Conductivity of CH\(_3\)COOH decreases on dilution.
The conductivity of acetic acid (CH₃COOH) decreases on dilution because it is a weak electrolyte that ionizes only partially in solution. This behavior can be understood by looking at the properties of weak electrolytes.
Acetic acid is a weak electrolyte, meaning that it does not fully ionize in solution. The ionization of acetic acid in water is represented as: \[ \text{CH}_3\text{COOH} (aq) \rightleftharpoons \text{CH}_3\text{COO}^- (aq) + \text{H}^+ (aq) \] This reaction is reversible, and only a small fraction of acetic acid molecules dissociate into ions.
When acetic acid is diluted with water, the concentration of both dissociated ions (\( \text{CH}_3\text{COO}^- \) and \( \text{H}^+ \)) decreases. Since conductivity (\( \kappa \)) depends on the number of ions in the solution, the conductivity decreases as the solution becomes more dilute.
In the case of weak electrolytes like acetic acid, dilution shifts the equilibrium of the ionization reaction to the right, meaning more acetic acid dissociates. However, despite the increase in ionization, the overall number of ions still decreases with dilution, which results in a net decrease in conductivity.
As the solution is diluted, the number of dissociated ions decreases, reducing the electrical conductivity. This is because the conductivity is directly proportional to the concentration of ions in solution. Therefore, even though the ionization of acetic acid might increase slightly upon dilution, the overall ion concentration is lower, leading to reduced conductivity.
In summary, dilution of acetic acid decreases the conductivity because it is a weak electrolyte that ionizes partially. As the concentration of ions decreases, the overall conductivity of the solution decreases as well.
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(i) (CH3 )2CHNH2 (ii) CH3 (CH2 )2NH2 (iii) CH3NHCH(CH3 )2
(iv) (CH3 )3CNH2 (v) C6H5NHCH3 (vi) (CH3CH2 )2NCH3 (vii) m–BrC6H4NH2
Give one chemical test to distinguish between the following pairs of compounds.
(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
(iv) Aniline and benzylamine
(v) Aniline and N-methylaniline
Account for the following:
(i) pKb of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
(v) Aniline does not undergo Friedel-Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines. (vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.