Question

In qualitative analysis, $H{g}^{2+}$ can be separated from $C{u}^{2+}$ by the addition of $3MHN{O}_3$ because:

• A. $Cu(N{O}_3{)}_2$ is less soluble than $Hg(N{O}_3{)}_2$
• B. $Hg(N{O}_3{)}_2$ is less soluble than $Cu(N{O}_3{)}_2$
• C. $HgS$ is less soluble than $CuS$ in $3MHN{O}_3$
• D. $CuS$ is less soluble than $HgS$ in $3MHN{O}_3$

Answer 1

The correct option is C $HgS$ is less soluble than $CuS$ in $3MHN{O}_3$

By addition of $3M$ $HN{O}_3$, $H{g}^{2+}$ can be departed from $C{u}^{2+}$. This means that $H{g}^{2+}$ can be precipitated from the solution by addition of $3M$ $HN{O}_3$. Hence, the solubility product of $HgS$ and ultimately it's solubility is less than $CuS$ in $3M$ $HN{O}_3$.

Hence, $H{g}^{2+}$ can be departed from $C{u}^{2+}$ because of its less solubility in $3M$ $HN{O}_3$.