Question

Traces of aniline can be determined by reaction with an excess of electrolytically generated Br₂:

${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2+3{\mathrm{Br}}_2\to {\mathrm{C}}_6{\mathrm{H}}_2{\mathrm{Br}}_3{\mathrm{NH}}_2+3{\mathrm{H}}^{+}+3{\mathrm{Br}}^{-}$

The polarity of the working electrode is then reversed, and the excess bromine is determined by a coulometric titration involving the generation of Cu(I):

${\mathrm{Br}}_2+2{\mathrm{Cu}}^{+}\to 2{\mathrm{Br}}^{-}+2{\mathrm{Cu}}^{2+}$

Suitable quantities of KBr and copper(II) sulfate were added to a 25.0-mL sample containing aniline. Calculate the mass in micrograms of C₆H₅NH₂ in the sample from the accompanying data:

Short answer

The mass of C₆H₅NH₂ in micrograms is $542\mathrm{\mu g}$.

Step-by-step solution

Given Information

To calculate the mass of C₆H₅NH₂ in micrograms.

Explanation

The reaction which takes place is:

${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2+3{\mathrm{Br}}_2\to {\mathrm{C}}_6{\mathrm{H}}_2{\mathrm{Br}}_3{\mathrm{NH}}_2+3{\mathrm{H}}^{+}+3{\mathrm{Br}}^{-}$........ (I)

As from reaction (I) 1 molC₆H₅NH₂ is equal to 3 mol of Br₂ and 6 mol of Br^-.

From reaction (I), each Br₂ required 2e^-, therefore the 3 mol of Br₂ required 6 mole e^-.

The expression of number of moles of C₆H₅NH₂ is:

${\mathrm{n}}_{{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2=}\frac{\mathrm{I}({\mathrm{t}}_{\mathrm{anode}}-{\mathrm{t}}_{\mathrm{cathode}})}{\mathrm{nF}}\left(1\mathrm{mol}\mathrm{of}{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2\right)$......... (II)

Here, the electric current is I, the time taken in anode is t_anode and time taken in cathode is t_cathode, the number of moles is nand the Faraday constant is F.

The expression of mass of C₆H₅NH₂ is:

${\mathrm{m}}_{{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2}={\mathrm{n}}_{{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2}\times {\mathrm{m}}_{\mathrm{molar}\mathrm{of}{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2}$........ (III)

Here, the molar mass of C₆H₅NH₂ is .

Substitute 1.61 mA for I 3.76 min for t_anode, 0.27 min for t_cathode , 6 mole e^- for n and 96485 C/mole e^- for F in Equation (II).

$$\begin{gathered} {\mathrm{n}}_{{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2}=\frac{(1.61\mathrm{mA})(3.76\min -0.27\min )(1\mathrm{mol}\mathrm{of}{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2)}{(6\mathrm{mole}{\mathrm{e}}^{-})(96485\mathrm{C}/\mathrm{mole}{\mathrm{e}}^{-})} \\ =\frac{(1.61\mathrm{mA})\left({\displaystyle \frac{1\mathrm{A}}{1000\mathrm{mA}}}\right)\left(3.49\min \right)\left({\displaystyle \frac{60\mathrm{s}}{1\min }}\right)(1\mathrm{mol}\mathrm{of}{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2)}{(578910\mathrm{C})\left({\displaystyle \frac{1\mathrm{As}}{1\mathrm{C}}}\right)} \\ =\frac{0.337134\mathrm{mol}\mathrm{of}{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2}{\left(578910\right)} \\ =5.824\times {10}^{-7}\mathrm{mol}\mathrm{of}{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2 \end{gathered}$$

Substitute 5.824 x 10^-7 mol of C₆H₅NH₂ for nC₆H₅NH₂ and 93.13 g/mol of C₆H₅NH₂ for ${\mathrm{m}}_{\mathrm{molar}\mathrm{of}{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2}$in Equation (III).

$$\begin{gathered} {\mathrm{m}}_{{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2}=\left(5.824\times {10}^{-7}\mathrm{mol}\mathrm{of}{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2\right)\times \left(9.13\mathrm{g}/\mathrm{mol}\mathrm{of}{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2\right) \\ =\left(5.42\times {10}^{-7}\mathrm{g}\right)\left(\frac{1\mathrm{\mu g}}{{10}^{-5}\mathrm{g}}\right) \\ =542\mathrm{\mu g} \end{gathered}$$