Correct Answer - C
An aromatic `-NH_(2)` group `(K_(a)=10^(-5))` and hence cannot appreciably neutralize the strongly acidic `-SO_(3)H` group as in sulphanilic acid In other words aliphatic `NH_(2)` is sufficiently basic to accept an `H^(+)` from `COOH` The `COOH` is not strong enough to donate `H^(+)` to the weakly basic `ArNH_(2)` but `SO_(3)H` is a sufficiently strong acid to do so .
