2. 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.

3. 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.

4. 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.

5. 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

6. Lab.3 Three component system
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.

7. 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.

8. 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).

9. 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.

10. 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.

11. 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

12. 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?

13. 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.

14. 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.

15. 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?

16. 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.

17. 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

18. 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.

19. 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?

20. 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:
AB (solid) A− (solution) + B+(solution) ………(1)
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

21. 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:
Ksp = Ag + Cl -
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:
AgCl Ag+ + CL-
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.

22. 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
KHT HT
Flask no.(1) :Ksp= [HT-] [K+]
in other flasks : Ksp = [HT-] [K+ + K+ from KCl]
KCl K+ + CL-

23. 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

24. 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+] Ksp
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 .