Strong Acid
Determination of heat of neutralization between a strong base and strong acid.
PRINCIPLE:
Heat of neutralization 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+ + OH-
Therefore, the neutralization between strong acid and strong base
may also be called as formation of one mole of water. So, the heat change
accompanying this reaction is constant.
In order to obtain the heat of neutralization according to the
quantitative definition,
Let,
M2 = Mass
of mixture.
M1 = Mass of calorimeter (with stirrer and
thermometer)
S2 = Specific heat of mixture .
S1 = Specific
heat of calorimeter
T1 =
Initial temperature .
T2 = Final temperature .
If V ml of HCl with strength ‘S’ is neutralized by NaOH, the heat
of neutralization is as –
1
Δ H = --------- { (M1
S1 + M2 S2) } (T2 - T1 )
} Cal
V x S
APPARATUS:-
a) Calorimeter, thermometer and stirrer.
b) Beaker 250 ml
c) Conical flask
d) Measuring cylinder 100ml and 10ml
e) Volumetric flask
f) Burette, Pipette.
g) Stop watch.
REAGENT:
i) Oxalic acid
ii)HCl, H2SO4
, HNO3
iii)Phenolphthalein
iv) NaOH or
KOH
v)Methyl
orange
vi)
Distilled water.
PROCEDURE:
ii)
Find out the exact concentration of supplied acid and base and suppose that
they are (a) and (b) respectively.
100
100
iii)
Take -------- ml of the supplied acid
and --------- ml of supplied
a
b
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 of reaction.
vi)
Find out the correct final temperature and calculate heat of neutralization
according to the equation.
Table for Oxalic acid and NaOH :-
No.of expt
|
Volume of Oxalic acid
|
S1(N) of Oxalic acid
|
IBR
|
F.B.R
|
Difference
|
Mean
|
1
|
10
|
|||||
2
|
10
|
|||||
3
|
10
|
CALCULATION-
V1
S1 = V2 S2 V1 = Volume of Oxalic acid
S1
= Strength of Oxalic acid
V1 S1 V2
= Volume of NaOH
Or, S2
= ---------------- S2 = Strength of NaOH
V2
S2 = ----------------
S2 =
(N)
Table for HCl and NaOH:-
No.of expt .
|
Volume of NaOH
|
Strength of NaOH
|
IBR
|
F.B.R
|
Difference
|
Mean
|
1
|
10
|
|||||
2
|
10
|
|||||
3
|
10
|
CALCULATION-
V1
S1 = V2 S2 V1 = Volume of NaOH
S1 = Strength of NaOH
V1 S1 V2 =
Volume of HCl
Or, S2
= ---------------- S2
= Strength of HCl
V2
S2 = ----------------
S2 = (N)
Then,
Where,
V1 S1 = V S V
= Volume of HCl
= V x S= Strength of HCl
V1 = Volume
of NaOH
V = -------- S1 = Strength of NaOH
=
Now,
M1 = Mass of calorimeter + stirrer + thermometer
=
M3 = Mass of calorimeter + stirrer + mixture =
Mass of mixture M2
=
S1 = Specific heat of calorimeter =
S2 = Specific heat of mixture =
T1 = Initial temperature =
T2 = Final temperature =
Time
|
½
|
1
|
1½
|
2
|
2 ½
|
3
|
3½
|
4
|
4½
|
5
|
5½
|
6
|
6½
|
7
|
7½
|
Temp
|
|||||||||||||||
Time
|
8
|
8½
|
9
|
9½
|
10
|
10½
|
11
|
11½
|
12
|
12½
|
13
|
13½
|
14
|
14½
|
15
|
Temp
|
1
Δ H = --------- { (M1
S1 + M2 S2) } (T2 - T1 )
} Cal
V x S
=
=
=
RESULT: -
Heat of
neutralization of HCl = Kcal
Strong Acid
Reviewed by M H Islam
on
6:17 AM
Rating:
Reviewed by M H Islam
on
6:17 AM
Rating:
