Question

Sodium Carbonate cannot be used in place of $(NH_4)_2CO_3$ for the identification of $Ca^{2+},\, Ba^{2+}$ and $Sr^{2+}$ ions (in group V) during mixture analysis because :

• A. $Mg^{2+}$ ions will also be precipitated.
• B. Concentration of $CO_3^{2-}$ ions is very low.
• C. Sodium ions will react with add radicals.
• D. Na+ions will interfere with the detection of $Ca^{2+},\, Ba^{2+},\, Si^{2+}$ ions.

Answer 1

The Correct Option is A

To understand why Sodium Carbonate (\(Na_2CO_3\)) cannot be used in place of Ammonium Carbonate (\((NH_4)_2CO_3\)) for the identification of \(Ca^{2+},\, Ba^{2+}\), and \(Sr^{2+}\) ions in qualitative analysis, we need to consider the reactions involved in the precipitation process in group V cations.

The qualitative analysis involves the following: 

1. Ammonium Carbonate is used to precipitate Group V cations such as \(Ca^{2+}\), \(Ba^{2+}\), and \(Sr^{2+}\) by forming their respective carbonates:

\(Ca^{2+} + CO_3^{2-} \rightarrow CaCO_3 \, (\text{solid})\\ Ba^{2+} + CO_3^{2-} \rightarrow BaCO_3 \, (\text{solid})\\ Sr^{2+} + CO_3^{2-} \rightarrow SrCO_3 \, (\text{solid})\)

1. Using \(Na_2CO_3\) will not only precipitate \(Ca^{2+}\), \(Ba^{2+}\), and \(Sr^{2+}\) ions, but it will also precipitate \(Mg^{2+}\) ion:

\(Mg^{2+} + CO_3^{2-} \rightarrow MgCO_3 \, (\text{solid})\)

1. This unwanted precipitation of \(MgCO_3\) occurs because \(CO_3^{2-}\) ions from \(Na_2CO_3\) can also react with \(Mg^{2+}\) ions which are generally not meant to be identified during the Group V analysis.
2. Ammonium Carbonate works effectively without bothering Groups IV and V cations because ammonia (from dissociation) suppresses the precipitation of \(Mg^{2+}\) ions by the common ion effect, while ensuring effective precipitation of the Group V ions.

Therefore, \(Na_2CO_3\) is unsuitable since \(Mg^{2+}\) ions will also be precipitated, leading to confusion in identification. Hence, the correct answer is:

\(Mg^{2+}\) ions will also be precipitated.