Determination of titration endpoint by Hahn's method
Hahn's method is one of the best methods used to determine titration endpoint (EP). It is characterized by
simplicity and high reproducibility of the results. The method can only be applied to experimental data if
titration reagent (titrant) was dosed in portions of equal volumes (at least in the neighbourhood of EP).
To determine EP by Hahn's method, one needs to find the biggest increase of the measured signal (SEM, pH,
etc.) ΔEmax and the next two largest increases in the measured signal; ΔE1 and ΔE2. ΔEmax lies between ΔE1
and ΔE2 and ΔEmax >ΔE1>ΔE2.
Based on these values a correction factor a is calculated using the following formula:
a=Δ V⋅qa
where:
Δ V - volume of titrant portion
qa
- symmetry factor calculated using following formula:
q a=
Δ E2
2⋅Δ E 1
Hahn's method will provide accurate results if two conditions are fulfilled: q a⩾0,25
and
Q=
Δ E max
⩾2,5 , while the symmetry factor q a is of greater importance.
Δ E1
The first step in determination of EP is saving experimental data in the right form. Table 1 contains data
obtained during potentiometric determination of chlorides, in this particular example 0.6 cm 3 of the titrant
has been added to the sample and then the sample was titrated using 0.1 cm 3 portions of the titrant. The first
column of the table contains total volume of titrant added, in the second column measured signal (sliver
electrode potential) corresponding to the given volume of titrant has been recorded. The third column
contains potential increases caused by addition of each portion of titrant. Please note that the values in this
column are written between rows of first and second column. Data from Table 1 is also shown in Figure 1.
Table 1 Potentiometric titration of chlorides (ΔE1 ahead of ΔEmax)
Prepared by PhD Andrzej Wasik, based on; A. Cygański, Podstawy metod elektroanalitycznych, WNT, Warszawa, 1999.
The second step is to calculate symmetry factor qa:
Δ E2
21
=
=0.175
2⋅Δ E 1 2⋅60
400
not satisfy the basic condition to obtain an accurate
340
result by Hahn's method. If the volume of dispensed
320
portion of titrant is relatively small (as in this case) it is
often sufficient to use every second point of the
experimental data and calculate symmetry factor again
300
280
260
(see Table 2). Note that you can do it in two ways,
240
starting from 0.6 or from 0.7 cm3.
220
Table 2 Potentiometric titration of chlorides (ΔE1 after ΔEmax)
ΔEmax
360
EAg [mV]
Unfortunately, the value of qa is only 0.175 so it does
ΔE2
380
ΔE1
q a=
200
180
160
0,6 0,7 0,8 0,9
1 1,1 1,2 1,3 1,4
VAgNO3 [ml]
Figure 1 Potentiometric titration curve (ΔE1 ahead of ΔEmax).
Lets calculate qa:
q a=
Δ E2
26.5
=
=0.442
2⋅Δ E 1 2⋅30
and Q:
Q=
Δ E max 155
=
= 5.2
Δ E1
30
Now qa value fulfils the condition: q a⩾0,25 and Q value fulfils the condition: Q ≥ 2.5 so we can calculate
correction factor a:
Δ E2
26.5
a=Δ V⋅qa =0.2⋅
=0.2⋅
=0.0884
2⋅Δ E 1
2⋅30
Note that now ΔV equals 0.2 cm3.
The third step is to find the volume of the titrant (V) lying between ΔE1 and ΔEmax. In our example (Table 2)
V = 1.0 cm3.
The last step is to calculate volume of the titrant corresponding to titration EP (V EP). To do so we need to
correct volume V using correction factor a. We need to add correction factor a to volume V if ΔE1 is ahead of
ΔEmax (as in Table 1) or subtract a from V if ΔE1 follows ΔEmax (as in Table 2):
V EP = V −a = 1.0−0.0884 = 0.9116 cm
3
Prepared by PhD Andrzej Wasik, based on; A. Cygański, Podstawy metod elektroanalitycznych, WNT, Warszawa, 1999.