Question

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

Hi, this problem can be solved using single point calibration method. We will assume that the ratio of signal to concentration is a constant for both the readings. The equation that we can use here is:

Here x is for the sample and s for standard.

the term is the new concentration after the addition of 0.5 mL standard solution. It is:

which is

this value can be put in the above equation to get

now we can put all the values given in the question to get:

solving this equation will give = 1.05 mg/L fluoride in the original water sample. Now we know that the atomic weight of fluorine is 19 while the formula weight of sodium fuoride is 42, so the amount of sodium fluoride in water is .

I hope it helps.