Acid-Base Equilibria (Buffers ) Green & Damji Chapter 8, Section 18.2 Chang Chapter 16 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

A buffer solution is a solution of: 1.A weak acid or a weak base and 2.The salt of the weak acid or weak base Both must be present! A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. 16.3

Add strong acid: H + ions react with conj base… H + ions are therefore removed from solution, pH increases back to near its original level. H + (aq) + CH 3 COO - (aq) CH 3 COOH (aq) Add strong base: OH - (aq) reacts with undissociated acid… OH - (aq) are therefore removed from solution, pH decreases back to near its original level. OH - (aq) + CH 3 COOH (aq) CH 3 COO - (aq) + H 2 O (l) Consider an equal molar mixture of CH 3 COOH and CH 3 COONa CH 3 COOH (aq) CH 3 COO - (aq) + H + (aq)

16.3 Add strong acid: H + ions react with base… H + ions are therefore removed from solution, pH increases back to near its original level. NH 3 (aq) + H + (aq) NH 4 + (aq) OH - (aq) + NH 4 + (aq) H 2 O (l) + NH 3 (aq) Consider an equal molar mixture of NH 3 and NH 4 + (from ammonium chloride) NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) Add strong base: OH - (aq) reacts with ammonium ions… OH - (aq) are therefore removed from solution, pH decreases back to near its original level.

16.3

Which of the following are buffer systems? (a)KF/HF (b) KBr/HBr, (c) Na 2 CO 3 /NaHCO

Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na 2 CO 3 /NaHCO 3 (a) HF is a weak acid and F - is its conjugate base buffer solution (b) HBr is a strong acid not a buffer solution (c) CO 3 2- is a weak base and HCO 3 - is its conjugate acid buffer solution 16.3

Chemistry In Action: Maintaining the pH of Blood 16.3 CO 2 (g) + H 2 O (g) H + (aq) + HCO 3 - (aq) Maintaining the pH of blood (7.4) is critical because enzymes only function effectively over a limited pH range. The buffer system in action here is carbonic acid and sodium bicarbonate… H 2 CO 3 (g) H + (aq) + HCO 3 - (aq) The buffer system is typically written as:

Chemistry In Action: Maintaining the pH of Blood 16.3

To make a buffer – Method 1 Mix a soln of a weak acid (HA) and a salt of the weak acid (NaA) –Ex: ethanoic acid and sodium ethanoate Mix a soln of a weak base and a salt of the weak base –Ex: ammonia and ammonium chloride

To make a buffer – Method 2 Mix a small amount of a strong base to an excess of weak acid –Ex: sodium hydroxide with excess ethanoic acid Mix a small amount of a strong acid to an excess of weak base –Ex: hydrochloric acid to excess ammonia

Calculations involving buffers: For buffers systems involving weak acids and their conjugate base, we can use the formula for the acid dissociation constant: We can assume that the [HA] is the concentration of the acid and the [A-] is the concentration of the conjugate base… and rearrange to this form… K a = [H + ][A - ] [HA] [H + ] = K a [HA] [A - ]

Calculations involving buffers: Or this form: The pH of a buffer system depends on… K a of the weak acid ratio of conj base to acid – pH of a buffer does not change with dilution… but is less effective when buffer components decrease pH = pK a + log [A - ] [HA] Henderson-Hasselbach equation Or…. pH = pK a - log [HA - ] [A - ]

Calculations involving buffers: For buffer systems involving a weak base and its conjugate acid… we can use the formula for the base dissociation constant: We can assume that the [B] is the concentration of the base and the [BH + ] is the concentration of the conjugate base… and rearrange to this form… K b = [BH + ][OH - ] [B] [OH - ] = K b [B] [BH + ]

Calculations involving buffers: Or this form: pOH = pK b + log [BH + ] [B] Henderson-Hasselbach equation Or…. pH = pK a - log [B] [BH + ]

Effective buffers Concentration of the weak acid & its salt or weak base & its salt must be >> than the strong acid/base added Conc of weak acid = conc of conj base pH = pKa Effective buffer range of any weak acid/base is in the range of pKa + 1

Solid sodium ethanoate is added to M ethanoic acid until it reaches a concentration of M. Calculate the pH of the buffer solution formed. [Ka for ethanoic acid is 1.74 x 10-5 M.] Assume no volume change on dissolving the solid. Try this….

Solid sodium ethanoate is added to M ethanoic acid until it reaches a concentration of M. Calculate the pH of the buffer solution formed. [Ka for ethanoic acid is 1.74 x M.] Assume no volume change on dissolving the solid. Method 1: [H + ] = 1.74 x x (0.200/0.0500) = 6.95 x M pH = - log [H + ] = -log (6.95 x M) = 4.16 [H + ] = K a [HA] [A - ]

Solid sodium ethanoate is added to M ethanoic acid until it reaches a concentration of M. Calculate the pH of the buffer solution formed. [Ka for ethanoic acid is 1.74 x M.] Assume no volume change on dissolving the solid. Method 2: pH = log (1.74 x ) – log (0.200/0.0500) = 4.76 – 0.6 = 4.16 pH = pK a - log [HA] [A - ]

Calculate the pH of the 0.30 M NH 3 /0.36 M NH 4 Cl buffer system. NH 4 + (aq) H + (aq) + NH 3 (aq) pH = pK a + log [NH 3 ] [NH 4 + ] pK a = 9.25 (you’d have to look this up) pH of the buffer = log [0.30] [0.36] =

Textwork: 18.2 Read Section 18.2 on pp Do Ex 18.2 # 1-6 Due: _______________________