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(A) When a sodium salt containing the bicarbonate ion $(HCO_3^-)$ is added to $MgCl_2$ solution,no precipitate is formed at room temperature because m

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$HCO_3^-$

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(A) When a sodium salt containing the bicarbonate ion $(HCO_3^-)$ is added to $MgCl_2$ solution,no precipitate is formed at room temperature because magnesium bicarbonate is soluble in water. Upon heating,the bicarbonate ion decomposes to form carbonate ions $(CO_3^{2-})$: $2HCO_3^- \xrightarrow{\Delta} CO_3^{2-} + H_2O + CO_2$ The resulting carbonate ions react with $Mg^{2+}$ ions to form a white precipitate of magnesium carbonate $(MgCO_3)$: $Mg^{2+} + CO_3^{2-} \rightarrow MgCO_3 \downarrow$ Therefore,the anion is $HCO_3^-$.

$A$ sodium salt gives a white precipitate with $MgCl_2$ only upon heating. What is the anion of the sodium salt?

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