Question

Sodium Carbonate cannot be used in place of $(N{H}_4{)}_{2}C{O}_3$ for the identification of $C{a}^{2+},B{a}^{2+}$ and $S{r}^{2+}$ ions (in group V) during mixture analysis because :

• A. $M{g}^{2+}$ ions will also be precipitated.
• B. Concentration of $C{O}_3^{2-}$ ions is very low.
• C. Sodium ions will react with add radicals.
• D. Na+ions will interfere with the detection of $C{a}^{2+},B{a}^{2+},S{i}^{2+}$ ions.

Answer 1

Answer: (A)

If $N{a}_{2}C{O}_3$ is used in place of $(N{H}_4{)}_{2}C{O}_3$. It will precipitate group $V$ radicals as well as magnesium radicals. The reason for this is the high ionization of $N{a}_{2}C{O}_3$ in water into $N{a}^{+}$ and $C{O}_3^{--}$. Now the higher concentration of $C{O}_3^{--}$ is available which exceeds the solubility product of group $V$ radicals as well as that of magnesium radicals.