Buffers Buffers Weak acidconjugate base mixtures OR weak

Buffers

Buffers Weak acid/conjugate base mixtures OR weak base/conjugate acid mixtures “buffers” or reduces the affect of a change in the p. H of a solution Absorbs slight changes in p. H resulting from the addition of small acid/base amounts to water.

Buffer Formation Mix a weak acid and its conjugate base Huge amounts of both weak acid and weak base in solution Example 1: HOAc + H 2 O(l) H 3 O+ + OAc – [HOAc] and [OAc] ions must be greater than amount of acid/base added to maintain p. H

Example 1: (cont. ) HOAc + H 2 O(l) H 3 O+ + OAc – What happens if an acid is added? ? ? Reacts with OAc ion [HOAc] increases slight, [OAc] decreases slightly, ratio mostly the same No p. H change

Example 1: (cont. ) HOAc + H 2 O(l) H 3 O+ + OAc – What happens if a base is added? ? ? Reacts with HOAc ion More OAc ion formed, removes excess OHfrom solution No p. H change

Buffer Capacity How much strong acid/base can be added to a buffer solution without changing the p. H drastically.

Types of buffers 1) Acidic Buffers Formed from mixing a weak acid and its conjugate base p. H < 7 Ex. HOAc and OAc – 2) Basic Buffers Formed from mixing a weak base and its conjugate acid p. H > 7 Ex. NH 3 and NH 4+

How do we tell acidic vs. basic buffers? Compare Ka and Kb from acid-conjugate base pair Ka > Kb, (generally > 1 x 10 -7) ---- acidic buffer Ka < Kb, (generally > 1 x 10 -7) ---- basic buffer

Calculating the p. H of buffers 1) Method applying “Common Ion Effect” 2) Henderson-Hasselbalch equation

Henderson-Hasselbalch Equation Easier method p. H = p. Ka + log[A-]/[HA] p. OH = p. Kb + log[HB+]/[B] = molarity of weak base [HB+] = molarity of conjugate acid Assumption: weak acids and conjugate bases do NOT change concentration with equilibrium.

Example 1: Find the p. H of a buffered solution created by mixing 0. 15 mol NH 4 NO 3 with 0. 65 L of a 0. 25 M NH 3 solution. Assume that the volume change is negligible. (Kb = 1. 8 x 10 -5)

Example 2: If 0. 02 moles of HCl were added to 1. 0 L of the buffered solution from example 1, what would be the new p. H? Assume that the volume does not change.

Buffer Details p. H = p. Ka when conjugate base and acid concentrations equal. Formation of buffers from VERY weak acids and their salts (conjugate bases)------high p. H value Formation of buffers from strong weak acids and their salts (conjugate bases)------low p. H value

Buffer Details (cont. ) What makes the best buffer? Acid and conjugate base have ~equal concentrations More acid/base can be added to buffers with more concentrated components. What is the p. H range where a buffer is most effective? One p. H unit ± p. H = p. Ka Example 3: NH 3/NH 4+ buffer p. Ka = 9. 26 for NH 4+ Buffer p. H range = 8. 26 --10. 26

Buffer Preparation Use the Henderson-Hasselbalch equation Must determine the concentration of acid and conjugate base to add in order to create buffer with a certain p. H

Example 3: What concentration of acetate ion in 0. 500 M CH 3 COOH produces a buffer solution with p. H= 5. 00

Buffer Biological Application 1) Enzymes Active only at an optimal p. H range Reactions using enzymes rely on the maintenance of a certain p. H range to function 2) Fluids within the body Very narrow p. H ranges (blood p. H 7. 36— 7. 42) 3 buffer systems Example: bicarbonate(HCO 3 -)/carbonic acid(H 2 CO 3) buffering system maintains blood p. H