Question

A 0.400-g sample of toothpaste was boiled with a 50-mL solution containing a citrate buffer and NaCl to extract a fluoride ion. After cooling, the solution was diluted to exactly 100 mL. The potential of ISE with Ag-AgCl (sat) reference electrode in a 25.0-mL aliquote of the sample was found to be -0.1823 V. Addition of 5.0 mL of a solution containing 0.00107 mg F-/mL caused the potential to change to -0.2446 V. Calculate the percentage of F- in the sample.

Answer 1

Explanation:

It is given that 400 mg sample is prepared into 100 ml of solution.

And, from 100 ml solution 25 ml is used as aliquote and potential found is -0.1823 V.

Therefore, for the addition of ($5 \times 0.00107$ = 0.0053 mg) $F^{-}$ potential will change into -0.2446 V. Hence, potential created by 0.00535 mg $F^{-}$ is as follows.

                  (0.2446 - 0.1823)

                 = 0.0623 V

Therefore, -0.1823 V potential will be caused as follows.

                ($\frac{0.00535 \times 0.1823}{0.0623}$)

               = 0.0156 mg of $F^{-}$

Hence, 25 ml of aliquote contains 0.0156 mg of $F^{-}$. And, 100 ml of aliquote contains :

               $0.0156 \times 4 = 0.0626 mg of F^{-}$

That is, 400 mg of sample contains 0.0626 mg of $F^{-}$. Therefore, % of $F^{-}$ in the sample is calculated as follows.

                      $\frac{0.0626}{400} \times 100$

                    = 0.0156%

Thus, we can conclude that the percentage of $F^{-}$ in the sample is 0.0156%.