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:

∴[H2 S][H+]2[ S2−]=1×10−7×1.2×10−13[0.1][0.2]2[ S2−]=1.2×10−20[ S2−]=3×10−20

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Ionisation of acids and bases

An aqueous solution contains 0.10 {~M~H}_{2} {~S} and 0.20 {~M

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

5 \times 1 0^{- 8}

3 \times 1 0^{- 20}

6 \times 1 0^{- 21}

5 \times 1 0^{- 19}

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Updated on: Mar 19, 2023

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Sebastian Discussed

An aqueous solution contains

0.10 M H_{2} S

and

0.20 M HCl

. If the equilibrium constants for the formation of

HS^{-}

from

H_{2} S

1.0 \times 1 0^{- 7}

and that of

S^{2 -}

from

HS^{-}

ions is

1.2 \times 1 0^{- 13}

then the concentration of

S^{2 -}

ions in aqueous solution is:

11 mins ago

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11 mins ago

Text solutionVerified

∴ \frac{[ H ^{+} ] ^{2} [ S ^{2 -} ]}{[ H _{2} S ]} = 1 \times 1 0^{- 7} \times 1.2 \times 1 0^{- 13}

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

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

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Question Text	An aqueous solution contains $0.10 M H_{2} S$ and $0.20 M HCl$ . If the equilibrium constants for the formation of $HS^{-}$ from $H_{2} S$ is $1.0 \times 1 0^{- 7}$ and that of $S^{2 -}$ from $HS^{-}$ ions is $1.2 \times 1 0^{- 13}$ then the concentration of $S^{2 -}$ ions in aqueous solution is:
Updated On	Mar 19, 2023
Topic	Equilibrium
Subject	Chemistry
Class	Class 11
Answer Type	Text solution:1 Video solution: 6
Upvotes	714
Avg. Video Duration	6 min

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

Text solutionVerified

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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:

Text solutionVerified

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