1. Meaning of smallest $pK_b$

For a base, $pK_b = -\log K_b$ So, smaller $pK_b$ means larger $K_b$, i.e. stronger base.

Hence, we need to identify the strongest base in aqueous solution among the given amines.

2. Given amines

• A: $(CH_3)_3N$ = trimethylamine (tertiary amine)
• B: $C_6H_5NH_2$ = aniline
• C: $(CH_3)_2NH$ = dimethylamine (secondary amine)
• D: $CH_3NH_2$ = methylamine (primary amine)

3. Basic strength in aqueous solution

In aqueous solution, basic strength depends on:

1. $+I$ effect of alkyl groups, which increases electron density on nitrogen and increases basicity.
2. Solvation of the conjugate acid formed after protonation.

For aliphatic amines in aqueous solution, the usual order is: $2^\circ > 1^\circ > 3^\circ > NH_3$ This happens because secondary amines have strong electron-donating effect and still get solvated well enough. Tertiary amines, though electron-rich, are less solvated due to steric hindrance.

So among the methyl-substituted amines: $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N$

4. Compare with aniline

Aniline, $C_6H_5NH_2$, is much less basic because the lone pair on nitrogen is delocalized into the benzene ring by resonance: $C_6H_5NH_2 \leftrightarrow C_6H_5NH_2^+\text{-like resonance forms}$ Thus, the lone pair is less available for protonation.

So, $C_6H_5NH_2$ is weaker than the aliphatic amines.

5. Overall order of basic strength

Therefore, in aqueous solution: $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > C_6H_5NH_2$

So the strongest base is: $(CH_3)_2NH$

Hence it has the smallest $pK_b$.

6. Option check

• A: $(CH_3)_3N$ — not the strongest in aqueous solution
• B: $C_6H_5NH_2$ — weakest due to resonance
• C: $(CH_3)_2NH$ — strongest, hence smallest $pK_b$
• D: $CH_3NH_2$ — less basic than secondary amine

7. Final answer

The amine with the smallest $pK_b$ is: $\boxed{(CH_3)_2NH}$ which corresponds to Option C.