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Question

Given the equation representing a reversible reaction: NH3(g)+H2O()NH4 +(aq)+OH(aq)\mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftarrows \mathrm{NH}_{4} \ ^{+}(a q)+\mathrm{OH}^{-}(a q) According to one add-base theory, the reactant that donates an H+\mathrm{H}^{+} ion in the forward reaction is: (1) H2O()\mathrm{H}_{2} \mathrm{O}(\ell) (2) NH3(g)\mathrm{NH}_{3}(g) (3) NH4+(aq)\mathrm{NH}_{4}^{+}(a q) (4) OH(aq)\mathrm{OH}^{-}(a q).

Solution

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The reaction

NH3(g)+H2O(l)NH4+(aq)+OH(aq)\mathrm{ NH_3(g) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq) }

Here we can see that the reactant that donates an H+^+ ion (in the forward reaction) is $\text{\color{#4257b2}(1) H2O(l)\mathrm{ H_2O(l) }}$

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