`Na_(2)CO_(3)` cannot be used in place of `(NH_(4))_(2) CO_(3)` for the precipitation of fifth group radicals, since `:`

Options

A. Magnesium will be precipitated
B. Concentration of carbonate ions is very low
C. Sodium ions will react will acidicc radicals
D. `Na^(+)` ions interfere with the detection of fifth group

Answer

A

Explanation

`Na_(2)CO_(3)` will yeild large concentration of ions to exceed the solubility product of `MgCO_(3)` and it will be precipitated along with carbonates of `Ca^(2+), Sr^(2+)` and `Ba^(2+)` in fifth group. Further, in absence of common ion `NH_(4)OH` will ionize more to give sufficient `[OH^(-)]` to exceed the solubility product of `Mg(OH)_(2)` and its ppt.