Correct Answer - C
`CO_(2)` si only slightly soluble in water. Consequently, the concentration of `H_(2)CO_(3)` is very low. Moreover, since `H_(2)CO_(3)` is a very weak acid, the concentration of `HCO_(3)^(-)` in the reaction mixture is extremely low:
`CO_(2)+H_(2)OhArrH_(2)CO_(3)`
`H_(2)CO_(3)hArrH^(+)+HCO_(3)`
`NH_(3)` combines with `H^(+)` ion forming `NH_(4)^(+)` ion. As a result the equilibrium shifts in the forward direction to produce a sufficient amoun of `HCO_(3)^(-)` ion which enables sparingly soluble `NaHCO_(3)` to precipitate out.
