A solution is 0.020 M in Zn^2+ and is saturated with H2S ([H2S] = 0.10 M). For H2S take K_a1K_a2 = 1.0x10^-21 and for ZnS take K_sp = 4.0x10^-22. Find the pH of the solution at which ZnS just begins to precipitate. Give the answer to two decimal places.

JEE Advanced levelNumerical answerDifficulty 3 of 3qc165

A solution is 0.020 M in $Zn X^{2 +}$ and is saturated with $H X_{2} S$ ( $[H X_{2} S] = 0.10$ M). For $H X_{2} S$ take $K_{a 1} K_{a 2} = 1.0 \times 1 0^{- 21}$ and for $ZnS$ take $K_{s p} = 4.0 \times 1 0^{- 22}$ . Find the pH of the solution at which $ZnS$ just begins to precipitate. Give the answer to two decimal places.

Numerical answer type, tolerance 0.01.

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

. Precipitation starts when

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

, so

[S X^{2 -}] = 4.0 \times 1 0^{- 22} /0.020 = 2.0 \times 1 0^{- 20}

M. Then

[H X^{+}]^{2} = 1.0 \times 1 0^{- 21} \times 0.10/ (2.0 \times 1 0^{- 20}) = 5.0 \times 1 0^{- 3}

, so

[H X^{+}] = 0.0707

M and pH = 1.151, i.e. 1.15. In 0.3 M HCl (pH about 0.5) the pH is lower, so ZnS does not precipitate in group II.