Which of the following have the lowest value of \(pK_b\) in their aqueous solution?

\((CH_3)_2NH\)

The compound with the lowest value of \(pK_b\) in its aqueous solution is 1. \((CH_3)_2NH\).

The \(pK_b\) of a compound is a measure of its basicity in aqueous solution. A lower \(pK_b\) value indicates a stronger base. The basicity of a compound depends on the ability of its lone pair of electrons to accept a proton from water.

In the case of the four compounds listed, \(C_6H_5NH_2\) (aniline) is the least basic because the nitrogen atom has a partial positive charge due to the inductive effect of the phenyl group. This partial positive charge makes it less likely for the nitrogen atom to attract a proton from water. \(NH_3\) is more basic than \(C_6H_5NH_2\) because the nitrogen atom in \(NH_3\) does not have any inductive effects that would make it less likely to attract a proton from water.

\((CH_3)_2NH\) is more basic than \(NH_3\) because the methyl groups have a slight inductive effect that makes the nitrogen atom more likely to attract a proton from water.

\((CH_3)_3N\) is the most basic of the four compounds because the three methyl groups have a significant inductive effect that makes the nitrogen atom very likely to attract a proton from water.

Therefore, the compound with the lowest value of \(pK_b\) in its aqueous solution is \((CH_3)_2NH\).