A solution is 0.010 M in Mn^2+ and is saturated with H2S ([H2S] = 0.10 M, K_a1K_a2 = 1.0x10^-21). K_sp(MnS) = 2.5x10^-13. Find the pH (two decimals) above which MnS begins to precipitate. Use log 2 = 0.301.

JEE Advanced levelNumerical answerDifficulty 3 of 3ia206

A solution is $0.010 M$ in $Mn X^{2 +}$ and is saturated with $H X_{2} S$ ( $[H X_{2} S] = 0.10 M$ , $K_{a 1} K_{a 2} = 1.0 \times 1 0^{- 21}$ ). $K_{s p} (MnS) = 2.5 \times 1 0^{- 13}$ . Find the pH (two decimals) above which $MnS$ begins to precipitate. Use $lo g 2 = 0.301$ .

Numerical answer type, tolerance 0.01.

MnS

starts when

[S X^{2 -}] = 2.5 \times 1 0^{- 13} /0.010 = 2.5 \times 1 0^{- 11} M

. Then

[H X^{+}]^{2} = K_{a 1} K_{a 2} [H X_{2} S] / [S X^{2 -}] = 1 0^{- 22} / (2.5 \times 1 0^{- 11}) = 4 \times 1 0^{- 12}

, so

[H X^{+}] = 2 \times 1 0^{- 6} M

pH = 6 - 0.301 = 5.70