Question:
A sodium salt of an unknown anion when treated with $ MgCl_2 $ gives white precipitate only on boiling. The anion is
A sodium salt of an unknown anion when treated with $ MgCl_2 $ gives white precipitate only on boiling. The anion is
Updated On: Jun 14, 2022
- $ SO_4^{2-} $
- $ HCO_3^- $
- $ CO_3^{2-} $
- $ NO_3^- $
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The Correct Option is B
Solution and Explanation
$ Mg(HCO_3)_2 $ on boiling decomposes to give white precipitate of $ MgCO_3 $ as:
$ Mg(HCO_3)_2 (aq) \, \, _{\, \, \rightarrow}^{Heat} MgCO_3 \downarrow + H_2O + CO_2 \uparrow $
$ Mg(HCO_3)_2 (aq) \, \, _{\, \, \rightarrow}^{Heat} MgCO_3 \downarrow + H_2O + CO_2 \uparrow $
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