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Volumetric Notes | NEB Grade 12 Notes | Physical Chemistry | Nepali Educate


Volumetric Analysics Class 12 NEB notes

Unit 1 Volumetric Analysis

Qualitative analysis: 

The study which deals with the qualitative properties of substances like color, surface tension, viscosity etc. is called Qualitative analysis. 

Quantitative analysis: 

The study which deals about the information related to mass volume, number of moles of substances is known as Quantitative analysis. 

Types of quantitative analysis: 

         i.Volumetric analysis 

       ii.Gravimetric analysis 

     iii.Instamental analysis 

Volumetric Analysis: 

The process of determination of concentration of unknown solution by finding out the titrating volume with the help of standard solution is called Volumetric Analysis. 

Types of volumetric analysis: 

        i.Acid Base Titration: 

The process of determination of strength of unknown solution of acid by titrating against the standard solution of base or strength of unknown solution of base by titrating against the standard solution of acid using a special indicator is called acid base titration. 

Example: Titration of HCl Vs N/10 Na 2CO3 Solution. 

       ii.Acidimetry: 

The process of determining the strength of an unknown solution of acid by titrating against the standard solution of base using a special indicator is called acidimetry. 

Example: Determination of concentration of unknown solution of HCl with 1/10 Na 2CO3. Alkalimetry: 

The process of determining the strength of an unknown solution of base by titrating against the standard solution of acid using a special indicator is called Alkalimetry. 

Example: Determination of concentration of unknown solution of NaOH with 1/10 Oxalic acid. 

     iii.Redox Titration: 

The titration in which reduction and oxidation occurs simultaneously is called redox titration. Example: Titration between acidified KMnO4 and Oxalic Acid. 

KMnO4 + 3H2SO4 +5(OH-O=C-C=O-OH) → K2SO4 + 2MnSO4 + 3H2O + 5[O]

 Gravimetric analysis: 

The process of determining the mass of constituents of the solution by finding out the mass of ppt obtained from reaction between the solution and precipitating agent is known as gravimetric analysis. 

Example:

Mass of NaCl dissolved in solution can be determined by titrating AgNO3(excess) solution. Gravimetrically, NaCl + AgNO3 → NaNO2 + AgCl 

Steps involved in Gravimetric analysis: 

        i.Properties of solution: 

Constituents are dissolved in water to make an unknown concentration of solution. 

       ii.Precipitation: 

Excess precipitating agent is added to the solution for obtaining maximum precipitant. 

     iii.Filtration: 

The ppt is filtered using filter paper of W. No. 41 and W. No. 42 with the help of suction pump. 

     iv.Drying: 

The ppt is dried in the desiccator and then it is burned along with the filter paper. 

       v.Weighing: 

The dried ppt is then weighed again and again till to get constant mass, then the mass  constituents are determined by using stoichiometric calculation. 

Concentration Of Solution: 

        I.Molarity: 

∙ The number of moles dissolved in 1000ml of known solution is called molarity. 

∙ It is denoted by M. 

∙ Mathematically, 

$\begin{array}{l}Molarity(M) = \frac{{No.{\rm{ }}of{\rm{ Moles}}}}{{Volume\;of{\rm{ Solution in ml}}}} \times 1000\\ \Rightarrow Molarity(M){\rm{ = }}\frac{{\frac{{Mass}}{{Molecular\;{\rm{Mass}}}}}}{{Volume\;of{\rm{ Solution in ml}}}} \times 1000\\ \Rightarrow Molarity(M){\rm{ = }}\frac{W}{{Molecular\;{\rm{Mass}} \times Volume\;of{\rm{ Solution in ml}}}} \times 1000\\ \Rightarrow Molarity(M){\rm{ = }}\frac{W}{{M \times V}} \times 1000\\\therefore Molarity(M){\rm{ = }}\frac{{gm/l}}{{Molecular\,Mass}} \times 1000\end{array}$

Types of Molar Solution: 

i. One Molarity: 

The Solution which contains 1 mole of substance dissolved in 1000 ml of solution is known as one molar solution and the concentration is known as 1 molarity. 

It is denoted by 1M. 

ii. Semi Molarity: 

The Solution which contains mole of substance dissolved in 1000 ml of solution is known as semi molar solution and the concentration is known as semi molarity. 

It is denoted by 0.5M. 

iii. Deci Molarity: 

The Solution which contains mole of substance dissolved in 1000 ml of solution is known as deci-molar solution and the concentration is known as Deci molarity. 

It is denoted by 0.1M. 

iv. Centi-Molarity: 

The Solution which contains mole of substance dissolved in 1000 ml of solution is known as centi-molar solution and the concentration is known as centi-molarity.

It is denoted by 0.01M. 

      II.Normality: 

The number of gram equivalent of a substance dissolved in 1000 ml of solution is known as normality. 

It Is denoted by N. 

Mathematically, 

$\begin{array}{l} \Rightarrow Normality(N) = \frac{{No.{\rm{ }}of{\rm{ gram Equivalent}}}}{{Volume\;of{\rm{ Solution in ml}}}} \times 1000\\ \Rightarrow Normality(N) = \frac{{Mass}}{{Equivalent{\rm{ Weight}} \times Volume\;of{\rm{ Solution in ml}}}} \times 1000\\\therefore Normality(N){\rm{ = }}\frac{{gm/l}}{{Equivalent{\rm{ Weight}}}} \times 1000\end{array}$

Types Of Normality: 

        i.1 Normality: 

The solution which contains 1 gram equivalent of substance dissolve in 1000 ml of solution is known as 1 normal solution and the concentration is known as 1 normality. 

It is denoted by 1N. 

       ii.Semi normality: 

The solution which contains gram equivalent of substance dissolve in 1000 ml of solution is known as semi normal solution and the concentration is known as semi normality. 

It is denoted by ½ N. 

     iii.Deci normality: 

The solution which contains gram equivalent of substance dissolve in 1000 ml of solution is known as decinormal solution and the concentration is known as deci-normality. 

It is denoted by 0.1N. 

     iv.Centi-normality: 

The solution which contains gram equivalent of substance dissolve in 1000 ml of solution is known as centi normal solution and the concentration is known as centi normality. 

It is denoted by 0.01N. 

Relationship between Molarity and Normality: 

We have, 

$\begin{array}{l}Molarity(M){\rm{ = }}\frac{{gm/l}}{{Molecular\,Mass}} \times 1000 -  -  -  -  -  - (i)\\Normality(N){\rm{ = }}\frac{{gm/l}}{{Equivalent{\rm{ Weight}}}} \times 1000 -  -  -  - (ii)\end{array}$

Dividing above Equation and we get:  

$\frac{{Normality}}{{Molarity}} = \frac{{Equivalent{\rm{ Weight}}}}{{Molecular\,Mass}}$ 

    III.Molality: 

The number of moles of substance dissolves in 1000 gm of solvent is known as molality.

Mathematically, 

$\begin{array}{l}Molality(m) = \frac{{No.{\rm{ }}of{\rm{ Moles of Substance}}}}{{Molecular{\rm{ Mass}} \times Mass{\rm{ of Solvent in gm}}}} \times 1000\\\therefore Molality(m) = \frac{{Mass}}{{Molecular{\rm{ Mass}} \times Mass{\rm{ of Solvent in gm}}}} \times 1000\end{array}$ 

Relationship Between Molality and Molarity:

$\frac{{Molarity}}{{Molality}} = \frac{{Mass{\rm{ }}of\;{\rm{solvent in KG}}}}{{Mass{\rm{ of solution in liter}}}} \times 1000$  

    IV.Formality: 

The number of formula mass of substance dissolve in 1000ml of solution is known as formality.

It is denoted by F. 

Mathematically,  

$Formality{\rm{  =  }}\frac{{No.\,{\rm{of formula mass}}}}{{Volume\;of{\rm{ Solution in ml}}}} \times 1000$  

      V.Gram Per Liter: 

The mass of substance in gm dissolve in 1l solution is called gram per liter. 

It is denoted by g/l. 

Mathematically, 

$\frac{g}{l} = \frac{{Mass{\rm{ of Substance}}}}{{Volume{\rm{ of solution in liter}}}}$

    VI.Percentage: 

The number of parts by mass or by volume of substance present in 100 parts by mass or volume of solution is known as percentage. 

Mathematically, 

$Percentage = \frac{{Mass{\rm{ of Substance}}}}{{Mass{\rm{ of Solution}}}} \times 100$  

  VII.Part Per Million: 

The number if part by weight or by volume of substance present in million parts by weight or by volume of substance is called part per billion. 

It is denoted by ppm. 

Mathematically, 

$ppm = \frac{{Mass{\rm{ of Substance}}}}{{Mass{\rm{ of Solution}}}} \times {10^6}$ 

VIII.Part Per Billion: 

The nu mb er if part by weight or by volume of substance present in billion parts by weight or by volume of substance is called part per billion. 

It is denoted by ppb. 

Mathematically, 

$ppb = \frac{{Mass{\rm{ of Substance}}}}{{Mass{\rm{ of Solution}}}} \times {10^9}$ 

Primary Standard Substance: 

The thermally stable substance which can be accurately weighed to prepare desired concentration of primary standard solution are known as primary standard substance. 

Example: 

Na 2CO3, 2H2O, Oxalic acid, etc.  

Secondary Standard Substance:

The thermally unstable substance which can’t be accurately weighed to prepare desired concentration of secondary standard solution are known as secondary standard substance. 

Example: 

NaOH, KOH, HCl etc. 

S.NO 

Primary Standard Substance 

Secondary Standard Substance

1. 

It is thermally stable substance that can be accurately weighed.

It is thermally unstable substance that can’t be accurately weighed.

2. 

It is non-hygroscopic, non-efflorescent, non-deliquescent. 

It is hygroscopic, efflorescent, deliquescent. 

3. 

It’s composition changes. 

Its composition doesn’t change.

4. 

It is required to prepare primary standard solution.

It is required to prepare secondary standard solution.

5. 

Their equivalent weight and molecular mass are comparatively higher.

Their equivalent weight and molecular mass may or may not be higher.

 

Primary Standard Solution: 

The chemical substance whose exact concentration solution can be prepared by directly weighing the sample and dissolving it in the definite volume of solution is called primary standard solution. 

Example: Na 2CO3, Oxalic acid. 

Secondary Standard Solution:  

The chemical substance whose standard solution cannot be prepared by directly weighing the sample and dissolving it in the definite volume of solution is called secondary standard solution

Example: NaOH, KOH. 

S.NO 

Primary Standard Solution 

Secondary standard solution

1. 

It can be prepared by directly weighing accurate amount of primary standard solution of primary standard substances.

It can’t be prepared by directly weighing accurate amount of primary standard solution of substances.

2. 

It is used to deter mine concentration of unknown solution.

Its concentration is determined by primary standard solution.

3. 

Its composition doesn’t change. 

It’s composition changes with time.

4. 

Example: N/10 Na 2CO3 solution 

Example: N/10 HCl solution.

 

Equivalent Weight: 

        i.Equivalent weight of element: 

The nu mb er of parts by weight of element that can displace or combine with 1.008 parts by weight of hydrogen, 8 parts by w eight of oxy gen and 35.5 parts by w eight of chlorine is called equivalent weight of element. 

Mathematically, 

$Equivalent{\rm{ Weight = }}\frac{{Mass{\rm{ of Element}}}}{{Mass{\rm{ of H displaced}}}} \times 1.008(at\,{\rm{NTP}})$  

       ii.Equivalent weight of acid:

The number of parts by weight of acid that contains 1.008 parts by weight of displaceable hydrogen is called equivalent weight of acid. 

Mathematically, 

$Equivalent{\rm{ Weight of Acid = }}\frac{{Weight\,{\rm{of Acid}}}}{{Basicity}}$  

Basicity: It is strength of acid that can neutralize 1gm equivalent of base. It is determined by taking number of displaceable hydrogen present in acid. 

     iii.Equivalent weight of base: 

The numb er of parts by weight of base that can neutralize 1gm equivalent of acid is called equivalent weight of base. 

Mathematically, 

$Equivalent{\rm{ Weight of Base = }}\frac{{Weight\,{\rm{of Acid}}}}{{Acidity}}$  

Acidity: It is strength of acid that can neutralize 1gm equivalent of acid. It is determined by taking number of displaceable hydroxyl ion (OH-)present in acid. 

     iv.Equivalent Weight of Oxidizing Agent: 

The amount of oxidizing agent that can produce 8 parts by mass of oxygen is known as equivalent weight of oxidizing agent. 

Mathematically, 

$Equivalent{\rm{ Weight = }}\frac{{Mass{\rm{ of Oxygen Produced}}}}{{{\rm{Mass}}\;{\rm{of Oxidizing Agent}}}} \times 8$  

       v.Equivalent Weight of Reducing Agent: 

The amount of r educing agent that can produce 1.008 parts by mass of hydrogen is known as equivalent weight of reducing agent. 

Relationship between valency atomic mass and equivalent weight: 

Let us consider A, E, V be atomic mass, equivalent weight and valency of element respectively. V*1.008 parts by weight of hydrogen combines with A parts by weight of the element. 

$1.008{\rm{ parts by weight of hydrogen combines with }}\frac{A}{{V \times 1.008}} \times 1.008 = \frac{A}{V}parts{\rm{ }}by{\rm{ weight}}$

$Equivalent{\rm{ Weight(E) = }}\frac{{Atomic{\rm{ Mass(A)}}}}{{Valency(V)}}$

Some Terminology: 

I. End Points: 

·         The point at which the Sharpe change in color of indicator appears. 

·         It Is the practical point. 

·         It can be observed in titration. 

·         It is completion point of titration. 

II. Equivalent Point:

·         The point at which equal number of gm of acid can neutralize same numb er of gram equivalent of base. 

·         It is theoretical point. 

·         It can be observed at titration. 

·         It is neutralization point.


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