Titrations
Strong Acid – Strong Base: the net reaction will be H+ + OH-  H2O.
Therefore, any amount of acid added will consume a stoichiometrically
equivalent amount of base. Let’s use 50.00 mL of 0.020 M KOH with 0.100
M HBr as an example.
(1) First calculate the volume of HBr required to reach the equivalence
Volume HBr Molarity of HBr   Volume KOH Molarity of KOH 
Vol 0.100 M   50.0 mL0.020 M  and the volume is thus 10.0 mL of HBr
point:
(2) Calculating the pH prior to the equivalence point: the pH will be
determined by the excess base (OH-) in solution. For example, when
3.00 mL of acid (HBr) has been added, we can calculate the pH via the
pOH.
 vol of HBr to equivalence  vol of acid added 
 initial OH vol 

initial base conc 

[OH ]  
vol of HBr to equivalence


 Total volume 
Fraction of OH- remaining
initial conc
dilution factor
 10.00  3.00 
 50.00 
[OH  ]  
0.020M 
  0.0132M
 10.00 
 50.00  3.00 
Calculate the pOH and subtract from 14 and you get a pH of 12.12
(3) Calculating the pH at the equivalence point: exactly enough acid has
been added to consume all of the base. Now the product (H2O) will
dissociate and you use Kw to find the H+ concentration. For a strong
acid-strong base, the pH will be 7.
(4) Calculating the pH after the equivalence point: after the equivalence
point, you are adding excess acid to the solution and the excess acid
will determine the pH. For example, after adding 10.50 mL of acid:
 vol of excess acid 

[ H ]  initial conc of H  
 total volume 
Dilution factor
0.50
4


[ H ]  0.100M 
  8.26 10 M
 50.00  10.50 
and pH is therefore 3.08
Weak Acid – Strong Base: the reaction is HA + OH-  A- + H2O. For this
illustration lets assume 50.00 mL of 0.0200 M HA (pKa = 6.15) and 0.100
M OH-.
(1) Prior to adding base, the solution only has HA in solution and the
equilibrium is HA
 H+ + A2
 6.15
[ H ][ A ]
x

 10
[ HA]
0.02  x
0.02-x
x
x
X is 1.19 x 10-4
Write the Ka,
and pH is 3.93.
(2) Prior to the equivalence point: as you add OH-, you will have a
mixture of HA and A- and you can use the Henderson-Hasselbalch
equation.
For example: 3.00 mL of OH- has been added and the equivalence
volume is 10.0mL. We can setup a table of relative quantities HA +
OH-  A- + H2O
Relative initial quantities (HA = 1)
1
3/10
--
-Relative final quantities
3/10 -Now use Henderson-Hasselbalch
pH  pKa  log
7/10 -[ A ]
3 / 10
 6.15  log
 5.78
[ HA]
7 / 10
At the ½ equivalence point, pH = pKa
(3) At the equivalence point: all of the HA has been consumed and we
have A- and H2O in solution. The reaction is now: A- + H2O  HA +
OH-. Find the Kb from the Ka value.
x
x
Write the Kb expression
[OH  ][ HA]
[ A ]
F’ –x
but note, the concentration of A-
 initial vol HA 

has changed since there was a dilution, thus F’ initial HA conc 
 Total volume 
50.00


and F’ is therefore 0.020M 
  0.0167 M . Substitute this into the
 50.00  10.00 
2
8
x
 Kb  1.43 10 and [OH-] =
Kb expression and solve for [OH ].
0.0167  x
-5
1.54 x 10 M. Find pOH and then pH = 9.18.
(3) After the equivalence point: let’s use a volume of 10.10 mL of base
added. Therefore, the pH is calculated by the excess base in solution.
 volume excess OH
[OH  ]  initial conc of OH 
 Total volume
pH is now 10.22

0.10
4


  0.100M 
  1.66 10
 50.00  10.10 

Weak Base – Strong Acid: the reaction is B + H+  BH+.
(1) Before adding acid, the solution only has weak base, B, in solution
(water). Thus the reaction is B + H2O  BH+ + OH-.
F-x
x
x
Where F is the initial concentration of base. Use the Kb
expression to calculate the OH- and then the pH.
(2) Prior to equivalence point, you have a mixture of B and BH+, which
is a buffer. Therefore, you can use Henderson-Hasselbalch
[ B]
pH  pKa( for BH  )  log
[ BH ]
Remember halfway to the equivalence point, pH = pKa.
(3) At the equivalence point, all of the B has been converted to BH+
and the reaction is
F’ – x
BH+  B + H+
x
x
Use the Ka expression to solve
for H+ and remember to calculate the concentration, F’, as indicated
above.
(3) After the equivalence point, you will have excess strong acid which
will determine the pH. Calculate the H+ by taking into account the
dilution similar to step 3 of weak acid- strong base