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

Short answer

(a) \(\mathrm{NH}_{3}(a q) + \mathrm{H}_{2}\mathrm{O}(l)\), equilibrium lies to the right. (b) \(\mathrm{CH}_{3}\mathrm{COOH}(a q)+\mathrm{H}_{2}\mathrm{O}(l)\), equilibrium lies to the left. (c) \(\mathrm{HF}(a q) + \mathrm{CO}_{3}^{2-}(a q)\), equilibrium lies to the right.

Step-by-step solution

Identify Acid and Base

In this reaction, we have \(\mathrm{NH}_{4}^{+}(a q)\) and \(\mathrm{OH}^{-}(a q)\). The \(\mathrm{NH}_{4}^{+}\) ion donates a proton and acts as an acid, while the \(\mathrm{OH}^{-}\) ion accepts the proton, functioning as a base.

Write the Reaction

Now we can write the reaction: \[ \mathrm{NH}_{4}^{+}(a q) + \mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{NH}_{3}(a q) + \mathrm{H}_{2}\mathrm{O}(l) \]

Compare Ka Values

To determine whether equilibrium favors the left or the right, we will compare the acid dissociation constants (Ka) of the participating acids. In this case, we need the Ka value of \(\mathrm{NH}_{4}^{+}\). The pKa of \(\mathrm{NH}_{4}^{+}\) is 9.25. The pKb of \(\mathrm{OH}^{-}\) is 0.74. Since the pKa of \(\mathrm{NH}_{4}^{+}\) is greater than the pKb of \(\mathrm{OH}^{-}\), equilibrium lies to the right. (b)

Identify Acid and Base

In this reaction, we have \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)\) and \(\mathrm{H}_{3} \mathrm{O}^{+}(a q)\). The \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) ion functions as a base, receiving a proton, while the \(\mathrm{H}_{3} \mathrm{O}^{+}\) ion donates a proton, functioning as an acid.

Write the Reaction

Now we can write the reaction: \[ \mathrm{CH}_{3}\mathrm{COO}^{-}(a q)+\mathrm{H}_{3}\mathrm{O}^{+}(a q)\rightleftharpoons \mathrm{CH}_{3}\mathrm{COOH}(a q)+\mathrm{H}_{2}\mathrm{O}(l) \]

Compare Ka Values

To determine whether equilibrium favors the left or the right, we will compare the acid dissociation constants of the acids. We need the Ka value of \(\mathrm{CH}_{3}\mathrm{COOH}\) and the Kb value of \(\mathrm{CH}_{3} \mathrm{COO}^{-}\). The Ka of \(\mathrm{CH}_{3}\mathrm{COOH}\) is \(1.8 \times 10^{-5}\), and the Kb of \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) is \(5.6\times 10^{-10}\). Since the Ka of \(\mathrm{CH}_{3}\mathrm{COOH}\) is greater than the Kb of \(\mathrm{CH}_{3} \mathrm{COO}^{-}\), equilibrium lies to the left. (c)

Identify Acid and Base

In this reaction, we have \(\mathrm{HCO}_{3}^{-}(a q)\) and \(\mathrm{F}^{-}(a q)\). The \(\mathrm{HCO}_{3}^{-}\) ion can donate a proton, acting as an acid, while the \(\mathrm{F}^{-}\) ion can accept a proton, functioning as a base.

Write the Reaction

Now we can write the reaction: \[ \mathrm{HCO}_{3}^{-}(a q) + \mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{HF}(a q) + \mathrm{CO}_{3}^{2-}(a q) \]

Compare Ka Values

To determine whether equilibrium favors the left or the right, we will compare the acid dissociation constants of the participating acids. In this case, we need the Ka value of \(\mathrm{HCO}_{3}^{-}\) and the Ka value of \(\mathrm{HF}\). The pKa of \(\mathrm{HCO}_{3}^{-}\) is 6.37, and the pKa of \(\mathrm{HF}\) is 3.17. Since the pKa of \(\mathrm{HCO}_{3}^{-}\) is greater than the pKa of \(\mathrm{HF}\), equilibrium lies to the right. In conclusion, the equilibria of the given reactions are as follows: (a) Equilibrium lies to the right (b) Equilibrium lies to the left (c) Equilibrium lies to the right

Key concepts

Equilibrium Position

In an acid-base reaction, equilibrium is the state where both the reactants and products are present in concentrations that do not change over time. The position of the equilibrium is determined by comparing the relative strengths of the acids and bases involved. When predicting which side of the reaction arrow the equilibrium will lie, we look to see which side contains the weaker acid and base. Equilibrium will favor the formation of the weaker acid-base pair.

Consider the reaction: \[\mathrm{NH}_{4}^{+}(a q) + \mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{NH}_{3}(a q) + \mathrm{H}_{2}\mathrm{O}(l)\]Here, the equilibrium position is determined by comparing the acidic strength of \(\mathrm{NH}_{4}^{+}\) with \(\mathrm{H}_{2}\mathrm{O}\), the acid formed in the products. If the acid in the reactants is weaker, then the equilibrium will lie to the right, favoring the products.

Understanding equilibrium positions in acid-base reactions helps predict which substances predominantly exist once equilibrium is reached. This concept is crucial in chemistry as it helps in understanding reaction outcomes and the feasibility of reactions under given conditions.

Proton Transfer

Proton transfer is fundamental to acid-base reactions, where acids donate protons (\(\text{H}^+\)) and bases accept them. In these transactions, an acid and a base are identifiable on both sides of the reaction—both in the reactants and the products.

For instance, in the reaction: \[\mathrm{CH}_{3}\mathrm{COO}^{-}(a q)+\mathrm{H}_{3}\mathrm{O}^{+}(a q) \rightleftharpoons \mathrm{CH}_{3}\mathrm{COOH}(a q)+\mathrm{H}_{2}\mathrm{O}(l)\]Proton transfer occurs from \(\mathrm{H}_{3}\mathrm{O}^{+}\) to \(\mathrm{CH}_{3}\mathrm{COO}^{-}\), converting \(\mathrm{CH}_{3}\mathrm{COO}^{-}\) into \(\mathrm{CH}_{3}\mathrm{COOH}\) and \(\mathrm{H}_{3}\mathrm{O}^{+}\) into \(\mathrm{H}_{2}\mathrm{O}(l)\).

Understanding proton transfer enables chemists to predict the outcome of reactions and develop mechanisms that explain how reactions proceed at a molecular level. It is crucial in fields ranging from biochemistry, where enzyme mechanisms often involve proton transfers, to industrial chemistry, where it is important for designing efficient processes.

Acid Dissociation Constant

The acid dissociation constant, \(K_a\), is a quantitative measure of the strength of an acid in solution. It indicates the degree to which an acid dissociates into its ions in water. A higher \(K_a\) value signifies a stronger acid, which dissociates more completely.

In the context of the reaction: \[\mathrm{HCO}_{3}^{-}(a q) + \mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{HF}(a q) + \mathrm{CO}_{3}^{2-}(a q)\]The dissociation constants, \(K_a\) of \(\mathrm{HCO}_{3}^{-}\) and \(\mathrm{HF}\) are compared to predict equilibrium position. With \(\mathrm{HF}\) having a lower \(pK_a\), indicating a stronger acid compared to \(\mathrm{HCO}_{3}^{-}\), the reaction favors the formation of \(\mathrm{HF}\) and \(\mathrm{CO}_{3}^{2-}\).

Understanding \(K_a\) values helps chemists assess reaction tendencies and design chemical processes with desired properties, such as specific pH levels. It also assists in predicting reaction directions and conditions under which reactions proceed most efficiently.

pKa and pKb Comparisons

The comparison of \(pK_a\) and \(pK_b\) values is instrumental in determining the direction of acid-base reactions and thus the position of equilibrium. \(pK_a\) represents the negative logarithm of the acid dissociation constant \(K_a\), while \(pK_b\) is the same for the base dissociation constant \(K_b\). Lower values of \(pK_a\) or \(pK_b\) indicate stronger acidic or basic character, respectively.

In the previously provided reactions:

• For \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{OH}^{-}\), the \(pK_a\) of \(\mathrm{NH}_{4}^{+}\) is 9.25, which is greater than the \(pK_b\) of \(\mathrm{OH}^-\) (0.74). Thus, equilibrium lies right.
• For \(\mathrm{CH}_{3}\mathrm{COOH}\) and \(\mathrm{CH}_{3} \mathrm{COO}^{-}\), the \(pK_a\) of \(\mathrm{CH}_{3}\mathrm{COOH}\) (4.76) is greater than the \(pK_b\) of \(\mathrm{CH}_{3} \mathrm{COO}^{-}\), so the equilibrium lies to the left.
• For \(\mathrm{HCO}_{3}^{-}\) and \(\mathrm{HF}\), the \(pK_a\) of \(\mathrm{HCO}_{3}^{-}\) (6.37) is higher than that of \(\mathrm{HF}\) (3.17), thus favoring rightward equilibrium.

Comparing these values is essential as they succinctly reveal which substances are stronger acids or bases in a given reaction. This comparison aids significantly when deciding equilibrium positions and predicting outcomes in both theoretical and practical chemical analyses.