Sulphanilic acid is soluble in dil. NaOH but not in dil. HCl. Explain.

Explanation

Sulphanilic acid exists as a zwitter ion, `NH_(3)^(+)` `SO_(3)^(-)`. In the presence of dil. NaOH, the weakly acidic `NH_(3)^(+)` group transfers its `H^(+)` to `OH^(-)` to form a soluble p- `NH_(2)C_(6)H_(4)SO_(3)^(-)Na^(+)`. On the other hand, `SO_(3)^(-)` group is very weak base and therefore, does not accept a proton from dil. HCl to form p- `NH_(3)^(+)C_(6)H_(4)SO_(3)H`. Hence, it does not dissolve in dil. HCI.