CHM 2046C    Sample Module 12   Name: _________________

 

Part C:  Derivation of Henderson-Hasselbalch Equations     10 points

 

1.     Given the ionization of weak acid, HA in water, derive the Henderson-Hasselbalch equation for a weak acid.

 

Start with writing the ionization reaction of the Weak Acid and setup the equilibrium constant expression for Ka:

 

HA + H2O ß à  H3O1+  +  A1-

 

           [A-1]1 [H3O+1]1

Ka = ---------------------  

                [HA]1

 

Through algebraic manipulation of the equilibrium constant

 expression:

 

 

                                        [HA]

[H3O+1] =   Ka x  ----------- 

                                        [A-1]

 

Now take the negative log of each side of the equation

 

                                                                 [HA]

- log [H3O+1] =  - log (K a x  --------- )

                                                                 [A-1]

 

 

 

 

The log The log of a product of two quantities is the sum of two logs or the difference when applying the negative log:

 

 

 

 

                                                                             [HA]

- log [H3O+1] =  - log (K a ) –  log  (--------)

                                                                             [A1-]

 

The - log [H3O+1] The - log [H3O+1] is pH, while - log (K a ) is pKa and you may change the –Log of ratio of the weak acid concentration to the concentration of its conjugate base to its inverse:

 

                                           [A1-]

pH =  pK a  +  log  (--------)

                                          [HA]

 

 

 

 

 

 

which is The Henderson-Hasselbalch equation

for a weak acid:

 

 

                              [conjugate base]

pH =   pKa + log (-------------------------)

                                  [weak acid]

 

 

 

2.     Given the ionization of base acid, BOH in water, derive the Henderson-Hasselbalch equation for a weak base.

 

Start with writing the ionization reaction of the weak Base and setup the equilibrium constant expression for Kb:

 

BOH  +  HOH  ß à B+1   +   OH-1

 

           [B1+]1 [OH1-]1

Kb = --------------------

                [BOH]1

 

 

 

 

 

 

 

 

Through algebraic manipulation of the equilibrium constant expression:

 

                                     [BOH]

[OH1-] =   Kb x  ----------- 

                                     [B1+]

 

Now take the negative log of each side of the equation

 

                                                           [BOH]

- log [OH1-]  =  - log (K b x  --------- )

                                                            [B1+]

 

The log of a The log of a product of two quantities is the sum of two logs or the difference when applying the negative log:

 

                                                                         [BOH]

- log [OH1-]  =  - log (K b ) –  log  (--------- )

                                                                           [B1+]

 

The - log [OH1The - log [OH1-] is pOH, while - log (K b ) is the pKb and you may change the –Log of ratio of the weak base concentration to the concentration of its conjugate acid to its inverse:

 

                                               [B1+]

pOH =  pK b  +  log  (---------)         

                                             [BOH]

 

 

 

 

 

 

 

                               [conjugate acid]

pOH =  pKb + log (------------------------)

                                  [weak base]

 

This is the Henderson-Hasselbalch Equation