Problem 25 - Chemistry: The Central Science

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) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\)

Step-by-Step Solution

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 1: (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.

Step 2: (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.

Step 3: (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: \[\mathrm{O}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons 2\ \mathrm{OH}^{-}(a q)\]

Step 4: (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.

Step 5: (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.

Step 6: (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.

Step 7: (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: \[\mathrm{CH}_{3}\mathrm{COOH}(a q)+\mathrm{HS}^{-}(a q) \rightleftharpoons \mathrm{CH}_{3}\mathrm{COO}^{-}(a q)+\mathrm{H}_{2}\mathrm{S}(a q)\]

Step 8: (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.

Step 9: (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.

Step 10: (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.

Step 11: (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: \[\mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{HNO}_{2}(a q)+\mathrm{OH}^{-}(a q)\]

Step 12: (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.