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 reaction arrow: (a) \(\mathrm{O}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) (b) \(\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{HS}^{-}(a q)\) (c) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\)

Short answer

(a) Products: \(\mathrm{OH}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\); Equilibrium lies to the right. (b) Products: \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{2} \mathrm{S}(a q)\); Equilibrium lies to the left. (c) Products: \(\mathrm{HNO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\); Equilibrium lies to the right.

Step-by-step solution

Identify the acidic and basic components

In this reaction, \(\mathrm{O}^{2-}\) is the base as it can accept a proton, and \(\mathrm{H}_{2}\mathrm{O}\) is the acid, as it can donate a proton.

Predict the products of the reaction

The base, \(\mathrm{O}^{2-}\), will accept a proton from \(\mathrm{H}_{2}\mathrm{O}\), resulting in the formation of \(\mathrm{OH}^{-}\) and \(\mathrm{HOH}\) which is another \(\mathrm{H}_{2}\mathrm{O}\). The balanced reaction is as follows: \[ \mathrm{O}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{OH}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \]

Predict the direction of equilibrium

To predict the direction of equilibrium, we need to compare the relative strengths of the acids and bases involved. The stronger the base, the larger the equilibrium constant, and the reaction will be directed toward products. \(\mathrm{O}^{2-}\) is a stronger base than \(\mathrm{OH}^{-}\). Therefore, the equilibrium will lie to the right of the reaction arrow. (b) \(\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{HS}^{-}(a q)\)

Identify the acidic and basic components

In this reaction, \(\mathrm{CH}_{3} \mathrm{COOH}\) is the acid, and \(\mathrm{HS}^{-}\) is the base.

Predict the products of the reaction

The base, \(\mathrm{HS}^{-}\), will accept a proton from the acid, \(\mathrm{CH}_{3} \mathrm{COOH}\), resulting in the formation of \(\mathrm{H}_{2}\mathrm{S}\) and \(\mathrm{CH}_{3} \mathrm{COO}^{-}\). The balanced reaction is: \[ \mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{HS}^{-}(a q) \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{2} \mathrm{S}(a q) \]

Predict the direction of equilibrium

To predict the direction of the equilibrium, we compare the relative strengths of the acids and bases involved. Acetic acid (\(\mathrm{CH}_{3} \mathrm{COOH}\)) is a weaker acid than hydrosulfide (\(\mathrm{HS}^{-}\)), so the equilibrium will lie to the left of the reaction arrow. (c) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\)

Identify the acidic and basic components

In this reaction, \(\mathrm{NO}_{2}^{-}\) is the base, and \(\mathrm{H}_{2}\mathrm{O}\) is the acid.

Predict the products of the reaction

The base, \(\mathrm{NO}_{2}^{-}\), accepts a proton from the acid, \(\mathrm{H}_{2}\mathrm{O}\), resulting in the formation of \(\mathrm{HNO}_{2}\) and \(\mathrm{HOH}\) which is another \(\mathrm{H}_{2}\mathrm{O}\). The balanced reaction is: \[ \mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HNO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \]

Predict the direction of equilibrium

To predict the direction of equilibrium, we compare the relative strengths of the acids and bases involved. Nitrite (\(\mathrm{NO}_{2}^{-}\)) is a weaker base than hydroxide (\(\mathrm{OH}^{-}\)), and \(\mathrm{H}_{2}\mathrm{O}\) is a weaker acid than \(\mathrm{HNO}_{2}\). Therefore, the equilibrium will lie to the right of the reaction arrow.

Key concepts

Acid-Base Reaction Prediction

Understanding how to predict acid-base reactions is crucial for mastering chemistry. In these reactions, an acid donates a proton (a hydrogen ion, or H+) to a base, which accepts it. The key is to identify the acidic and basic components of the reactants.

For instance, in the reaction \(\mathrm{O}^{2-}(aq) + \mathrm{H}_2\mathrm{O}(l) \rightleftharpoons\), the oxide ion (\(\mathrm{O}^{2-}\)) acts as a base because it can accept a proton, while water (\(\mathrm{H}_2\mathrm{O}\)) is the acid, able to donate a proton. When these two react, the oxide ion accepts a hydrogen ion from water, forming hydroxide (\(\mathrm{OH}^{-}\)) and leaving behind another water molecule.

To successfully predict reaction products, always identify the prospective acid and base, and then determine the resulting products after they exchange the hydrogen ion.

Equilibrium Direction Prediction

Equilibrium direction in an acid-base reaction can be predicted by evaluating the relative strengths of the acids and bases involved. When a strong base reacts with a weaker acid, the equilibrium will favor the right side, forming more products.

For example, in the reaction involving \(\mathrm{CH}_{3}\mathrm{COOH}\) (acetic acid) and \(\mathrm{HS}^{-}\) (hydrosulfide), since acetic acid is a weaker acid than the conjugate acid of the hydrosulfide (\(\mathrm{H}_2\mathrm{S}\)), the equilibrium position is towards the reactants, or the left side.

Understanding how to predict the direction of an acid-base reaction's equilibrium is essential in determining the predominance of products or reactants after the reaction has reached its stable state.

Acid-Base Reaction Products

The products of an acid-base reaction are typically a conjugate base of the acid and a conjugate acid of the base. During the reaction, the acid loses a proton, becoming a conjugate base, and the base gains a proton, becoming a conjugate acid.

For instance, in the reaction between nitrite (\(\mathrm{NO}_2^{-}\)) and water, nitrite is the base and accepts a proton to become nitrous acid (\(\mathrm{HNO}_2\)), while water is the acid that donates a proton, essentially remaining as water. These reactions allow for the formation of new substances, and predicting the products involves understanding both the initial reactants and the concept of proton transfer in acid-base reactions.