Acid base equilibria Physical and inorganic chemistry 1

Acid – base equilibria Physical and inorganic chemistry 1 st Year Clinical pharmacy

Acids and bases in aqueous solution: p. H n The concentration of H 3 O+ ions in solution is usually denoted by a p. H value n p. H = - log [H 3 O+] n n If the p. H is known, then the hydrogen ion concentration can be found from this

p. H calculation for strong acid n Calculate the p. H of hydrochloric acid solution of concentration 5. 0 x 10 -2 M

p. H calculation for weak acid Formic acid has a pka of 3. 75. What is the p. H of an aqueous solution of concentration 5. 0 x 10 -3 M p. Ka = - log Ka = 3. 75 ka = 1. 8 x 10 -4 [H 3 O+] [HCOO-] Ka = ------------ = 1. 8 x 10 -4 [HCOOH] But [H 3 O+] = [HCOO-] [H 3 O+] 2 ------- = ka [HCOOH] , : , [H 3 O+]2 = ka x [HCOOH]

(continued) [H 3 O+] = √ ka x [HCOOH] = √ 1, 8 x 10 -4 x 5 x 10 -3 = 9. 5 x 10 -4 p. H = - log [H 3 O+] = - log 9. 5 x 10 -4 = 3. 02 (3. 07 more accurate) n n A general equation for calculation of the p. H of weak acid solution p. H = ½ pka + ½ p. A where p. A = - log acid concentration

Some examples of aqueous acids and bases covering the p. H range (0 – 14).

p. H calculation of strong base n Strong base is completely ionized in aqueous solution. For example 0. 1 M of 0. 1 M Na. OH will ionized to give 0. 1 M Na+ and OH-; Na. OH Na+ + OH 0. 1 M p. OH = – log [OH-] = – log 10 -1 = 1 p. H = pkw – p. OH = 14 – 1 = 13

p. H calculation for weak base Methyl amine (CH 3 NH 2) is a wek base pkb of 3. 34. What is the p. H of an aqueous solution of concentration 0. 1 M CH 3 NH 2 + H 2 O CH 3 NH 3+ + OHp. Kb = - log Kb = 3. 34 kb = 4. 6 x 10 -4 [CH 3 NH 3+] [OH-] Ka = ------------ = 4. 6 x 10 -4 [CH 3 NH 2] [OH-] 2 ------- = 4. 6 x 10 -4 [CH 3 NH 2] , : , But [OH-] = [CH 3 NH 3+] [OH-]2 = kb x [CH 3 NH 2]

continue [OH-] = √ kb x [CH 3 NH 2] = √ 4, 6 x 10 -4 x 0. 1 = 6. 8 x 10 -3 M p. OH = – log [OH-] = – log 6. 8 x 10 -3 = 2. 18 x (2. 17 more accurate) n A general equation for calculation of the p. H of weak base solution (B): p. OH = ½ pkb + ½ p. B p. H = pkw – p. OH p. H = pkw – (½ pkb + ½ p. B) p. H = pkw – ½ pkb – ½ p. B

Buffer solutions n n Buffer solution is the solution which resist the change in p. H upon the addition of small amount of acid or base. A buffer solution usually consists of an aqueous solution of a weak acid and a salt of that acid (e. g. acetic acid and sodium acetate) or of a weak base and its salt. To provide the maximum buffering capacity, the relative concentrations of the weak acid and its salt (or the weak base and its salt) must be 1 : 1. Buffers are extremely important in living organisms where a constant p. H is essential; human blood has a p. H of 7. 4 and is naturally buffered

Example: (acetate buffer) n n Consider a solution that is made up of aqueous acetic acid and sodium acetate. CH 3 COOH is partially dissociated while its sodium salt is fully dissociated.

n n The combination of these two solutions produces a solution with a buffering effect. Because CH 3 COOH is a weak acid, most of it is in the undissociated form. The amount of CH 3 COO- from the acid is negligible with respect to that originating from sodium acetate. If a small amount of acid H 3 O+, is added, it will be consumed by CH 3 COO- and form more CH 3 COOH in the back reaction Any OH- added to the solution will be neutralized by H 3 O+, and equilibrium will shift to the righthand side.

Buffer action A buffer solution is able to react with either acid (H 3 O+) or base (OH-) ions, whichever is added. n Thus a buffer solution resists changes in p. H upon the addition of a modest amount of strong acid or base. How ? n A buffer solution contains a conjugate acid-base pair both the acid and the base in a reasonable concentration. n The acidic component reacts with added strong bases, n While the basic component reacts with added strong acids. n

Mechanism of buffer action 1) Acidic buffer solution a) upon addition of strong acid H+ + CH 3 COO- CH 3 COOH b) upon addition of strong base OH- + CH 3 COOH CH 3 COO- + H 2 O n The strong acid is converted to a weakly dissociated acid, n while in (b) the added OH- is converted to undissociated H 2 O, n thus no change in the p. H occurs.

Mechanism of buffer action 2) Alkaline buffer solution (ammonia buffer) a) upon addition of strong acid : H+ + NH 4 OH NH 4+ + H 2 O b) upon addition of strong base: OH- + NH 4+ NH 4 OH n Here the added acid is converted to H 2 O, n while in (b) the added OH- is converted to a weakly dissociated ammonia, n thus no change in the p. H occurs.

Calculation of the p. H of buffer solution n For acid type buffer: [A-] p. Ka = p. H – log -----[HA] HA H+ + A[A-] [H+] Ka = ------[HA] [A-] p. H = p. Ka + log ------[HA] n n This final equation is called “Henderson. Hasselbalch equation”. It should be noted that p. H of the buffer solution is n n dependent only on the ratio between the molar concentration of the acid and the salt, but not the absolute concentration of the acid and the base.

Henderson–Hasselbalch equation n for acidic type buffer n [salt] p. H = pka + log -----[acid] for alkaline type buffer will be [salt] p. H = pkw - p. Kb – log -----[base]

General equation This is the general Henderson–Hasselbalch equation and can be applied to solutions consisting of a weak acid and its salt, or a weak base and its salt.

Example -1

Example - 2

Example Calculate the p. H of the solution when 0. 20 ml of HNO 3 (0. 50 M is added to : a) 100 ml of water. b) a buffer solution consisting of 50 ml 0. 045 M aqueous CH 3 COOH (p. Ka = 4. 77) and 50 ml 0. 045 M Na[CH 3 COO]. n The total volume before addition is 100 ml, and the addition of the HNO 3 causes little increase in volume; let us assume that the total volume remains 100 ml. n

a) The case of water At first we should calculate the final concentration of the acid after dilution. n Concentration of nitric acid after dilution with 100 ml water will be n n [H+] = = = 0. 001 p. H = -log [H+] = - log 0. 001 = 3. 00 n If the nitric acid had simply been added to 100 ml of water, the p. H would have been 3. 00. n

b) If nitric acid is added to buffer n n The initial p. H of the buffer solution (before addition of extra acid) is found from the Henderson–Hasselbalch equation, and because the concentrations of acid and base are equal: Since the concentration of both acetic

Now consider the HNO 3. It is a strong acid, so fully dissociated in aqueous solution. The H 3 O from 0. 20 ml of a 0. 50 M solution of HNO 3 reacts completely with CH 3 COO- in the buffer solution: n HNO 3(aq) + CH 3 COO- (aq) CH 3 COOH(aq) + NO 3 -(aq) n n Therefore, some CH 3 COO- is consumed from the buffer solution, and an equal amount of CH 3 COOH is produced. -

n Concentration of CH 3 COOH and sodium acetate after addition of nitric acid will be [CH 3 COOH] = 0. 0235 [CH 3 COO-] =0. 0215 p. H = p. Ka + log 4. 77 + log 0. 915 = 4. 77 + log =

Conclusion Adding the HNO 3 to the buffer solution will result in only a small change in p. H (4. 77 to 4. 73) on n While if the nitric acid had simply been added to 100 ml of water, the p. H would have been 3. 00. (change from p. H 7. 00 to p. H 3. 00) n

p. H of strong acid / base solution Calculate the p. H of 0. 1 M hydrochloric acid. Hydrochoric acid is a strong acid which is completely ionized in water HCl + H 2 O H 3 O+ + Cl 0. 1 M p. H = - log [H 3 O+] = - log [0. 1] = - log [10 -1] = 1 Calculate the p. H of a solution which HNO 3 concentration is 0. 05 M.