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 < pKa < 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 Ka

equation

– Ka = [H+][A-]/[HA] = [H+][A-]o/[HA]o

(always valid) (valid for traditional buffer)

– But log version more common

– pH = pKa + 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 (HONH3+Cl-) needs

to be added to 500 mL of 0.020 M HONH2 to obtain a buffer solution with a pH of 6.20? The pKa for HONH3

+ is 5.96.

– What is the pH of a solution made from mixing 400 mL of 0.018 M CH3CO2H (pKa = 4.75) with 100 mL of 0.024 M NaOH?

Acid – Base Equilibria

• Last Example Problem:– How many mL of 0.0500 M NaOH should

be added to 50.0 mL of 0.00850 M methyl ammonium chloride (CH3NH3

+Cl-) in order to make a buffer with a pH of 10.80? The pKa for CH3NH3

+ is 10.645.

Acid – Base Equilibria

• Additional Questions:– Which of the following will result in a

traditional buffer?• 0.00100 M HNO3 + 0.00200 M NaNO3

• 0.010 M NH4Cl + 0.003 M NH3

• 1.0 x 10-5 M CH3CO2H + 0.010 M NaCH3CO2

• 0.0020 M HCl + 0.010 M CH3CO2H

• 0.0020 M HCl + 0.010 M NaCH3CO2