iterative method of calculation in excel

Laboratoire de chimie analytique
Study of an ion-selective electrode for
the determination of fluoride ions
I - Principle.
We consider two solutions (1) and (2), containing a common ion X, separated by a
selective membrane. As activities in (1) and (2) are different, it appears a membrane
potential. We cannot measure this potential alone; we have to build a complete
electrochemical chain :
electrode A
solution (1)
Electrode A Potential
solution (2)
electrode B
Membrane potential
Electrode B potential
So that the potential difference measured between the electrodes A and B depends
only on the membrane potential, it is necessary that the electrode potentials of A and
B remain constant, this means that A and B are reference electrodes.
The electrochemical chain will be formed as follows:
fixed ax
with membrane
Test solution
Commercial electrode
It goes without saying that the quality of the electrode is essentially dependent
on the quality of the membrane. In the case of the fluorine selective electrode, the
membrane consists of a single crystal of lanthanum fluoride, LaF 3.
Under these conditions, the measured potential difference is related to the activity of
X in solution by the relation:
 E = A (T) + (RT / zF) * ln ( ax) (1) with z = charge of X with its sign
Cautions to take during measures with the fluoride electrode for reducing its
response time:
- Check for bubbles in the electrode (same thing using pH electrode)
- Do not wipe the electrode (modification of the crystal surface)
- Soak 30 s between 2 measurements in a beaker of deionized water
- Start with the more dilute solutions
Laboratoire de chimie analytique
II - Study of the characteristics of the fluoride sensitive electrode.
II - 1: Slope of the electrode - Calibration.
Refer to the simplified instructions of the potentiometer on the bench to understand
the concept of stability criteria and the procedure for the calibration.
Preparation of solutions of constant ionic strength from 0.1 mol.L -1 NaF solution in
NaCl medium by serial dilutions.
Solution pF = 1
pF = 2
pF =3
pF = 4
pF =6
solution NaF 0.1 mol.L -1 NaF solution 0.1 mol L -1
5 mL of pF=1 + NaCl 0.1 mol.L -1 qsp 50.mL
5 mL of pF=2 + NaCl 0.1 mol.L -1 qsp 50.mL
5 mL of pF=3 + NaCl 0.1 mol.L -1 qsp 50.mL
5 mL of pF=4 + NaCl 0.1 mol.L -1 qsp 50.mL
5 mL of pF=5 + NaCl 0.1 mol.L -1 qsp 50.mL
Verify that these solutions are at constant ionic strength: this allows to fix the value of
activity coefficients and to have a solution sufficiently conductive to reduce response
i i where c i is the concentration
Reminder: the ionic strength is defined by I 2
of the ion i and z i its charge number.
Show that under these conditions the relation (1) can be written E =  (T) + (RT
ln(10)/ F) * pF
For each solution, starting with the more diluted, measure the potential
difference (in mV) between the ion-selective electrode and the external
Use the B method and press if necessary on pH / mV to see the potential
difference  E.
Plot the curve  = f (pF). Search the linear part of the curve. (Linear
regression) and verify that the experimental slope of the linear part is RTln
(10) / F.
We can now calibrate the instrument. Take two solutions located in the linear
part, as for instance the vials pF = 2 and pF = 5.
. The% slope must be close to 100%.
We will now check that the calibration is correct. Prepare by appropriate dilution a
solution of 5.10 -5 mol.L -1 of F - in NaCl 0.1 mol L -1. Measure the pF of this solution
and compare with the expected value of pF. We will carry out three times the
measure so far to check the repeatability of the measurement.
Do not discard solutions.
II-2: pH of the test solution
Laboratoire de chimie analytique
In acidic medium, free [F -] decreases by the formation of HF. But the electrode is
only sensitive to free F -. In basic medium, the electrode is attacked by OH -.
We will study at a constant concentration of total fluorides, the influence of pH on the
fluoride electrode response, this for pH values close to 3, 4, 5, 7 and 9.
Prepare the solutions by adding:
- 2.5 mL of a 0.01 mol L -1 fluoride solution
- 10 mL of the corresponding Britton-Robinson buffer
- 10 mL of NaCl 0.1 mol L -1
- Deionized water, qsp 25 mL.
For each solution, measure :
Its pH, please refer to the corresponding part in the simplified instructions :
proceed to the calibration of the device (use the pH electrode)
The potential difference  (fluoride selective electrode), refer to the manual.
Plot the response curve  E) = f (pH), conclusion?
III - Fluoride content in mineral water.
One could, for this measure, consider implementing a calibration method as was
seen in § II-1.
This could be realized by making all the calibration solutions in the same conditions
of ionic strength of the sample water. However it’s difficult to perform given the wide
differences in mineral content that may exist between two waters.
On the other hand, if we want to determine the whole of fluoride present, we must
take into account those who may be engaged in complexes with trivalent metals
such as Fe 3 + and Al 3 +. Finally the pH conditions should be optimal (see § II-2). To
meet these requirements, it is better to implement a method of standard additions in
a TISAB medium (Total Ionic Strength Adjustment Buffer).
Appendix: composition of TISAB: for 1 L of aqueous solution:
57 mL glacial acetic acid
58 g NaCl
4 g DCTA: diamino-1 ,2-cyclohexane tetraacetic acid
- Adjust to pH 5 with NaOH 5 mol L -1
The water to be analyzed, once degassed, will be diluted to half in the medium
WATER / TISAB volume to volume (50% vol TISAB) .
Measure the pH of the water + TISAB to verify that the pH conditions are
Laboratoire de chimie analytique
 Measure the potential
difference between the fluoride sensitive electrode and
the external reference in the water medium / TISAB volume to volume for the
following solutions in a beaker:
1) 19 ppm F - in the medium WATER / TISAB volume to volume.
2) 1.9 ppm F - in the medium WATER / TISAB volume to volume.
Determine the slope of the electrode in the medium WATER / TISAB volume to
volume (We can write E = A + slope *log [F-] ).
Prepare a 100 mL beaker containing V 0 = 50.0 mL of sample (sample water
diluted 50% vol TISAB). Measure the potential difference (denoted  E 1)
Add a volume V a = 1.0 mL of a fluoride solution concentration C a = 0.01 mol
L -1. Measure the potential difference (denoted  E 2)
We then have E  1 -  E 2 = slope * log  * (c i + c a V a) / (c i V i)] (2)
with  =(V i +V TISAB )/ (V i +V TISAB + V a ), Vi is the volume of the analyzed water,
VTISAB is the volume of the TISAB solution added.
Determine the fluoride concentration c i of the water analyzed. For this, demonstrate
the relation (2) and use the iterative method of calculation in excel presented below.
The result is expressed as ppm F -.
It is desirable to perform at least three (3) times this determination, in order to
determine the confidence interval on the result.
In a cell (for instance A1), enter any value for c i. It is in this cell that c i will be read after the
iterative calculation. In a second cell (for instance A2) enter the formula "= slope * log  * (c
i V i + Va c a) / (c i V i)]"
In the menu "tool" (« outil » ) choose application "target value" (« valeur cible » ) and fill in:
"Set cell" (« Cellule à définir » ) put the cell of the mathematical function (A2 in our example)
"Target value" (« valeur cible »): value for the desired function (eg 10 mV)
" changing cell" (« cellule à modifier » ) cell where c i is displayed (A1 in this example).
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