BORN HABER CYCLE OF ALUMINIUM OXIDE

Construct Born Haber Cycle Aluminium Oxide

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https://youtu.be/yoFNQnYS5_8

Born Haber Cycle

 Born Haber Cycle for Aluminium Floride

 Check the video :
https://youtu.be/O4d47THj7RY

Reaction of Organic compound

reaction of Amine with Nitrous Acid

Reaction of amine with Nitrous Acid

For further explaination, check the video 

https://www.youtube.com/watch?v=uaKTbgUm8Mg&t=34s

CHEMICAL TEST IN ORGANIC COMPOUND

https://drive.google.com/file/d/1NHJbsJSmBO1lzhu7CeEQPLZvaUIjAeFM/view?usp=sharing

Mechanism Halogenation of Benzene

check the video:

Mechanism in Alkene

Mechanism free radical substitution

Lets check the video:

BASIC BUFFER SOLUTION

	Basic Buffer Solution (Weak base + conjugate acid)
Preparation	By adding a weak base to a salt which contains its conjugate acid.
Example	NH3 and NH4Cl
Reaction	(W.B) : NH3 + H2O   NH4+   + OH- (C.A):  NH4Cl  → NH4+  + Cl-
Henderson-Hasselbalch’s equation	pH = - log Kb  +  log  [ conjugate acid ]                                     [ weak base ]
Addition of a small amount of strong acid	When a small amount of acid is added (H+), NH3(aq) +  H+(aq) →NH4+(aq) Acid (H+) form strong acid reacts with NH3to form NH4+. Acid added is consumed. Concentration of [NH4+] increase, but concentrations of [NH3] decrease. As a result, pH of the solution is not much affected.
Addition of a small amount of strong acid	When a small amount of base is added (OH-), OH-(aq) + NH4+ (aq) →NH3 (aq) + H2O(l) Base (OH-) form strong base reacts with NH4+ to form NH3 Base added is consumed. Concentration of [NH3] increase, but concentrations of [NH4+] decrease. As a result, pH of the solution is not much affected.
NOTE: The pH of a buffer solution can always be maintained no matter a strong acid or a strong base is added

Acidic Buffer solution

Definition:  Buffer solution is a solution which has the ability to maintain its pH when a small amount of strong acid or strong base is added.

  Two types of buffer solution

i)        Acidic Buffer – contains weak acid and its conjugate base (pH < 7)

ii)      Basic Buffer – contains weak base and its conjugate acid (pH > 7)

	Acidic Buffer Solution (Weak acid + conjugate base)
Preparation	By adding a weak acid to a salt which contains its conjugate base.
Example	CH3COOH  and  CH3COONa
Reaction	(W.A) : CH3COOH(aq)  + H2O(l)   CH3COO-(aq)  +   H3O +(aq) (C.B) :  CH3COONa(aq) →    CH3COO-(aq)   +   Na+(aq)
Henderson-Hasselbalch’s equation	pH = pKa  +  log  [ conjugate base ]                                 [ weak acid ] pH = - log Ka  +  log  [ conjugate base ]                                        [ weak acid ]
Addition of a small amount of strong acid	When a small amount of acid is added (H+), H+(aq) + CH3COO-(aq) → CH3COOH(aq) Acid (H+) form strong acid reacts with CH3COO- to form CH3COOH. Acid added is consumed. Concentration of [CH3COOH] increase, but concentrations of [CH3COO-] decrease. As a result, pH of the solution is not much affected.
Addition of a small amount of strong acid	When a small amount of base is added (OH-), OH-(aq) + CH3COOH (aq) →CH3COO- (aq) + H2O(l) Base (OH-) form strong base reacts with CH3COOH to form CH3COO-. Base added is consumed. Concentration of [CH3COO-] increase, but concentrations of [CH3COOH] decrease. As a result, pH of the solution is not much affected.
NOTE: The pH of a buffer solution can always be maintained no matter a strong acid or a strong base is added

Calculate pH for weak acid.

Example: (PSPM 2010/2011)

An amine, RNH2 is a weak base with a dissociation constant, Kb of 1.8 x 10-5. It was dissolved in water to form a dilute solution.

a) Write the dissociation equation for the base.

b) Calculate the pH of a 0.4 M aqueous solution of the amine. 

                                                                                                                                 
►►►Hint:
Step 1 : Weak base dissociation equation

                      RNH2(aq) + H2O(l)        →⟶   RNH3⁺(aq) + OH^‐(aq)

Step 2 : Derive the dissociation constant

Step 3: Write ICE table

	RNH2(aq) +	H2O(l)	RNH3+(aq)	OH‐(aq)
Initial (M)	0.4	-	0	0
Change (M)	-x	-	+x	+x
Equilibrium (M)	0.4-x	-	+x	+x

Step 3 : Solve to find [OH-]. Include any assumptions made

1.8 x 10^-5 =    (x)(x)

                     (0.4-x)

Checking = √Kb x 100   < 5%

                      [ ]i 

Kb << 1, thus 0.4-x = 0.4

1.8 x 10^-4 =    (x)(x)

                     (0.4)

x = 2.68 x 10^‐3M

[x] =[ OH^-] = 2.68 x 10^‐3M

Step 4 : Find the pH

pOH = ‐log [OH‐] = 2.57

pH = 14 ‐ pOH

= 11.43

Guidelines for calculating pH of strong acid and strong base

Strong acid: Acid that completely ionised to form H⁺ or H₃O⁺ ions. Example HCl, H₂SO₄

Strong base: Base theat completely ionised to form OH^- ions. Example KOH, Mg(OH)₂

Guidelines for calculating pH value of strong acid and base

Le Chatellier's Principle

 Concept of excercise reaction:               

      The reaction between NO and O₂ is exothermic:

→          2NO (g)  +  O₂ (g)          NO₂ (g)

       

      State the effects on equilibrium 

                  

1. O₂ is added
2. NO₂ is taken out
3. The temperature is increase
4. The total pressure is decreased
5. Neon gas is added at constant volume
6. Argon gas is added at constant pressure

                   Answer:

                          

                     1.    When O₂ gas is added

·         Equilibrium position shift to the right in order to reduce the added O₂ gas

·         As a result, concentration of NO₂ gas increase

·         Concentration of NO and O₂ gas is decrease

2.    When NO₂ is taken out

·         Equilibrium position shift to the right in order to increase NO₂ gas

·         As a result, concentration of NO₂ gas increase

·         Concentration of NO and O₂ gas is decrease

                     3.    The temperature is increased

·         Equilibrium position shift to the left in order to reduce the increase in temperature

·         As a result, concentration of NO₂ gas decrease

·         Concentration of NO and O₂ gas is increase

                    4.    The total pressure is decreased

·         Equilibrium position shift to the left (higher number of mole) in order to increase in pressure

·         As a result, concentration of NO₂ gas decrease

·         Concentration of NO and O₂ gas is increase

                    5.    Neon gas is added at constant                               volume

·         Partial pressure of each reacting gas remains constant

·         Equilibbrium position remain unchange

6.    Argon gas is added at constant pressure

·         Partial pressure of each reacting gasses decrease

·         System will respond by increasing pressure

·         Equilibrium position shifts to the side with higher number of moles

·         So equilibrium position shift to the left

·         As a result, concentration of NO₂ gas decrease

·         Concentration of NO and O₂ gas is increase