Problem 28 - 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) \(\mathrm{NH}_{4}{\underline{\phantom{xx}}}^{+}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons\) (b) \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons\) (c) \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons\)

Step-by-Step Solution

Answer

In summary, the products and the direction of the equilibrium for the given acid-base reactions are: (a) \(NH_{4}^{+}(aq) + OH^{-}(aq) \rightleftharpoons NH_{3}(aq) + H_{2}O(l)\) (equilibrium lies to the right) (b) \(CH_{3}COO^{-}(aq) + H_{3}O^{+}(aq) \rightleftharpoons CH_{3}COOH(aq) + H_{2}O(l)\) (equilibrium lies to the right) (c) \(HCO_{3}^{-}(aq) + F^{-}(aq) \rightleftharpoons CO_{3}^{2-}(aq) + HF(aq)\) (equilibrium slightly lies to the right)

Step 1: Determine the acid and the base

In this reaction, \(NH_{4}^{+}\) is an acidic species (a proton donor) and \(OH^{-}\) is a basic species (a proton acceptor).

Step 2: Predicting the products

When \(NH_{4}^{+}\) donates a proton, it will become its conjugate base \(NH_{3}\). When \(OH^{-}\) accepts a proton, it forms its conjugate acid \(H_{2}O\). Therefore, the products are \(NH_{3}\) and \(H_{2}O\). The equation becomes: \[NH_{4}^{+}(aq) + OH^{-}(aq) \rightleftharpoons NH_{3}(aq) + H_{2}O(l)\]

Step 3: Predicting the equilibrium direction

Since \(OH^{-}\) is a strong base and it is reacting with \(NH_{4}^{+}\), a weak acid, the equilibrium will lie to the right. The reaction favors the formation of weak acid (\(H_{2}O\)) and weak base (\(NH_{3}\)). (b) Predicting the products and the equilibrium direction: \(CH_{3}COO^{-}(aq) + H_{3}O^{+}(aq)\rightleftharpoons\)

Step 1: Determine the acid and the base

In this reaction, \(H_{3}O^{+}\) is an acidic species and \(CH_{3}COO^{-}\) is a basic species (a proton acceptor).

Step 2: Predicting the products

When \(H_{3}O^{+}\) donates a proton, it will become its conjugate base \(H_{2}O\). When \(CH_{3}COO^{-}\) accepts a proton, it forms its conjugate acid \(CH_{3}COOH\). Therefore, the products are \(CH_{3}COOH\) and \(H_{2}O\). The equation becomes: \[CH_{3}COO^{-}(aq) + H_{3}O^{+}(aq) \rightleftharpoons CH_{3}COOH(aq) + H_{2}O(l)\]

Step 3: Predicting the equilibrium direction

Since a strong acid (\(H_{3}O^{+}\)) is reacting with a weak base (\(CH_{3}COO^{-}\)), the equilibrium will lie to the right. The reaction favors the formation of the weak acid (\(CH_{3}COOH\)) and water. (c) Predicting the products and the equilibrium direction: \(HCO_{3}^{-}(aq) + F^{-}(aq) \rightleftharpoons\)

Step 1: Determine the acid and the base

In this reaction, \(HCO_{3}^{-}\) is an acidic species and \(F^{-}\) is a basic species.

Step 2: Predicting the products

When \(HCO_{3}^{-}\) donates a proton, it will become its conjugate base \(CO_{3}^{2-}\). When \(F^{-}\) accepts a proton, it forms its conjugate acid \(HF\). Therefore, the products are \(CO_{3}^{2-}\) and \(HF\). The equation becomes: \[HCO_{3}^{-}(aq) + F^{-}(aq) \rightleftharpoons CO_{3}^{2-}(aq) + HF(aq)\]

Step 3: Predicting the equilibrium direction

Since a weak acid (\(HCO_{3}^{-}\)) is reacting with a weak base (\(F^{-}\)), it is necessary to compare their respective \(K_a\) and \(K_b\) values to predict the direction of the equilibrium. \(HCO_{3}^{-}\) has a larger \(K_a\) value than \(HF\), and \(F^{-}\) has a larger \(K_b\) value than \(CO_{3}^{2-}\). Thus, the equilibrium will slightly lie to the right, favoring the formation of \(CO_{3}^{2-}\) and \(HF\).