Physical Pharmacy-I lab (With values) MANIK

Department of Pharmacy
University of Rajshahi
Rajshahi-6205, Bangladesh

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LABORATORY MANUAL
FOR
HYSICAL · HARMACY
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Date Page no.Name of the experimentSerial no.
INDEX
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b) Secondary Standard Substance: Secondary standard substances are the substances from
which standard solution cannot be prepared by directly weighting and then dissolving in suitable
solvent. Examples are:
iv. The gram equivalent weight is large.
iii. The substances are stable both in solid or liquid state.
ii. The molecular weight of the substance corresponds to the chemical formula.
i. They are chemically pure substance.
Criteria of Primary Standard substance:
I t
a) · Primary Standard substance: Primary standard substances are substances from which
standard solution can be prepared directly by taking weight at an electrical balance and then
dissolving that substance in distilled water or any other suitable solvent. Examples are:
Standard solution: In the case of volumetric analysis, the solution whose strength is known
is called standard solution. Standard solution can be prepared directly from primary standard
substance by taking exact weight and then dissolving in distilled water.
PRINCIPLE:
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j I EXPERIMENT - 1:
/ ~ a) Preparation of pritnary standard solution of Sodium thiosulfate (Na2S03), Oxalic acid
{(COOH)z. 2H20}, Potassium dichromate (K2Cr201).
b) Preparation of secondary standard solution of Sulfuric acid (H2S04), Hydrochloric acid
(HCl) and Acetic acid (CH3COOH).
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Secondarystandard substance:
i. Sulfuric acid (H:~S04)
ii. Hydrochloric acid (HCI}
iii. Nitric acid (HNQ3)
iv. Acetic acid (CH3COOH)
v. Sodium hydroxide (NaOH)
vi. Distilled water (H20)
iv. Potassium dichromate (K2Cr201)
iii. Sodium oxalate (NaOOC-COONa)
ii. Oxalic acid (HOOC-COOH)
i. Sodium carbonate (Na2C03)
Primarystandardsubstance:
Reagent:
Concentrationterm:Normal solution.
S2 = Strength of secondary standard solution
V2 = Volume of secondary standard solution ·
S1 =Strength of primary standard solution
V1 = Volume of primary standard solutionWhere,
In this case, the appropriate weight of the substance under test is taken and then dissolved in
distilled water and dil~te solution is prepared. This solution is then standardized with primary
standard solution and then its strength is determined by using the following equation:
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v. Preparation of standard O.l(N) Sodium thiosulfate solution: The gram equivalent
weight of this substance is 158. So, to prepare 100ml of 0.1 (N) solution of Na2S203, we need
(158 + 100) = l.58gm ofNa2S203.
iv. Preparation of standard O.l(N) Potassium dichromate (K2Cr201) solution: The gram
equivalent weight of potassium dichromate is _f;j ~-b6 = 49. So, to prepare 1OOml of 0.1 (N)
potassium dichromate solution we need (49 + I00) = 0.49gm K2Cr201.
Now weigh 0.49gm of K2Cr207 in the electrical balance and dissolved it in 50ml of distilled
water in lOOml volumetric flask and shake gently. Then make the volume up to the mark with
distilled water.
iii. Preparation of standard O.l(N) Sodium oxalate solution: The gram equivalent weight
of this substance is 170 + 2 = 85. So, to prepare IOOml ofO.l(N) solution of sodium oxalate, we
need (85 + 100) = 0.85gm of sodium oxalate.
Now weigh 0.85gm of sodium oxalate in the electrical balance and dissolved it in 50ml of
distilled water in 1OOml volumetric flask and shake gently. Then make the volume up to the
mark with distilled water.
ii. Preparation of standard O.l(N) Oxalic acid solution: The gram equivalent weight of
oxalic acid is 126 + 2 = 63. So, to prepare IOOml of 0. l(N) solution of oxalic acid, we need (63 +
100) = 0.63gm of oxalic acid.
Now weigh out 0.63gm of oxalic acid in the electrical balance and dissolved it in 50ml of
distilled water in lOOml volumetric flask and shake gently. Then make the volume up to the
mark with distilled water. ·
Apparatus:
i, Balance
ii. Measuring cylinder
iii. Funnel
iv. Volumetric flask
Procedure:
a) Primary Standard Substance:
i. Preparation of standard O.l(N) Na2C03 solution: The gram equivalent weight of
Na2C03 is 53.50. 53 + 100 = 0.53 gram ofNa2C03 is needed to prepare O. l(N) lOOml solution of
Na2C03. ·..t ·~__:---~----------- -------------·----------------------···---------_,
Now 0.5fgm of Na2C03 was weighed in the electrical balance and dissolved it in 50ml of~-----------------:-;:_,,___ --------------·--------- -....
distilled water in. l_OOml volumetric flask and shake gently. Then make the volume up to the
mark with distilled water.
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Mass = volume x density
Thus density ofHCI is 1.055. So, to prepare lOOml O. l(N) solution of CH3COOH the needed
volume of CH3COOH is (0.6 + 1.055) = 0.56ml = 0.6ml.
Now~ take lOOml volumetric flask, which contains 50ml of distilled water and 0.6ml of
CH3COOH is added slow,Iy. Then make the volume up to the mark with distilled water.
Page 6 of 23
iii. Preparation of standard O.l(N) Acetic acid (CH3COOH) solution: The gram
equivalent weight of CH3COOH is 60. So, to prepare lOOml of O. l(N) acetic acid solution (60 +
100) = 0.6gm of 100% CH3COOH is needed. But CH3COOH is liquid and secondary standard
substance. So its volumetric measurement is needed which may be obtained by the following
equation:
Thus density ofHCI is 1.18. So, to prepare lOOml O. l(N) solution of HCI the needed volume
of32% HCI is (1.14 + 1.18) = 0.966ml = lml.
Now take lOOmlvolumetric flask, which contains 50ml of distilled water and I ml of HCl is
added slowly. Then make the volume up to the mark with distilled water.
ii. Preparation of standard O.l(N) HCI solution: The gram equivalent weight of HCl is
36.5. So, to prepare lOOmlof O. l(N) HCI solution (36.5 + 100) = 0.365gm of 100% HCI acid is
needed. Since HCl is not 100% pure, it is 32% pure. So to prepare lOOml of O. l(N) HCI solution
{(100 x 0.365) + 32} = l.14gm HCI is needed. But HCl is liquid and secondary standard
substance. So its volumetric measurement is needed which may be obtained by the following
equation:
b) Secondarystandard substance
i. Preparation of standard O.l(N) Sulfuric acid solution: The gram equivalent weight of
H2S04 is (98 + 2) = 49. So, to prepare lOOml ofO.l(N) H2S04 solution (49 + 100) = 0.49gm of
100% H2S04 acid is needed. Since H2S04 is not 100% pure, it is 98% pure. So to prepare 1OOml
ofO. I(N) H2S04 solution {(100 x 0.49) -'!,_ 98} = 0.5gm H2S04 is needed. But H2S04 is liquid and
secondary standard substance. So its volumetric measurement is needed which may be obtained
by the following equation:
Thus density of H2S04 is 1.89. So, to prepare lOOml 0. l(N) H2S04 the needed volume of
98% H2S04 is 0.5 + 1.89 = 0.2645ml
Now take lOOml volumetric flask, which contains 50ml of distilled water and 0.2645ml of
H2S04 is added slowly. Then make the volume up to the mark with distilled water.
Now weigh l.58grrt of Na2S203 in the electrical balance and dissolved it in 50ml of distilled
water in lOOmlvolumetric flask and shake gently. Then make the volume up to the mark with
distilled water.
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1L
Precaution:
i. Acid should be poured very slowly to the water. If acid is poured to water very rapidly
then the flask will be heated and may burst.
ii. Concentrated acid should be measured with measuring cylinder pipette as accurately as
possible.
iii. The surface of the funnel should be washed into flask with distilled water so that no acid
loss occurs.
ii. Preparationof standardO.l(N) KOHsolution: The gram equivalent weight of KOH is
56. So, to prepare lOOml ofO. l(N) KOH solution, we need (56 -i- 100) = 0.56gm of KOH.
Now take 50ml of distilled water in a conical flask and shake gently>until the KOH granules
dissolved completely. Then pour this solution to 1OOml volumetric flask and rinse the conical
flask into the volumetric flask and make the volume up to the mark with distilled water.
iv. Preparation of standard O.l(N) NaOH solution: The gram equivalent weight of NaOH
is 40. So, to prepare lOOmlofO. l(N) NaOH solution, we need (40 + 100) = 0.4gm ofNaOH.
Now take lOOml volumetric flask, which contains 50ml of distilled water and 0.4gm of
NaOH is added slowly and shake gently until the NaOH granules dissolved completely. Then
make the volume up to the mark with distilled water.
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/'., Procedure:
i. Filling the burette: A clear and dried burette is filled with O. l(N) acid solution and mark
is adjusted with respect to zero.
ii. Placement of O.l(N) acid solution in conical flask by pipette: A clear and dried conical
flask is used to take 0.l(N) Na2C03 solution from the volumetric flask to the conical flask. Take
lOml of O. l(N) Na2C03 solution in a conical flask, add IOml of distilled water and 2-3 drops of
Apparatus: '-i
a) Conical flask e) Volumetric flask
b) Pipette f) Beaker
c) Burette g) Measuring cylinder
d) Funnel
Chemicals:
1-ar-b. l(N) H2S04 (secondary standard solution)
L-bt--0.1 (N) HCl (secondary standard solution)
/? c) O. I(N) CH3COOH (secondary standard solution)
Ld) O. l(N) Na2C03 (primary standard solution)
e) Indicator - methyl orange (for H2S04, HCl)
-Phenolphthalein (for CH3COOH)
S2 =Strength ofbase solution
V2 =Volume of base solution
S1 = Strength of acid solution
V1 =Volume of acid solutionWhere,
Principle:
This can be considered as the neutralization reaction between a strong base {e.g. 0.1 (N)
Na2C03} and strong acid {e.g. 0.1(N) HCl, 0.1(N) H2S04 etc.} of course CH3COOH is a weak
and organic acid. The following reactions take place between them.
H+. + OH- ~ H20
To determine the strength of the acid the following equation is used:
V1S1 = V2S2
Experiment -1/:
Determination of ltrength of O. l(N) H2S04, O. l(N) HCl, O. l(N) Acetic acid (CH3COOH)
solution by standard Od(N) sodium carbonate (Na2C03) solution.
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(N)
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V1 =Volume of acid =
S1 = Strength ofacid =
V2 =Volume ofbase =
S2 = Strength of base =
Where,
Solution no. Volume, of base Initial burette reading F.B.R Difference Mean
2 io 0 0·'2. ~'L.
2 10 ~ ·'L .~· b 9·lt ~·~
2 to ti· (:, 2-1· b 9'·~'
Where, V1 =Volume ofacid = <6· b ml"'?
S1 = Strength of acid = 1. (N)
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V2 =Volume ofbase = Q mlr
'--- S2 =Strength ofbase = C) r 1 (N)
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We know, V1S1 = V2S2
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b· 2..i (N)r 1--- = =
' v: <6· l 4I
-,
Thus, strength ofacid (H2S04) = O•Z~ (N)·/ t-
~· '7I ~<~
Table for O.l(N) HCI · e
Solution no. Volume of base Initial burette reading F.B.R Difference Mean
I IO 0 <6·1 ~·t
1 10 g.t_ b·~ cg. '2. g. l(,
I 10 (,.':) 2-~·a g.'l_ --
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methyl orange. The colour becomes yellow. Add 1 Oml distilled water and 3-4....~~
phenolphthalein for CH]COOH. This colour becom_!s red.
·- iii. Wipe out the bottom of the conical flask and place a piece ofwhite paper below it. Take
initial burette reading. Now add drops of acid solution from the burette to the conical flask along
with gentle shaking. A drop of acid solution will make it orange colour but this colour will fade
away rapidly. When the end point reaches, then the colour become orange. Taken the final
burette reading and fill the table for each acid.
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i. Taken necessary step to prevent from the stop cork of burette.
ii. Acid solution from burette should be added slowly.
iii. Not more than 3-4 drops of methyl orange should be used.
iv. Difference of acid solution between two titrations should not vary more than 0.2ml. If it
is high, repeat the titration.
v. In case of titration of CH3COOH used phenolphthalein indicator.
--
Precaution:
(N)
(N)
(N)
e t'2. ~
O· 07
0·';>1-
../Result:
a) Strength ofH2S04 {O. l(N)} =
b) Strength ofHCI {O. l(N)} =
c) Strength ofCH3COOH {O.l(N)} =
I -1
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Where, V1 = Volume of acid = 7.-, ml
S1 =Strength of acid = 41 (N)
V2 =Volume of base = t0 ml
S2 =Strength of base = o.t (N)
0 S - V2S2 toi--0·1
0·~1-r, 1--v,- = = (N)
-1-·?I
Thus, strength of acid (CH3COOH) = 0·1~7- (N)/L
3 10 C) ::;. i--,:.
3 10 1- -.
'lt·~ 1--·? i·~
3 10 ~·~ 2·<'h 1·~
L
Thus, strength of ad~ (HCl) = a· l 0 7-- (N)
tTableforO.t(N)C~,coo?r--
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= o- l01- (N)
to~ ·'.
=
9. ':l
We know,
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ii. Placement of O.l(N) standard oxalic acid solution in conical flask by pipette: A clear
and dried conical flask is used to take O. l(N) oxalic acid solution from the volumetric flask. Add
- Procedure:
i. Filling the burette: A clear and dried burette is filled with 0.1 (N) basic solution and
lower meniscus of the solution is adjusted to zero.
e) Volumetric flask
f) Stand clamp
g) Measuring cylinder
Apparatus:"
a) Conical flask
b) Pipette
c) Burette ·
d) Funnel
e) Indicator - Phenolphthalein
O. l(N) KOH (secondary standard solution)
O, l(N) NaOH (secondary standard solution)
O. l(N) NH40H (secondary standard solution)
O. l(N) 'oxalic acid (primary standard solution)
Chemicals:
a)
b)
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S2 = Strength of acid solution
V2 =Volume of acid solution
S1 =Strength of base solution
V1 =Volume of base solutionWhere,
Ir + OH-~ H20
acid base water
The calculation is performed by using the following equation:
V1S1 = V2S2
Principle:
Standardization of strong base by oxalic acid can be considered as neutralization reaction
between strong base and weak acid. Phenolphthalein should be used as indicator. The following
reaction takes place.
Experiment - 3: '
Determination of strength of O. l(N) Potassium hydroxide (KOH), O. l(N) Ammonium
hydroxide (NH40H), 0.1 {N) Sodium hydroxide (NaOH) by standard oxalic acid.
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O·~~ ml
1 (N)
V1 =Volume of base =
S1 =Strength of base =
Where,
Solution no. Volume of acid Initial burette reading F.B.R Difference Mean
2 ·- 10 C) lO•i IO· I
2 10 1.0·t 2· 1i lO·~~
2 10 2-t I 1 1>·0 0·0)
l_
Where, V1 =Volume ofbase = g.1 ml
S1 =Strength ofbase = '2. (N)
V2 = Volume of acid = LO ml~
S2 = Strength of acid = Q., (N)
o s - v2s2 l0-1- ().I
a.,i. ~r 1--.-. = = (N), v: <l.. II
Thus, strength of base (KOH) = 0-~:9 (N)
Table for O.l(N) NaOH
I IO 0 <3·1- ~·
1 10 t·1. l~· '?> ~-2- s.t
1 10 ((;·~ l~·~ g.o
Table for O.l(N) KOH
iii. Titration: .
a) The bottom ofthe conical flask is wiped and a piece of white paper is placed below it.
b) Take initial burette reading.
c) Add drop by drop NaOH solution from the burette to the conical flask along with gentle
shaking. Each drop produces a pink colour which disappears rapidly.
d) At the end point the colour of the solution becomes pink which remain for long time.
e) Take the final burette reading and fill the table, from which we can calculate the strength
of solution.
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lOml of distilled wafer and 3-4 drops of phenolphthalein solution to it. The solution remains
colourless.L
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i. The necessary step should be taken to prevent the breaking of stop cork of the burette.
ii. Acid solution from burette should be added slowly.
iii. Not more than 3-4 drops of phenolphthalein should be used.
iv. Difference of acid solution between two titrations should not vary more than 0.2ml. If it
'L..
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0 S - V2S2
01-0.'
o. _'2.'5' (N)r, i-V = g =
1
Thus, strength of base (NH40H) = O·'_'L~ (N)
.>
Result:
.>a) Strength ofH2S04 {O.l(N)} = 0·'l'3 (N)
b) Strength ofHCI {O. I(N)} = C·OOJ1. (N)
~
c) Strength of CH3COOH {0.1 (N)} = CJ ·l'L~ (N)
Precaution:
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Where, V1 =Volume of base = s ml
Si= Strength of base = 'l (N)~
V2 =Volume of acid = 0 ml
S2 = Strength of acid = ().I (N)
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3 10 0 ~-9 1-· °)
3 10 ~·~ b <8· I g
3 10 (, ~~ s
'r '--
Table for O.l(N) NH40H
V2 = Voh.ime of acid = 0 ml
t.., S2 = Strength of acid = 0.1 (N)
-
We know, ViS1 = V2S2
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Oi' S _ V2S2 01-0·1
r 1--- = = 0- o~ 9-. (N)
' v. le)·~'?I
Thus, strength of base (NaOH) = 0 · OC{)f (N)
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1 10 C) ~'2. t~
1 10 '2. 2-15 i 12.·~'?
1 10 'L*'! ~ 1--- ..2_
Table for O.l(N) NaOH
•
i. Determination of standard 0.1 (N) oxalic acid solution.
ii. Determination of strength of 0.1 (N) NaOH solution by oxalic acid solution.
iii. Calculation ofstrength of 0.1 (N) NaOH by titration.
iv. Standardization of 0.1(N) HCl with standardized NaOH solution.
v. Calculation ofstrength ofO. l(N) HCI by titration.
e) Volumetric flask
f) Stand clamp
g) Measuring cylinder
h) Beaker
i. O. l(N) Potassium hydroxide (KOH)
ii. O. l(N) Hydrochloride acid (HCl)
iii. O. l(N) Oxalic acid solution
iv. Indicator - methyl orange and phenolphthalein
Apparatus:
a) Conical flask
b) Pipette
c) Burette
d) Funnel
Procedure:
H+ +OH.~ H20
As both acid and alkali are secondary standard substance, so anyone of them must be
standardized by primary standard solution.
Calculation is done by the following equation:
VS= V1S1
Where, S = Strength ofacid
V = Volume of acid
S1 =Strength of base
V1 =Volume of base
Chemicals:
Principle:
During the standardization of a strong base with a strong acid neutralization reaction takes
place by this way:
Experiment - 4:
The standardization ofstrong base {secondary standard solution O. l(N) NaOH} with strong
acid {secondary standard sdlution O. l(N) HCl}
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Precaution:
i, Take necessary step to prevent breaking the stop cork of the burette.
ii. Acid solution from the burette is slowly added.
iii. Not more than 3-4 drops of phenolphthalein should be used.
iv. Difference of basic solution between two titrations should not vary more than 0.2ml. If it
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(N)
vs ~™-0-1
Or, S1 = ~
2
= • ~ . {, - .>"l 0 ~
Thus, strength of acid (HCI) =1C (N) /
Result. /
i. Strength ofNaOH {O.l(N)} = 0 · 0<6 (N) "' /
ii. Strength ofHCl {O. l(N)} = 6. tO~ (N) / _
' '
~' '---
r 1
L
We know,L
Where, V1 =Volume of base = 0 ml
S1 = Strength of base =
o.. ' (N)
V2 =Volume of acid = 9. (, ml
S2 = Strength of acid = 'l (N)
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3 IO C) ~·~ 9·~
3 IO ~·Lt & . 0) ~-? ~.{:,
3 IO ~· 0) 1-~·1-- 9.g
TableforO.l(N) HCI
Where, V1 = Volume ofacid = 0 ml;
'-- S1 = Streng!h ofacid = 0· (N)
V2 = Volurrie of base = 12..'?1 ml
S2 =Strengih of base = ·'l (N)
~-
o s - v;s1
01'0·
r, 2-V =
'2-·~'.
= . (N)
2
Thus, strength of base (NaOH) = 0. 0<6 ( (N)
T.
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Now let the amount of solid MH taken for the experiment is n gm of molecular weight M
Page 16 of 23
:. Total heat absorbed= (M1S1 - M2S2) (T2 - T1) cal= Q cal (say)
Where, M1 =mass of the solution
M2 = mass ofthe calorimeter + stirrer + thermometer
S1 = specific heat of solution
S2 = specific heat of glass
T1 = Initial temperature
T2 = Final temperature
Theory:
Since the heat which is evolved or absorbed on dissolving a substance depends on amount of
water or other solvent employed, the statement of the heat of solution has· a definite meaning.
Only when the concentration of the solution formed is given. If the dilution is so great that
further dilution is unaccompanied by any heat effect, then the heat measured per mole ofsolute
is known as the heat of solution at infinite dilution, usually, however, it will not be possible to
determine this heat of solution directly and one therefore must state the number of molesof
water in which one mole of solute is dissolved. Further a clear definition must be made between
the quantity known as the integral heat of solution (which is the heat obtained when I mole is
dissolved in X moles of solvent) and the different heat of solution. The latter is the heat change
per mole of the solute when an infinitesimal amount of solute is dissolved in a large amount of
solution of stated concentration.
The dissolution of most of the solid salt (MA) in water is an endothermic process. This
means that a definite amount of heat energy is necessary for the decomposition of MA into M
1
and A- ions and hence to go into solution. The heat absorbed in the reaction must be equal to the
heat loss due to the decrees of temperature of the solution in the calorimeter and other glass and
with the calorimeter through (T2 - T1)°C. this is exactly equal to the sum of the masses of the
parts multiplied by their specification.
Principle:
When solids are dissolved into H20 or any other solvent heat is either absorbed or evolved.
Heat of the solution is the heat content change of the solvent when one mole ofa solute is
dissolved in it. It depends on the concentration of the resulting solution. The integral heat of a
solution is the total heat content change for dissolving one mole completely that is when the
solution is complete.
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Experiment- 5:
~ Determination of integral heat of solution of Potassium chloride (KCI), Potassium nitrate
(KN03), Ammonium chloride (NH4CI) and Sodium chloride (NaCl).
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/ Page 17 of 23
Mass of calorimeter + Stirrer + Thermometer (M2) =
Molecular weight of KCI (M) = .f.. Lt 1 ~
i_(), ~ gm
Calculation:
ForKCI
Mass of KCl (W) =
Procedure:
i. 200ml of distilled water is taken into a pre-weighed calorimeter with stirrer and
thermometer.
ii. The temperature of the contents of the calorimeter was noted and recorded.
iii. 1 Ogm of supplied solid sample was weighed using a rough balance.
iv. Supplied sample added quickly to the calorimeter.
v. The temperature recorded after 30 seconds interval at the time the solution was stirred
quickly. The temperature time recording continued for 15 minutes.
vi. The final weight of calorimeter with its whole content was taken. The differences
between the weight of calorimeter and final weight give the weight of the solution.
vii. The temperature time curve was drawn and corrected final temperature was taken from
the graph in case of each supplied sample.
iii. Solid NH4CJ
iv. Solid N&CI
L
L
.f
L Chemicals:
i. Solid KCI
ii. Solid KNQ3
iv. Rough balance
v. Measuring flask
vi, Stop watch
Apparatus:
i. Calorimeter
ii. Stirrer
iii. Thermometer
w:. Number of moles = -
M
For the dissolution of moles of the salt MA, the amount of heat absorbed Q cal (say).
w:. For one mole of MA, the heat absorbed H = Q x -
M
H = -1-{M (M1S1 +M2S2)(1~ -1;)}Kcal
1000 w -
i
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Md.
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Manik

·- -~)
Page 18 of 23
Specific heat of solution (S1) = 1
Specific heat of glass (S2)= 0.16
Mass of solution (M1) =f~3-~~)m {~g +. ~ ~1oi r
gmMass of solution + Calorimeter + Stirrer+ Thermometer (M3) =
Mass of calorimeter + Stirrer + Thermometer (M2) = ~ ~ ? gm
Molecular weight of NH4CI (M) = '5'3 ·'i"
ForNH.,CI
Mass of NH4Cl (W) = ! 0 gm
The total heat absorbed:
H= _l_ {M(M1 S1 +M~S2 )(T2 -1;)}Kcal
1000 fV_ -
~ 1 ·) ~f-t·'i (2.0lS·(.)(l +(i73f..-O·I~) (Z·1- - 21J1K...u-'
1000 o·~ ~
-z - L_. C. 2.. ~(.o..l
Time (min) 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 LU 14 14.5 15
Temp(°C)
1Yt 21·1- l·1- 2.·f- 2l ·7- 2·1 2Y1- 2·l- 2Yl- 21·1 2·'.1- 219- 2·/ 21·+ 21·J
Time (min 'C 0.5 I 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5
Tcmp(0C)
~ 22-"i 21·} 2FI- 2.1·1- Z1- .2..1·°7- 2Y7- 21-7 2.F1- 21·7- 2J·1- '2 .-,.. 2. ·7- 2Ft 2-·~
Final temperature (T2) = '2.~•1- °C
Initial temperature (T1) = · ?.~ °C
Specific heat of solution (S1) =
Specific heat of glass (S2).= 0.16
Mass of solution (M1) = 2.0&.<, gm
L
gmMass of solution + Calorimeter + Stirrer+ Thermometer (M3) = C ~ ' 'L
L
Md.
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Manik

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Page 19 of 23
Time (min) 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15
Temp (°C)
lS 2..·.) 1·) 2.:) 2.·5" lS l6° u.s 21.r 2:5 2·S 2·J 2.1 ·J l·5' ll·!>
Time (min e 0.5 l 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5" 7 7.5
Temp (°C)
~ 1-1- 2·) 2tS 2..-S- 2,s 2.'iS' 2·~ 1·S- 2~ 21·5' 2u 'l·) 2J·j 2·) -z'..r
~1I
I
lI 1
gm
ForKN03
Mass ofKNOJ (W) = i~ gm
Molecular weight ofKNOJ (M) = 1.0t ~ . ·
Mass of calorimeter + Stirrer + Thermometer (Ma) = G (p· ' gm
Mass gfsehrtron + Calorimeter + Stirrer + Thern<ometer (~ 'i41· '2-.
Mass of solution (M1) = 2.0~· gm
Specific heat of solution (S1) = 1
Specific heat of glass (S2) = 0.16
Initial temperature (T1) = .2.~
0
°C
Final temperature (T2) = 2-1-) °C
I
L
H = -1-{M (M1S1 +M,S,)(T2 - t, )}Kcal
1000 w ..
.L
Time (min) 8 8.5 9· 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15
Temp (°C)
9·f ~·S" ~-S ,,.5 ¥!)·5 9·5 '1>5 I"). S" i<}f ~.s ~'5" ~1:) 1'3·f 0-5 1':5
Time (min 6 0.5 1 ·- t.s 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5
Temp (°C) .-i
?J.. 20·5 lO·S 20 '3·5 !l)tS' tgS ¥?>-.f '!>·) '9-5" gs i~s·19-5' l~J t~·S"":.;,
Final temperature (T2) = 1 ~ ·) °C
Initial temperature (T1) = 2 ~ °CI~
1
!
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Manik

·.~
Page 20 of 23
Result:
Heat ofKCl solution absorbed per mole=
Heat ofNH4Cl solution absorbed per mole=
Heat of KN03 solution absorbed per mole=
Heat of NaCl solution absorbed per mole =
Kcal
Kcal
Kcal
Kcal
- l..· (,L_
-S- - 2- b
-7· I~
~ ,.(:,)
H= -1-{M (M1S1 +M2S~)(1~ -li)}Kcal
1000 JV -
t i~ 5~: (Li 5·~ 7 I-t /.ii l.·'!l >< O :I 4) ( 31 - ->tf U.-
-z - 1· ~ S' ~~G-'
L·i
Time (min e o.s 1 1.5 2 2.5 3 3.5 4 4,5 5 5.5 6 6,5 7 7.5
Temp (°C)
31. ~IS ; ~' .,,' ~ ~ ~ ~I ~" ~' '3' ~I 3 1 ~1
Time (min) 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 l3 13.5 14 14.5 15
Temp (°C) ,, ')' ~' ~' ~' ~) ">l '3 I ~' '3' ~' "31 ~' ~1 "S
gm
For NaCl
Mass of NaCl (W) = i 0 gm
Molecular weight of NaCl (M) = 5<6·)
Mass of calorimeter + Stirrer + Thermometer (M2) = 4.t b· ~ gm
Mass of solution+ Calorimeter+ Stirrer+ Thermometer (M3) = C~t · S-
Mass of solution (M1) = 21r· C. gm
Specific heat of solution (S1) = I
Specific heat of glass (S2) = 0.16
Initial temperature (T1) = ~ 'l °C
Final temperature (T2) = 31 °C
~[!· I
'
Ll
:1f .
H= lO~O {: (M,S, +M2S,)(7;-1;)}Kcal
e: _i_ ~ to1 ~(1,t>~., "I,..- 4,~t·t~ ().
.1000 lo
z._ - .;. • t b l(tc.
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Page-21 of 23
Procedure:
i. Preparation of solution of 0.1 (N) oxalic acid, H2S04, HN03, HCl and NaOH.
ii. Find out the exact concentration of supplied acid and base and suppose that they are (a)
and (b) respectively.
iv. NaOH or KOH
v. Methyl orange
vi. Distilled water
Reagent:
i. Oxalic acid
ii. HCl, H2S04, HN03
iii. Phenolphthalein
f) Volumetric flask
g) Stop watch
h) Beaker 250ml.
Apparatus:
a) Conical flask
b) Pipette
c) Burette
d) Calorimeter, thermometer and stirrer
e) Measuring cylinder 1OOml and 1 Oml
Principle:
Heat of neutralization is defined as heat evolved when one gram equivalent of an acid is
completely neutralized by a base.
In dilute solution strong acid and strong base and the salt produced remain completely
dissolved. So the neutralization reaction may be represented as:
H+ + A" + B+ + Off = H20 + B+ + A
Therefore, the neutralization between strong acid and strong base may also be called as
formation of one mole ofwater. So, the heat change accompanying this reaction is constant.
In order to obtain the heat of neutralization according to the quantitative definition:
Let, W 1 = mass of mixture
W2 =mass of calorimeter with stirrer and thermometer
S1 =specific heat of mixture
S2 = specific heat of calorimeter
T1 = initial temperature
T2 = final temperature
lfVml ofHCl with strength Sis neutralized by NaOH, the heat of neutralization is as:
£H= v~s {cw1s1 +w2s2)(T2-1;)}ca1
Experiment - 6:
Determination of heat of neutralization between a strong base and strong acid.
~
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Md.
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of 23
t 0 ml
O·()i (N)
9,4 ml
1 (N)
'
V1 =Volume ofNaOH =
S1 =Strength NaOH =
V2 =Volume ofHCI =
52 =Strength ofHCl =
Where,
0
Number of Volume of Strength of I.B.R F.B.R Difference Mean
experiment Na OH NaOH
1 10 c~oi 0 0·1t 9·~
2 10 o o i 9,Lt &·~ 0. S"' 9· f:,
3 10 C)· Qi I~~ 2J~; 1-- 0·&
Number of Volume of Strength of LB.R F.B.R Difference Mean
experiment oxalic acid oxalic acid
I 10 0·{ 0 I 'L l i._
-··
2 10 6·1 .z. 2) 1') I~.')'
3 10 6·1 'l '5 )1- 'L
Table for oxalic acid and NaOH
iii. Take 100/a ml of the supplied acid and 100/b ml of supplied base in two separate flask.
iv. Pour one of these solutions into the previously cleaned and weighed calorimeter and note
down the temperature at a regular interval of time.
v. Plot the graph temperature versus time in order to find out the true final temperature after
the correction for heat loss bf reaction.
vi. Find out the correct final temperature and calculate heat of neutralization according to the
equation.L
t
·L
Where, V1 =Volume of oxalic acid = 10 ml- S1 =Strength of oxalic acid = (.). (N)
V2 =Volume ofNaOH = 1-:>~ ml
S2 = Strength of NaOH = 'l (N)
0 S - V.S1
O"'- O· .
o .()rg_.r 2--- =
'2 ·~'> = (N)
' v2
Thus, strength of base (NaOH) = () -~~1 (N)
-:Table for O.l(N) HCI
Md.
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of 23
Result: Heat of neutralization ofHCl = ~- 2 6 Cal
. :. L.H = - I~· '& f, Cal
I
i
Thus, heat ofneutralization: L.H = -.-1- {CWiSI + w2s2 )(T2 - Ti)}Cal
f· xS
Or L.H= 1 ) (2..._i:i<. +~o?• 2~D·9 (~Q·S"-J.o)
• 1·t<?>·I> Lr)( a.o'IS~ 1 L
Time (min) 8 8.5 9 9.5 10 10.5 l l 11.5 12 12.5 13 13.5 14 14.5 15
Temp(°C) >o·) 'JO.S 3l>T -so.r YiS 30.J 30.S "},l•S" ~us ~OS '1~.S- 30.S '30.5" 3) S'"' 10-')
Time(min 0 0.5 l l.5 2 2.5 " 3~ 3.5., 4 4.5 5 5.5 6 6.5 7 7.5
Tempec>
- ,
~ '}O·J' ~()·S- '30·l ~o·) 10.s ~-d·J ~u<v 3()·!$" ';l.O·s ~-S- 3V·f' ':O·S ltl·~ ~Cl-S- <3.<l·.J
t-
gm
Now, Mass of calorimeter + Stirrer+ Thermometer (W'!) =
Mass of solution+ Calorimeter+ Stirrer+ Thermometer (W3) =
~~on(Wi)= 2J·i gm
Specific heat of solution (S1) ". 1
Specific heat of glass (S2) = _ · 0.16
Initial temperature (Ty = ':>() °C
Final temperature (T2) = )Cl -:S-°C -· ,.r- .,,,,,.,,..
Let, volume'V' of exact Q.l(N) HCl solution was,,____ '--v-->
Vx 0-0B~ = lQlJj._ C>-1
Or, V=t~-O~ml
,(1() )()O·) -:-·
0- ~~·· (
V1S1 =VS -..
V =Volume ofi':JaOH = r l.q&- ~t ml
S =Strength NaOH = a.oil (N)
V2 =Volume ofHCl = lQ.2.-- ml
S2 =Strength ofHCl = b·i (N)
Then,
Where,
(N)
= (N)
V1S1 = V2S2
0 . s - i'iS1
r1 2-V
. 2
Thus, strength ofacid (HCI) =
We know,
-
·'
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Md.
Imran
Nur
Manik

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Physical Pharmacy-I lab (With values) MANIK

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