Question

$ZnS{O}_4$ and $Pb(N{O}_3{)}_2$ have similar powdery white appearance. So, which of the given options help us to identify $Z{n}^{2+}$ and $P{b}^{2+}$ when both are treated with $NaOH$ and $N{H}_{4}OH$?

• A. Formed precipitate $Zn(OH{)}_2$ is soluble in excess $NaOH$
• B. Formed precipitate $Zn(OH{)}_2$ is insoluble in excess $N{H}_{4}OH$
• C. Formed precipitate $Pb(OH{)}_2$ is insoluble in excess $NaOH$
• D. Formed precipitate $Pb(OH{)}_2$ is insoluble in excess $N{H}_{4}OH$

Answer 1

The correct option is C Formed precipitate $Pb(OH{)}_2$ is insoluble in excess $NaOH$

$\begin{array}{ccc}\text{For}ZnS{O}_4& \text{NaOH}& N{H}_{4}OH\\ \text{Colour}& \text{Gelatinous ppt.}& \text{Gelatinous ppt.}\\ \text{Solubility}& \text{Precipitate formed will be}& \text{Precipitate formed will be}\\ & \text{soluble in excess NaOH}& \text{insoluble in excess}N{H}_{4}OH\end{array}$

$\begin{array}{ccc}\text{For}Pb(N{O}_3{)}_2& \text{NaOH}& N{H}_{4}OH\\ \text{Colour}& \text{Chalky white ppt}& \text{Chalky white ppt.}\\ \text{Solubility}& \text{Precipitate formed will be}& \text{Precipitate formed will be}\\ & \text{soluble in excess NaOH}& \text{insoluble in excess}N{H}_{4}OH\end{array}$