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.

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 ( = 0.0053 mg) potential will change into -0.2446 V. Hence, potential created by 0.00535 mg is as follows.

                  (0.2446 - 0.1823)

                 = 0.0623 V

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

                ()

               = 0.0156 mg of

Hence, 25 ml of aliquote contains 0.0156 mg of . And, 100 ml of aliquote contains :

That is, 400 mg of sample contains 0.0626 mg of . Therefore, % of in the sample is calculated as follows.

                    = 0.0156%

Thus, we can conclude that the percentage of in the sample is 0.0156%.

2b2t

explanation:

2b2t