration of salt and acid provides a greater alkaline acid reserve.

Approximate formula to calculate buffer capacity was introduced by Van Slyke is:

∆B

β=------------------

PH

May be used, in which delta (∆) has its usual meaning, a finite change, and B is the small increment in gm equivalent/liter of strong base.

Various buffer systems have been suggested for different pharmaceutical solutions:

1-Sorensen phosphate: buffer which consists of two stock solutions M /15 sodium acid phosphate and M/15 disodium phosphate, which are mixed in various volumes to obtain a desired PH

2-Acetate buffer: consist of 0.2 M solution of acetic acid (A) and0.2M solution of sodium acetate (B) by mixing x ml of A and Y ml of B and enough water to produce 100 ml.

Experimental work:

Procedure:

1-prepares acetate buffer solution according to the following table then measure the PH in each case:

Solution (A) ml + Solution (B) ml D.W to 100ml pH

46.3 3.7

30.5 19.5

25.5 24.5

20 30

14.8 35.2

4.8 45.2

The PH of each solution is measured by using:

1. Paper indicator: by immersing a strip of wide range PH paper into small quantity of buffer solution and observing the color changes of the paper which changes according to the PH value of the buffer.

2. Liquid universal indicator: which consists of several indicators e.g.

a mixture of methyl yellow, methyl red, bromothymol blue ,thymol blue and phenolphthalein which covers PH range (1-11).

The PH of buffer solution is measured by the addition of 2 drops of universal indicators to 10 ml buffer solution ,then compare the color result with color found on the bottle of liquid universal indicator .each color represents certain value of PH.

Method (1) and (2) are called colorimetric methods for the determination of PH. they are probably less accurate and less convenient but also less expensive than electrometric method (by using PH meter) . They are used in the determination of PH of aqueous solutions which are not colored or turbid.

PH indicators: indicators may be considered as weak acids or weak bases which exhibit color changes as their degree of dissociation varies with PH. Indicators therefore offer a convenient alternative method to electrometric techniques for the determination of PH of solution. The color of an indicator is a function of the PH of the solution.

3- PH meter :

1 .Put the electrode of the PH meter in the buffer solution & read the PH

Note: a- the PH meter should be opened before reading the PH about 30 min .

b- The electrode should always be immersed in D.W when it's not used

c- Standardization for instrument should be done using standard buffer solution PH 7 &PH 4 when our samples are acidic, or PH 7 & PH 10 when our samples are basic .

2-a. Take a certain volume of acetate buffer solution, add to it 0.0004 M sodium hydroxide portions (0.1 ml of 0.1 M). B- Measure the PH. C- Calculate the buffer capacity.

Materials ,equipment , and instruments

-acetic acid, sodium acetate, water, liquid universal indicator, paper indicator .

-solution: 0.2 M HAc, 0.2 M NaAc, 0.1 M NaOH.

-conical flask, beaker, pipette, volumetric flasks .

-PH meter .

Home work :

1-explain how buffer resist the changes in PH by using equations? 2-derive buffer equation for acidic buffer? 3-How can you differentiate between buffer and non buffer system in lab?

Lab .8

Determination of solubility product constant of slightly soluble salts

When slightly soluble electrolytes are dissolved to form saturated solution, the solubility is described by a special constant, known as solubility product Ksp.

Theory: in case of inorganic salt, when solute goes into solution such as AB, we then write the following equation for solution product:

A− (solution) + B+(solution) ………(1) AB (solid)

There is an equilibrium between the solute and the saturated solution

phases, the law of mass action defines as equilibrium constant, Keq as :

a A− (solution) × a B+(solution)

Keq = ---------------------------------------- …………….(2) a AB(solid)

Where a A− (solution) , a B+

(solution) and a AB(solid) are the activities of A and B in solution and AB in solid phase . Since the activity of a solid is defined as unity, and that in dilute solution (e.g. where we have slightly soluble salt) , concentration may be substituted for activities ; equation

(2) then becomes:

Keq = CA . CB

Where CA and CB are the concentration of A and B in solution .Keq in this situation has a special name , the solubility product , Keq , thus :

Ksp = CA . CB ……….(4)

This equation will hold true theoretically only for slightly soluble salt .

As an example of this type of solution, consider the solubility of silver chloride, as we can write:

] Cl [- ]Ag +[Ksp =

Where the brackets [ ] designate concentration in molar /litter (molar concentration).

So, the Ksp is the multiplication of molar conc. of two ions. If the silver ion concentration is increased by the addition of soluble silver salt, the chloride ion concentration must decrease until the product of the concentration is again numerically equal to the solubility product.

In order to affect the decrease in chloride ion concentration, silver chloride is precipitated, and hence its solubility is decreased .in a similar manner an increase chloride ion concentration by the addition of soluble chloride salt, a decreased in the silver ion conc. until the numerical value of the solubility product is attained. Again this decrease in silver ion conc. Is brought about by the precipitation of silver chloride.

The solubility of AgCl in saturated aqueous solution of the salt may be calculated by assuming that the concentration of silver ion is the same as the conc. Of chloride ion , both expressed in gm mol/l and that the conc. of dissolve silver chloride is numerically the same since each silver chloride molecule gives rise to one silver ion and one chloride ion , since:

Ag+ + CL-AgCl

Procedure-:

1-into five clean dry conical flasks, add 1gm of KHT (potassium acid tartar ate) +40 ml of different molarities of KCL .

a- In the first flask add 40 ml D.W. b- In the second flask add 40 ml of 0.01 M KCl. c- In the third flask add 40 ml of 0.02M KCl. d- In the fourth flask add 40 ml of 0.03 M KCl. e- In the fifth flask add 40 ml of 0.05 M KCl .

Note : you are provided with 0.1 M KCl e.g. in order to prepare 40 ml of 0.01 M KCl and complete the volume with water to 40 ml depending on the dilution equation.

C1 ×V1 = C2 ×V2 and so on.

2 -Shake for 10 min, leave 15 min for equilibration. 3-filter, rinse the flask with first portion of the filtrate (e.g. 1ml) , complete the filtrate . 4-take 10 ml of the filtrate, titrate against M/50 NaOH using phenolphthalein as indicator. 5-calculate the solubility product of potassium acid tartar ate .

Calculation:

O O

C OH C OH

H C OH H C OH

H C OH H C OH

C OK C OH

O O

HT KHT

Flask no.(1) :Ksp= [HT-] [K+] in other flasks : Ksp = [HT-] [K+ + K+ from KCl]

KCl K+ + CL-

in titration:

HT + NaOH NaHT + H2O

1 eq.wt of HT- ≡ 1 eq.wt of NaOH

1M.wt of HT- ≡ 1 M.wt of NaOH

1L 1M NaOH ≡ 1M.wt of HT- =149 gm

1ml 1M NaOH ≡ 149 / 1000 gm

1 149 1

1ml ------- M NaOH ≡ ------------ × ----------- gm

50 1000 50

149 1

For flask no .1 E.P1×-----------×---------------- = gm HT/10ml

1000 50

149 1 100

E.P1×-----------×---------------- ×------------- = mole /liter

1000 50 149

So E.P 1 /500 = molar concentration of HT = molar conc. of K+

E.P1 E.P1

KSP for flask (1) = [-------] [--------]

500 500

E.P2 E.P2

KSP for flask (2) = [-------] [-------- + 0.01]

500 500

E.P3 E.P3

KSP for flask (3) = [-------] [-------- + 0.02]

500 500

E.P4 E.P4

KSP for flask (4) = [-------] [-------- + 0.03]

500 500

E.P5 E.P5

KSP for flask (5) = [-------] [-------- + 0.04]

500 500

Tabulate your result as follows:

Flask no. E.P M KCL [HT-] [K + ] K sp

Materials and equipments:

-potassium acid tartarate , potassium chloride , sodium hydroxide . –solution: 0.1 M KCl , M/50 NaOH, phenolphthalein indicator. – Conical flask, pipette , burette , filter paper , measuring cylinder .