Preparation of solutions
1. Saturated solution
2. Standard solution
3. Percent solution
Solvent:It is a liquid in which solute goes into solution
Solute:It is a solid chemical to be dissolved in a solution
The crystalline chemical when dissolve in a liquid the chemical is referred as the solute and the
liquid is called as solvent.
I .Saturated solution:A saturated solution is prepared by continuously dissolving this solute in a
solvent is visible in the solution.
Eg: Ammonium sulphate crystal is added in a small portion to a fixed amount of water with
constant stirring.The addition is continued till some crystals are left undissolved.The clear
solution obtained is reffered as saturated ammonium sulphate solution.
II.Percent solution:Percent solution is prepared by dissolving gram of substance in 100ml
percent solution is expressed as weight i.e., if known weight of solute is dissolved in solvent
or by Volume Volume
Volume i.e., a known volume of solvent dissolved in another volume of solvent.
Eg: 1.Preparation of 0.9% Nacl solution , in this case 0.9g Nacl is dissolved in 100ml of solution.
Method of preparation : weigh exactly 0.9g of Nacl or 900mg Nacl and mix with 80 ml of H2o
until it dissolves completely. Then transfer the contents to 100ml volumetric flask or a
measuring cylinder with stopper.The volume is made upto 100ml with distiiled water. The
cylinder is stoppered and mixed properly it is then transfer to reagent bottle and labelled.
2.Eg: A percent solution Volume
Volume
Eg: Preparation of 2% HCL Molecular weight -36.46
Note: stock HCl is provided by the company is 35-38%.
Formula-V1P1=V2P2
V1-How much volume?
V2-100
P2-2
P1-35
= 200
35
=5.71 ml of concentrated HCl
Make up to 100ml of water
Method of preparation: Approximate 50 ml of water is taken in a 100ml volumetric flask.
1) Concentrated HCl is drawn in to the pipette with help of rubber bulb exactly 5.7 ml is
transferred to the flask containing water with the constant mixing.Then water is added upto the
mark.The flask is stoppered and solution is mixed properly it is then transferred to reagent bottle
and labelled.
III Normal solution: A gram equivalent weight of a substance dissolved in 1 litre of solution is
called 1 normal solution.It is denoted as N.
Eg: If the equivalent weight of a substance is 40g this is dissolved in 400ml of water and then
make upto volume 1 litre(1Normality).Like wise 80g of a substance if it is dissolved in 1 litre is
named as 2 normality solution.If 20g of substance is added then its 0.5 Normality.4g of
substance is dissolved in 1 litre then it is 0.1N solution.
To prepare a normal solution equivalent weight of a substance is calculated as follows.
Equivalent weight of an acid: Acid is the one which has ionisable or replaceable hydrogen
ion.If an acid has 1 replaceable hydrogen ion it is called monobasic acid if 2 replaceable
hydrogen ion then dibasic if 3 replaceable hydrogen ion then tribasic.
Equivalent weight of an acid: Molecular weight
Number of replaceable hydrogen ion
Ex: Molecular weight of HCL is 36.46
replaceable hydrogen ion =1
2)Eg: equivalent weight of H2SO4 (sulphuric acid)
Molecular weight
Number of replaceable hydrogen ion
Equivalent weight = 98
2 = 49
1 molar H2SO4(sulphuric acid) = 2 normality
2) To find out the weight to be taken for a known volume of V ltrs of acid of a strength N of
equivalent weight E.
3) Weight of V ltrs of N normal= E×N×V
E-Equivalent weight
N-Normality
V-volume
4) The equivalent weight of a base is taken by molecular weight of a particular base di
vided by number of replaceable hydrogen ions.
5) Eg.1) NaoH – it has 1 replaceable hydrogen ions.So its equivalent weight is
Equivt wt is = Molecular weight
1
=40
6) Eg.2) Calcium hydroxide Ca(OH2)2 it has 2 replaceable hydrogen ion. its equivalent
weight is
74 =37 eqvt wt of Ca (OH2)2
2
The equivalent weight of salt such as silver nitrate (AgN03)& KmNo4 are determined by number
of electrons which they give or take during a reaction.
Ex : (AgN03) it gives 1 electron therefore its equivalent weight (AgN03) is
= Molecular weight
1
Approximate normality of liquids can be calculated using the following formula
Percent of purityof liquid × Specific gravity × 10
Equivalent weight
Eg: HCl: % purity of HCL is 37,specific gravity is 1.18kg,equivalent weight is 36.4
37 × 1.18 × 10
36.4
= 11.9 Normality
To get 1N HCl ,10 ml of concentrated HCl made upto 100ml deionized water
Or
12 ml of concentrated HCl + 108 ml water gives 1 Normality
N1V1=N2V2
N1 =11.9
V1=?
N2 =1N
V2=how much ml i.e., 100
N1V1=N2V2
11.9× V1×1N×100ml
100
11.9
=8.33ml
Preparation of normal solution:Exact normal solution can be prepared only when a chemical
available in its pure state. Moreover correct weighing is possible if chemical doesnot absorb or
loose water on exposure.On the other hand liquid like acid as not pure as supplied commercially
so as exact solution is possible to prepare by titration and filtration.
Sodium carbonate and oxalic acids are available in pure form.So it is easy to prepare exact
normal solution by weight.These are called primary standards.NaOH when exposed it absorbs
water and liquefies.Sodium carbonate can be used as a primary base,where as oxalic acid is a
primary acids.To standardization of other normal solution these acids or bases can be used.
Ex: preparation of primary standard solution sodium carbonate solution of 0.1N.
Formula=gram/litre= E×N×V
1000
53×0.1N×100
1000
=0.53g
0.53g sodium carbonate dissolved in 80ml water then make upto 100ml
Method of preparation: transfer a 550mg of sodium carbonate crystals into clear 100ml of
flask,use clean funnel,mix properly to dissolve the chemical.Add distill water upto the mark,
stopper the flask and mix again ,transfer to reagent bottle and label it.
Eg: preparation of 0.01N oxalic acid solution.
Molecular weight of oxalic acid is 126.067 Molecular weight
1
126.067
2
Equivalent wt=63.033
gram/litre= 63.033×0.1N×100
1000
=0.6303g
Accurate 0.6303g of oxalic crystals are weighed and prepared 100ml solution.
