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

The products and equilibrium directions of the given acid-base reactions are: (a) \(2 \mathrm{OH}^{-}(a q)\); equilibrium lies to the right. (b) \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q) + \mathrm{H}_{2} \mathrm{S}(a q)\); equilibrium lies to the right. (c) \(\mathrm{HNO}_{2}(a q) + \mathrm{OH}^{-}(a q)\); equilibrium lies to the right.

Step 1: A. Identify the Acids and Bases in (a)

In the given reaction, \(\mathrm{O}^{2-}(a q)\) is the base and \(\mathrm{H}_{2} \mathrm{O}(l)\) is the acid.

Step 2: A. Determine the Conjugate Acid and Base and Predict Products in (a)

The conjugate acid of \(\mathrm{O}^{2-}\) is \(\mathrm{OH}^{-}\) while the conjugate base of the \(\mathrm{H}_{2} \mathrm{O}\) is \(\mathrm{OH}^{-}\). So, the predicted products of the reaction are \(\mathrm{OH}^{-}\) and \(\mathrm{OH}^{-}\). The reaction becomes: \(\mathrm{O}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons 2 \mathrm{OH}^{-}(a q)\).

Step 3: A. Determine the Equilibrium Direction in (a)

Since \(water\) is a weaker acid than \(\mathrm{OH}^{-}\) and \(\mathrm{O}^{2-}\) is a strong enough base to remove a proton from water, the equilibrium will lie to the right.

Step 4: B. Identify the Acids and Bases in (b)

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

Step 5: B. Determine the Conjugate Acid and Base and Predict Products in (b)

The conjugate base of \(\mathrm{CH}_{3} \mathrm{COOH}\) is \(\mathrm{CH}_{3} \mathrm{COO}^{-}\), and the conjugate acid of \(\mathrm{HS}^{-}\) is \(\mathrm{H}_{2} \mathrm{S}\). The predicted products of the reaction are \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) and \(\mathrm{H}_{2} \mathrm{S}\). The reaction becomes: \(\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{HS}^{-}(a q) \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}(a q) + \mathrm{H}_{2} \mathrm{S}(a q)\).

Step 6: B. Determine the Equilibrium Direction in (b)

Since \(\mathrm{CH}_{3} \mathrm{COOH}\) is a weaker acid than \(\mathrm{H}_{2} \mathrm{S}\) and \(\mathrm{HS}^{-}\) is a stronger base than \(\mathrm{CH}_{3} \mathrm{COO}^{-}\), the equilibrium will lie to the right.

Step 7: C. Identify the Acids and Bases in (c)

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

Step 8: C. Determine the Conjugate Acid and Base and Predict Products in (c)

The conjugate acid of \(\mathrm{NO}_{2}^{-}\) is \(\mathrm{HNO}_{2}\), and the conjugate base of \(\mathrm{H}_{2} \mathrm{O}\) is \(\mathrm{OH}^{-}\). The predicted products of the reaction are \(\mathrm{HNO}_{2}\) and \(\mathrm{OH}^{-}\). The reaction becomes: \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HNO}_{2}(a q) + \mathrm{OH}^{-}(a q)\).

Step 9: C. Determine the Equilibrium Direction in (c)

Since \(\mathrm{HNO}_{2}\) is a weaker acid than water, and \(\mathrm{NO}_{2}^{-}\) is a stronger base than \(\mathrm{OH}^{-}\), the equilibrium will lie to the right.

Key Concepts

Equilibrium Direction

In acid-base reactions, equilibrium direction is influenced by the relative strengths of the acids and bases involved. Understanding equilibrium is crucial as it tells us which side of the reaction is favored, resulting in the predominant products. In general, an equilibrium will favor the side with the weaker acid and weaker base. This occurs because weaker acids do not donate protons as easily, and weaker bases do not accept protons as readily.

Take for instance, reaction (a) where the oxygen ion \((\mathrm{O}^{2-})\) reacts with water \(\mathrm{H}_{2} \mathrm{O}\). Here, water acts as the acid, and the equilibrium lies to the right because water is a weaker acid than hydroxide ions \(\mathrm{OH}^{-}\), making the reaction favorable towards the formation of \(2\mathrm{OH}^{-}\). Similarly, in (b), the equilibrium favors the right since the acetic acid \(\mathrm{CH}_{3} \mathrm{COOH}\) is weaker compared to \(\mathrm{H}_{2} \mathrm{S}\). Comprehending equilibrium direction allows you to predict the dominant species in a solution at equilibrium, giving insight into the solution's behavior.

Conjugate Acids and Bases

In every acid-base reaction, substances transform into their conjugate acids and bases. A conjugate acid is formed when a base gains a proton, and a conjugate base is made when an acid donates a proton.

For instance, in reaction (a), the base \(\mathrm{O}^{2-}\) becomes \(\mathrm{OH}^{-}\), its conjugate acid, after accepting a proton from water. Meanwhile, water, acting as the acid, transforms into its conjugate base \(\mathrm{OH}^{-}\) upon donating a proton.

In reaction (b), \(\mathrm{CH}_{3} \mathrm{COOH}\) donates a proton to form \(\mathrm{CH}_{3} \mathrm{COO}^{-}\), its conjugate base, while \(\mathrm{HS}^{-}\) becomes \(\mathrm{H}_{2} \mathrm{S}\), its conjugate acid. Recognizing the connections between acids/bases and their conjugates helps to understand the mechanism of proton transfer during the reaction and in predicting the products of acid-base reactions.

Proton Transfer

Proton transfer is a fundamental aspect of acid-base reactions, where a proton \(\mathrm{H}^{+}\) is transferred from an acid to a base. This process illustrates how acidic or basic the substances involved are.

In reaction (a), a \(\mathrm{H}^{+}\) is transferred from water to the oxide ion \(\mathrm{O}^{2-}\), highlighting its strong basic nature as it readily accepts a proton. This results in the formation of two \(\mathrm{OH}^{-}\) ions.

Contrast this with reaction (b), where the acetic acid \(\mathrm{CH}_{3} \mathrm{COOH}\) donates a proton to the bisulfide ion \(\mathrm{HS}^{-}\), forming \(\mathrm{H}_{2} \, \, \mathrm{S}\). Here, proton transfer demonstrates that \(\mathrm{HS}^{-}\) is strong enough to accept a proton from \(\mathrm{CH}_{3} \mathrm{COOH}\).

• Proton transfer dictates the eventual equilibrium position, with efficient proton acceptors/bases and donors/acids determining the favored reaction pathway.
• It helps predict the location of equilibrium and understand acidity/basicity comparisons across different reactions.

Understanding how protons are transferred is essential in predicting the behavior of acids and bases in a reaction.