A solution is 0.10 M in H2S and 0.20 M in HCl (so [H+] = 0.20 M). For H2S, K_a1 = 1.0x10^-7 and K_a2 = 1.2x10^-13. Taking [H2S] = 0.10 M, the concentration of S^2- (in M) is

JEE Main levelSingle correct optionDifficulty 2 of 3ia024

A solution is 0.10 M in $H X_{2} S$ and 0.20 M in $HCl$ (so $[H X^{+}] = 0.20$ M). For $H X_{2} S$ , $K_{a 1} = 1.0 \times 1 0^{- 7}$ and $K_{a 2} = 1.2 \times 1 0^{- 13}$ . Taking $[H X_{2} S] = 0.10$ M, the concentration of $S X^{2 -}$ (in M) is

For the overall equilibrium

H X_{2} S 2 H X^{+} + S X^{2 -}

K = K_{a 1} K_{a 2} = 1.2 e - 20

[S X^{2 -}] = K_{a 1} K_{a 2} [H X_{2} S] / [H X^{+}]^{2} = 1.2 e - 20 \times 0.10/0.040 = 3.0 e - 20

M. Forgetting to square

[H X^{+}]

or to include

[H X_{2} S]

gives the other choices.