A solution is 0.040 M in Zn^2+ and saturated with H2S ([H2S] = 0.10 M). Take K_a1K_a2 = 1.1x10^-21 for H2S and K_sp(ZnS) = 2.5x10^-22. Calculate the minimum [H^+] in mol L^-1 needed to just prevent precipitation of ZnS (answer to two decimal places).

JEE Advanced levelNumerical answerDifficulty 3 of 3qc110

A solution is $0.040 M$ in $Zn X^{2 +}$ and saturated with $H X_{2} S$ ( $[H X_{2} S] = 0.10 M$ ). Take $K_{a 1} K_{a 2} = 1.1 \times 1 0^{- 21}$ for $H X_{2} S$ and $K_{s p} (ZnS) = 2.5 \times 1 0^{- 22}$ . Calculate the minimum $[H X^{+}]$ in $mol L^{- 1}$ needed to just prevent precipitation of $ZnS$ (answer to two decimal places).

Numerical answer type, answer in mol/L, tolerance 0.01.

No precipitation requires

[Zn X^{2 +}] [S X^{2 -}] \leq K_{s p}

, so

[S X^{2 -}]_{ma x} = 2.5 \times 1 0^{- 22} /0.040 = 6.25 \times 1 0^{- 21} M

. Since

[S X^{2 -}] = K_{a 1} K_{a 2} [H X_{2} S] / [H X^{+}]^{2}

[H X^{+}]^{2} = 1.1 \times 1 0^{- 21} (0.10) /6.25 \times 1 0^{- 21} = 0.0176

. Hence

[H X^{+}] = 0.1327

, i.e.

0.13 M