In the reaction: H_2S rightleftharpoons 2H^{+ | vedclass

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(D) The dissociation of $H_2S$ is given by: $H_2S \rightleftharpoons 2H^{+} + S^{2-}$.When $NH_4OH$ is added,it dissociates as: $NH_4OH \rightleftharp

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Concentration of $S^{2-}$ increases

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(D) The dissociation of $H_2S$ is given by: $H_2S \rightleftharpoons 2H^{+} + S^{2-}$.When $NH_4OH$ is added,it dissociates as: $NH_4OH \rightleftharpoons NH_4^{+} + OH^{-}$.The $OH^{-}$ ions react with $H^{+}$ ions from $H_2S$ to form water: $H^{+} + OH^{-} \rightleftharpoons H_2O$.According to Le Chatelier's principle,the removal of $H^{+}$ ions shifts the equilibrium of $H_2S$ dissociation to the right.Therefore,the concentration of $S^{2-}$ increases.

In the reaction: $H_2S \rightleftharpoons 2H^{+} + S^{2-}$,when $NH_4OH$ is added,then

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