Question

[Ion selective electrode] If some of your F^- was bound to Ca²⁺, how would your measured voltage signal change?

Background Info (might be useful):

For this lab, you will determine the fluoride content in commercial mouthwashes using an ion selective electrode. This lab uses the direct potentiometry method, which assumes that matrix interferences are not a problem for determining the concentration of fluoride in toothpaste. To combat possible matrix problems, the total ionic strength adjustment buffer (TISAB) solution used in the experiment contains 1,2-cyclohexylenedinitrilotetraacetic acid (CDTA), a complexing agent. The presence of the strong CDTA ligand ensures that cations of metals such as iron and copper in the sample will not complex the fluoride ion. The TISAB will also adjust and stabilize the ionic strength of the measured solutions. Ionic strength can be a large factor affecting accuracy in ion selective electrode (ISE) measurements.

Answer 1

The relation between measured potential with fluoride activity is described by Nernst Equation.

E = E^o - RT ln a_F-

E is the measured potential

E^o is the standard potential

R is gas constant

T = temperature

a_F- is the activity of fluoride ion.

According to this equation, as the fluoride activity will change, the measured potential will also change. So, it is actually dissociated F- ions. We should be sure that the F- ions are not in complex forms or undissociated forms, otherwise it will affect the measured potential of the cell.

If Ca2+ is bound with F-, it will form CaF2, which means that F- ions are not free to carry current in the cell. The fluoride ion activity will decrease and thus measured potential signal will also change.