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Bronsted Lowry Theory

H2Ol+S2-⇌HS-+...

Question

$H_{2} O (l) + S^{2 -} ⇌ {H S}^{-} + {O H}^{-}$
In the equation for the reaction represented above, the species acting as acids (proton donors) are:

A

H_{2} O (l)

and

{O H}^{-}

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B

H_{2} O (l)

and

{H S}^{-}

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C

{O H}^{-}

and

{H S}^{-}

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D

{H S}^{-}

and

S^{2 -}

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E

H_{2} O (l)

and

S^{2 -}

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Solution

The correct option is B $H_{2} O (l)$ and ${H S}^{-}$
$H_{2} O (l) + S^{2 -} ⇋ H S^{-} + O H^{-}$
In forward reaction, $H_{2} O$ donates a proton and forms $O H^{-}$ ion.
In backward reaction, $H S^{-}$ donates a proton to form $S^{2 -}$ ion.
Hence, the species acting as acids (proton donors) are $H S^{-}$ and $H_{2} O (l)$ .

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Similar questions

H S^{-} + O H^{-} \to S^{2 -} + H_{2} O

In the system indicated by the equation above, brosnted acids are:

Q. What is the conjugate base in this reaction?

H_{2} S_{(g)} + {H_{2} O}_{(l)} ⇌ {H_{3} O}_{(a q)}^{+} + {H S^{-}}_{(a q)}

Q. In the following reaction acts as Bronsted-Lowry acid and acts as Bronsted-Lowrybase:

{NH}_{3} (aq) + H_{2} O (l) \to {NH}_{4}_{4}^{+} (aq) + {OH}^{-} (aq)

K_{a 1}, K_{a 2}

K_{a 3}

are the respective ionisation constants for the following reactions;

H_{2} S ⇌ H^{+} H S^{-}

H S^{-} ⇌ H^{+} S^{2 -}

H_{2} S ⇌ 2 H^{+} S^{2 -}

The correct relationship between

K_{a 1}, K_{a 2}

K_{a 3}

Q. Which choice or choices demonstrate amphoterism?
i.

H C l + H_{2} O \to H_{3} O^{+} + C l^{-}

H_{2} O + N_{3} \to O H^{-} + N H_{4}^{-}

H S^{-} + H C l \to C l^{-} + H_{2} S

H S^{-} + N H_{3} \to N H_{4}^{+} + S^{2 -}

H C l + N a O H \to N a C l + H_{2} O

N a C l + H_{2} O \to H C l + N a O H

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