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 × 1 0 - 7 and that of S 2 - from H S - ions is 1.2 × 1 0 - 13 then the concentration of S 2 - ions in aqueous solution is

H 2 S ( a q ) ( 0.1 - x ) = 0.1 ⇌ 2 H + ( 2 x + 0.20 ) ≃ 0.2 + S 2 - x , K a = K a 1 = 1.2 × 1 0 - 20 1.2 × 1 0 - 20 = ( 0.2 ) 2 × [ S 2 - ] 0.1 [ S 2 - ] = 3 × 1 0 - 20 '

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ChemistryChemical EquilibriumMiscellaneous/MixedEasy2 minPYQ_2020

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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 $H S^{-}$ 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

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Answer:

Solution:

$\underset{\underset{= 0.1}{(0.1 - x)}}{H_{2} S (a q)} ⇌ \underset{\underset{≃ 0.2}{(2 x + 0.20)}}{2 H^{+}} + \underset{x}{S^{2 -}},$
$K_{a} = K_{a_{1}} = 1.2 \times 1 0^{- 20}$
$1.2 \times 1 0^{- 20} = \frac{(0.2)^{2} \times [S^{2 -}]}{0.1}$
$[S^{2 -}] = 3 \times 1 0^{- 20}$ '

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:Chemical EquilibriumSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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