A solution is 0.020 M in Zn^2+ and 0.50 M in HCl, and is saturated with H2S (0.10 M). Take K = [H+]^2[S^2-]/[H2S] = 1.0x10^-21 and K_sp(ZnS) = 3.0x10^-22 (neglect H+ from H2S). The correct conclusion is

no ZnS precipitates, so Zn^2+ passes into the filtrate and is found in group IV

ZnS precipitates and Zn^2+ is found in group II

no ZnS precipitates, and Zn^2+ is then found in group III as Zn(OH)2

ZnS precipitates and Zn^2+ is found in group IV

JEE Advanced levelSingle correct optionDifficulty 3 of 3qc042

A solution is $0.020 M$ in $Zn X^{2 +}$ and $0.50 M$ in $HCl$ , and is saturated with $H X_{2} S$ ( $0.10 M$ ). Take $K = [H X^{+}] X^{2} [S X^{2 -}] / [H X_{2} S] = 1.0 \times 1 0^{- 21}$ and $K_{s p} (ZnS) = 3.0 \times 1 0^{- 22}$ (neglect $H X^{+}$ from $H X_{2} S$ ). The correct conclusion is

Ano $ZnS$ precipitates, so $Zn X^{2 +}$ passes into the filtrate and is found in group IV
B $ZnS$ precipitates and $Zn X^{2 +}$ is found in group II
Cno $ZnS$ precipitates, and $Zn X^{2 +}$ is then found in group III as $Zn (OH) X_{2}$
D $ZnS$ precipitates and $Zn X^{2 +}$ is found in group IV

[S X^{2 -}] = (1.0 \times 1 0^{- 21}) (0.10) / (0.50)^{2} = 4.00 e - 22 M

. Ionic product

= 0.020 \times 4.00 e - 22 = 8.00 e - 24

, which is below

K_{s p} = 3.0 \times 1 0^{- 22}

, so no precipitation.

Zn X^{2 +}

remains in the group II filtrate and is later precipitated as white

ZnS

H X_{2} S

in ammoniacal medium (group IV);

NH X_{4} Cl

NH X_{4} OH

does not precipitate

Zn (OH) X_{2}