BDL 11/22/2013
Algebra for Modelling Titration Curves
Or: Where the @#$^! Does  come from?
Interested in how we get from charge balance to phi (fraction of titration, ) for the
titration of a diprotic weak acid with a strong base? Remember, phi tells us where we are
relative to the nth equivalence point in a titration. For example,  = 1 means we are at the 1st
equivalence point, = 1.5 means we have passed the 1st equivalence point and are midway to
the second equivalence point, if there is one.
For a long time, I just accepted that the phi function was reasonable, but now I want to
know how it is derived. The Harris textbook starts the derivation, but then skips a lot of algebra
and jump to the result, something I’ve become uncomfortable with! Since the algebra to
determine phi may not be obvious, here it is! I show the derivation for the titration of a weak
diprotic acid with a strong base, NaOH. The derivation for other acid/base combinations is
similar. Note that terms highlighted in yellow will be cancelled in the subsequent step.
In this system we have the following reactions and equilibria to consider:
H2A + 2 NaOH → Na2A + 2H2O
H2A ⇌ HA- + H+
HA-⇌ A2- + H+
H2O ⇌ H++ OH-
Begin with charge balance
H   Na   HA   2A   OH 



2

At any time in the titration, the total acid concentration, FA and the sodium concentration [Na+]
are:
FA 
 
Ca Va
Cb Vb
and Na  
Va  Vb
Va  Vb
Where Ca and Cb are the intital concentration of the weak acid and NaOH, Va is the volume of
acid solution used and Vb is the volume of titrant solution added. Inserting into the charge
balance:
H   Na   


HA 

FA  2 A 2 FA  OH

and
H   VC VV

b
a
b
b
 C V 
 C V 
 HA   a a   2 A 2  a a   OH
 Va  Vb 
 Va  Vb 


Multiplying both sides by (Va+Vb) and simplifying:
Va  Vb  H  




 C V 
 C V 
Cb Vb 
  Va  Vb  HA   a a   2 A 2  a a   OH 


Va  Vb 
 Va  Vb 
 Va  Vb 






 C V 
 C V 
 Cb Vb 
  Va  Vb   HA   a a    Va  Vb  2 A 2  a a    Va  Vb  OH




 Va  Vb  
 Va  Vb  
 Va  Vb 


Va  Vb H   Va  Vb 

H V  H V


a
b




 Cb Vb   HA  Ca Va   2 A 2 Ca Va   OH Va  OH Vb

BDL 11/22/2013
Collecting “a” terms on one side and “b” terms on the other:
 




 




 
Cb Vb  H Vb  OH Vb   HA  Ca Va   2 A 2 Ca Va   OH Va  H Va
Dividing both sides by CaVa:
 
Cb Vb H Vb OH Vb  HA  Ca Va  2 A 2 Ca Va  OH Va H Va






Ca Va Ca Va
Ca Va
Ca Va
Ca Va
Ca Va
Ca Va
Simplifying:
 



  
Cb Vb H Vb OH Vb
OH
H


  HA   2 A 2 

Ca Va Ca Va
Ca Va
Ca
Ca
Now we have something that has (CbVb/CaVa) in it, but we still have Vb appearing in other terms
and we need to somehow get rid of it in all terms except for (CbVb/CaVa).
I struggled with this for a while, but it turns out that we need to factor (CbVb/CaVa) out of the left
side (This part isn’t very obvious, but we need to get to the correct form for phi). Factoring the
left side and simplifying:
Cb Vb
Ca Va
 



  

H Vb OH Vb 
OH
H
1 
   HA   2 A 2 


Cb Vb
Cb Vb 
Ca
Ca

Cb Vb
Ca Va
  

H
OH
1 

Cb
Cb

  


HA
 2 A 2 

OH   H 

Ca
Rearranging:
Cb Vb

Ca Va
 HA   2 A 2 
OH   H 

Ca

H
OH
1 

Cb
Cb

  



Ca


Which can be rewritten to match the format in the text:
Cb Vb

Ca Va
 HA   2 A 2 
H   OH 


Ca

H  OH 
1 

Cb


  




Ca