`Na_(2)CO_(3)` cannot be used in place of `(NH_(4))_(2)CO_(3)` for precipitation of group V cations because

Options

A. `Na^(+)` interferes in the detection of group V
B. concentration of `CO_(3)^(2-)` is very low
C. `Na^(+)` will react with acid radicals
D. `Mg_(2+)` will precipitate as `MgCO_(3)` il present in the mixture

Explanation

`Na_(2)CO_(3)` is highly ionizable electrolyte which produces very high concentration of `CO_(3)^(2-)` ions. As a result ionic product of `MgCO_(3)` may increase its K, and it may get precipitated along with the radicals of group V.
Hence, (D) is the correct answer.