Question

The fluoride ISE is used routinely for measuring fluoridated water and fluoride ion in dental products such as mouthwash. A 50 mL aliquot of water containing sodium fluoride is analyzed using a fluoride ion electrode and the MSAs. The pH and ionic strength are adjusted so that all fluoride ion is present as free F- ion. The potential of the ISE/reference electrode combination in a 50 mL aliquot of the water was -0.1805 V. Addition of 0.5 mL of a 100 mg/L F- ion standard solution to the beaker changed the potential to -0.3490 V. Calculate the concentration of (1) fluoride ion and (2) sodium fluoride in the water sample.

Answer 1

concentration of F- standard solution = 100 mg/19 g/mol x 1000 = 5.26 x 10^-3 M

concentration of F- in 50.5 ml standard = 5.26 x 10^-3 x 50.5/1000 = 2.66 x 10^-4 M

Standard F- solution gave potential reading of = (-0.3490 - (-0.1805) = -0.1685 V

Potential for F- in 50 ml solution = 0.1685 x 50.5/50 = 0.1702 V

Potential reading due to F- in water sample = -0.1895 V

concentration of F- ion in water sample = 2.66 x 10^-4 x 0.1895/0.1702 = 2.96 x 10^-4 M

Concentration of sodium fluoride in water sample = 2.96 x 10^-4 M x 0.05 L/41.99 g/mol = 3.52 x 10^-7 M