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Showing posts with label standard solution. Show all posts
Showing posts with label standard solution. Show all posts

Sunday, August 9, 2020

Primary and Secondary Standards Definition, Properties and Uses

                                                          

 Standard Solution

Solutions of accurately known strength are called standard solutions. A standard solution contains a known weight of reagent in a definite volume of solution.

 A standard solution is prepared by dissolving an accurately weighed quantity of a highly pure substance in a definite volume of solvent. 

Types of Standard Solution

1. Primary Standard 
2. Secondary Standard

Primary Standard

A primary standard is a chemical or reagent which has certain properties such as

1. It is extremely pure
2. Highly stable
3. It is anhydrous
4. It is less hygroscopic
5. Has very high molecular weight
6. Can be weighed easily
7. Should be ready to use and available
8. Should be preferably non toxic
9. Should not be expensive

1. It is Extremely Pure

Primary standard material should be extremely pure which means that, it should be a chemical of high grade of purity, preferably 99.98%.
 

2. Highly Stable

It should be highly stable which means it usually does not react easily when kept in its pure form or in other words it should have very low reactivity. 
Importance:
Because if a reagent reacts easily with atmospheric oxygen or water or changes its property over time then it is unreliable chemical to use as primary standard. 

3. It is Anhydrous

It should be anhydrous which means that it does not contain any water molecule in its molecular structure. 
e.g. magnesium sulphate (MgSO4), is found with formula MgSO4.7H2O. Therefore, to prepare  primary standard solution of MgSO4, anhydrous MgSO4 preferably of analytical reagent grade with purity greater than 99.98% will be required. 

4. It is Less Hygroscopic


The chemical preferably should be less hygroscopic that is on opening the container it should not absorb water molecules from atmosphere. 

5. Has Very High Molecular Weight


It should have very high molecular weight to minimize weighing errors.

Examples of Primary Standard

These are materials which, after drying under the specified conditions, are recommended for use as primary standards in the standardization of volumetric solutions.

Acid - Base Titrations: Potassium Hydrogen Phthalate (C8H5KO4), Anhydrous Sodium Carbonate (Na2CO3)
Redox titrations: Arsenic Trioxide (As2O3), Sodium Oxalate (Na2C2O4), Potassium bromate (KBrO3)
Precipitation titrations: Sodium Chloride (NaCl)

Uses of Primary standard

1. Used to standardize a volumetric solutions.

2. They are used for standardization of titration solutions. 

3. They are used for calibration of secondary standards.

4. Used as reference to determine unknown concentrations or to calibrate analytical instruments.

Secondary Standard

A secondary standard is a solution which contain exactly known amount of the substance in unit volume of the solution which is determined by titrating against a primary standard.   
OR
A secondary standard solution is a solution in which the concentration of dissolved solute has been determined by reaction (titration) with a primary standard solution. 

Secondary Standard Properties

A secondary standard is a chemical or reagent which has certain properties such as: 
1. It has less purity than primary standard 
2. Less stable and more reactive than primary standard but its solution remains stable for a long time 
3. Titrated against primary standard 

Uses of Secondary Standards

Secondary standards are commonly used to calibrate analytical equipment and analytical techniques. 
 Used in titration. 
                                                      
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Wednesday, June 6, 2018

Normality and Gram Equivalent Weight


Normality

Normality is defined as the number of gram equivalents of a solute contained in 1 litre of a solution. Thus a solution which contains one gram equivalent of the solute per litre of the solution is called normal solution. It is represented as 1 N. 

Gram equivalent is the weight of substance in grams numerically equivalent to its equivalent weight.

Equivalent weights of Acids and Bases:


The equivalent weight of an acid is that weight of it which contains 1.008 gm of replaceable hydrogen. 
The equivalent weight of a base is that weight of it contains one replaceable hydroxyl group. 

For example, HCl and NaOH have only replaceable H⁺ and OH-hence the equivalent weights  of them are equal to their molecular weights. 

Equivalent weight of HCl
                        Molecular weight of HCl  =  H(1 X 1) + Cl(35.5 X 1)
                                                                  = 1+35.5 
                                                                  = 36.5
            Number of replaceable ions (H⁺)  = 1
                           Gram equivalent weight = molecular weight/number of replaceable ions
                           Gram equivalent weight = 36.5 gm

Equivalent weight of NaOH

                      Molecular weight of NaOH = Na (23 X 1) + O (16 X 1) + H (1 X 1)
                                                                   = 23 + 16 + 1
                                                                   = 40 gm    
            Number of replaceable ions (OH-) = 1
                             Gram equivalent weight = molecular weight/number of replaceable ions
                            Gram equivalent weight  = 40 gm
                                       

Sulphuric acid (H₂SO₄) and Oxalic acid (HOOC-COOH) each have two replaceable hydrogen, hence their equivalent weights are equal to half of their respective molecular weights.


Equivalent weights of some common Acids:


Acidity: No. of replaceable OH- ions.

 Basicity : No. of replaceable H⁺ ions.

Acid
Molecular weight
Basicity
Equivalent weight
Hydro Chloric Acid, HCl
36.46
1
36.46
Nitric Acid, HNO₃
63.01
1
63.01
Sulphuric Acid, H₂SO₄
98.07
2
49.03
Acetic Acid, CH₃COOH
60.05
1
60.05
Boric Acid, H₃BO₃
61.83
3
20.61
Oxalic Acid (dihydrate), (COOH)₃,2H₂O
126.07
2
63.03

Equivalent weights of some common Bases:

Base
Molecular weight
Acidity
Equivalent weight
Sodium Hydroxide, NaOH
40.00
1
40.00
Potassium Hydroxide, KOH
56.11
1
56.11
Sodium Carbonate, NaCO
105.99
2
53.00
Sodium bicarbonate, NaHCO
84.01
1
84.01
Potassium Carbonate, KCO
138.21
2
69.10
Potassium bicarbonate, KHCO
100.12
1
100.12



Specific gravities, percentage and approximate Normalities of some common Acids:

Acid
Specific gravity
Percentage
Approximate normality
Concentrated sulphuric acid
1.84
Not less then 97.0 w/w HSO
36N
Concentrated hydrochloric acid
1.18
35.0 – 38.0 w/w HCL

11.5N
Concentrated nitric acid
1.42
69.0 – 71.0 w/w HNO
16N
Acetic acid (glacial)
1.05
Not less then 99.0 w/w CHO
17.5N


Tuesday, June 5, 2018

How to Prepare 0.5 N and 1 N HCl

Concentrated HCl

      Concentrated HCl has strength of approximately 31-36%,in case of 36% means 360 ml HCl is present  per 1000 ml of solution for 32 % HCl, 320 ml of HCl is present per 1000 ml of solution. 
      For preparing dilute normal solutions from concentrated HCl we need to calculate its volume according to its percent strength required to mix with water. To calculate required volume of concentrated HCl to prepare different normal solutions  follow below mentioned steps.  normally HCl found in chemistry labs is of 36% so let us take example of 36% HCl for calculation.

Step 1

Calculate weight of HCl in 360 ml of HCl by multiplying it with specific gravity of 36% HCl i.e. 1.1789

360 x 1.178=424.08 gm
 means 360 ml of HCl = 424.08 gm

Step 2

Calculate no. of moles in 424.08 gm of HCl by using formula

                                   No. of moles =given weight/molecular mass of HCl
                                                     N=424.08/36.5 =11.6 moles
as HCl has only one replaceable Hydrogen ion its gram equivalent weight and molcular weight is same and hence Normality and Molarity is same as well. Means no. of moles is equal to gram equivalents which ultimately is its Normailty.
Hence, Normality of 36% HCl = No.of moles = 11.6 N

Step 3

Prepare 1N HCl by using Normality equation- 

let us calculate volume of concentrated HCl required to prepare 100 ml 1 N HCl.


(Conc. HCl)N1 V1= N2V2 (1N HCl)
11.6NV1= 1N x 100ml
V1 =1 x 100/11.6
V1  = 8.6 ml

So to prepare 100 ml 1 N HCl from concentrated HCl  8.6 ml of concentrated Acid is diluted up to 100 ml with water.

Different normal solutions can be prepared  by using the above equation by changing the value of required volume and normality of the required solution .e.g.

To prepare 500 ml of 0.5 N HCl

(Conc. HCl)N1 V1= N2V2 (1N HCl)
11.6 N x V1= 0.5 N  x 500 ml
V1 =0.5 x 500/11.6
V1  = 21.5 ml
So to prepare 500 ml of 0.5 N HCl, 21.5 ml of concentrated Acid is diluted up to 500 ml.
It is not necessary to calculate every time no. of moles to calculate volume if u remember Normality of concentrated HCl i.e. 11.6 N, then volume can be easily calculated by direct from normality equation for different volume and normality.