In the part B of the synthesis 2.4 g of 2,6-dimethylaniline free base is dissolv

Question

In the part B of the synthesis 2.4 g of 2,6-dimethylaniline free base is dissolved in 20 ml of acetic acid to react with alpha-chloroacetyl chloride. The pKa value of 2,6-dimethylanilinium is 3.9, the pKa value of carboxylic acid is 4.75. Taking into account that the density of 2,6-dimethylaniline is 0.98 g/ml, molecular weight is 121g/mol, density of acetic acid is 1.0 g/ml, and molecular weight of acetic acid is 60 g/mol calculate the concentration of 2,6-dimethylanilinium cation in mole per liter (protonated 2,6-dimethylaniline) in the solution. Round your answer to the nearest hundredth.

Not quite sure how to go about this particular question, thank you in advance!

Explanation / Answer

The reaction is-

2,6-dimethylaniline +CH3COOH(acetic acid)--->2,6-dimethylanilium ion + acetate ions(CH3COO-)

pka of 2,6-dimethylanilium<pka of acetic acid so 2,6-dimethylanilium(more acidic) is strongly deprotonated than acetic acid.

density of acetic acid=1.0g/ml

So mass of acetic acid in 20 ml=1.0g/ml*20ml=20g=20g/60g/mol=0.333 moles

moles of 2,6-dimethylaniline=2.4g/121g/mol=0.0198 moles

As the moles of acetic acid is high ,so 2,6-dimethylaniline will be the limiting reagent.

[2,6-dimethylaniline]=[2,6-dimethylanilium cation]=0.0198 moles/20ml=0.0198moles/0.02L=0.992 mol/L

[

2,6-dimethylanilinium cation=[2,6-dimethylanilinium cation]

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