In the presence of 0.3 M HCl, H2S precipitates CuS but not ZnS. The correct reason is

the acid suppresses ionisation of H2S, so [S^2-] is small and the ionic product of ZnS stays below its larger K_sp

ZnS reacts with HCl to form a soluble complex [ZnCl4]^2-

Zn^2+ is oxidised to Zn^3+ in acidic medium

the acid suppresses ionisation of H2S, so [S^2-] is small and the ionic product of ZnS stays below its larger K_sp

H2S does not ionise at all in acidic solution, so no sulphide can form

JEE Main levelSingle correct optionDifficulty 2 of 3qc171

In the presence of $0.3 M$ HCl, $H X_{2} S$ precipitates $CuS$ but not $ZnS$ . The correct reason is

A $ZnS$ reacts with HCl to form a soluble complex $[ZnCl X_{4}] X^{2 -}$
B $Zn X^{2 +}$ is oxidised to $Zn X^{3 +}$ in acidic medium
Cthe acid suppresses ionisation of $H X_{2} S$ , so $[S X^{2 -}]$ is small and the ionic product of ZnS stays below its larger $K_{s p}$
D $H X_{2} S$ does not ionise at all in acidic solution, so no sulphide can form

Common-ion effect of

H X^{+}

H X_{2} S 2 H X^{+} + S X^{2 -}

keeps

[S X^{2 -}]

near

1 0^{- 21} M

. Sulphides with very small

K_{s p}

(CuS, PbS, CdS) still exceed their

K_{s p}

, but ZnS, whose

K_{s p}

is much larger, does not.