# practical205

```Zagazig University
Faculty of pharmacy
Clinical pharmacy program
Practical
Quantitative Analytical Chemistry 1
Pc 205
First level – Second Semester
2013
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Photo
Name:
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Serial No:
Group:
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Section:
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Attendance Sheet
Week
Marks
Signature
1st week
2nd week
3rd week
4th week
5th week
6th week
7th week
8th week
9th week
10th week
Total marks
10
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Analytical Chemistry Department
Faculty of Pharmacy
Student name:
Student number:
Section:
Group:
Final examination 2013
First sample:
NO:
Second sample
NO:
Results and calculation
Final Marks
4
Part 1
Acid – base titration
Standard solution
Standard solution may be generally defined as “a solution of exactly
known concentration”
In titrimetric analysis the standard solution is often the titrating agent.
It is more usual to describe such standard solutions either as empirical;
molar or normal.
1.
Molar standard solutions:
A molar solution (M or M/1) is one, which contains the gram molecular
weight of the substance in one liter of solution.
The molar concentration or molarity of a solution is stated by a
number indicating how many moles or what fraction of a mole is
present per liter of solution.
2.
Normal standard solutions
A normal solution (N or N/1) is a solution, which contains one
equivalent of substance per liter of solution. Any convenient multiple
or fraction of the gram – equivalent weight may be used when a
normal is too weak or too strong for the purpose at hand.
5
3.
Empirical standard solutions
These solution are prepared in such a concentration that 1 ml of it
react with some definite, convenient quantity of another substance,
as for example with 5 gm. This often desirable when the solution is to
be used in the determination of only one substance, but different
empirical solutions bear no relation to each other
In the determination of zinc a standard solution of potassium
ferrocyanide is prepared so that 1 ml of which reacts with 5 gm of zinc
according to the following equations
3 Zn2+ + 2 K4 (Fe(CN)6) = K2Zn3[Fe(CN)6] + 6 K+
 2 K4(Fe(CN)6). 3 H2O = 3 Zn2+
i.e. 2 x 422.3 gm K4(Fe(CN)6). 3 H2O / 1 = 3 x 65.38 mg Zn/ml
And if 21.2 gm K4(Fe(CN)6). 3 H2O are dissolved in a liter of its solution,
1 ml = 5 mg Zn
It is evident that this relation does not hold well if the standard
ferrocyanide solution is used for the determination of substances other
than zinc.
Equivalent weight of acids and bases:
The equivalent weight is determined by “the number of replaceable
hydrogen atoms in the acid molecule”. Thus to make a normal
solution of the mono-basic hydrochloric, hydrobromic, hydroiodic,
nitric or acetic acids, it is necessary to have a molecular weight in
grams (1mole) of the acid dissolved in a liter of solution. To make one
liter of normal solution of the diabasic sulphuric acid, only half mole of
the acid is necessary. Phosphoric acid has three equivalent weights
6
depending on the number of its hydrogen, which enter into the
reaction
H3PO4 + OH– = H2O + H2PO4–
equivalent weight =
H3PO4
1
H3PO4 + 2 OH– = 2H2O + HPO4–2 equivalent weight =
H3PO4
2
H3PO4 + 3 OH– = 3 H2O + PO4–3 equivalent weight =
H3PO 4
3
The normal weight of a base is determined by the number of
replaceable hydroxide in the base molecule. Thus of potassium
hydroxide, KOH, sodium hydroxide, NaOH and ammonium hydroxide,
NH4OH, 1 mole per liter makes a normal solution. Of barium hydroxide
Ba(OH)2, calcium hydroxide Ca(OH)2 and strontium hydroxide,
Sr(OH)2, only half mole per liter is required.
Salts of weak acids and strong bases have a basic reaction. The
equivalent weight of such basic salts, for example sodium carbonate,
is determined in terms of “the weight which reacts with one hydrogen”
with methyl orange as indicator, one mole of sodium carbonate
reacts with 2 moles of hydrochloric acid, that is 2 hydrogen; hence,
the equivalent weight is half mole of sodium carbonate. With
phenolphthalein as indicator, however, the end point is reached with
1 mole of hydrochloric acid that is one hydrogen; in this case the
equivalent weight of sodium carbonate is its molecular weight.
Preparation of standard solutions:
There are two general methods for the preparation of a standard
solution:
1. Direct method: This method consists of dissolving the equivalent
weight, or a definite fraction or multiple of a substance of known purity
in the solvent, usually water, and making uniform after dilution to the
proper final volume.
7
2. Indirect method: By this method a solution of approximately the
desired normality is first prepared, then standardized against a
standard of exactly known concentration or purity.
The following are some important substances, which may be
obtained rather easily in a condition of suitable purity for the direct
preparation of standard solution, and are called, therefore, primary
standards: benzoic acid, potassium acid phthalate, sodium
carbonate, potassium bormate, potassium iodate, potassium
dichromate, arsenious oxide, iodine, potassium chloride, sodium
chloride and silver nitrate.
Primary standard should have the following characteristics in order
to be ideally satisfactory:1. easy to obtain, purity and dry without change in composition
2. Capable of being tested for impurities by simple qualitative tests of
known sensitivity.
3. stable in contact with air of average humidity over a long periods
4. Having a high equivalent weight in order that weighing errors may
have a small relative effect.
5. Readily soluble under the conditions in which it is employed.
6. React with the other substances rapidly and stoichiometrically.
Preparation of standard acid:
Hydrochloric and sulphuric acids are widely employed in the
preparation of standard solutions of acids. Both of these are
commercially available as concentrated solution; conc. HCl is about
10 to 12 N, and conc. H2SO4 is about 36 N. By suitable dilution, solution
of any desired approximate strength may be readily prepared.
Hydrochloric acid is more convenient, since most chlorides are soluble
in water. Nitric acid is rarely employed, because it almost invariably
contains a little nitrous acid, which has a destructive action upon
many indicators.
8
Preparation of 0.1 N HCl and standardization:
Measure, by means of a graduated cylinder or a burette, 9 ml of pure
conc. HCl, pour the acid into a liter measuring flask containing about
500 ml distilled water. Dilute to the mark with distilled water and mix.
This will give a solution approximately 0.1 N solution.
If 1.0 N HCl is required, use 90 ml of the conc. HCl. Approximately 0.1
N H2SO4 is similarly prepared from 3 ml. of pure conc. H2SO4.
Standardization with anhydrous sodium carbonate:
The sodium carbonate required for this purpose must be free from
chloride, sulphate and from insoluble impurities. Analytical reagent
quality (A.R.) sodium carbonate of 99.9 % purity is obtainable
commercially. This contains a little moisture and must be dehydrated
by heating at 260 to 270 oC. For half an hour and allowed to cool in a
desiccator before use.
Alternatively, pure Na2CO3 may be prepared by heating pure
NaHCO3 to 260 to 270 oC for 1 to 1.5 hours. The temperature must not
may lose CO2. As small quantity, 3 to 4 g of pure NaHCO3 is placed in
a porcelain crucible and pressed against the walls of crucible so as to
form uniformly thick layer. Insert the crucible in a sand bath, immerse
the bulb of a 300oC thermometer in the sand bath with a free flame,
and maintain the temperature at 260 to 270oC for 1 hour; stir the mass
and allow the crucible to cool in a desiccator.
Procedure:
Weight out accurately about 0.1 gm of the pure Na2CO3 into 25 ml.
indicator. Titrate with the acid in the usual manner. Repeat the
titration with two other portions of Na2CO3. From the weights of
Na2CO3 and the volumes of HCl solution employed the strength (and
9
the normality) of an acid may be computed for each titration. The
results should agree to about 0.2 %. The mean of the results is taken as
the strength of the solution.
Results
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Preparation of standard alkali
The hydroxides of sodium, potassium and barium are used for this
purpose, since they are strong bases and are readily soluble in water
sodium hydroxide is most commonly used because of its cheapness.
None of these hydroxides can be obtained pure, a certain amount of
water and alkali carbonates being always present. A standard
solution cannot, therefore, be prepared by directly dissolving a known
weight in a definite volume of water. It is therefore, necessary to
prepare a solution of approximately the proper concentration and
then to standardize this alkali against hydrochloric acid of known
normality, or against any of the primary standards previously
mentioned, as, for example potassium acid phthalate.
Preparation of carbonate free sodium hydroxide:
Dissolve 50 g of sodium hydroxide in 50 ml. of water in a conical flask.
Transfer the solution to a 75 ml- test tube, and insert a well – fitting
stopper covered with the foil. Allow standing until the undissolved
Na2CO3 has settled. The clear solution is practically free from
carbonate, and is about 15 N. To prepare an approximately 0.1 N
10
NaOH carefully with draw 6.5 ml through a graduated pipette into a
liter flask and dilute quickly to 1 liter with CO2 – free water.
Standardization of the approximately 0.1 NaOH
If the sodium hydroxide solution contains carbonate, methyl orange
must be used in standardization against hydrochloric acid of known
normality. Phenolphthalein and similar indicators, which affected by
CO2 cannot be used. In the case of carbonate – free sodium
hydroxide, phenolphthalein or thymol blue may also be used as the
indicator, and standardization may be made with standard
hydrochloric acid or potassium hydrogen phthalate.
Results
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Quiz 1
Define:
Normal solution:
Molar solution:
Equivalent weight (EW) = Molecular weight (MW) / Valencey. (Z)
Normality (N) = Equivalent weight (EW) / Volume of Solution (1000 ml.)
Meq = Actual weight (W) / Equivalent weight (EW)
Example: Molecular wt of K2Cr2O7 is 294.148 gm
Valency = 6 (because of Cr -6)
EW = 294.148/6 = 49.025
N = ………………………………….
To make 0.25 N solution = ………………….
Molarity = Molecular Weight, grams / One Liter of Distilled Water
Example:
1 M NaCl = ……………………………………………
0.5 M NaCl = …………………………………………
12
1-
Determination of Boric acids
Principle:
Boric acid act as a weak monobasic acid (Ka = 5.8 x 10-10), and,
therefore, produces no change in colour in methyl orange. If
phenolphthalein is used, the end point appears very rapidly owing to
hydrolysis of the sodium borate formed.
NaBO2 + 2 H2O ⇄
NaOH + H3BO3
If sufficient polyhdroxy compound, e.g. glycerol, is added the
reaction can be made quantitatively, presumably owing to the
formation of a much stronger complex acid of the form (C3H5O2OH)
B(OH). The correct end point is reached when the pink colour no
longer disappears on adding more glycerol.
Procedure:
1Pipette 10 ml sample solution (1 gm in 100 ml water)
2Add 10 ml glycerol previously neutralised and titrate with 0.1 N
NaOH using phenolphthalein as indicator
N.B.
 The quantity of glycerol must be 50 – 30% of the total volume of the
solution. At the end point the volume of glycerol must not fall below
1/3 of the total volume.
 Mannitol is more effective (solid). Dextrose and invert sugar can
also be used.
NaOH + H3BO3 = NaBO2 + 2 H2O
1 ml 0.1 N NaOH = 0.006184 gm H3BO3
Results
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2.
Determination of borax
Principle:
When borax is dissolved in water. It is hydrolyzed into
Na2B4O7 + 7 H2O ⇄4 H3BO3 + 2 NaOH
If the aqueous solution is titrated with 0.1 N HCl using M.O. as
indicator, it is the NaOH that is actually titrated; boric acid being of no
effect on the indicator, and the net reaction is;
Na2B4O7 + 5 H2O + 2 HCl ⇄ 4 H3BO3 + 2 NaCl
Na2B4O7 . 10 H2O = 2 HCl
The residual solution can be titrated for the remaining boric acid with
0.1 N NaOH after adding glycerol and using ph.ph as indicator. The
reaction would be:
Na2B4O7 + 5 H2O + 2 HCl ⇄ 4 H3BO3 + 2 NaCl
4 H3BO3 + 4 NaOH  4 NaBO2 + 8 H2O
 Na2B4O7. 10 H2O = 4 NaOH
In the other words, when a pure sample of borax containing no free
acid is titrated, the volume of standard alkali used would be exactly
double the volume of standard acid.
14
If the borax solution is treated directly with glycerol and titrated
against 0.1 N NaOH, the solution could be graded as boric acid that
is half neutralised.
Procedure:
A) To 10 ml sample (2.0gm in 100 ml distilled water) add 3 drops methyl
orange and titrate with 0.1 N HCl
B) To the titrated solution add an equal volume of glycerol and titrate
with 0.1 N NaOH using phenolphthalein as indicator. the volume of 0.1
N NaOH must be double that of 0.1 N HCl, if the sample of borax is
pure
1 ml 0.1 N HCl= Na2B 4 O7. 10H2 O  0.01907 gm Na2B4O7.10 H2O
2 x 10 x 100
Results
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Quiz 2
1) Calculate the pH of (a) 0.055 M HNO3 (b) 4.55 M HCl.
2) Calculate the pH of 0.10 M NaOH
3) A solution has a pH of 6.88. What is [H3O+]? [OH-]?
4)
Very concentrated nitric acid has a pH of -1.2. What is [H3O+]?
5) Calculate the pH at t = 0, at 5.00 mL, at 10.00 mL, at the
equivalence point and at 40.00 ml in a titration of 25.0 mL of
0.120 M formic acid with 0.105 M NaOH.(kA of formic acid = 1.8
E-4)
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6) Calculate [H+] in a solution with a volume of 1.000 L and which
contains 2.70 g of HCN and 2.45 g of NaCN. HCN has a Ka of
4.0 x 10-10.
7) Determine the pH of 0.30 M acetic acid (HC2H3O2) with
the Ka of 1.8x10-5.
8) Determine the pH of a 0.010 M solution of Ba(OH)2.
9) Determine the pH of 0.15 M ammonia (NH3) with a Kb=1.8x10-5.
10) Find the pH of a solution formed by dissolving 0.100 mol of
HC2H3O2 with a Ka of 1.8x10-8 and 0.200 mol of NaC2H3O2 in a
total volume of 1.00 L.
17
3. Determination of alkaline earth salts of strong acids:
(e.g. BaCl2 .2H2O)
Principle:
The alkaline earth metal (e.g. Ba2+) is precipitated as its insoluble
carbonate by addition of a known excess of 0.1 N Na2CO3. The
solution is then boiled, cooled to about 0oC and the excess sodium
carbonate remaining is titrated with 0.1 N HCl using phenolphthalein
as indicator, multiply the volume of acid by 2.
The volume used in precipitating the metal ion is obtained by
difference.
Procedure:
1- Pipette 20 ml sample into a beaker (1 gn in 100 ml distilled water)
2- Dilute with 150 ml water
3- Heat to about 70oC and slowly run 25 ml of 0.1 N Na2CO3 solution
4- Boil for 2 min. cool to about 0oC and titrate with 0.1 N HCl using
phenolphthalein as indicator
Na2CO3 + BaCl2 = BaCO3 + 2 NaCl
Na2CO3
= BaCl2 . 2 H2O
= 244.36 gm BaCl2 . 2 H2O
244 .36
1 ml of 0.1 N Na2CO3 =
= 0.012218 g BaCl2.2 H2O
2 x10 x1000
4. Determination of Mercury oxide
Principle:
HgO dissolves readily in KBr or KI solutions with the production of
potassium mercuric bromide or iodide and an equivalent quantity of
18
KOH. This can be titrated with 0.1 N HCl using methyl orange or
phenolphthalein as indicator.
Procedure:
To 0.2 gm of HgO, accurately weighed in an Erlenmeyer flask, add 7
gm of KI and 20 ml water. Shake until completely dissolved and titrate
the solution with 0.1 N HCl, using methyl orange or phenolphthalein as
indicator
HgO + 4 KI + H2O = K2 (HgI4) + 2 KOH
2 KOH + 2 HCl = 2 KCl + 2 H2O
2 HCl = HgO
1 ml 0.1 N HCl=
HgO
216.61

 0.010831 gm HgO
2x10x1000 2x10x1000
Results
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5. Determination of ammonium salts
(e.g. NH4Cl)
A)
Indirect method:
When a known excess of standard NaOH is added to an ammonium
salt (other than carbonate or bicarbonate) and the mixture boiled,
NH3 is removed and an equivalent amount of the alkali is used up.
19
Procedure:
1. To 20-ml sample (1 gm in 100 ml distilled water) solution add 1 drop
of methyl orange and neutralize the solution if necessary.
2. Add 25 ml of 1 N NaOH solution and boil, placing a funnel in the
neck of the flask to minimize loss by evaporation. Continue boiling till
no more NH3 is driven off. Cool and titrate with 1 NHCl using methyl
orange as indicator.
NH4Cl + NaOH = NaCl + NH3 + H2O
NH 4 Cl
1 ml 1 N NaOH =
= 0.0535 gm NH4Cl
1000
B)
Formal titration:
When formaldehyde is added to ammonium salt solution,
hexamethylene tetramine (CH2)6N4 is produced together with an
amount of acid equivalent to the amount of ammonium salt used. The
liberated acid is titrated with standard alkali using phenolphthalein as
indicator.
Procedure:
To 20 ml sample in a beaker add 5 ml of formaline (previously
neutralised to phenolphthalein) and after 5 min. titrate with 0.1 N
NaOH, using phenolphthalein as indicator.
4 NH4Cl + 6 HCHO = (CH2)6N4 + 4 HCl + 6 H2O
4 NH4Cl = 4 HCl = 4 NaOH
 NH4Cl = NaOH
53.5
1 ml 0.1 N NaOH =
= 0.00535 gm NH4Cl
10 x 1000
Results
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Quiz 3
Write on the following
1. Formol titration
2. Back titration with example
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6. Determination of ammoniated mercury
Principle:
The determination depends upon the reaction, previously used for
the determination of mercury oxide:
NH2HgCl + 2 KI + 2 H2O = NH4OH + KOH + KCl + HgI2
The HgI2 dissolves in the excess KI producing a soluble complex;
HgI2 + 2 KI = K2HgI4 (soluble)
The alkali liberated is titrated with standard acid
Procedure:
To 20 ml sample ( add 4 gm KI and 5 drops methyl red and titrate with
0.1 N HCl until the solution is just pink
Results
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22
7. Determination of Potassium Persulphate
(e.g. K2S2O8)
Principle:
When a salt of potassium persulphate is boiled in presence of a
suitable catalyst, the salt decomposes as follows:
2 K2S2O8 + 2 H2O = 4 KHSO4 + O2
the resulting KHSO4 may be titrated with standard alkali using methyl
orange as indicator.
Procedure:
To 20 ml sample (4.0 gm in 50 ml cold water) add methyl orange and,
if the indicator goes red, bring back the yellow colour with 0.1 N
NaOH. This is a measure of KHSO4 in K2S2O8.
Now add about 5 drops 0.1 N AgNO3 and boil for 30 minutes, cool,
add methyl orange and titrate with 0.1 N NaOH.
K2S2O8 + 2 H2O = 4 KHSO4 + O2
4 NaOH = 4 KHSO4 = 2 K2S2O8
K 2S 2 O 8
270 .2
1 ml 0.1 N NaOH=

 0.01351 gm K2S2O8
2 x 10 x 1000 2 x 10 x 1000
Results
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Quiz 4
(1)
A buffer solution was prepared which had a concentration of 0.20
mol dm-3 in ethanoic acid and 0.10 mol dm-3 in sodium ethanoate.
If the Ka for ethanoic acid is 1.74 x 10-5 mol dm-3, calculate the
theoretical hydrogen ion concentration and pH of the buffer
solution.
(2)
In what ratio should a 0.30 mol dm-3 of ethanoic acid be mixed
with a 0.30 mol dm-3 solution of sodium ethanoate to give a buffer
solution of pH 5.6?
24
(3)
What is the pH of a buffer solution made from dissolving 2.0g of
benzoic acid and 5.0g of sodium benzoate in 250 cm3 of water?
(Ka benzoic acid = 6.3 x 10-5 mol dm-3, Ar's: H = 1, C = 12, O = 16, Na =
23)
(4)
Calculate the pH of a buffer made by mixing 100 cm3 of a 0.40 M
sodium propanoate and 50 cm3 of 0.2 M propanoic acid solution.
(Ka propanoic acid = 1.3 x 10-5 mol dm-3, total volume of buffer = 150
cm3)
(5)
Calculate the pH of buffer solution made by mixing together 100
cm3 of 0.100M ethanoic acid and 50 cm3 of 0.400M sodium
ethanoate, (given that Ka for ethanoic acid is 1.74 x 10-5 mol dm-3)
25
8. Determination of mixture of Carbonate and Bicarbonate
(e.g. Na2CO3 and NaHCO3)
Principle:
Two methods may be used:
A) The total alkali (carbonate and bicarbonate) is determined by
titration with standard acid, using as indicator methyl orange, or
bromophenol blue.
The carbonate content is determined by a similar titration but using
phenolphthalein as indicator. The volume of standard acid used in the
titration is multiplied by two to correspond to the carbonate content
in the aliquot taken.
B) The total alkalinity is determined as in method (A). the bicarbonate
content is determined by transforming into carbonate by addition of
a very slight excess of standard NaOH, the total carbonate
precipitated with BaCl2, and the residual NaOH titrated with standard
acid, using phenolphthalein as indicator.
Procedure (A)
120 ml samples (1 gm in 200 water) diluted with 50-ml water and
titrated with 0.1 N HCl using methyl orange as indicator.
This is a measure of carbonate and bicarbonate.
2Another 20 ml sample diluted with water and titrated with 0.1 N
HCl using phenolphthalein as indicator. Multiply the volume of
standard acid by two, and this would be a measure of the carbonate
content in 20 ml.
3Subtract volume of standard acid equivalent to carbonate from
that used in first step and this would be a measure of the bicarbonate
content in 20 ml.
26
Procedure (B)
1the total alkalinity as under procedure A
2Measure 20 ml sample solution and add 0.1 N NaOH (free from
carbonate) equivalent to ml of 0.1 N HCl used in step 1 (assuming that
the solution contains NaHCO3 only). Dilute with water to about 150 ml,
heat to about 70 oC, and precipitate the carbonate original and
produced- by adding BaCl2 solution (about 0.1 N) in slight excess. Boil
for 2 min. cool to room temperature and immediately titrate with 0.1
N HCl using phenolphthalein as indicator (or thymol blue) as indicator
until the pink colour is almost discharged (or the blue colour changes
into yellow).
The difference between ml 0.1 N HCl used in 1 and that used in 2 is a
rough measure of the bicarbonate or, in other words, of the volume
of 0.1 N NaOH required to transform the bicarbonate into carbonate.
3The process is repeated until the volume of 0.1 N NaOH added
to transform the bicarbonate into carbonate is reduced to a
minimum, indicated by the fact that when the pink colour produced
on addition of phenolphthalein after precipitation with BaCl 2 solution
is discharged by not more than 3 – 4 drops of 0.1 N HCl.
40.1N HCl used to discharge the pink colour after addition of
phenolphthalein is subtracted from ml 0.1 N NaOH added, and the
difference is a measure of the NaHCO3. subtract that difference from
total 0.1 N HCl used in 1 and the balance is equivalent to Na2CO3
Na2 CO3
106

 0.0053 gm Na2CO3
2 x 10 x 100 2 x 10 x 1000
NaHCO 3
84
1 ml 0.1 N HCl =

 0.0084 gm NaHCO3
10 x 1000 10 x 1000
1 ml 0.1 N HCl =
Results
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9. Determination of mixture of calcium oxide and calcium
carbonate
Principle:
The estimation is based on the facts that:
1CaO suspended in water is alkaline to phenolphthalein.
2Phenolphthalein loses its colour when (H+) is lower than that
required attacking the precipitate of CaCO3.
3CaO is more soluble in sucrose solution than in water, the
complex calcium saccharate being equally alkaline as CaO.
Procedure:
1- To 20 ml sample solution (0.5 gm of mix., 2ml alcohol add 100 ml
10% sucrose solution) add 10 drops phenolphthalein and titrate with
0.5 N HCl solution.
The number of ml of 0.5 N acid ia a measure of CaO
2- To another 20 ml sample in beaker add known excess of 0.5 N HCl
sufficient to convert all carbonate to oxide. Boil off Co 2, cool and
titrate the excess acid with 0.5 N NaOH using phenolphthalein as
indicator.
3- Subtract the volume of 0.5 N HCl used in steps 1 from that
consumed by the mixture in step 2, the difference is equivalent to
CaCO3 in 0.5 gm of the sample.
CaO + 2 HCl = CaCl2 + H2O
28
2 HCl = CaO
1 ml 0.5 N HCl =
CaO
 0.02502 gm CaCO3
2x 2 x 1000
CaCO3 + 2 HCl = CaCl2 + CO2 + H2O
1 ml 0.5 N HCl =
CaCO 3
 0.01402 gm CaCO3
2x 2 x 1000
Results
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10.
Determination of mixture of boric acid and borax
Principle:
Solutions of alkali borate may be titrated with standard acid using
methyl orange as indicator. They react towered this indicator as if they
were solutions of alkali hydroxides. They behave as dinormal bases
when titrated with acids.
Na2B4O7. 10 H2O + 2 HCl = 4 H3BO3 + 2 NaCl + 5 H2O
While the liberated boric acid consumes 4 molecules of NaOH when
titrated with alkali, using phenolphthalein as indicator in presence of
glycerol.
29
Procedure:
1. Titrate 20 ml sample solution with 0.1 N HCl using methyl orange as
indicator. (this is a measure of the borax)
2. To the same solution add glycerol previously neutralised to
phenolphthalein (equivalent to the bulk of the solution) and titrate the
boric acid –original and resultant – with 0.1 N NaOH, using
phenolphthalein as indicator.
3. Multiply the number of ml 0.1 N HCl used with methyl orange by two
and subtract the product from ml of 0.1 N alkali used with
phenolphthalein. The difference corresponds to boric acid originally
present is the free state
Results
…………………………………………………………………………………….
…………………………………………………………………………………….
…………………………………………………………………………………….
…………………………………………………………………………………….
…………………………………………………………………………………….
…………………………………………………………………………………….
Quiz 5
Write on the following
Types of buffer solutions
30

The following titration curve is the kind of curve expected for the
titration of a ____ acid with a ____ base.
(a) strong, strong
(b) weak, strong
(c) strong, weak
(d) weak, weak
(e) none of these


Consider the titration of 30.0 mL of 0.20 M nitrous acid by adding
0.0500 M aqueous ammonia to it. The pH at the equivalence point is
_____. (Note: This is the titration of a weak acid with a weak base.)
(a) greater than 7
(b) equal to 7
(c) less than 7
(d) cannot be determined without more data (not including
Ka and Kb)
(e) is impossible to predict
The pH of a 0.02 M solution of an unknown weak acid is 3.7. what
is the pKa of this acid?
(a) 5.7
(b) 4.9
(c) 3.2
(d) 2.8
(e) 3.7
31
Part 2
Gravimetery
1- determination of calcium as calcium oxalate
monohydrate
Principle
The calcium is precipitate as calcium by treating hot hydrochloric
acid solution with ammonium oxalate and slowly neutralizing with
ammonium hydroxide. The precipitate is washed with dilute
ammonium oxalate solution and then weighed in the following forms:
1. As CaC2O4. H2O by drying at 105o C- 110oC.
2. As CaCO3 by drying at 475 - 525 oC.
3. As CaO by igniting at 1200OC.
Procedure:
1. to 20 ml sample in beaker add 3 drops M.O
2. Add NaOH dropwise till yellow colour
4. Add in excess till faint acidic colour (pink).
5. Heat to boil, add slowly 10 ml boiled freshly prepared saturated
ammonium oxalate solution and heat to boil to accommodate the
ppt. for 2 minute
6. Leave the precipitate for one hour.
7. Filter the precipitate and test the solution with few drops of
ammonium oxalate solution for complete precipitation.
8. Wash the precipitate twice with 1% ammonium oxalate solution.
9. Transfer the precipitate to the Gooch with water.
10. Dry the Gooch at 105 - 110oC and weight.
32
11.
Repeat drying and weighing until constant weight
Calculation
CaCO3 + 2 HCl = CaCl2 + H2O
CaCl2 + H2O + C2O4H2 = C2O4 Ca.H2O + 2 HCl
i.e. CaCO3 = C2O4 Ca.H2O
100.09 gm = 146.1162 gm
Convertion factor =
CaCO3
= 0.685
C2O4 Ca.H2O
Results
…………………………………………………………………………………….
…………………………………………………………………………………….
…………………………………………………………………………………….
…………………………………………………………………………………….
…………………………………………………………………………………….
…………………………………………………………………………………….
33
2 Determination of Nickel with dimethyl glyoxime
Principle
When dimethylglyoxime is added to an alkaline solution of a nickel
salt, at a pH range 7.5 - 8.1, a bright red insoluble complex is
produced. The precipitate is filtered, dried at 120oC and weight.
Procedure:
1. To 10 ml sample add water to 100 ml, followed by 2 drop dil HCl
and 15 ml 1 % DMG.
2. Add ammonia till ammonia odour and leave to one hour.
3. Decant clear supernatant and wash the precipitate with cold
water.
4. Filter and dry the precipitate at 120oC for 1 hour then weight.
5. Repeat drying and weighing till constant weight
Calculation:
NiSO4. 7 H2O + 2 C2H8O2N2 = Ni (C4H7O2N2)2 + H2SO4.
i.e. NiSO4. 7 H2O = Ni (C4H7O2N2)2
280.8934 gm = 288.91 gm
Convention factor = NiSO4. 7 H2O = 1.0285
Ni (C4H7O2N2)2
Results
…………………………………………………………………………………….
…………………………………………………………………………………….
…………………………………………………………………………………….
…………………………………………………………………………………….
…………………………………………………………………………………….
…………………………………………………………………………………….
34
Quiz 6
Write on the following:
Fundamental Requirements for the precipitation process.
Types of contamination:
35
Part 3
Precipitmetric determination
1. Determination of Zinc salts (e.g. ZnO)
Principle:
Zinc ions in neutral or acid medium react with ferrocyanide forming
a precipitate of potassium zinc ferrocyanide:
2 K4Fe(CN)6 + 3 Zn2+ = K2Zn3(Fe(CN)6)2 + 6 K+
The end point is detected by an external indicator such as uranyl
acetate, ammonium molybdate or ferric chloride. An internal
indicator is preferentially used, e.g. diphenylamine
Procedure:
1- To 10 ml sample add 1 gm NH4Cl and dilute with 50 ml water.
2- Add 1 drop of ferric sulphate solution and 2 - 3 drop
diphnylamine indicator
3- Awit till the blue violet colour develops and titrate with 0.05 N
potassium ferrocyanide solution with vigorous shaking till the blue
colour changes to a permanent pale green.
Calculation
2 K4Fe(CN)6 + 3 Zn2+ = K2Zn3(Fe(CN)6)2 + 6 K+
2 K4Fe(CN)6 + 3 Zn
K4Fe(CN)6 = 3/2 Zn
36
1 ml 0.05 N K4Fe(CN)6 =
3 Zn
= 3 x 65.38
2 x 20 x 1000
2 x 2000
Results
…………………………………………………………………………………….
…………………………………………………………………………………….
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2- Determination of mercuric chloride
Principle:
when HgCl2 solution is added drop by drop to a KI solution, a red
precipitate of HgI2 mercuric iodide is formed when the two liquids first
come in contact with one another, but on stirring it redissolves in
excess KI owing to the formation of soluble complex iodide.
HgCl2 + 2 KI = HgI2 + 2 KCl
(red ppt.)
HgI2 + 2 KI = K2HgI4
(soluble)
The total reaction is, therefore:
HgCl2 + 4 KI = K2HgI4 + 2 KCl
37
After all the potassium iodide has been converted into potassium
mercuric iodide, the addition of a drop excess of HgCl 2 produces a
permanent brilliant red colour, due to the formation of the insoluble
HgI2, which even in the smallest quantity communicates its tint in the
liquid.
HgCl2 + K2HgI4 = 2 HgI2 + 2 KCl
Procedure:
To 10 ml 0.1 N KI in conical flask, titrate with HgCl2 sample (in burette)
drop by drop with constant stirring till brilliant red colour.
Calculation
HgCl2 = 4 KI
1ml 0.1 N KI =
HgCl2
4x10x1000
= 271.52 = 0.006788 gm HgCl2
4000
Results
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38
3 Determination of mixture of chloride iodide
(e.g. KCl and KI)
Principle:
These two ions differ in the ease which they are adsorbed on the
corresponding silver halide. This makes it possible to select adsorption
indicators, which will permit of the estimation of chloride and iodide
in the presence of one another. Thus, the iodide may be determined
with eosin by titration with standard AgNO3, and the total iodide
chloride by similar titration with fluorescein; chloride is obtained by
difference.
Procedure:
1- 20 ml sample + 50 ml water + 1 ml fluorescein indicator, titrate
with 0.05N AgNO3 solution - the end point red colour on the
precipitate (Total)
2- New 20 ml sample + 50 ml water + 1 ml eosin titrate , with 0.05 N
AgNO3 until red colour is formed on the precipitate
e.p.1 - e.p.2 = Cl
Results
…………………………………………………………………………………….
…………………………………………………………………………………….
…………………………………………………………………………………….
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39
Quiz 7
Write on the following

Solubility product constant:
If the solubility of AgCl is 0.0015 g/l what is the solubility product.
(143 is Molecular weight)

Calculate the solubility of silver sulphide in pure water.
Ag2S ⇋ 2 Ag+ + S2
40
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