A solution is 0.010 M in each of Zn^2+ and Cu^2+ and is kept saturated with H2S ([H2S] = 0.10 M). For H2S, K = ([H+]^2[S^2-])/([H2S]) = 1.0 x 10^-21. Given K_sp(ZnS) = 4.0 x 10^-22 and K_sp(CuS) = 1.0 x 10^-36. As the pH of the solution is gradually raised from 0, at what pH does ZnS start to precipitate, and what is [Cu^2+] at that stage?

pH = 1.30; [Cu^2+] = 2.5 x 10^-17 M

pH = 0.30; [Cu^2+] = 2.5 x 10^-15 M

pH = 0.80; [Cu^2+] = 2.5 x 10^-17 M

pH = 1.30; [Cu^2+] = 1.0 x 10^-34 M

JEE Advanced levelSingle correct optionDifficulty 3 of 3ia069

A solution is $0.010 M$ in each of $Zn X^{2 +}$ and $Cu X^{2 +}$ and is kept saturated with $H X_{2} S$ ( $[H X_{2} S] = 0.10 M$ ). For $H X_{2} S$ , $K = \frac{[ H X ^{+} ] ^{2} [ S X ^{2 -} ]}{[ H X _{2} S ]} = 1.0 \times 1 0^{- 21}$ . Given $K_{s p} (ZnS) = 4.0 \times 1 0^{- 22}$ and $K_{s p} (CuS) = 1.0 \times 1 0^{- 36}$ . As the pH of the solution is gradually raised from 0, at what pH does $ZnS$ start to precipitate, and what is $[Cu X^{2 +}]$ at that stage?

ZnS

starts when

[S X^{2 -}] = 4.0 \times 1 0^{- 22} /0.010 = 4.0 \times 1 0^{- 20} M

. Then

[H X^{+}]^{2} = 1.0 \times 1 0^{- 21} \times 0.10/ (4.0 \times 1 0^{- 20})

, so

[H X^{+}] = 5.0 \times 1 0^{- 2} M

and pH = 1.30. At this

[S X^{2 -}]

[Cu X^{2 +}] = 1.0 \times 1 0^{- 36} / (4.0 \times 1 0^{- 20}) = 2.5 \times 1 0^{- 17} M

(CuS precipitates almost completely before ZnS starts).