Select the correct statements about the use of HCl and H2S in the group II separation.

Added H+ lowers [S^2-] by shifting H2S 2H+ + S^2- to the left.

ZnS remains dissolved because its K_sp is much larger than that of CdS.

CdS still precipitates at this low [S^2-] because its K_sp is very small.

Added H+ lowers [S^2-] by shifting H2S 2H+ + S^2- to the left.

ZnS remains dissolved because its K_sp is much larger than that of CdS.

CdS still precipitates at this low [S^2-] because its K_sp is very small.

A large excess of concentrated HCl is preferred because it makes precipitation of CdS more complete.

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3qc035

Select the correct statements about the use of $HCl$ and $H X_{2} S$ in the group II separation.

AAdded $H X^{+}$ lowers $[S X^{2 -}]$ by shifting $H X_{2} S 2 H X^{+} + S X^{2 -}$ to the left.
B $ZnS$ remains dissolved because its $K_{s p}$ is much larger than that of $CdS$ .
C $CdS$ still precipitates at this low $[S X^{2 -}]$ because its $K_{s p}$ is very small.
DA large excess of concentrated $HCl$ is preferred because it makes precipitation of $CdS$ more complete.

Statements 1 to 3 describe the common-ion control of

[S X^{2 -}]

and the relative

K_{s p}

values correctly. Statement 4 is wrong: too much acid pushes

[S X^{2 -}]

so low (and

CdS

even dissolves in strong acid) that

CdS

precipitation becomes incomplete.