Buffer 1 p H l p H log

Buffer 1

p. H l p. H = - log [ H+] or p. H = - log [ H 3 O+ ] Example I What is the p. H of solution with [ H+] = 32 X 10 -5 M/L ? p. H = - log [ H+] p. H = - log 32 X 10 -5 p. H = 3. 495 2

Example II The hydronium ion concentration of 0. 1 M solution was found to the 1. 32 X 10 -3 M, What is the p. H of the solution? p. H = - log [ H 3 O+ ] p. H = - log [ 1. 32 X 10 -3 ] p. H = 2. 88 Example III If the p. H of a solution is 4. 72, what is the hydrogen ion concentration? p. H = - log [ H+ ] = 4. 72 log [ H+ ] = - 4. 72 [ H+ ] = 1. 91 X 10 -5 take anti-log for both side 3

l p. H = 0 p. H = 1 p. H = 2 p. H = 3 l Note : the change in one p. H unit means 10 fold change in [H+]. l p. H + p. OH = 14 l l l H+ = 1 M/l = 1 X 100 = 1 H+ = 1 X 10 -1 = 0. 1 M/L H+ = 1 X 10 -2 = 0. 01 M/L H+ = 1 X 10 -3 = 0. 001 M/L 4

Introduction When a minute trace of hydrochloric acid is added to pure water, a significant increase in hydrogen-ion concentration occur immediately. l 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. l 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 p. H. l Therefore, it is said to be unbuffered. l E. g. CO 2 + H 2 O H 2 CO 3 decrease p. H from 7 to 5. 8 These change of p. H are of great concern in pharmaceutical preparation also Na. Cl solution ability to resist change of p. H. To ensure stability and solubility, we used to control p. H by using a buffer 5 l

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

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

Adjustment of p. H by the buffer l 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, p. H 3. 5 - 7 2 - strong acid, p. H 0 -3. 5 3 - weak base, p. H 7 -10 4 - strong base, p. H 10. 5 -14 l 0 3. 5 Strong acid 7 Weak acid 10. 5 Weak base 14 Strong base 8

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

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. 10

Selection of buffer system depends on: 1. p. H rang. 2. Buffer capacity desired. 3. The purpose for which it is required. 4. Compatibility with active ingredients. 11

Method of preparation of buffer 1. Buffer equation. 2. Buffer table. 12

l 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 CH 3 COOH CH 3 COONa l Weak base + its salt e. g. Ammonia + Ammonium chloride l NH 3 + NH 4 Cl 13

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: 14

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

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

l 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. l However, remember that the salt of a weak acid is a base (and the salt of a weak base is an acid). l 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-). l Here it is: l Remember this: the base is the one WITHOUT the proton and the acid is the one WITH the proton. 17

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] p. H = p. Kw – p. Kb + log -----[ Salt] 18

An alternate form of the Henderson. Hasselbalch Equation l The last discussion used p. H and p. Ka. There is a alternate form of the Henderson-Hasselbalch Equation using p. OH and p. Kb. [Salt] p. OH = p. Kb + log -----[Base] 19

Mechanism of action of buffer 1 – Acid / Its salt e. g. CH 3 COOH/CH 3 COONa. CH 3 COOH + OH- CH 3 COO- + H 2 O This means acid will react with base (OH- ) to neutralize it. CH 3 COO-/Na+ + H 3 O+ CH 3 COOH + H 2 O This means salt of weak acid will react with acid ( H 3 O ) to neutralize it. 20

Mechanism of action of buffer… cont 2 – Base / Its salt e. g. NH 3/NH 4 Cl. NH 3 + H 3 O+ NH 4+ + H 2 O This means base will react with acid to neutralize it. NH 4+ + OH- NH 3 + H 2 O This means salt of weak base will react with base (OH- ) to neutralize it. 21

p. H + p. OH = 14 For acid & its salt : [Salt] p. H = p. Ka + log -----[Acid] For base & its salt : [Salt] p. OH = p. Kb + log -----[Base] or [Base] p. H = p. Kw – p. Kb + log -----[ Salt] 22

Example I What is the molar ratio of salt/acid required to prepare an acetate buffer p. H = 5 ? Ka = 1. 8 X 10 -5 p. Ka = - log Ka p. Ka = - log 1. 8 X 10 -5 = 4. 75 p. H = p. Ka + log salt/acid 5 = 4. 75 + log salt/acid = 5 – 4. 75 = 0. 25 take anti-log for both side salt/acid = 1. 78 So, Ratio of salt/acid = 1. 78/1 Mole fraction of acid = 1 / ( 1. 78 + 1) =0. 3597 X 100 = 35. 97 % Mole fraction of salt = 1. 78 / ( 1. 78 + 1) =0. 6403 X 100 = 64. 03 % mole fraction multiply by 100 to get of mole % 23

Example II Prepare 200 ml of acetate buffer p. H = 6 with molar conc. = 0. 4 M & Ka = 1. 8 X 10 -5 ? p. Ka = - log Ka p. Ka = - log 1. 8 X 10 -5 = 4. 75 p. H = p. Ka + log salt/acid 6 = 4. 75 + log salt/acid = 6 – 4. 75 = 1. 25 salt/acid = 17. 78/1 take anti-log for both side Molecular weight of acetic acid ( CH 3 COOH ) = 60 Molecular weight of sodium acetate ( CH 3 COONa ) = 82 Weight of acid = mole fraction X Conc. X M. wt X V (L) = 1/18. 78 X 0. 4 X 60 X 0. 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 X 0. 2 = 6. 21 g 24

Example III If we add 0. 1 M sodium acetate to 0. 09 M acetic acid. What is the p. H if you know Ka = 1. 8 X 10 -5 ? p. Ka = - log Ka p. Ka = - log 1. 8 X 10 -5 = 4. 75 p. H = p. Ka + log salt/acid p. H = 4. 75 + log 0. 1/0. 09 p. H = 4. 796 25

Example IV What is the p. H of of a solution containing 0. 1 mole of ephedrine base and 0. 01 mole of ephedrine HCl / liter of solution ? p. Kb of ephedrine = 4. 64 p. H = p. Kw - p. Kb + log base/salt p. H = 14 - 4. 64 + log 0. 1/0. 01 p. H = 9. 36 + log 10 = 10. 36 Or p. OH = p. Kb + log salt/base p. OH = 4. 64 + log 0. 01/0. 1 p. OH = 4. 64 + log 0. 1 = 3. 64 p. H + p. OH = 14 p. H = 14 – p. OH p. H = 14 – 3. 64 = 10. 36 26

Change in p. H with addition of an acid or base Calculate the change in p. H of a buffer solution with the addition of a given amount of acid or base in the following example : Example Calculation the change in p. H 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 p. Ka value of acetic acid is 4. 76 at 25 OC. The p. H of the buffer solution is calculated by using the buffer equation p. H = p. Ka + log salt/acid p. H = 4. 76 + log 0. 2/0. 2 p. H = 4. 76 27

l 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 p. H = p. Ka + log ------------acid - base 28

salt + base p. H = p. Ka + log -----------acid - base 0. 2 + 0. 04 p. H = 4. 76 + log ---------0. 2 - 0. 04 p. H = 4. 76 + 0. 1761 = 4. 9361 = 4. 94 Because the p. H before addition sodium hydroxide was 4. 74, the change in p. H = 4. 94 – 4. 76 = 0. 18 unit. 29

Buffer Capacity l The ability of buffer solution to resist change in p. H upon addition of acid or base. K a. [H+] Buffer Capacity “ B “ = 2. 303 X C X --------( K a + [H+] )2 30

Example I At hydrogen ion conc. of 1. 75 X 10 -5 (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. 1 + 0. 1 = 0. 2 mol/L K a. [H+] Buffer Capacity “ B “ = 2. 303 X C X --------( K a + [H+] )2 (1. 75 X 10 -5 ). (1. 75 X 10 -5 ) Buffer Capacity “ B “ = 2. 303 X 0. 2 X --------------------[(1. 75 X 10 -5 ) + (1. 75 X 10 -5 )]2 B = 0. 115 31

Thank you 32