Question

Question: The measurement of ${\mathbf{F}}^{\mathbf{-}}$ion concentration by ion-selective electrodes at 25.00⁰C obeys the equation

${\mathbf{E}}_{\mathbf{meas}}\mathbf{=}{\mathbf{E}}_{\mathbf{ref}}\mathbf{-}\frac{\mathbf{0}\mathbf{.}\mathbf{0591}}{\mathbf{n}}\mathbf{log}\left[F^{-}\right]$

a. For a given solution,is 0.4462 V. If is 0.2420 V, what is the concentration of${\mathbf{F}}^{\mathbf{-}}$in the solution?

b. Hydroxide ion interferes with the measurement of. Therefore, the response of a fluoride electrode is

${\mathbf{E}}_{\mathbf{meas}}\mathbf{=}{\mathbf{E}}_{\mathbf{ref}}\mathbf{-}\frac{\mathbf{0}\mathbf{.}\mathbf{0591}}{\mathbf{n}}\mathbf{log}\left(\left[F^{-}\right]+k\left[{\mathrm{OH}}^{-}\right]\right)$
where $\mathit{k}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{1}}$and is called the selectivity factor for the electrode response. Calculate for the data in part a if the pH is 9.00. What is the percent error introduced in the ${\mathbf{F}}^{\mathbf{-}}$if the hydroxide interference is ignored?
c. For the ${\mathbf{F}}^{\mathbf{-}}$in part b, what is the maximum pH such that $\raisebox{1ex}{$\left[F^{-}\right]$}\!\left/ \!\raisebox{-1ex}{$\mathbf{k}\left[{\mathrm{OH}}^{-}\right]$}\right.\mathbf{=}\mathbf{50}$.?
d. At low pH,${\mathbf{F}}^{\mathbf{-}}$is mostly converted to HF. The fluoride electrode does not respond to HF. What is the minimum pH at which 99% of the fluoride is present as ${\mathbf{F}}^{\mathbf{-}}$and only 1% is present as HF?

e. Buffering agents are added to solutions containing fluoride before making measurements with a fluoride-selective electrode. Why?

Short answer

1. The concentration of $\left[{\mathrm{F}}^{-}\right]$is $3.534x{10}^{-4}\mathrm{M}$.
2. The concentration of $\left[{\mathrm{F}}^{-}\right]$in this case is $2.534x{10}^{-4}\mathrm{M}$. The percentage error will be 40%
3. The maximum pH value will be $7.70$
4. The minimum pH value will be $5.14$
5. A buffer of $\mathrm{pH}=6.0$should be used.

Step-by-step solution

Determine concentration of F-  in the solution

Given Data

$$\begin{gathered} {\mathrm{E}}_{\mathrm{meas}}={\mathrm{E}}_{\mathrm{ref}}-\frac{0.0591}{\mathrm{n}}\log \left[{\mathrm{F}}^{-}\right] \\ {\mathrm{E}}_{\mathrm{meas}}=0.4462\mathrm{V} \\ {\mathrm{E}}_{\mathrm{ref}}=0.2420\mathrm{V} \end{gathered}$$

Plugging the respective data in the given equation the following can be obtained

$\begin{array}{rcl}0.4662& =& 0.2420-\frac{0.0591}{1}\log \left[{\mathrm{F}}^{-}\right]\\ \log \left[{\mathrm{F}}^{-}\right]& =& \frac{0.2420-0.4662}{0.0591}\\ \log \left[{\mathrm{F}}^{-}\right]& =& -3.4517\\ \left[{\mathrm{F}}^{-}\right]& =& 3.534\times {10}^{-4}\\ & & \end{array}$

Determine the percentage error

Given data

$$\begin{gathered} {\mathrm{E}}_{\mathrm{meas}}={\mathrm{E}}_{\mathrm{ref}}-\frac{0.0591}{\mathrm{n}}\log \left[{\mathrm{F}}^{-}\right] \\ {\mathrm{E}}_{\mathrm{meas}}=0.4462\mathrm{V} \\ {\mathrm{E}}_{\mathrm{ref}}=0.2420\mathrm{V} \\ \mathrm{k}=1.00\times {10}^1 \\ \mathrm{pH}=9.00 \end{gathered}$$

Therefore,$\mathrm{pOH}=5.00,\left[{\mathrm{OH}}^{-}\right]=1\times {10}^{-5}\hspace{0.17em}\mathrm{M}$

Plugging the respective data in the given equation the following can be obtained

$\begin{array}{rcl}0.4662& =& 0.2420-0.0591\log \left(\left[{\mathrm{F}}^{-}\right]+10.0\times \left(1\times {10}^{-5}\right)\mathrm{M}\right)\\ \log \left(\left[{\mathrm{F}}^{-}\right]+\left(1\times {10}^{-4}\right)\mathrm{M}\right)& =& \frac{0.2420-0.4662}{0.0591}\\ \log \left(\left[{\mathrm{F}}^{-}\right]+\left(1\times {10}^{-4}\right)\mathrm{M}\right)& =& -3.452\\ \left[{\mathrm{F}}^{-}\right]+\left(1\times {10}^{-4}\right)\mathrm{M}& =& 3.534\times {10}^{-4}\mathrm{M}\\ \left[{\mathrm{F}}^{-}\right]& =& 2.534\times {10}^{-4}\mathrm{M}\\ & & \\ & & \end{array}$

True value for $\left[{\mathrm{F}}^{-}\right]$is $2.534\mathrm{x}{10}^{-4}\mathrm{M}$. Ignoring $\left[{\mathrm{OH}}^{-}\right]$the value $\left[{\mathrm{F}}^{-}\right]$of will be $3.534\mathrm{x}{10}^{-4}\mathrm{M}$.

Therefore, percentage error $=\frac{3.534\mathrm{x}{10}^{-4}\mathrm{M}-2.534\mathrm{x}{10}^{-4}\mathrm{M}}{2.534\mathrm{x}{10}^{-4}\mathrm{M}}\times 100\%=40\%$

Determine maximum pH

From part (b) it was obtained that the true concentration foris

Given,

$\frac{\left[\mathrm{F}\right]}{\mathrm{k}\left[\mathrm{OH}\right]}=50$

Plugging respective values in the above equation the following can be obtained

$$\begin{gathered} \frac{2.534\mathrm{x}{10}^{-4}\mathrm{M}}{1\times {10}^1\left[{\mathrm{OH}}^{-}\right]}=50 \\ \left[{\mathrm{OH}}^{-}\right]=\frac{2.534\mathrm{x}{10}^{-4}\mathrm{M}}{1\times {10}^1\times 50} \\ \left[{\mathrm{OH}}^{-}\right]=5.0\times {10}^{-7} \\ \mathrm{pOH}=6.30 \end{gathered}$$

Therefore,$\mathrm{pH}=7.70$

Determine minimum pH

$$\begin{gathered} \mathrm{HF}\stackrel{}{\rightleftharpoons }{\mathrm{H}}^{+}+{\mathrm{OH}}^{-} \\ {\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{F}}^{-}\right]}{\left[\mathrm{HF}\right]}=7.2\times {10}^{-4} \end{gathered}$$

If 99% is $\left[{\mathrm{F}}^{-}\right]$and 1% is $\left[\mathrm{HF}\right]$then

$\frac{\left[{\mathrm{F}}^{-}\right]}{\left[\mathrm{HF}\right]}=99$

Given,

$$\begin{gathered} \frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{F}}^{-}\right]}{\left[\mathrm{HF}\right]}=7.2\times {10}^{-4} \\ \left[{\mathrm{H}}^{+}\right]\times 99=7.2\times {10}^{-4} \\ \left[{\mathrm{H}}^{+}\right]=\frac{7.2\times {10}^{-4}}{99} \\ \left[{\mathrm{H}}^{+}\right]=7.3\times {10}^{-6} \\ \mathrm{pH}=5.14 \end{gathered}$$

Explanation regarding part e

Buffer solutions are used to control the pH of a reaction. As a result, there is little HF present and there is little response to $\left[{\mathrm{OH}}^{-}\right]$. A buffer of $\mathrm{pH}=6.0$ should be used.