Preparation of
primary standard solution of sodium thiosulfate (Na2S2O3),
Oxalic acid [ ( COOH )2.2H2O ], Potassium dichromate (K2Cr2O7).
Preparation of secondary standard solution of Sulfuric acid (H2SO4),
Hydrochloric acid (HCl), Acetic acid (CH3COOH).
PRINCIPLE :
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 .
(A) PRIMARY STANDARD SUBSTANCE : Primary standard substance are substance
from which standard solution can be
prepared directly by taking weight at a electrical balance and then dissolving
that substance in distilled water or any other
suitable solvent . Examples are-
| |
COOH COONa
CRITERIA OF PRIMARY STANDARD SUBSTANCE :
(I) They are chemically
pure substance .
(II) The molecular weight of the substance
corresponds to the chemical
formula .
(III) The substance are stable
both in solid or liquid state .
(IV) The gram equivalent weight is large .
(B) SECONDARY STANDARD SUBSTANCE : Secondary standard substance are the substance
from which standard solution cannot be prepared by directly weighting and then
dissolving in suitable solvent.
Examples are :- NaOH, KOH, H2SO4,
HCl, CH3COOH, KMnO4 etc .
In this case , the appropriate weight of
the substance under test is taken and then dissolved in distilled water and
dilute solution is prepared. This solution is then standardized with primary
standard solution and then its strength is determined by using the following equation
--
V1S1 =
V2S2
S1
= Strength of Primary standard solution
V1
= Volume of Primary standard solution
S2
= Strength of Secondary standard solution (?)
V2 = Volume of Secondary standard
solution
CONCENTRATION TERM : Normal solution
REAGENT :-
Primary standard substance :
I) Sodium
carbonate (Na2CO3) = 106 ÷ 2 = 53
II) Oxalic
acid (HOOC-COOH).2H2O = 126 ÷ 2 = 63
III) Sodium
oxalate(NaOOC-COONa).2H2O=170÷ 2=85
IV) Potassium dichromate (K2Cr2O7)
=149 ÷ 6= 24
V) Distilled
water (H2O)
Secondary standard substance :
I) Sulfuric
acid (H2SO4)
II) Hydrochloric acid (HCl)
III) Nitric
acid (HNO3)
IV) Acetic
acid (CH3COOH)
V) Sodium
hydroxide (NaOH)
VI) Distilled water (H2O)
APPARATUS :
i) Balance
ii) Conical flask
iii) Measuring cylinder
iv) Funnel
v) Volumetric flask.
PROCEDURE :
A) Primary standard substance :
(i) Preparation
of standard 0.1(N) Na2CO3 solution : The gram equivalent weight of Na2CO3 is 53.50. 53 ÷ 100= 0.53 gm. of Na2CO3
is needed to prepare 0.1(N) 100ml solution of Na2CO3 .
Now 0.53gm. of Na2CO3 was weighted in the electrical balance and
dissolved in 50ml of distilled water in 100ml volumetric flask and shake
gently. Then make the volume up to the mark with distilled water.
(ii) Preparation of
standard 0.1(N) Oxalic acid solution: The gram equivalent weight of oxalic acid is 126 ÷ 2= 63 .
So, to prepare 100ml of 0.1(N) solution of oxalic acid we need (63 ÷ 100)= 0.63gm of oxalic
acid .
Now weight out 0.63gm of
oxalic acid in electrical balance and dissolved in 50ml of distilled water in a 100ml volumetric flask
and shake gently. Then make the volume up to the mark with distilled water.
(iii) Preparation of
standard 0.1(N) Sodium Oxalate
solution: The gram equivalent weight of
this substance is 170 ÷ 2= 85.
So , to prepare 100ml of 0.1 (N) solution of sodium Oxalate, we need
(85 ÷ 100)=
0.85gm of sodium Oxalate.
Now weight out 0.85gm of sodium
Oxalate in electrical balance and dissolved it in 50ml of distilled water in a 100ml volumetric flask and shake
gently. Then make the volume up to the mark with distilled water.
(iv) Preparation of standard
0.1(N) Potassium dichromate (K2Cr2O7)
solution : The gram equivalent weight of Potassium dichromate is 149 ÷ 6= 49 .
So , to prepare 100ml of 0.1 (N) Potassium dichromate solution we need
(49 ÷ 100)=
0.49gm of K2Cr2O7
.
Now weight out 0.49gm of K2Cr2O7 in electrical balance and dissolved it
in 50ml of distilled water and shake
gently. Then make the volume up to the mark with distilled water.
(v) Preparation of standard
0.1(N) Sodium thiosulfate
solution: The gram equivalent weight of Na2S2O3 is 158 .
So, to prepare 100ml of 0.1 (N) solution of sodium Oxalate, we need
(158 ÷ 100)=
1.58gm of Na2S2O3.
Now weight out 1.58gm of Na2S2O3
in electrical balance and dissolved it in 50ml of distilled water in a 100ml volumetric flask and shake
gently. Then make the volume up to the mark with distilled water.
(B) SECONDARY STANDARD SUBSTANCE :
(i) Preparation of standard
0.1(N) Sulfuric acid (H2SO4) solution: The gram equivalent weight
of H2SO4 is (98 ÷ 2)= 49 .So , to prepare 100ml of 0.1 (N) H2SO4 solution
49 ÷ 100=0.49gm of 100% H2SO4 acid is
needed. Since H2SO4 is not 100% pure,it is 98% pure. So to prepare 100ml of 0.1(N) H2SO4
solution [ (100 x 49) ÷ 98 ] = 0.5gm H2SO4
acid is needed. But H2SO4 is liquid and Secondary
standard substance. So its volumetric measurement may be obtained by the
following equation.
mass = volume x density.
Thus density of H2SO4
is 1.89 so to prepare 100ml 0.1(N) H2SO4 the needed
volume of 98% H2SO4
is 0.5 ÷ 1.89=0.282ml=0.3ml.
Now take 100ml volumetric flask, which
contains 50ml of distilled water and 0.282ml of H2SO4 is
added slowly. Then make the volume up to the mark with distilled water.
(ii) Preparation of standard 0.1(N) HCl solution:-
The gram equivalent
weight of HCl is 36.5 .So, to prepare 100ml of
0.1 (N) HCl solution 36.5 ÷ 100 = 0.36gm of 100% HCl acid is needed .Since HCl
is not 100 pure,it is 32% pure.So
to prepare 100ml of 0.1(N) HCl solution of HCl [(100 x 0.365) ÷ 32] = 1.4gm HC acid is needed. But HCl is liquid and
Secondary standard substance. So volumetric measurement is needed which may be
obtained by the following equation.
mass = volume x density.
Thus density of HCl is 1.18 so to prepare 100ml 0.1(N) solution of HCl
the needed volume of 32% HCl is (1.140 ÷ 1.18)= 0.966ml= 1ml .
Now take 100ml volumetric flask,
which contains 50ml of distilled water and 1ml of HCl is added slowly. Then
make the volume up to the mark with distilled water.
(iii) Preparation of standard
0.1(N) Acetic acid (CH3COOH) solution: The gram equivalent
weight of CH3COOH is 60. So , to prepare 100ml of 0.1 (N) Acetic acid solution we need (60 ÷ 100)= 0.6gm of CH3COOH
which is 100% pure . But
it is liquid and Secondary standard substance.
So volumetric measurement is needed which may be obtained by the following equation.
mass = volume x density.
Thus density of CH3COOH is 1.055 .So to prepare 100ml 0.1(N) CH3COOH
is (0.6 ÷ 1.055)=
0.56ml= 0.6ml .
Now
take 100ml volumetric flask, which contains 50ml of distilled water and
0.6ml of CH3COOH is added slowly. Then make the volume up to the
mark with distilled water.
(iv) Preparation of standard
0.1(N) Sodium hydroxide (NaOH) solution : The gram equivalent
weight of NaOH is 40 . So , to prepare 100ml of 0.1 (N) NaOH solution we need (40 ÷ 100)= 0.4gm of NaOH .
Now take 50ml of distilled water in a conical
flask (volume is 100ml) and Shake gently until the NaOH granules dissolved
completely. Then make the volume up to the mark with distilled water.
(v) Preparation of standard
0.1(N) Potassium hydroxide (KOH) solution : The gram equivalent weight
of KOH is 56 . So , to prepare
100ml of 0.1 (N) KOH solution we need
(56 ÷ 100)= 0.56gm
of KOH .
Now take 50ml of distilled water in a conical
flask (volume is 100ml) and Shake gently until the KOH granules dissolved
completely. Then Pour this solution to 100ml volumetric flask and rinse the
conical flask into the volumetric flask and make the volume up to the mark with
distilled water.
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 .
Potassium
Reviewed by M H Islam
on
12:32 PM
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