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%.

Answer 2

We have that for the Question "Calculate the percentage of F- in the sample."

It can be said that

• The percentage of $F^-$ in the sample = $0.0156\%$

From the question we are told

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

For addition of

$5 * 0.00107 = (0.00535mg) F^-$ potential changed to $-0.2446v$

So,$0.00535mg F^-$ causes

$0.2446 - 0.1823 = 0.0623v$ of potential

Therefore, $-0.1823v$ potential caused by

$\frac{0.00535*0.1823}{0.0623}\\\\=0.01565mg of F^-$

so,

25ml aliquote contains $0.0156mg of F^-$

100ml aliquote contains $0.01565*4$

$=0.0626mg of F^-$

that is,

400mg of sample contains 0.0626mg of F^-

therefore $\% of F^-$ in the sample

$= \frac{0.0626}{400}*100\\\\= 0.0156\%$

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