# Titrations Help File

### Calculations

Here is an example of a titration curve for a difunctional acid (H2A)
titrated against KOH.

From the lines from positions A,B,C and D, the concentration of the acid
and its pK values can be determined.

### Determination of Concentration

The reactions which occur during the titration are:

H2A -> HA- + H+
HA- -> A2- + H+

Point B, represents the position of the end-point of the reaction
between H2A and HA-, that is, when one mole of protons has been removed
from H2A by reaction with one mole of base and the point where HA- now
fully exists.

If you used the default concentration of KOH (0.100 mol dm-3) then the
number of moles of base is:

0.100 x B / 1000
= B x 10-4

If we let the concentration of the acid be represented by X and given
that the volume of acid added to the beaker is 16 cm3, then the number
of moles of acid is given by:

X x 16 / 1000

Since at point B the number of moles of base is equal to the number of
moles of acid, these two expressions are equivalent and by reading off
the volume at B from the curve it is possible to calculate X.

For example, if B=16 cm3, then X=0.100 mol dm-3.

Point D represents the position of the end-point of the reaction between
HA- and A2-. This must be at twice the volume at B, since both protons
have now been removed and each would require the same amount of base to
neutralise them.

### Determination of pK values

The calculations needed to determine the pK values can best be seen by
writing the equilibrium expressions for the reactions above in terms of
pK and pH.

For the first step:

Ka = [H+].[HA-]/ [H2A]
or [H2A]/[HA-] = [H+]/Ka
hence log10 [H2A]/[HA-] = pKa - pH

When the concentrations of H2A and HA- are equal, then the left hand
side of this expression becomes zero (log 1=0.0) and at that point the
pKa1 is equal to the pH.

From the titration curve, this is represented by point A where the
volume is half that of B. At this point, half of the original H2A is
converted to HA- and there is a 50:50 mixture of both species present in
the solution.

Similarly the point C, which is 1.5 times the volume at point B and 3
times the volume at A, can be used to determine pKa2 since at this point
there is a 50:50 mixture of the species HA- and A2- and so again the pK
can be read off the pH axis.

In the example given, pK1 is =1.6 and pK2 =3.9.

Further information on pH and
logarithms is available.

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May-96, rjl