BUFFER SOLUTION

Topics cover :
(1) DEFINITION
(2)TYPE OF BUFFER SOLUTION
(3) pH OF ACIDIC BUFFER
(4) pH OF BASIC BUFFER
(5) DILUTION OF BUFFER
(6) BUFFER CAPACITY OR BUFFER  INDEX
(7) IDEAL BUFFER SOLUTION

(1) DEFINITION:
The aqueous electrolyte solution which resist the any change in pH even after addition of small amount of strong acid or strong base  called  buffer Solution.

(2)TYPE OF BUFFER SOLUTION:
(1) Simple Buffer Solution or Neutral buffer
(2) Mixed Buffer Solution

(3) pH OF ACIDIC BUFFER:

(4) pH OF BASIC BUFFER:

(5) DILUTION OF BUFFER:

(6) BUFFER CAPACITY OR BUFFER  INDEX:

(7) IDEAL BUFFER SOLUTION:

ILLUSTRATIVE EXAMPLE (1):
Pkb(NH4OH) is 5 and a buffer solution contains 0.1M NH4OH and 0.1M (NH4)SO4 calculate pH of this buffer solution ?
(Ans-5.3 )

ILLUSTRATIVE EXAMPLE (2):
Pls of HX is 4.7 , (1) find the pH of Solution having 0.5 M HX and KX 0.25M ? (2) pH of this solution if it is diluted 10000 times ?
(Ans- (1) 4.4 (2) )

ILLUSTRATIVE EXAMPLE (3):
Calculate pH of acidic buffer mixture containing 1.0 M HA (Ka=1.5x10-1) and 0.1M NaH .
(Ans- 0.824)

ILLUSTRATIVE EXAMPLE (4):
Calculate the mass of NH3 and NH4Cl required to prepared a buffer solution of pH  =9.0 when total  concentration of buffering reagents is 0.6 mole /litre .( Pkb for NH3 is 4.7, log2 is 0.3010)
(Ans- a=0.2 mole ,b=0.4 mole)

ILLUSTRATIVE EXAMPLE (5):
One liter buffer solution is prepared by mixing of 1.0 mole HCOOH (formic acid) and 1 mole HCOONa . (Given Pka HCOOH =4.0) then calculate
(i) pH of buffer
(ii) pH of Solution if 1/3 mole of HCl is added
(iii) pH of Solution if 1/3 mole of NaCl is added
(iv) pH of Solution if it diluted to 10 liter
(v) pH of Solution if it diluted to 1000 liter
(Ans- i-4.0 , ii- 3.7 , iii-4.3010 , vi- 4.0 v- 4.07)

ILLUSTRATIVE EXAMPLE (6):
Calculate the pH of an aqueous solution originally containing 0.4 M acetic acid and 0.2 M sodium acetate (ka CH3COOH= 1.8x10-5 ).
(Ans- 4.4)

ILLUSTRATIVE EXAMPLE (7):
Calculate pH of Solution Originally having 0.2 M (NH4)SO4 and 0.4 M NH4OH (given Kb NH4OH =1.8x10-5)
(Ans- 9.26)
ILLUSTRATIVE EXAMPLE (8):
In 100 ml of 0.4M C6H6COOH Solution,0.1M C6H6COONa is added , calculate pH of resulting solution (given Ka C6H6COOH is 4x10-5)
(Ans-4.1)

ILLUSTRATIVE EXAMPLE (9):
A solution contains 0.2 mole acetic acid and 0.10 mole of sodium acetate ,made up to 10 liter volume , calculate the pH of Solution ( given Ka CH3COOH is 1.8x10-5)
(Ans-4.44)
ILLUSTRATIVE EXAMPLE (10):
What mass of , in gram ,of NaNO2 must added to 700 ml of 0.165 M HNO2 to produce a Solution with pH of 3.50 ? ( Ka HNO2 is 6.0x10-4)
(Ans- 15.1gm)

ILLUSTRATIVE EXAMPLE (11):
In 500 ml of a buffer solution containing 0.8 M CH3COOH and 0.6 M CH3COONa , 0.2 M HCl is added. Calculate the pH of Solution before and after adding HCl. (Ka CH3COOH is 1.8x10-5) .
(Ans- i - 4.62  ii- 3.96)

ILLUSTRATIVE EXAMPLE (12):
A mixture of 0.2 mole RNH2 and 0.4 mole RNH3Cl is  mixed .the volume of Solution prepared is 10 liter  (given Kb RNH2 is 10-5) calculate.
(i) pH of resulting solution
(ii) pH of Solution if diluted to 1000 litres
(iii) pH of Solution if 200 ml buffer is mixed with 2 milimoles of H+
(iv) pH of Solution if 200 ml buffer is mixed with 2 milimoles of OH-
(Ans- i-  5.30 , ii- 5.31 , ii- 8.3 , iv -9.0)

BRONSTED LOWERY ACID-BASE CONCEPT

According to Bronsted theory the species which donate protons (H+) in any medium is consider as acid and the species which accept proton is consider as base.

Acid and base characters are realised in the presence of each other.
For example

CONJUGATE ACID-BASE PAIRS
(1) conjugate pair is acid-base pair differing in single proton (H⁺)
(2) conjugate acid is written by adding  H⁺ and conjugate is written by removing H⁺ .

(3) Strong acid has weak conjugate base and vice versa. Similarly strong base has weak conjugate and vice versa.

(4) Equilibrium always moves from strong acid to weak acid and strong base to weak base.

(5) Conjugate acid - base pair differ by only one proton. Reaction will always proceed from strong acid to weak acid or from strong base to weak base.

MERITES OF BRONSTED CONCEPT:
(1) the role of solvent clearly defined.
(2) the acidic and basic  character may be observed in non aqueous medium also .
(3) the acidic ,basic or Amphoteric nature of most of the substance may be defined.
(4) the acid having greater tendency to donate protons are stronger acid and base  having greater tendency to accept protons are stronger base .
(5)In conjugate pair ,if one is strong then other must be weak .
The weak acid or base are normally determined by comparing the the stability of different acid or base
DEMERITES OF BRONSTED CONCEPT:
(1) Proton is a nuclear particle hence reaction should not explained in term of proton.
(2) the neutralized process becomes multiples step process.
(3) Most of the Amphoteric solvent become Amphoteric.

AMPHOTERIC SPECIES (Amphiprotic):
The species which have a tendency to donate proton as well as accept proton (H⁺) such species are known as Amphoteric species.
For example H₂O,NH₃ HS^- ,HPO₃^- ,HC₂OO₄^- , H₂O₄ etc

ILLUSTRATIVE EXAMPLES:
(1)The conjugate base of HCO₃ is –
  (A) H₂CO₃        (B) CO₂             (C) H₂O      (D) CO₃^-   
(2) The conjugate acid of HSO₃^- is -
  (A) SO₃^2-          (B) SO₄^2-                  (C) H₂SO₄    (D) H₂SO₃

(Ans: 1-D 2-D)

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