Acid – Base Equilibria Buffer Solutions: –Question: Was the ICE Problem set up needed? –Answer: No. The assumption of x << [HA], [A - ] is valid for all “traditional” buffers –Traditional Buffer Weak acid (3 < pK a < 11) Ratio of weak acid to conjugate base in range 0.1 to 10 mM concentration range

Acid – Base Equilibria Buffer Solutions: –Since ICE not needed, can just use K a equation –K a = [H + ][A - ]/[HA] = [H + ][A - ] o /[HA] o (always valid) (valid for traditional buffer) –But log version more common –pH = pK a + log([A - ]/[HA]) –Also known as Henderson-Hasselbalch Equation

Acid – Base Equilibria Buffer Solutions: –Ways to make buffer solution: Mix weak acid and conjugate base Add strong base to weak acid (weak acid must be in excess) – this converts some of the weak acid to its conjugate base Add strong acid to weak base (weak base must be in excess) – this converts some of weak base to its conjugate acid

Acid – Base Equilibria Example Problems: –How many moles of hydroxyl ammonium chloride (HONH 3 + Cl - ) needs to be added to 500 mL of M HONH 2 to obtain a buffer solution with a pH of 6.20? The pK a for HONH 3 + is –What is the pH of a solution made from mixing 400 mL of M CH 3 CO 2 H (pK a = 4.75) with 100 mL of M NaOH?

Acid – Base Equilibria Last Example Problem: –How many mL of M NaOH should be added to 50.0 mL of M methyl ammonium chloride (CH 3 NH 3 + Cl - ) in order to make a buffer with a pH of 10.80? The pK a for CH 3 NH 3 + is

Acid – Base Equilibria Additional Questions: –Which of the following will result in a traditional buffer? M HNO M NaNO M NH 4 Cl M NH x M CH 3 CO 2 H M NaCH 3 CO M HCl M CH 3 CO 2 H M HCl M NaCH 3 CO 2