Question

(A) gives white ppt (B) with NaOH, but (B) dissolves in excess of NaOH forming (C). (C) gives white ppt (D) when $H_{2}S$ gas is passed into it. (A) also gives white ppt (E) with $BaC{l}_2$. (E) is insoluble in conc. $HN{O}_3$. Identify (A), (B), (C), (D) and (E).

Answer 1

The compound (A) is $ZnS{O}_4$ .
It gives white ppt (B) of $Zn(OH{)}_2$ with NaOH,
$ZnS{O}_4+2NaOH\to Zn(OH{)}_2\downarrow (whitheppt)+N{a}_{2}S{O}_4$
But (B) dissolves in excess of NaOH forming (C) which is $N{a}_{2}Zn{O}_2$.
$Zn\left(O{H}_2\right)\stackrel{excessNaOH}{\to }N{a}_{2}Zn{O}_2$
(C) gives white ppt (D) (ZnS) when $H_{2}S$ gas is passed into it.
$N{a}_{2}Zn{O}_2\stackrel{H_{2}S}{\to }ZnS\downarrow \left(whiteppt\right)$
(A) also gives white ppt (E) with $BaC{l}_2$.
$ZnS{O}_4+BaC{l}_2\to BaS{O}_4\downarrow \left(whiteppt\right)+ZnC{l}_2$
(E) $\left(BaS{O}_4\right)$ is insoluble in conc. $HN{O}_3$.

A	$ZnS{O}_4$
B	$Zn(OH{)}_2$
C	$N{a}_{2}Zn{O}_2$
D	$ZnS$
E	$BaS{O}_4$