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

Let the new concentration after addition of 0.5mL of 100mg /L of F- = Cstandard

The concentration before addition of 0.5mL of 100mg / L of F- = Csample

Therefore

C2 X Vfinal = Csample X 50mL + 100 X 0.5 mL

Vfianl = 50.5 mL

Cstandard X 50.5 = Csample X 50 + 100 X 0.5 ......(1)

Now we know that

electrode potential can be related to concentration as

Esample / Cstandard = Estandard/ C standard

Esample = Potential of sample = -0.1805 V

Csample = concentration of sample

Estandard = Potential of standard after addition of 0.5mL of 100mg/L F-

Putting values from (1)

-0.1805 / Csample = 50.5 X Estandard / Csample X 50 + 50

-0.1805 / Csample = 50.5 X (-0.3490) / Csample X 50 + 50

On solving Csample = 1.05 mg / L NaF

We know that atomic weight of F- = 19 g / mole and molecular weight of NaF = 42g / mole

So

in 42g of NaF there are19g of F-

if 1 g of F- there will be 42/ 19 g of NaF

So if 1.05mg of F-is presnet the amount of NaF = 42X 1.05 / 19 mg = 2.32 mg / L