1. Buffer

2. pH Example I pH = - log [ H+] or pH = - log [ H3O+ ]
What is the pH of solution with [ H+] = 32 X M/L ?
pH = - log [ H+]
pH = - log 32 X 10-5
pH = 3.495

3. Example II
The hydronium ion concentration of 0.1 M solution was found to the 1.32 X 10-3 M, What is the pH of the solution?
pH = - log [ H3O+ ]
pH = - log [ 1.32 X 10-3 ]
pH = 2.88
Example III
If the pH of a solution is 4.72, what is the hydrogen ion concentration?
pH = - log [ H+ ]
- log [ H+ ] = 4.72
log [ H+ ] = take anti-log for both side
[ H+ ] = 1.91 X 10-5

4. pH = H+ = 1 M/l = 1 X 100 = 1
pH = H+ = 1 X 10-1 = 0.1 M/L
pH = H+ = 1 X 10-2 = 0.01 M/L
pH = H+ = 1 X 10-3 = M/L
Note : the change in one pH unit means 10 fold change in [H+].
pH + pOH = 14

5. Introduction
When a minute trace of hydrochloric acid is added to pure water, a significant increase in hydrogen-ion concentration occur immediately.
In a similar manner, when a minute trace of sodium hydroxide is added to pure water, it cause correspondingly large increase in the hydroxyl-ion concentration.
These change take place because water alone cannot neutralize even trace of acid or base, i.e. it has no ability to resist change in hydrogen-ion concentration or pH.
Therefore, it is said to be unbuffered.
E.g. CO2 + H2O H2CO3 decrease pH from 7 to 5.8
These change of pH are of great concern in pharmaceutical
preparation also NaCl solution ability to resist change of pH.
To ensure stability and solubility, we used to control pH by using
a buffer

6. Buffer :
Compound or a mixture of compound which by presence in solution to resist change in pH up of addition of small quantities of acid, base or a solvent.

7. Adjustment of pH by the buffer
pH important for stability & solubility, therefore should be adjusted pH by buffer .
Also pH play important role in :
1-Parenteral dosage form.
2-Eye drops.
3-Nasal drops.

8. Adjustment of pH by the buffer
Degree of acidity and alkalinity depends on the relative concentration of H+ ion and OH- ion.
if H+ > OH- = acidic
H+ = OH- = neutral
H+ < OH- = alkaline
Acidity and alkali may be strong or weak:
1- weak acid, pH strong acid, pH 0-3.5
3- weak base, pH strong base, pH
Strong base
Weak base
Weak acid
Strong acid

9. The product of H+ ion and OH- ion in the any aqueous liquid is constant i.e. Kw = [ H+] [ OH- ]
Increase of one tend to decrease of another.

10. Some notes about buffer:
1-Buffers solution should be prepared using freshly boiled and cooled water.
2- Buffers solution should be stored in containers of Alkali-free glass.
3-Buffers solution should be discarded no later than three months from the date of manufacture.

11. Selection of buffer system depends on:
pH rang.
Buffer capacity desired.
The purpose for which it is required.
Compatibility with active ingredients.

12. Method of preparation of buffer
Buffer equation.
Buffer table.

13. Buffer solution consist of mixture of weak acid and its conjugate base or weak base and its conjugate acid.
Weak acid + its salt
e.g. acetic acid + sodium acetate
CH3COOH CH3COONa
Weak base + its salt
e.g. Ammonia + Ammonium chloride
NH NH4Cl

14. Henderson-Hasselbalch Equation buffer equation for weak acid and its salt
Dissociation Constant of weak acid is given by the equation : Where : A- = Salt HA = Acid we isolate the H+ and put it on the left-hand side of the equation:

15. take the negative log of each of the three terms in the last equation, they become: - log [H+]  this is the pH - log Ka  this is the pKa - log ([HA] / [A-])  to get rid of the negative sign + log ([A-] / [HA]) Inserting these last three items (the pH, the pKa and the rearranged log term), we arrive at the Henderson-Hasselbalch Equation:

16. common way the Henderson-Hasselbalch Equation is presented
in a textbook explanation:

17. Remember that, in a buffer, the two substances differ by only a proton
Remember that, in a buffer, the two substances differ by only a proton. The substance with the proton is the acid and the substance without the proton is the salt.
However, remember that the salt of a weak acid is a base (and the salt of a weak base is an acid).
Consequently, another common way to write the Henderson-Hasselbalch Equation is to substitute "base" for "salt form" (sometimes you will see "conjugate base" or "base form"). This is probably the most useful way to decribe the interactions between the acidic form (the HA) and the basic form (the A-).
Here it is:
Remember this: the base is the one WITHOUT the proton and the acid is the one WITH the proton.

18. Buffer equation of weak base & its salt
Dissociation Constant of weak acid is given by the equation :
( B + ) ( OH- )
Kb =
( BOH)
Where : B + = Salt
BOH = Base
[Base]
pH = pKw – pKb + log
[ Salt]

19. An alternate form of the Henderson-Hasselbalch Equation
The last discussion used pH and pKa. There is a alternate form of the Henderson-Hasselbalch Equation using pOH and pKb.
[Salt]
pOH = pKb + log
[Base]

20. Mechanism of action of buffer
1 – Acid / Its salt e.g. CH3COOH/CH3COONa.
CH3COOH + OH-  CH3COO- + H2O
This means acid will react with base (OH- ( to neutralize it.
CH3COO-/Na+ + H3O+  CH3COOH + H2O
This means salt of weak acid will react with acid ( H3O ) to neutralize it.

21. Mechanism of action of buffer… cont
2 – Base / Its salt e.g. NH3/NH4Cl.
NH3 + H3O+  NH4+ + H2O
This means base will react with acid to neutralize it.
NH4+ + OH-  NH3 + H2O
This means salt of weak base will react with base (OH- ) to neutralize it.

22. pH + pOH = 14
For acid & its salt :
[Salt]
pH = pKa + log
[Acid]
For base & its salt :
pOH = pKb + log or
[Base]
pH = pKw – pKb + log
[ Salt]

23. log salt/acid = 5 – 4.75 = 0.25 take anti-log for both side
Example I
What is the molar ratio of salt/acid required to prepare an acetate buffer pH = 5 ?
Ka = 1.8 X 10-5
pKa = - log Ka
pKa = - log 1.8 X = 4.75
pH = pKa + log salt/acid
5 = log salt/acid
log salt/acid = 5 – 4.75 = take anti-log for both side
salt/acid = 1.78
So, Ratio of salt/acid = 1.78/1
Mole fraction of acid = 1 / ( ) = X 100 = %
Mole fraction of salt = 1.78 / ( ) = X 100 = %
mole fraction multiply by 100 to get of mole %

24. Example II
Prepare 200 ml of acetate buffer pH = 6 with molar conc. = 0.4 M & Ka = 1.8 X 10-5 ?
pKa = - log Ka
pKa = - log 1.8 X = 4.75
pH = pKa + log salt/acid
6 = log salt/acid
log salt/acid = 6 – 4.75 = take anti-log for both side
salt/acid = 17.78/1
Molecular weight of acetic acid ( CH3COOH ) = 60
Molecular weight of sodium acetate ( CH3COONa ) = 82
Weight of acid = mole fraction X Conc. X M.wt X V (L)
= 1/18.78 X 0.4 X 60 X0.2
= 0.26 g
Weight of salt = mole fraction X Conc. X M.wt X V (L)
= 17.78/18.78 X 0.4 X 82 X0.2
= 6.21 g

25. Example III
If we add 0.1 M sodium acetate to 0.09 M acetic acid .What is the pH if you know Ka = 1.8 X ?
pKa = - log Ka
pKa = - log 1.8 X = 4.75
pH = pKa + log salt/acid
pH = log 0.1/0.09
pH = 4.796

26. Example IV What is the pH of of a solution containing 0
Example IV What is the pH of of a solution containing 0.1 mole of ephedrine base and 0.01 mole of ephedrine HCl / liter of solution ? pKb of ephedrine = 4.64
pH = pKw - pKb + log base/salt
pH = log 0.1/0.01
pH = log 10 = 10.36 Or
pOH = pKb + log salt/base
pOH = log 0.01/0.1
pOH = log 0.1 = 3.64
pH + pOH = 14
pH = 14 – pOH
pH = 14 – 3.64 = 10.36

27. Change in pH with addition of an acid or base
Calculate the change in pH of a buffer solution with the addition of a given amount of acid or base in the following example :
Example
Calculation the change in pH after adding 0.04 mol of sodium hydroxide to a liter of a buffer solution containing 0.2 M conc. of sodium acetate and acetic acid. The pKa value of acetic acid is 4.76 at 25 OC.
The pH of the buffer solution is calculated by using the buffer equation
pH = pKa + log salt/acid
pH = log 0.2/0.2
pH = 4.76

28. pH = pKa + log ------------------------
The addition of 0.04 mol of sodium hydroxide converts 0.04 mol of acetic acid to 0.04 mol of sodium acetate. Consequently, the conc. Of the acetic acid is decrease and the conc. Of the sodium acetate is increase by equal amounts, according to the following equation :
salt + base
pH = pKa + log
acid - base

29. salt + base
pH = pKa + log
acid - base
pH = log
pH = = =
Because the pH before addition sodium hydroxide was 4.74, the change in pH = 4.94 – 4.76 = 0.18 unit.

30. Buffer Capacity
The ability of buffer solution to resist change in pH upon addition of acid or base.
Ka . [H+]
Buffer Capacity “ B “ = X C X
( Ka + [H+] )2

31. Example I
At hydrogen ion conc. of 1.75 X (4.76), what is the capacity of a buffer containing 0.1 mole each of acetic acid and sodium acetate / liter of solution ?
Ka = 1.75 x 10-5
Total C = [acid] + [ salt] = = 0.2 mol/L
Ka . [H+]
Buffer Capacity “ B “ = X C X
( Ka + [H+] )2
(1.75 X 10-5 ) . (1.75 X 10-5 )
Buffer Capacity “ B “ = X 0.2 X
[(1.75 X 10-5 ) + (1.75 X 10-5 )]2
B = 0.115

32. Thank you