An aqueous solution contains 0.10 M H2S and 0.20 M HCl. If the equilibrium constants for the formation of HS- from H2S is 1.0× 10- 7 and that of S2 - from HS- ions is 1.2× 10- 13 then the concentration of S2 - ions in aqueous solution is

Question

An aqueous solution contains 0.10 M H2S and 0.20 M HCl. If the equilibrium constants for the formation of HS- from H2S is 1.0× 10- 7 and that of S2 - from HS- ions is 1.2× 10- 13 then the concentration of S2 - ions in aqueous solution is (A) 5× 10- 19 (B) 5× 10- 8 (C) 3× 10- 20 (D) 6× 10- 21

Tardigrade Admin · Accepted Answer

underset underset= 0.1(. 0.1 - x .)H2 S (. a q .)leftharpoons underset underset∼ eq 0.2(. 2 x + 0.20 .)2 H++ undersetxS2 -, Ka=Ka1=1.2× 10- 20 1.2 × 10-20=((0.2)2 ×[ S 2-]/0.1) [.S2 -].=3× 10- 20 '

Q. An aqueous solution contains 0.10 M $H_{2} S$ and 0.20 M HCl. If the equilibrium constants for the formation of $H S^{-}$ from $H_{2} S$ is $1.0 \times 1 0^{- 7}$ and that of $S^{2 -}$ from $H S^{-}$ ions is $1.2 \times 1 0^{- 13}$ then the concentration of $S^{2 -}$ ions in aqueous solution is

$5 \times 1 0^{- 19}$

$5 \times 1 0^{- 8}$

$3 \times 1 0^{- 20}$

$6 \times 1 0^{- 21}$

$= 0.1 (0.1 - x) H_{2} S (a q) ⇌ \sim e q 0.2 (2 x + 0.20) 2 H^{+} + x S^{2 -},$

$K_{a} = K_{a_{1}} = 1.2 \times 1 0^{- 20}$

$1.2 \times 1 0^{- 20} = \frac{( 0.2 ) ^{2} \times [ S ^{2 -} ]}{0.1}$

$[S^{2 -}] = 3 \times 1 0^{- 20}$ '

Q. An aqueous solution contains 0.10 M H2​S and 0.20 M HCl. If the equilibrium constants for the formation of HS− from H2​S is 1.0×10−7 and that of S2− from HS− ions is 1.2×10−13 then the concentration of S2− ions in aqueous solution is

5×10−19

5×10−8

3×10−20

6×10−21