Question

When $H_{2}S$ gas is passed through the $HCl$ containing aqueous solution of $CuC{l}_2,HgC{l}_2$, $BiC{l}_3$ and $CoC{l}_2$, it does not precipitate out

• A. $CuS$
• B. $HgS$
• C. $B{i}_2{S}_3$
• D. $CoS$

Answer 1

Answer: (D)

When $H_{2}S$ gas is passed through the $HCl$ containing aqueous solution of $CuC{l}_2,HgC{l}_{2,}BiC{l}_3$ and $CoC{l}_2$ then
$CuC{l}_2+H_{2}S⟶\underset{\text{black ppt}}{\mathrm{CuS}}\downarrow +2HCl$
$HgC{l}_2+H_{2}S⟶HgS\downarrow +HCl$
$2BiC{l}_3+3H_{2}S⟶B{i}_2{S}_3+6HCl$
These are the alkaline radical of II A group. These radical precipitate in the presence of dil. $HCl$ and $H_{2}S$. Co is the alkaline radical of IV group which is precipitated in the presence of $H_{2}S$ gas and $N{H}_{4}OH$. Its reason that the solubility product of sulphide of II nd group is lesser than the solubility product of sulphide of IV group. So, $CoS$ will not precipitate.