Estimation of Aniline

Aim: Estimation of Aniline compound by brominating method.

Required: 0.1 N Na₂S₂O₃.5H₂O solution, 0.1 N Brominating solution, 10% KI solution, con. HCl

Procedure: Take 25 ml of given Aniline solution and dilute it up to 250 mL using distilled water in 250 mL conical flask take 10 mL of diluted solution in a stoppered bottle add 25 mL distilled water and 5 mL con. HCl then add 25 mL of brominating solution shake and allow to stand in dark for 15 minutes then add 10 mL KI solution and again keep it for 5 minutes. Then titrate the liberated I₂ with 0.1 N Na₂S₂O₃ solution using starch as indicator. Also perform a blank reading without aniline calculate the amount of aniline.

Result:

1. Amount of aniline in 250 ml Solution = ____________
2. Amount of Aniline in 1000 mL solution = ___________

Reactions:

Calculation:

Indicator: Starch

Color change: Blue to colorless

Blank Reading (V1): _____mL

Sample Reading (V2): ____mL

V1 – V2 = ____mL

3Br₂ = 3I₂ = 6Na₂S₂O₃ = 93.0 g Aniline

6000 mL 1 N Na₂S₂O₃ = 93.0 g Aniline

1 mL 0.1 N Na₂S₂O₃ = 0.00155 g Aniline

V1-V2 mL 0.1 N Na₂S₂O₃ = (?) g Aniline

V1-V2 mL × 0.00155 = ________g Aniline (in diluted solution)

In 250 mL diluted solution = ______× 25

                                                =______g Aniline

In 1000 mL solution = ______× 40

                                    =_______g/lit Aniline