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) \(\mathrm{NH}_{4}{ }^{+}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons\) (b) \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons\) (c) \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons\)

Short answer

In summary, we have the following reactions and their equilibrium directions: (a) \(\mathrm{NH}_{4}{ }^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\), equilibrium lies to the right. (b) \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \longleftarrow \mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\), equilibrium lies to the left. (c) \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) \longleftarrow \mathrm{CO}_{3}^{2-}(a q)+\mathrm{HF}(a q)\), equilibrium lies to the left.

Step-by-step solution

a) Identifying the Acid and Base

In the given reaction: \(\mathrm{NH}_{4}{ }^{+}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons\) The acid is \(\mathrm{NH}_{4}{ }^{+}\) because it can donate a proton (H+). The base is \(\mathrm{OH}^{-}\) because it can accept a proton (H+).

a) Predicting the Products

In an acid-base neutralization reaction, the acid donates a proton (H+) to the base, forming water and the conjugate base of the acid. In this case, the products will be \(\mathrm{NH}_{3}\) (the conjugate base of \(\mathrm{NH}_{4}{ }^{+}\)) and \(\mathrm{H}_{2} \mathrm{O}\). The overall reaction is: \(\mathrm{NH}_{4}{ }^{+}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

a) Determining the Equilibrium Direction

To determine the equilibrium direction, we need to consider the relative strengths of the acid and base and their conjugates. Since \(\mathrm{NH}_{4}{ }^{+}\) is a weak acid and \(\mathrm{OH}^{-}\) is a strong base, the equilibrium will lie to the right, favoring the formation of the products. Therefore, the reaction can be better represented as: \(\mathrm{NH}_{4}{ }^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

b) Identifying the Acid and Base

In the given reaction: \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons\) The acid is \(\mathrm{H}_{3} \mathrm{O}^{+}\) because it can donate a proton (H+). The base is \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) because it can accept a proton (H+).

b) Predicting the Products

In this acid-base neutralization reaction, the acid donates a proton (H+) to the base, forming water and the conjugate acid of the base. In this case, \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CH}_{3} \mathrm{COOH}\) will be formed. The overall reaction is: \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

b) Determining the Equilibrium Direction

Here, \(\mathrm{H}_{3} \mathrm{O}^{+}\) is a strong acid, and \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) is the conjugate base of a weak acid (\(\mathrm{CH}_{3} \mathrm{COOH}\)). This means the equilibrium will lie to the left, favoring the reactants. Therefore, the reaction is better represented as: \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \longleftarrow \mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

c) Identifying the Acid and Base

In the given reaction: \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons\) The acid is \(\mathrm{HCO}_{3}^{-}\) because it can donate a proton (H+). The base is \(\mathrm{F}^{-}\) because it can accept a proton (H+).

c) Predicting the Products

In this acid-base reaction, \(\mathrm{HCO}_{3}^{-}\) donates a proton (H+) to \(\mathrm{F}^{-}\), forming \(\mathrm{CO}_{3}^{2-}\) and \(\mathrm{HF}\). The overall reaction is: \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{CO}_{3}^{2-}(a q)+\mathrm{HF}(a q)\)

c) Determining the Equilibrium Direction

Here, both \(\mathrm{HCO}_{3}^{-}\) and \(\mathrm{F}^{-}\) are weak acids' conjugate bases (\(\mathrm{H}_{2} \mathrm{CO}_{3}\) and \(\mathrm{HF}\)). To determine the equilibrium direction, we can consider their \(K_{a}\) values. The \(K_{a}\) of \(\mathrm{H}_{2} \mathrm{CO}_{3}\) is around \(4.45 \times 10^{-7}\), and the \(K_{a}\) of \(\mathrm{HF}\) is around \(6.76 \times 10^{-4}\). Since \(\mathrm{HF}\) is a stronger acid than \(\mathrm{H}_{2} \mathrm{CO}_{3}\), the equilibrium will lie to the left, favoring the formation of the reactants. Therefore, the reaction is better represented as: \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) \longleftarrow \mathrm{CO}_{3}^{2-}(a q)+\mathrm{HF}(a q)\)