Course Content
CHAPTER 3: CLASSIFICATION OF ELEMENTS
Section Name Topic Name 3 Classification of Elements and Periodicity in Properties 3.1 Why do we Need to Classify Elements ? 3.2 Genesis of Periodic Classification 3.3 Modern Periodic Law and the present form of the Periodic Table 3.4 Nomenclature of Elements with Atomic Numbers > 100 3.5 Electronic Configurations of Elements and the Periodic Table 3.6 Electronic Configurations and Types of Elements: s-, p-, d-, f – Blocks 3.7 Periodic Trends in Properties of Elements
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CHAPTER 7: EQUILIBRIUM
Section Name Topic Name 7 Equilibrium 7.1 Equilibrium in Physical Processes 7.2 Equilibrium in Chemical Processes – Dynamic Equilibrium 7.3 Law of Chemical Equilibrium and Equilibrium Constant 7.4 Homogeneous Equilibria 7.5 Heterogeneous Equilibria 7.6 Applications of Equilibrium Constants 7.7 Relationship between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G 7.8 Factors Affecting Equilibria 7.9 Ionic Equilibrium in Solution 7.10 Acids, Bases and Salts 7.11 Ionization of Acids and Bases 7.12 Buffer Solutions 7.13 Solubility Equilibria of Sparingly Soluble Salts
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CHAPTER 10: S-BLOCK ELEMENTS
Section Name Topic Name 10 The s-Block Elements 10.1 Group 1 Elements: Alkali Metals 10.2 General Characteristics of the Compounds of the Alkali Metals 10.3 Anomalous Properties of Lithium 10.4 Some Important Compounds of Sodium 10.5 Biological Importance of Sodium and Potassium 10.6 Group 2 Elements : Alkaline Earth Metals 10.7 General Characteristics of Compounds of the Alkaline Earth Metals 10.8 Anomalous Behaviour of Beryllium 10.9 Some Important Compounds of Calcium 10.10 Biological Importance of Magnesium and Calcium
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CHAPTER 12: CHARACTERIZATION OF ORGANIC COMPOUND
Section Name Topic Name 12 Organic Chemistry – Some Basic Principles and Techniques 12.1 General Introduction 12.2 Tetravalence of Carbon: Shapes of Organic Compounds 12.3 Structural Representations of Organic Compounds 12.4 Classification of Organic Compounds 12.5 Nomenclature of Organic Compounds 12.6 Isomerism 12.7 Fundamental Concepts in Organic Reaction Mechanism 12.8 Methods of Purification of Organic Compounds 12.9 Qualitative Analysis of Organic Compounds 12.10 Quantitative Analysis
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Class 11th Chemistry Online Class For 100% Result
About Lesson

Protic and non- Protic acids

  • Protic acids: Are those which have Proton.

Example: HClàH+ + Cl

  • Non-Protic acid: A acid which has no Proton.

Example: carbon dioxide

  • Protic solvent: A solvent with proton like water.
  • Non- Protic solvent: A solvent with no proton like carbon tetrachloride etc.
  • Di-Protic acid: Which has two protons like H2SO4?
  • Tri-Protic acid: Which has three protons like phosphoric acid?

The method of calculating their Ka value is given below:

Let see,

H2SO4 àH+ + SO42-, let’s say for first reaction K=Ka1

HSO4àH+ + SO42-, let’s say for this K=Ka2

Please note Ka1 is always greater than Ka2 because as acid keeps on loosing proton ,its acidic strength keeps on decreasing .

Then for the complete reaction Ka =Ka1 x Ka2

Similarly, for bases we have: Kb = Kb1 x Kb2

equilibrium notes class 11

Hydrolysis of salt

It is the “interaction of anion and cation of salt with water to produce an acidic or basic solution”.

That is : Salt + water –>acid + base

Or

BA + H2OàHA + BOH

We can also say: “Hydrolysis as reverse of neutralization reaction”.

If salt has cation and anion, it reacts with water and result in acidic or basic solution (depending upon which is strong acid or base).

Like strong acid + weak base à acidic solution

Suppose we have salt of strong acid and weak  base:

NH4Cl +H2OàNH4OH + HCl

The resulting solution is acidic because NH4OH is a weak base and HCl is strong acid.

On the other hand, if we have salt of strong base and weak acid à basic solution

For example: CH3COONa +H2Oà NaOH +CH3COONa

The resulting solution is basic solution as NaOH is strong base. Like, we did Ka ,Kb, Kw .Similarly, we have Kh that is hydrolysis constant and degree of hydrolysis as ‘h’.

Different kind of salts

  1. Salt of weak acid and strong base
equilibrium notes class 11

Buffer solution

They are those solutions which has a constant value of pH , no matter what we add in  to them .

Example: The pH of Blood is 7.4-7.6. Blood act as a buffer. As u all know, in spite of eating so many types of food items, the pH of blood remains same.

The buffer solution is defined as: A solution which resists the change in hydrogen ion concentration on addition of a small amount of acid or a base in to it.

  • Buffer action: It is the ability to resist the change in pH on addition of acid or base.

There are two types of buffers

  • Acidic buffer
  • Basic buffer

Acidic buffer: It contains equimolar concentration of weak acid and its salt with a strong base. For example:

  • We take any weak acid say, acetic acid. To it, if we add equimolar concentration of sodium acetate, then the concentration of acetate ion increases. This acetate ion will neutralize the effect of hydrogen ion and the pH remains same.

Basic buffer: It is equimolar concentration of solution of weak base and its salt with a strong acid.

  • Let’s say, ammonium hydroxide dissociates to ammonium ion and hydroxide ion. It has ammonium chloride which also gives ammonium ion. So, this ammonium ion combines with hydroxide ion and neutralizes its effect. Therefore, pH remains same.

  A equation related to it was given by Henderson equation:

Solubility product

It is the amount of moles of solute dissolved in given amount of solvent.

Solubility of salt depends upon

  • Lattice enthalpy
  • Hydration enthalpy

If lattice enthalpy is low and hydration enthalpy is high, than the salt easily dissolves.

equilibrium notes class 11

2.To predict the precipitate  of salt

For example :

AgClàAg+ +Cl

K=[Ag+][Cl]/[AgCl]         (equation 1 )

Or, K[AgCl]= [Ag+][Cl]

Or, Ksp=[Ag+][Cl]             (equation 2 )

If we compare equation 1 and equation 2 ,we get :

  • If value of K>Ksp, then the precipitation occurs .
  • If value of K<Ksp, then the precipitation doesn’t occurs.

Numerical problems:

Problem 1:-

What is the Kc for the following equilibrium when the equilibrium concentration of each substance is :

               2SO2 + O2   <–>      2SO3

Given [SO2]=0.60M,[O2]=0.82M and [SO3]=1.90M?

Answer 1: The Kc =[SO3]2/[SO2]2[O2]

Substituting the values, we get :

Kc=(1.90)2/(0.60)2(0.82)=12.229 M-1

Problem 2:-

At certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of iodine atoms :

I  <–>2I .Calculate Kp for equilibrium?

Answer 2: Total pressure of the mixture at equilibrium =105 Pa

                   Partial pressure of iodine atoms = 40/100 x 105 Pa

                                                                              = 0.4 x 105 Pa

                   Partial pressure of I2 = 60/100 x 105 Pa

                                                          = 0.60 x 105 Pa

                                                       Kp=(pI)2/pI2

                                                          = (0.4 x 105)2/0.60 x 105=2.67 x 104

Problem 3:

Find out the value of Kc for each of the following from the value of Kp:

          2NOCl   <–>        2NO + Cl2  ,Kp= 1.8 x 10-2 at 500k

        CaCO3         <–>        CaO + CO ,K p = 167 at 1073 K

Answer 3:- Kp and Kc are related to each other by the relation :

Kp=KcRTchange In gaseous moles

  • Change in gaseous moles =3-2=1 ,Kp=kc(RT)1

or

 Kc= 1.8 x 10-2/0.0831 x 500=4.33 x 10-4

  • Change in gaseous moles = 1-0=1 , Kc=Kp/RT=167/0.0831 x 1073 =1.87

Problem 4:-

Reaction between N2 and O2 takes place as follows :

2N2 + O2   <–>         2N2O.

If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc= 2.0 x 10 -37,determine the composition of equilibrium mixture?

Answer 4:- let x moles of nitrogen gas take part in the reaction .Then according to the equation ,x/2moles of O2 will react giving x moles of N2O.

                                 2N2  +   O2      <–>         2 N2O

Initial                     0.482 mol       0.933mol              0

At equilibrium: 0.482-x                0.933-x/2           x/10

Molar concentration: 0.482-x/10   0.933-x/2/10     x/10

As Kc =2.0 x 10 -37 is very small ,this shows that the amount of nitrogen gas and oxygen gas reaction is to a very small extent and it can be neglected .Hence ,at equilibrium ,we have :

[N2]=0.0482 mol/L

[O2]=0.0933 mol/L

[N2O]=0.1x

Substituting the values, we get :

Kc= (0.1x)2/(0.0482)2(0.0933)=2.0 x 10-37

 x= 6.6 x 10 -20

On solving :

N2O=0.1x=6.6 x 10-21 mol/L

                     [N2]=0.0482 mol/L

                     [O2]=0.0933 mol/L

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