
Which radicals are precipitated in \[{(N{H_{4}})_2}C{O_3}\] in presence of alkali?
(A) $C{a^{ + 2}}, B{a^{ + 2}}, S{r^{ + 2}}$
(B) Mg
(C) Both
(D) None
Answer
161.1k+ views
Hint: \[{(N{H_{4}})_2}C{O_3}\] in presence of a base having common ion is used to precipitate group $5$ cations in qualitative inorganic analysis of cations. The base used is \[N{H_{4}}OH\] in presence of \[N{H_{4}}Cl\].
Complete Step by Step Solution:
the precipitation can be understood with help of following$O{H^ - }$ reaction:
\[N{H_{4}}Cl \to N{H_4}^ + + C{l^ - }\]
\[N{H_{4}}OH\overset {} \leftrightarrows N{H_4}^ + + O{H^ - }\]
\[{(N{H_{4}})_2}C{O_3}\overset {} \leftrightarrows 2N{H_4}^ + + C{O_3}^{2 - }\]
The concentration of \[C{O_3}^{2 - }\] decreases and concentration of $N{H_4}^ + $ increases.
The concentration of \[C{O_3}^{2 - }\]is sufficient enough to cause precipitation of carbonates of$C{a^{ + 2}},B{a^{ + 2}},S{r^{ + 2}}$ but the ionic product of magnesium carbonate or magnesium hydroxide is not sufficient enough to Enough to exceed the solubility product and hence magnesium does not get precipitated in presence of ammonium ions and remain soluble in the solution.
If sodium carbonate is used instead of ammonium carbonate, then due to being highly ionised electrolyte the concentration of carbonate is so high that it is sufficient to precipitate magnesium in group $5$itself. Thus, sodium carbonate is not used.
Sodium hydrogen carbonate or ammonium hydrogen carbonate is not used because of being water soluble.
A large amount of ammonium salts is avoided because it will enhance the solubilities of carbonate salts in ammonium salts.
Note: Group $5$ cations are tested in the order $B{a^{ + 2}}$ followed by $S{r^{ + 2}}$ and $C{a^{ + 2}}$. Tests for $S{r^{ + 2}}$ and $C{a^{ + 2}}$ are given by $B{a^{ + 2}}$. Similarly, tests for $C{a^{ + 2}}$ are given by $S{r^{ + 2}}$. Therefore, before confirming $S{r^{ + 2}}$ it is important to show the absence of $B{a^{ + 2}}$ or eliminate it. The value of solubility product is very high for strontium and calcium hence they are not precipitated on adding potassium chromate. Only barium is precipitated in the form of barium chromate.
Complete Step by Step Solution:
the precipitation can be understood with help of following$O{H^ - }$ reaction:
\[N{H_{4}}Cl \to N{H_4}^ + + C{l^ - }\]
\[N{H_{4}}OH\overset {} \leftrightarrows N{H_4}^ + + O{H^ - }\]
\[{(N{H_{4}})_2}C{O_3}\overset {} \leftrightarrows 2N{H_4}^ + + C{O_3}^{2 - }\]
The concentration of \[C{O_3}^{2 - }\] decreases and concentration of $N{H_4}^ + $ increases.
The concentration of \[C{O_3}^{2 - }\]is sufficient enough to cause precipitation of carbonates of$C{a^{ + 2}},B{a^{ + 2}},S{r^{ + 2}}$ but the ionic product of magnesium carbonate or magnesium hydroxide is not sufficient enough to Enough to exceed the solubility product and hence magnesium does not get precipitated in presence of ammonium ions and remain soluble in the solution.
If sodium carbonate is used instead of ammonium carbonate, then due to being highly ionised electrolyte the concentration of carbonate is so high that it is sufficient to precipitate magnesium in group $5$itself. Thus, sodium carbonate is not used.
Sodium hydrogen carbonate or ammonium hydrogen carbonate is not used because of being water soluble.
A large amount of ammonium salts is avoided because it will enhance the solubilities of carbonate salts in ammonium salts.
Note: Group $5$ cations are tested in the order $B{a^{ + 2}}$ followed by $S{r^{ + 2}}$ and $C{a^{ + 2}}$. Tests for $S{r^{ + 2}}$ and $C{a^{ + 2}}$ are given by $B{a^{ + 2}}$. Similarly, tests for $C{a^{ + 2}}$ are given by $S{r^{ + 2}}$. Therefore, before confirming $S{r^{ + 2}}$ it is important to show the absence of $B{a^{ + 2}}$ or eliminate it. The value of solubility product is very high for strontium and calcium hence they are not precipitated on adding potassium chromate. Only barium is precipitated in the form of barium chromate.
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