Problem 27 - 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) \(O^{2-}(aq) + H_2O(l) \rightleftharpoons OH^-(aq) + H_3O^+(aq)\); equilibrium to the right (b) \(CH_3COOH(aq) + HS^-(aq) \rightleftharpoons CH_3COO^-(aq) + H_2S(aq)\); equilibrium (c) \(NO_2^-(aq) + H_2O(l) \rightleftharpoons HNO_2(aq) + OH^-(aq)\); equilibrium

Step 1: Identify the acid and base in the reaction

In this reaction, oxide ion (O²⁻) is the base, and water (H₂O) is the acid.

Step 2: Predict the products of the reaction

The products of the acid-base reaction will be the conjugate acid of the base (O²⁻) and the conjugate base of the acid (H₂O). Since oxide ion accepts a proton, its conjugate acid will be hydroxide ion (OH⁻). And since water donates a proton, its conjugate base will be hydronium ion (H₃O⁺).

Step 3: Write the balanced equation

Using the reactants and products, we can write the balanced equation: \[ O^{2-}(aq) + H_2O(l) \rightleftharpoons OH^-(aq) + H_3O^+(aq) \]

Step 4: Determine the direction of equilibrium

Since water is amphoteric, it can act as an acid or a base. It varies in proton donation or acceptance depending on the context. As the oxide ion is a strong base and water is a weak acid, the equilibrium will lie to the right, favoring the formation of the conjugate acid (OH⁻) and conjugate base (H₃O⁺). (b)

Step 5: Identify the acid and base in the reaction

In this reaction, acetic acid (CH₃COOH) is the acid, and the hydrogen sulfide ion (HS⁻) is the base.

Step 6: Predict the products of the reaction

The products of the acid-base reaction will be the conjugate base of the acid (CH₃COOH) and the conjugate acid of the base (HS⁻). The conjugate base of CH₃COOH is CH₃COO⁻ and the conjugate acid of HS⁻ is H₂S.

Step 7: Write the balanced equation

Using the reactants and products, we can write the balanced equation: \[ CH_3COOH(aq) + HS^-(aq) \rightleftharpoons CH_3COO^-(aq) + H_2S(aq) \]

Step 8: Determine the direction of equilibrium

The equilibrium will lie in the direction of the weaker acid and base. Acetic acid is a weak acid, and the hydrogen sulfide ion is a relatively weak base. Since their strengths are comparable, the equilibrium lies neither to the left nor the right, and the reaction is considered to be in equilibrium. (c)

Step 9: Identify the acid and base in the reaction

In this reaction, nitrite ion (NO₂⁻) is the base, and water (H₂O) is the acid.

Step 10: Predict the products of the reaction

The products of the acid-base reaction will be the conjugate acid of the base (NO₂⁻) and the conjugate base of the acid (H₂O). The conjugate acid of NO₂⁻ is HNO₂, and the conjugate base of H₂O is OH⁻.

Step 11: Write the balanced equation

Using the reactants and products, we can write the balanced equation: \[ NO_2^-(aq) + H_2O(l) \rightleftharpoons HNO_2(aq) + OH^-(aq) \]

Step 12: Determine the direction of equilibrium

Nitrite ion (NO₂⁻) is a weak base, and water (H₂O) is a weak acid. The equilibrium will lie in the direction of the weaker acid and base. Since the nitrite ion and water have comparable strengths, the equilibrium lies neither to the left nor the right, and the reaction is considered to be in equilibrium.