5 p h,buffers

1

Self Ionization of Water. pH.

Water is weak, amphoteric, binnary electrolyte.

2

pH + pOH =143

4

5

Hydrolysis of Salts.

6

Salt hydrolysis may be defined as the reaction of the cation or anion of the salt with water to produce acidic or basic solution.Thus depending upon the relative strengths of the acid or base produced, the resulting solution is acidic, basic or neutral. There are four distinct types of hydrolytic behaviour of various salts. These are : 1.Salts of strong acids and strong bases. 2.Salts of strong acid and weak bases. 3.Salts of weak acids and strong bases. 4.Salts of weak acids and weak bases.

7

1. SALTS OF STRONG ACID AND STRONG BASE. Examples are NaCl, NaNO3 , Na2SO4 , KCl, KNO3 , K2SO4, etc.As an illustration, let us discuss the hydrolysis of NaCl. We may write

Thus it involves only ionisation and no hydrolysis. Further in the resulting solution [ H+] = [ OH−] So the solution is neutral. Hence it may be generalized that the salts of strong acids and strong bases do not undergo hydrolysis and the resulting solution is neutral , (pH=7), litmus-violet.

8

2. SALTS OF WEAK ACIDS AND STRONG BASES. Examples are : CH3COONa, Na2CO3 , K2CO3 , Na3PO4 etc.As an illustration , the hydrolysis of sodium acetate (CH3COONa) may be represented as follows: CH3COONa + H2O ↔ CH3COOH + NaOH or CH3COO− + Na+ + H2O ↔ CH3COOH + Na+ + OH−

or CH3COO− + H2O ↔ CH3COOH + OH−

As it produces OH− ions , the solution of such a salt is alkaline in nature (pH>7), litmus-blue.

9

3. SALTS OF STRONG ACIDS AND WEAK BASES. Examples are :NH4Cl, CuSO4 , NH4NO3 , AlCl3 , etc.As an illustration , the hydrolysis of NH4Cl may be represented as follows: NH4Cl + H2O ↔ NH4OH + HCl or NH4

+ + Cl− + H2O ↔ NH4OH + H+ + Cl−

or NH4+ + H2O ↔ NH4OH + H+

As it produces H+ ions , the solution of such a salt is acidic in character (pH<7), litmus-red.

10

4. SALTS OF WEAK ACIDS AND WEAK BASES Examples are CH3COONH4 , (NH4)2CO3 , AlPO4 etc.As an illustration, the hydrolysis of ammonium acetate may be represented as follows:

     CH3COONH4 + H2O↔ CH3COOH + NH4OHOr CH3COO− + NH4

+ + H2O ↔ CH3COOH + NH4OH

Thus it involves both anionic and cationic hydrolysis. The resulting solution may be neutral or slightly acidic or basic depending upon the relative degrees of ionisation of weak acid and weak base produced. Hence the resulting solution is almost neutral , (pH=7), litmus-violet.

11

Irreversible Hydrolysis.

Al2S3 + 6H2O = 2Al(OH)3↓ + 3H2S↑

12

salt

HYDROLYTIC CONSTANT (Kh) The general equation for the hydrolysis of a salt (BA) may be written as :               BA + H2O  HA + BOH

Applying the law of chemical equilibrium, we get :

13

K = the equilibrium constant.Since water is present in large excess in aqueous solution, its concentration [H2O] may be regarded as constant so that we have:

where K h is called hydrolysis constant.

14

DEGREE OF HYDROLYSIS (h) The degree of hydrolysis of a salt is defined as the fraction (or percentage) of the total salt hydrolysed. i.e.,

15

h →1

Salt hydrolysis:on An-

pH=7 + ½(pKa+logCsalt)

on Cat+

pH = 7-1/2(pKb+logCsalt)

16

17

Buffers.

18

Such solutions which oppose the change in

their pH on the addition of small amounts of

an acid or a base are called buffer solutions

or simply buffers. The buffer solutions have

thus reserved acidity & reserved alkalinity.

19

Characteristics of Buffer solutions:

•Its pH remains unsalted either on keeping the solution for long or on dilution•Its pH is very slightly changed by

the addition of small amount of strong base or an acid.

●It has a definite pH values

20

Acidic Buffer Solution.Mixture of weak acid + its conjugate baseEg. : aqueous mixture of ethanoic acid (CH3COOH )+ sodium ethanoate (CH3COONa)Sodium ethanoate dissociates fully in water.Ethanoic acid dissociates partially in water.An aqueous mixture of ethanoic acid and sodium ethanoate contains a large quantity of, Undissociated CH3COOH (the acid)

CH3COO- ions (the base conjugate)

21

Mixture of weak base + its conjugate acidEg. : aqueous mixture of ammonia NH4OH (the base) and ammonium chloride NH4Cl (the conjugate acid).Ammonium chloride dissociates fully in water.Ammonia dissociates partially in water.The aqueous mixture contains a large quantity of, Undissociates NH4OH (the base)

NH4+ (the conjugate acid)

Basic Buffer Solution.

22

Acidic Buffer Action.

weak a. c.b. strong

The presence of a common ions from highly ionized

a further suppresses the ionization of weak acid .

23

Thus even an adding HCl, a strong electrolyte does not produce an appreciable change in pH of the solution.

However, when a strong base is added, OH- ions are furnished by the base and are neutralized by and no change in pH is observed.

24

Basic Buffer Action.

Weak base. strong conjug.acid

Similary when a few drops of strong acid (HCl) is added, the H+ NH4OH ions combines with excess of

to form feebly ionized water molecules.

25

• In chemistry, the Henderson–Hasselbalch (often misspelled as Henderson–Hasselbach) equation describes the derivation of pH

• measure of acidity (using pKa, the acid dissociation constant) in biological and chemical systems.

• The equation is also useful for estimating the pH of a buffer solution and finding the equilibrium pH in acid-base reactions (it is widely used to calculate the isoelectric point of proteins).

• ( is the pH at which a particular molecule or surface carries no net electrical charge).

pH value of Buffer Solution(Henderson – Hasselbah’s

Equation).

26

Consider a buffer solution containing a weak acid

(CH3COOH) & its highly ionized salt (CH3COONa).

The dissociation of weak acid occurs as’

Applying Law Mass Action, we have

27

Taking logarithms:

28

for basic buffer:

29

Blood Buffer systems.•Hydrophosphate Buffer

•Hydrocarbonate Buffer:

30

•Hemoglobyne systems

Arterial blood Venous blood

31

Acid

Base

•Protein buffer system

+OH-

+H+

32

Buffers capacity (B) depends on the amount of acid & conjugate base in the solution. The best 1:1.

33

pH(weak acid) =1/2(pKa-logCa)

pH(weak base) =14-1/2(pKb-logCb)

Weak electrolytes

Hydrolizated salts

pH(salt) =7+1/2(pKa +logC(salt) )

pH(salt) =7-1/2(pKb +logC(salt) )

34

Thank You!