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

  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 pH Buffer Formation

  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 pH 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 OH - from solution  No pH change Example 1: (cont.)

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

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

  Compare Ka and Kb from acid-conjugate base pair  Ka > Kb, (generally > 1x10 -7 ) acidic buffer  Ka 1x10 -7 ) basic buffer How do we tell acidic vs. basic buffers?

 1)Method applying “Common Ion Effect” 2)Henderson-Hasselbalch equation Calculating the pH of buffers

  Easier method  pH = pKa + log[A - ]/[HA]  pOH = pKb + log[HB + ]/[B]  [B] = molarity of weak base  [HB + ] = molarity of conjugate acid  Assumption: weak acids and conjugate bases do NOT change concentration with equilibrium. Henderson-Hasselbalch Equation

  Find the pH of a buffered solution created by mixing 0.15mol NH 4 NO 3 with 0.65L of a 0.25M NH 3 solution. Assume that the volume change is negligible. (Kb = 1.8x10 -5 ) Example 1:

  If 0.02 moles of HCl were added to 1.0L of the buffered solution from example 1, what would be the new pH? Assume that the volume does not change. Example 2:

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

  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 pH range where a buffer is most effective?  One pH unit ± pH = pKa  Example 3: NH 3 /NH 4 + buffer pKa = 9.26 for NH 4 +  Buffer pH range = Buffer Details (cont.)

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

  What concentration of acetate ion in 0.500M CH 3 COOH produces a buffer solution with pH= 5.00 Example 4:

  1) Enzymes  Active only at an optimal pH range  Reactions using enzymes rely on the maintenance of a certain pH range to function  2) Fluids within the body  Very narrow pH ranges (blood pH 7.36—7.42)  3 buffer systems  Example: bicarbonate(HCO 3 - )/carbonic acid(H 2 CO 3 ) buffering system maintains blood pH Buffer Biological Application

  pp. 671 #81-83  p. 651 Read “Buffers in Blood” Homework