A solution is 0.010 M in each of Pb^2+, Cd^2+, Zn^2+ and Mn^2+ and is made 0.50 M in H+ and saturated with H2S ([H2S] = 0.10 M, K_1K_2 = 1.0x10^-21). Take K_sp: PbS 3x10^-28, CdS 8x10^-27, ZnS 3x10^-22, MnS 2.5x10^-13. Which statements are correct?

CdS precipitates and the Cd^2+ left in solution is 2.0x10^-5 M

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3qc179

A solution is $0.010 M$ in each of $Pb X^{2 +}$ , $Cd X^{2 +}$ , $Zn X^{2 +}$ and $Mn X^{2 +}$ and is made $0.50 M$ in $H X^{+}$ and saturated with $H X_{2} S$ ( $[H X_{2} S] = 0.10 M$ , $K_{1} K_{2} = 1.0 \times 1 0^{- 21}$ ). Take $K_{s p}$ : PbS $3 \times 1 0^{- 28}$ , CdS $8 \times 1 0^{- 27}$ , ZnS $3 \times 1 0^{- 22}$ , MnS $2.5 \times 1 0^{- 13}$ . Which statements are correct?

APbS precipitates
BCdS precipitates and the $Cd X^{2 +}$ left in solution is $2.0 \times 1 0^{- 5} M$
CZnS precipitates
DMnS precipitates

[S X^{2 -}] = K_{1} K_{2} [H X_{2} S] / [H X^{+}]^{2} = 1.0 \times 1 0^{- 21} \times 0.10/0.25 = 4.0 \times 1 0^{- 22} M

. Ionic product with

0.010 M

cation is

4.0 \times 1 0^{- 24}

, which exceeds

K_{s p}

of PbS and CdS but not of ZnS or MnS. Remaining

[Cd X^{2 +}] = 8 \times 1 0^{- 27} /4.0 \times 1 0^{- 22} = 2.0 \times 1 0^{- 5} M

. So options 1 and 2.