Question

A 0.400-g sample of toothpaste was boiled with a 50-mL solution containing a citrate buffer and NaCl to extract the fluoride ion. After cooling, the solution was diluted to exactly 100 mL. The potential of an ISE with a Ag/AgCl(sat’d) reference electrode in a 25.0-mL aliquot of the sample was found to be 20.1823 V. Addition of 5.0 mL of a solution containing 0.00107 mg F^-/mL caused the potential to change to 20.2446 V. Calculate the mass percentage of F^- in the sample.

Short answer

The mass percentage of F^- in the sample is $4.28\times {10}^{-4}\%$.

Step-by-step solution

Given information

Mass of toothpaste sample$=0.400\mathrm{g}$

Volume of solution$=50\mathrm{mL}$

Volume after dilution on cooling$=100\mathrm{mL}$

The potential of an ISE with a Ag/AgCl(sat’d) reference electroderole="math" localid="1649662232297" $=0.1823\mathrm{V}$

The potential change on adding 5.0 mL of $0.00107\mathrm{mg}$data-custom-editor="chemistry" ${\mathrm{F}}^{-}/\mathrm{mL}=-0.2446\mathrm{V}$

Ecell for the reaction

$$\begin{gathered} E_{cell}=K+\frac{0.0592}{1}{\mathrm{pF}}^{-} \\ {\mathrm{E}}_{\mathrm{cell}}=\mathrm{K}+\frac{0.0592}{1}\log \left[{\mathrm{F}}^{-}\right] \end{gathered}$$

If $C_x$is the concentration of diluted sample,

$-0.1823=K-0.0592\log C_x..........\left(1\right)$

Potential after addition of standard 

If the concentration of added standard is $C_s$and concentration of F- after addition of standard is $C_1$:

$$\begin{gathered} C_1=\frac{25.0C_x+5.00C_s}{30.0} \\ =0.833C_x+0.167C_s \end{gathered}$$

The potential after addition of standard:

$-0.2446=K-0.0592\log \left(0.833C_x+0.167C_s\right).........\left(2\right)$

Concentration of Cx

Subtracting equation (2) from equation (1) as follows:

$$\begin{gathered} 0.0623=-0.0592\log \frac{C_x}{\left(0.833C_x+0.167C_s\right)} \\ -1.0524=\log \frac{C_x}{\left(0.833C_x+0.167C_s\right)} \\ \frac{C_x}{\left(0.833C_x+0.167C_s\right)}=0.0886 \\ \frac{C_x}{\left(0.833C_x+0.167\times 0.00107\right)}=0.0886 \\ {C}_x=0.0738C_x+1.583\times {10}^{-5} \\ {C}_x=1.71\times {10}^{-5}\mathrm{mg}{\mathrm{F}}^{-}/\mathrm{mL} \end{gathered}$$

Mass Percentage of F- 

The mass percentage is as follows:

$$\begin{gathered} \%{\mathrm{F}}^{-}=\frac{1.71\times {10}^{-5}\mathrm{mg}{\mathrm{F}}^{-}/\mathrm{mL}\times {10}^{-3}\mathrm{g}/\mathrm{mg}}{0.400\mathrm{g}}\times 100\% \\ =4.28\times {10}^{-4}\% \end{gathered}$$