University of Sulaimani
Faculty of Pharmacy
General Chemistry
1st Stage
Practical General Chemistry
labs.
Fuad H.Shareef
2016-2017
Vinegar
is a versatile liquid that results from the fermentation of ethanol.
CH3CH2OH + O2 → CH3COOH + H2O
The key ingredient of vinegar is acetic acid, which gives it an acidic taste. pH ranges
= 2 - 3.5
Uses:* Food
preparation procedures, it is a multi-purpose product as an
condiment. ingredient and
* Outside of cooking,
* vinegar has medicinal,
* household cleaning, and
* agricultural applications.
Vinegar contains 4-5 % of acetic acid. The percentage of
acetic acid (w/w) % can be determined by titration of a solution
or dilute solution of vinegar against standardized NaOH solution
using phenolphthalein (Ph.Ph) as an indicator.
CH3COOH + NaOH → CH3COONa + H2O
The pH at end point (E.P) of the titration will be around
(8.7) Ph.Ph (color less to pink - red) .
Solution: a mixture consisting of a solute and a solvent
Solute: component of a solution present in the lesser amount
Solvent: component of a solution present in the greater amount
Concentration: amount of a solute present in a solution per standard amount
of solvent
(w/w) %, (w/v) % and (v/v) %: these variations on percentage concentration
are used in chemistry and biology.
Example: What is the w/w% concentration of sugar, if 5g of
sugar dissolved in 20g of water?
(w/v) %, used where a solid chemical is dissolved in liquid.
Example: dissolve 10 g of NaCl salt to make up a total volume of 100 mL of
solution.
(w/v) % = (10 g NaCl / 100 mL solution) × 100 = 10 w/v%
Procedure
Standardization of HCl by Standard borax
No. of mEq. Borax= No. of mEq. HCl
( N × V ) Borax= ( N × V ) HCl
A. Standardization of HCl by Standard NaOH
No. of mEq. NaOH= No. of mEq. HCl
( N × V ) NaOH = ( N × V ) HCl
1- No. of mEq. NaOH= No. of mEq. HAc
( N × V ) NaOH = (wt.(mg)/eq.wt.) CH3COOH , (eq.wt. = m.wt / 1 )
Wt. (mg)CH3COOH = N × V × m.wt
Wt. (mg)CH3COOH = N × V × m.wt × 100/10
Wt. (g)CH3COOH = N × V × m.wt × 100/10 × 10-3
(w/w) % = (wt CH3COOH / wt vinegar) × 100 = ??
(w/v) % = (wt CH3COOH / v vinegar) × 100 = ??
1- A 10 ml vinegar is
taken into a dry beaker
NaOH
standard
on sensitive balance.
2- transferee it into Vol.
flas complete to 100ml
3- take 10 ml then titrate
Ph.Ph
3 drops
against std NaOH
4- find w/w% and w/v%.
acetic acid in vinegar
10 ml
dilute
Vineger
Caustic soda
It is a white solid inorganic compound and highly
caustic metallic base.
Sodium hydroxide is used in many industries, mostly as a strong
chemical base in the manufacture of paper, textiles, soaps and
detergents and as a drain cleaner.
Sodium hydroxide NaOH is not used as a primary standard
solution Why?
2NaOH + CO2 → Na2CO3 + H2O
So, the commercial sample of solid sodium hydroxide
contains a quantity of Na2CO3
In the caring Acid/Base titration, what will happen when (NaOH and
Na2CO3) mixture are in the flask, titrate with HCl?
HCl + NaOH
Ph,Ph V1
NaCl + H2O
The first neutralization is: half neutralization
Na2CO3 + HCl
Ph,Ph V1
NaHCO3 + NaCl
The second neutralization is: half neutralization
NaHCO3 + HCl
M.orange V2
NaCl + CO2 + H2O
Method : Double indicator method
In the caring Acid/Base titration, what will happen when (NaOH
and Na2CO3) mixture are in the flask, titrate with HCl?
NaOH
Na2CO3
V1 HCl
Ph.Ph.
NaCl + H2O
1/2 NaCl + H2O + CO2
1/2 NaHCO3
V2 HCl
M. orange Ind.
NaCl + H2O + CO2
Data
V1= VHCl needed for {NaOH + ½ Na2CO3}
V2= VHCl needed for {½ Na2CO3}
V HCl for all Na2CO3 = 2 V2
V HCl for all NaOH = V1-V2
Calculation NaOH%
(V1-V2)×NHCl=(N×V)NaOH
Wt. (g/L) NaOH in mix.
= N NaOH × 40
Calculation Na2CO3
%
2V2×NHCl=(N×V)
Na2CO3
Wt (gm/L) Na2CO3 in mix.
procedure (Double indicator method)
1- Pipette 10ml of the mixture solution (NaOH + Na2CO3) to a
Conical flask; add 2drops of Ph.Ph indicator (pink).
2- Titrate against HCl until color changes (pink to colorless). Record
the volume of HCl added as (V1).
3- Add 3drops of M. orange indicator (yellow), complete the titration
with HCl until color changes (yellow to red). Record the volume of
HCl added as (V2). Calculate Na2CO3 % and NaOH % present in the
sample ?
HCl
standard
Ind.1
Ph.Ph (3 drops)
Ind.2
M.org. (3drops)
10 ml
mix.
Precipitation Titrations
1. Mohr method 2. Adsorption method 3. Volhard method
Depends on differential
precipitation
(with different
solubilities)
Depends on the adsorption
process
FAJANS method
Depends on Back titration
technique
 Precipitation titrations are based upon reactions that yield
ionic compounds of limited solubility.
 The most important precipitating reagent is silver nitrate.
 Titrimetric methods based upon silver nitrate are sometimes
termed argentometric methods.
Potassium chromate : end point indicator for Cl- , Br- 
CN- ions by reacting with silver ions to form a brick-red
silver chromate precipitate in the equivalence point region.
MOHR Method 
depends on differential precipitation (with
different solubility). Use to determines [Cl-] against (AgNO3 )
and using potassium chromate (K2CrO4) as an indicator.
During the titration:
Ag + + Cl- + AgCl¯ white ppt.
Ag + + Br- + AgBr¯ yellow ppt.
At End point:
2 Ag + + CrO 2- + Ag2CrO4¯
reddish brown ppt.
Chromate Indicator,
K2CrO4
How much do we need to indicate the precipitation of
Ag2CrO4 at the E.P?
At the E.P the only (Ag+) present is from the solubility of AgCl .
which is equal to (1.3 × 10-5 )
[Ag+]2[CrO42-] = Ksp (Ag2CrO4) to start the precipitation.
[1.3×10-5]2[CrO42-] =1.1×10-12
[CrO42-]=(1.1×10-12)/1.7×10-10=10-2, 0.01M
There is a systematic error (E) which comes from the
precipitation of the (reddish brown) ppt. of (Ag2CrO4) at the E.P, so
we must determine volume of titrant (AgNO3) required for this
precipitation which determined by BLANK titration, where :
{D.W+ some amount CaCO3 + same ind. (K2CrO4)} Titrate with AgNO3
A = O - E
Titration should be carried out in neutral medium or in a very faint
alkaline (base) solution (with pH range 6.5-9), because in acid
solution Chromic Acid will form.
2K2CrO4 + 2H+
2HCrO4-
2K2Cr2O7 + H2O
so the amount of indicator CrO4 will decrease in the solution and the ionic
product [Ag+][CrO42-] may be exceed Ksp of AgCrO4, so the ppt. will not
form at E.P. in alkaline solution, silver hydroxide (Ksp= 2.3×10-10) may be
precipitated.
2Ag+ + 2OH-
2AgOH
Ag2O + H2O
Precipitation Titrations
1. Mohr method 2. Adsorption method 3. Volhard method
Depends on differential
precipitation
(with different
solubilities)
Depends on the adsorption
process
FAJANS method
Depends on Back titration
technique
Adsorption method ( FAJANS method)
This method depends on the COLLOIDAL properties of 
the precipitates.
Colloid is a mixture in which one substance is divided into 
minute particles (called colloidal particles) and dispersed
throughout a second substance.
The mixture is also called a colloidal system, colloidal 
solution, or colloidal dispersion. Familiar colloids include
fog, smoke, homogenized milk, and ruby-colored glass.
Properties of colloids:

Diameters less than 10-4 cm. 
Solid particles

Surface activity 
Not settling down 
Large surface area
difficult to filter

Scatter light

The scattering of visible light by colloids, known as the Tyndall effect,
. e.g Starch
The particles of a colloid selectively absorb ions and acquire an electric charge.

If an electric potential is applied to a colloid, the charged colloidal particles move

toward the oppositely charged electrode; this migration is called electrophoresis.
If the charge on the particles is neutralized, they may precipitate out of the
suspension.

A colloid may be precipitated by adding another colloid with 
oppositely charged particles; the particles are attracted to
one another, coagulate, and precipitate out.
Addition of soluble ions may precipitate a colloid; the ions in 
seawater precipitate the colloidal silt dispersed in river water,
forming a delta. A method developed by F. G. Cottrell reduces
air pollution by removing colloidal particles (e.g., smoke, dust,
and fly ash) from exhaust gases with electric precipitators.
During the titration in the solution we have:
(1) Primary adsorbed layer
(2) Secondary layer
AgNO3
ClAgCl
+
-
-
+
-
(Na , Cl , NO3 , H , In )
Na+ Cl AgCl
Cl-
Na+
ClNa+
Na+ Cl-
The ppt. (AgCl) has
a tendency to adsorb
its own ions Ag+, Cl-
* Before E.P
we have an excess [Cl-]ions
so the secondary layer is
opposite ion which is [Na+]
The indicators (flourecein , Eosin) are weak acids : 
HIn
-
+
In + H
So [In-] ion can be adsorbed on the surface of the precipitate (before
E.P.) because of its (ve) charge.
The condition must be neutral so slightly alkaline to dissociate the
[HIn] especially when Fluerecein used.
At E.P. (when we have AgNO3) [excess Ag+ and Cl- consumed] So,
[In-] adsorbed on secondary layer, so the [In-] color will appear on
the surface of the precipitate (it is more strongly adsorbed than
NO3- ions) so there is a change in color at E.P. of titration.
(1) Primary adsorbed layer
(2) Secondary layer
AgNO3
ClAgCl
+
-
-
+
-
(Na , Cl , NO3 , H , In )
Na+ Cl AgCl
Cl-
Na+
ClNa+
Na+ Cl-
The ppt. (AgCl) has
a tendency to adsorb
its own ions Ag+, Cl-
* Before E.P
we have an excess [Cl-]ions
so the secondary layer is
opposite ion which is [Na+]
Conditions for adsorption indicators
1- The precipitate should separate as far as possible in the
colloidal condition.
2- The solution should not be very dilute as the amount of ppt.
formed will be small and the color change far from sharp.
3- The indicator ion [In-] must be opposite charge to the ion of
precipitating agent.
4- The [In-] ion should not be adsorbed before the ppt. has
been completely precipitated. But it should be strongly
adsorbed immediately after E.P…
All the changes in color depend on:
The adsorption of the halide ions [Cl- , Br- , I-] and [ In-]
of the indicator on the surface of the precipitate.
Indicators
Flourescein (w.acid Ka = 10-8) , condition : pH of (7 -10) optimum pH
Dichloroflourescein (stronger) , condition : pH – greater than 4.5
Eosin (Tetra bromo flourescein ) : (stronger acid ) , condition – pH (1-2)
Why Eosin cannot be used for detn. Of [Cl-] in this method? Because,
Eosin is so strongly adsorbed on silver halide ppt. and the [Eo-] ion can compete
with [Cl-] ion before (E.P) and thereby give a premature (early) indication of the
E.P .
Flouresein HFI
Eosin
EoH
FI- + H+ Flouresein (w.acid) but Eosin is stronger acid,
so for (FI) the condition must be slightly basic
- + to help the dissociation of it.
Eo + H
Experiment:12 Complexometric Titrations
A complexometric titration is the one in which the reaction between
the analyte and titrant involves the formation of a colored complex,
(which is stable and stays undissociated).
Therefore, controlling the pH is one major factor that affects
complexation.
EDTA (Ethylenediaminetetraacetic acid) is probably the most widely
used of these ligands.
The stability of the complexes can be attributed to the chelate
effect; when complexing a metal ion EDTA surrounds the metal
cation, forming a number of five-membered rings.
The key steps in designing a typical complexometric determination are:
• choosing a suitable complexing agent
• choosing a suitable method of detecting the end point
• choosing the experimental conditions that provide an optimum
titration Complexometric
All types of complex formation can not used for titration,
only the complexes which are:
Stable
.1
Fast complex formation
.2
There is a method for measuring end point.
.3
Most famous compound used is: EDTA (ethylene Diamine Tetra Acetic acid)
CH2COOH
HOOCH2C
NCH2CH2N
HOOCH2C
CH2COOH
This represented as H4Y
There is 6 sites for making coordination bonds as shown (four hydrogens and two
nitrogens) and can make 6 of five membered ring when complexed with metal
ions [M+]. So it can be combined with metal ions [M+] in 6 sites (more stable
complex). So EDTA called [hexadentate ligand].
EDTA is a weak acid because 4H+ can be replaced K1 (10-1), K2 (10-3), K3 
(10-6), K4 (10-8). . Any H+ can be replaced by [Na+, Li+, K+ and NH4+]
forming the salts of EDTA).
First and second (H) can be donated easily. Na2H2Y.2H2O (EDTA) 
(disodium salt) is used because it is more soluble than the acid (EDTA)
(H4Y) itself,
The famous salt is [Na2H2Y](disodium salt), or tetra sodium salt 
[Na4Y].
EDTA can be used as primary standard substance. 
But if the metal ion was more than monovalent, it will form 
complexes.
Stoichiometry: The stoichiometry of (M + EDTA) reaction is usually
1:1 and changing of metal ions will not affect, so we use it in the
calculations (M molarity = mole solute/ L soln.).
Properties of these reactions are:
1. 1:1 reaction.
2. Always produce same amount of hydrogen ions (2H+).
3. pH affected on titration results.
INDICATORS: Metal ion indicators or metalochromic indicators are
used:
H2In
Free Ind.
Color 1
+
M
=
MIn + 2H+
metal-Ind. comp.
color 2
These indicators have properties of acid-base indicators.
M+
+
M-In
H2In
+
=
EDTA
M-In + 2H+ (before starting the titration)
=
EDTA + 2H+ + M-In = H2In + M-EDTA
M-EDTA + In (during the titration)
(at the E.P. when last drop EDTA added)
Color 1
So M-EDTA complex more stable than M-In complex.
Hardness of water.
Hard water is water that contains cations with a charge of (+2) , 
especially Ca2+ and Mg 2+ (unit ppm CaCO3 = mg/ L)
Soft water only contains sodium cation. 
Types hardness of water.
Temporary hardness is due to the bicarbonate ion (HCO3- ) being present 
in the water. This type of hardness can be removed by boiling the water to
expel the CO2.
Permanent hardness is due to calcium and magnesium nitrate, sulfates 
and chloride…etc. this type of hardness cannot be eliminated by boiling.
when hard water heated Ca2+ react with bicarbonate ions to 
form insoluble CaCO3 (calcium carbonate)
A. Preparation and standardization of EDTA
solution
with std. 0.02M ZnSO4 solution: .
B. Determination of Mg2+ in unknown
solution.
C. Determination of hardness of water.
Total hardness
Pipette 25ml of tap water to a conical flask, add 0.5ml of buffer .1
pH=10, add 0.1gm of E.B.T indicator.
2. Titrate with EDTA solution from burette until the color change from
violet to blue. Record volume of EDTA as V1.
3.Calculate the total hardness of water as ppm CaCO3.
No. of mmol EDTA = No. of mmol CaCO3
M.V = Wt. (mg)CaCO3 / m.mass CaCO3 ,
m.mass CaCO3 = (100 g/mol)
Wt. (mg)CaCO3 in 25 mL = (M.V)EDTA* 100 / 25 mL
Wt. (mg)CaCO3 in 1 mL , 1 gm = (M.V)EDTA* 100 / 25 mL
No. of mmol EDTA = No. of mmol CaCO3
M.V = Wt. (mg)CaCO3 / m.mass CaCO3 ,
m.mass CaCO3 = (100 g/mol)
Wt. (mg)CaCO3 in 25 mL = (M.V)EDTA* 100 / 25 mL
Wt. (mg)CaCO3 in 1 mL , 1 gm = (M.V)EDTA* 100 / 25 mL
(1 g = 1000 mg)
Ppm = 1000,000
Wt CaCO3 in 1000,000 g water = (M.V)EDTA* 100 * 106 * 10-3/ 25 mL
ppm CaCO3 = (M.V1)EDTA* 105 / 25 mL
Permanent hardness
1. Place 250ml of tap water in a beaker, boil for 20-30min. filter to a 250ml vol.
flask.
2. Complete the volume by D.W.
3. Pipette 25ml of the water to a Conical flask, add 0.5ml of buffer pH=10, add
0.1gm of E.B.T. indicator.
4. Titrate with EDTA solution until the color change from violet to blue. Record vol.
of EDTA needed as V2.
Calculate the permanent hardness as ppm CaCO3.
Calculate the permanent hardness as ppm CaCO3.
Then calculate the temporary hardness as
ppm CaCO3. ppm CaCO3 = (M.V2)EDTA* 105 / 25 mL
Total hardness = Temporary hardness + permanent hardness