A solution is 0.015 M in Zn^2+ and is kept saturated with H2S ([H2S] = 0.10 M throughout). For the overall ionisation H2S 2H+ + S^2- take K = [H+]^2[S^2-]/[H2S] = 1.0x10^-21 and K_sp(ZnS) = 3.0x10^-22. Calculate the pH at which ZnS just begins to precipitate (above this pH it precipitates), correct to two decimal places.

JEE Advanced levelNumerical answerDifficulty 3 of 3qc034

A solution is $0.015 M$ in $Zn X^{2 +}$ and is kept saturated with $H X_{2} S$ ( $[H X_{2} S] = 0.10 M$ throughout). For the overall ionisation $H X_{2} S 2 H X^{+} + S X^{2 -}$ take $K = [H X^{+}] X^{2} [S X^{2 -}] / [H X_{2} S] = 1.0 \times 1 0^{- 21}$ and $K_{s p} (ZnS) = 3.0 \times 1 0^{- 22}$ . Calculate the pH at which $ZnS$ just begins to precipitate (above this pH it precipitates), correct to two decimal places.

Numerical answer type, tolerance 0.01.

At the onset of precipitation

[S X^{2 -}] = K_{s p} / [Zn X^{2 +}] = 3.0 \times 1 0^{- 22} /0.015 = 2.00 e - 20 M

. Then

[H X^{+}] X^{2} = K [H X_{2} S] / [S X^{2 -}] = (1.0 \times 1 0^{- 21}) (0.10) / (2.00 e - 20) = 5.000 e - 03

, so

[H X^{+}] = 0.0707 M

pH = - lo g 0.0707 = 1.151 \approx 1.15

. At the pH used for group II (about 0.5)

ZnS

therefore stays dissolved.