The following equilibrium constants have been determined for hydrosulfuric acid at \(25^{\circ} \mathrm{C}\) $$\begin{array}{l}\mathrm{H}_{2} \mathrm{~S}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{HS}^{-}(a q) \\\\\qquad \begin{aligned}K_{\mathrm{c}}^{\prime} &=9.5 \times 10^{-8} \\\\\mathrm{HS}^{-}(a q) \Longrightarrow \mathrm{H}^{+}(a q)+\mathrm{S}^{2-}(a q) \\\K_{\mathrm{c}}^{\prime \prime}=1.0 \times 10^{-19}\end{aligned}\end{array}$$ Calculate the equilibrium constant for the following reaction at the same temperature: $$\mathrm{H}_{2} \mathrm{~S}(a q) \rightleftharpoons 2 \mathrm{H}^{+}(a q)+\mathrm{S}^{2-}(a q)$$

Short Answer

Expert verified
The equilibrium constant for the reaction \(H_{2}S(aq) \rightleftharpoons 2H^{+}(aq) + S^{2-}(aq)\) at the same temperature is \( K_{c} = 9.5 \times 10^{-8} \times 1.0 \times 10^{-19} = 9.5 \times 10^{-27}\)

Step by step solution

01

Analyze the Given Reactions

Analyze the reactions that are given and write them down. The two reactions are \n\n1. \(H_2S(aq) \rightleftharpoons H^{+}(aq) + HS^{-}(aq)\) with \(K_{c}^{'} = 9.5 \times 10^{-8}\) \n\n2. \(HS^{-}(aq) \rightleftharpoons H^{+}(aq) + S^{2-}(aq)\) with \(K_{c}^{''} = 1.0 \times 10^{-19}\)
02

Sum Up the Reactions

Sum the two reactions to form the reaction for which we need to calculate the equilibrium constant. By adding the two reactions, we get the final reaction: \(H_{2}S(aq) \rightleftharpoons 2H^{+}(aq) + S^{2-}(aq)\)
03

Calculate the Equilibrium Constant of the New Reaction

The equilibrium constant, K, for the final reaction will be the product of the equilibrium constants of the two reactions. Thus, the equilibrium constant for the reaction is: \(K_{c} = K_{c}^{'} \times K_{c}^{''} = 9.5 \times 10^{-8} \times 1.0 \times 10^{-19}\)

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