Question

Predict the products of the following acid-base reactions, and predict whether the equilibrium lies to the left or to the right of the equation: (a) \(\mathrm{O}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) (b) \(\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{HS}^{-}(a q) \rightleftharpoons\) (c) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\)

Short answer

(a) The products of the reaction are \(2\ \mathrm{OH}^{-}(a q)\) and the equilibrium lies to the right. (b) The products of the reaction are \(\mathrm{CH}_{3}\mathrm{COO}^{-}(a q)\) and \(\mathrm{H}_{2}\mathrm{S}(a q)\) with the equilibrium lying close to the middle. (c) The products of the reaction are \(\mathrm{HNO}_{2}(a q)\) and \(\mathrm{OH}^{-}(a q)\) and the equilibrium lies to the left.

Step-by-step solution

(a) Identify the reactants

In reaction (a), the reactants are the oxide ion (\(\mathrm{O}^{2-}\)) and water (\(\mathrm{H}_{2}\mathrm{O}\)). We must determine which reactant acts as an acid and which as a base, predict the products, and evaluate their relative strengths.

(a) Identify the acid and base

The oxide ion (\(\mathrm{O}^{2-}\)) is a strong base due to its high charge density and its propensity to accept protons. Water (\(\mathrm{H}_{2}\mathrm{O}\)) can act as both an acid and a base, but in this case, it will act as an acid.

(a) Predict the products

The products of this acid-base reaction would be hydroxide ion (\(\mathrm{OH}^{-}\)) and hydroxide ion (\(\mathrm{OH}^{-}\)). The balanced equation would be: \[\mathrm{O}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons 2\ \mathrm{OH}^{-}(a q)\]

(a) Determine the equilibrium direction

Since \(\mathrm{O}^{2-}\) is a stronger base than \(\mathrm{OH}^{-}\), and since water is a weaker acid than the produced hydroxide ion, the equilibrium will lie to the right.

(b) Identify the reactants

In reaction (b), the reactants are acetic acid (\(\mathrm{CH}_{3}\mathrm{COOH}\)) and the hydrosulfide ion (\(\mathrm{HS}^{-}\)). We must again identify which reactant acts as an acid and which as a base, predict the products, and evaluate their relative strengths.

(b) Identify the acid and base

Acetic acid (\(\mathrm{CH}_{3}\mathrm{COOH}\)) is a weak acid, while the hydrosulfide ion (\(\mathrm{HS}^{-}\)) is a weak base.

(b) Predict the products

The products of this acid-base reaction would be acetate ion (\(\mathrm{CH}_{3}\mathrm{COO}^{-}\)) and hydrogen sulfide (\(\mathrm{H}_{2}\mathrm{S}\)). The balanced equation would be: \[\mathrm{CH}_{3}\mathrm{COOH}(a q)+\mathrm{HS}^{-}(a q) \rightleftharpoons \mathrm{CH}_{3}\mathrm{COO}^{-}(a q)+\mathrm{H}_{2}\mathrm{S}(a q)\]

(b) Determine the equilibrium direction

Since both the acid and base are relatively weak, the equilibrium will not favor either side strongly, and will lie close to the middle.

(c) Identify the reactants

In reaction (c), the reactants are the nitrite ion (\(\mathrm{NO}_{2}^{-}\)) and water (\(\mathrm{H}_{2}\mathrm{O}\)). As before, we will identify which reactant acts as an acid and which as a base, predict the products, and evaluate their relative strengths.

(c) Identify the acid and base

The nitrite ion (\(\mathrm{NO}_{2}^{-}\)) can act as a weak base, while water (\(\mathrm{H}_{2}\mathrm{O}\)) will again act as an acid.

(c) Predict the products

The products of this acid-base reaction would be nitrous acid (\(\mathrm{HNO}_{2}\)) and hydroxide ion (\(\mathrm{OH}^{-}\)). The balanced equation would be: \[\mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{HNO}_{2}(a q)+\mathrm{OH}^{-}(a q)\]

(c) Determine the equilibrium direction

Since the nitrite ion is a weaker base than the hydroxide ion, and water is a weaker acid than nitrous acid, the equilibrium will lie to the left.