Question

1)Ideally, what should the slope of the F- ISE’s calibration curve be? What should this value be for an H+ ISE (i.e., a pH electrode)? For a S2- ISE?

2)What is the composition of the F- ISE’s active surface? Why is it important to control the pH of the solution for the analysis with the F- ISE?

Answer 1

Ans: The electrode potential (E) is related to the concentration of the measured ion(M) by the Nernst equation :

E = E⁰ + 2.303(RT/nF) log[M]

= E⁰ + (0.059/n) log[M]

n = ion charge (including the sign)

Thus for an F- ISE , the slope of the calibration curve (E vs log [M]) will be negative (theoretically -0.059/1 or -0.059)

for an H-ISE, the slope of the calibration curve (E vs log [M]) will be positive(theoretically 0.059/1 or 0.059).

Similarly for an S2- ISE, the slope of the calibration curve (E vs log [M]) will be negative (theoretically -0.059/2 or -0.0295)

In a F- ion electrode, a LaF₃ crystal is doped with EuF₂ so that there are anion vacancies for the migration of F through the LaF₃ crystal. The composition of the active surface may be written as LaF3, F^-.

It is important to control the pH of the solution for the analysis with the F- ISE because at high pH interference occurs by the OH- ions and at low pH F- is converted to HF to which the electrode is sensitive.