Question

What would appear if the gas B were passed through an aqueous solution of $Pb(N{O}_3{)}_2$?

• A. A white precipitate soluble in hot dilute $HN{O}_3$
• B. A black precipitate soluble in hot dilute $HN{O}_3$
• C. A black precipitate insoluble in hot dilute $HN{O}_3$
• D. A yellow precipitate soluble in hot concentrated $HN{O}_3$

Answer 1

The correct option is B A black precipitate soluble in hot dilute $HN{O}_3$

Reaction:

$H_{2}S+Pb(N{O}_3{)}_2\to 2HN{O}_3+PbS(s)$ black ppt

One of the three colourless gases $-C{O}_2,S{O}_2$ and $H_{2}S-$ is generally produced by the action of dilute $H_{2}S{O}_4$ on a salt. As the gas turns acidified dichromate solution green along with the formation of a slightly coloured precipitate (may be S), the gas seems to be $H_{2}S$ and so the substance A seems to be a sulphide. Again, the action of $N{H}_3$ or an $NaOH$ solution on the solution of the substance in dilute $H_{2}S{O}_4$ suggests that the substance is a zinc salt. Thus, A is most likely $ZnS$. Let us go through the now,

$ZnS+2H^{+}\to$ $Z{n}^{2+}$ $+$ $H_{2}S\uparrow$

(white) (colourless) (colourless)

A C B

$C{r}_2{O}_7^{2-}+14H^{+}+3S^{2-}\to 2C{r}^{3+}+7H_{2}O+$ $3S$

$\left(H_{2}S\right)$ (green) (yellowish white)

B D

$S+O_2\to S{O}_2\uparrow$

D E

$2H_{2}S+S{O}_2\to 3S^{+}+$ $2H_{2}O$ $\underset{CuS{O}_4\left(white\right)}{\overset{anhydrous}{\to }}$ $CuS{O}_4.x{H}_{2}O$

D (colourless liquid) hydrated copper (II)

sulphur (blue)

$\underset{excessNaOH}{\overset{4N{H}_3}{\to }}\left[Zn\right(O{H}_3{)}_4{]}^{2+}+2O{H}^{-}$

$Z{n}^{2+}+2O{H}^{-}\to Zn\left(OH{)}_2\underset{excessN{H}_{4}OH}{\overset{4N{H}_3}{\to }}\right[Zn(O{H}_3{)}_4{]}^{2+}+2O{H}^{-}$

C $(N{H}_{4}OH$ (White) tetramminezinc

or $NaOH)$ zinc (II) ion

(soluble)