Answer to Question #156008 in General Chemistry for Sidney MUZYAMBA

Question

Amines form salts with picric acid and all amine picrates possess an absorption maximum at 359 nm with ɛ = 1.25 x 104

Firstly, 0.200 g of sample of aniline is dissolved in 500 mL of water. Next, a 25.0 mL aliquot is reacted with picric acid in 250 mL flask and diluted to volume. Finally, 10.0 mL aliquot of this is diluted to 100 mL and the absorbance read at 359 nm in 1 cm cuvette. If the absorbance was determined to be 0.425, compute the percentage purity of aniline.

Answer 1

The reaction between aniline and picric acid is:

C₆H₅NH₂ + C₆H₅OH → C₆H₅NH₃⁺C₆H₅O^- + H₂O

C₆H₅NH₂ =aniline

C₆H₅OH =phenol or picric acid

C₆H₅NH₃⁺C₆H₅​​​​​​​O^- =aniline picrate (it is the salt)

We preparate a solution with 0.2 g aniline in 500 mL water. We convert the grams into mmoles using the molar mass of aniline.

1 mole aniline = 93 g

We have:

$=\frac{0.2 \; g}{(93 \; g/mol} \\ = 2.15 \times10^{-3} \; moles \\ =2.15 \; mmoles \\ 1 \;mmole =10^{-3} \; moles$

This 2.15 mmoles are in 500 mL, in 25 mL we have:

$\frac{2.15 \; mmoles}{500 \; ml} \times 25 \; mL=0.1075 \; mmoles$

Then, we determine the amount of aniline reacted with picric acid. We determine the concentration of the final solution. To do so, we use the Lambert-Beer's law:

$A= Ɛ \times b \times c \\ Ɛ= molar \; absorptivity =1.25 \times 104 \; cm^{-1} \times mol^{-1} \times L$

b=optic path= 1 cm

c=concentration of the solution (in mol/L)

We determine the concentration:

$0.425 = 1.25 \times 104 \;cm^{-1} \times mol^{-1} \times L \times 1 \;cm \times c \\ c= \frac{0.425}{1.25 \times 104 \times 1} \\ c=3.4 \times 10^{-5} \; mol/L$

We convert the concentration into mmoles/L. We have:

$3.4 \times 10^{-5} \; mol/L \times 1 mmole/10^{-3} \; moles = 0.034 \; mmoles/L$

Then, we determine the amount of aniline picrate in the 100 mL. We have:

$\frac{0.034 \; mmoles}{1000 \; mL} \times 100 \; mL=0.0034 \; mmoles$

This 0.0034 mmoles came from the 10 mL aliquot, which was taken from a solution of 250 mL. Then, we have:

$\frac{0.0034 \;mmoles}{10 \; mL} \times 250 \; mL=0.085 \;mmoles$

Then, 0.085 mmoles of the salt are produced. According to the reaction's stochiometry, 1 mole of aniline reacts and produced 1 mole of the salt. Then, if 0.085 mmoles of the salt are produced, only 0.085 mmoles of the aniline in the original sample reacted.

The original sample is made up of 0.1075 mmoles of aniline. We determine the purity of the aniline sample as a fraction. We divide the mmoles of aniline reacted by the amount of aniline in the original sample. Thus, we have:

$\frac{0.085 \; mmoles}{0.1075 \; mmoles} = 0.79$

We multiply this decimal number by 100 to convert it into a percentage:

$0.79 \times 100 = 79 \%$

Answer: 79 %