Problem 26 - 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

(a) \(\mathrm{NH}_{4}^{+}(a q) + \mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{NH}_3 (a q) + \mathrm{H}_2 \mathrm{O}(l)\), equilibrium lies to the right. (b) \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}(a q) + \mathrm{H}_2 \mathrm{O}(l)\), equilibrium lies to the left. (c) \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{H}_2 \mathrm{CO}_3 (a q) + \mathrm{HF}(a q)\), equilibrium lies to the right.

Step 1: a) Reaction equation and identify acid/base reactants

First, we need to complete the reaction equation for the reactants given: \(\mathrm{NH}_{4}^{+}(a q) + \mathrm{OH}^{-}(a q) \rightleftharpoons\) Next, we need to identify the acid and base reactants. The \(\mathrm{NH}_{4}^{+}\) ion acts as an acid, as it donates a proton (H\(^+\)), and the \(\mathrm{OH}^{-}\) ion acts as a base, as it accepts a proton.

Step 2: a) Reaction products

Now that we've identified the acid and base reactants, we can predict the reaction products. In this case, the \(\mathrm{NH}_{4}^{+}\) ion donates a proton to \(\mathrm{OH}^{-}\), forming NH3 and H2O: \(\mathrm{NH}_{4}^{+}(a q) + \mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{NH}_3 (a q) + \mathrm{H}_2 \mathrm{O}(l)\)

Step 3: a) Equilibrium position

To predict if the equilibrium lies to the left or the right, we need to consider the strength of the acid and the base reacting and compare their conjugate acid-base pairs. The reaction between ammonium ions (\(\mathrm{NH}_{4}^{+}\)) and hydroxide ions (\(\mathrm{OH}^{-}\)) gives ammonia (\(\mathrm{NH}_3\)) and water (\(\mathrm{H}_2\mathrm{O}\)). Since the hydroxide ion is a stronger base than ammonia and the ammonium ion is a stronger acid than water, the equilibrium lies to the right due to the formation of the weaker acid and base.

Step 4: b) Reaction equation and identify acid/base reactants

First, we need to complete the reaction equation for the reactants given: \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons\) Next, we need to identify the acid and base reactants. The \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) ion (acetate ion) acts as a base, as it accepts a proton, and the \(\mathrm{H}_{3} \mathrm{O}^{+}\) ion (hydronium ion) acts as an acid, as it donates a proton.

Step 5: b) Reaction products

We can predict the reaction products by having the acetate ion accept a proton from the hydronium ion, forming acetic acid (CH3COOH) and water (H2O): \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}(a q) + \mathrm{H}_2 \mathrm{O}(l)\)

Step 6: b) Equilibrium position

When considering the strength of the acid and the base reacting and comparing their conjugate acid-base pairs, the equilibrium position is to the left because hydronium ion (\(\mathrm{H}_{3}\mathrm{O}^{+}\)) is a stronger acid than acetic acid (\(\mathrm{CH}_3 \mathrm{COOH}\)) and the acetate ion (\(\mathrm{CH}_{3} \mathrm{COO}^{-}\)) is a stronger base than water (\(\mathrm{H}_2\mathrm{O}\)).

Step 7: c) Reaction equation and identify acid/base reactants

First, we need to complete the reaction equation for the reactants given: \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons\) Next, we need to identify the acid and base reactants. The \(\mathrm{HCO}_{3}^{-}\) ion (bicarbonate ion) acts as an acid, as it donates a proton, and the \(\mathrm{F}^{-}\) ion (fluoride ion) acts as a base, as it accepts a proton.

Step 8: c) Reaction products

Now that we've identified the acid and base reactants, we can predict the reaction products. The bicarbonate ion donates a proton to the fluoride ion, forming carbonic acid (H2CO3) and hydrogen fluoride (HF): \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{H}_2 \mathrm{CO}_3 (a q) + \mathrm{HF}(a q)\)

Step 9: c) Equilibrium position

To determine the equilibrium position, we analyze the strength of the acid and the base reactants and compare their conjugate acid-base pairs. Bicarbonate ion (\(\mathrm{HCO}_{3}^{-}\)) is a weaker acid than hydrogen fluoride (\(\mathrm{HF}\)), and fluoride ion (\(\mathrm{F}^{-}\)) is a stronger base than the resulting carbonic acid (\(\mathrm{H}_2 \mathrm{CO}_3\)). Therefore, the equilibrium lies to the right due to the formation of the weaker acid and base.

Key Concepts

Equilibrium Position

Equilibrium position in a chemical reaction refers to the point at which the rates of the forward and backward reactions are equal. In acid-base reactions, this position can lean to the left or the right, depending on the strengths of the acids and bases involved. When strong acids and bases react to form weaker conjugates, the equilibrium shifts to the side with the weaker species.
For example, in the reaction between \(NH_4^{+}\) and \(OH^{-}\), the equilibrium lies to the right. This is because \(OH^{-}\) is a stronger base than ammonia (NH3), and \(NH_4^{+}\) is a stronger acid than water. As a result, the formation of the weaker acid (NH3) and base (H2O) dominates.
Understanding the equilibrium position is crucial for predicting the outcome of acid-base reactions, as it reveals which products are more likely to form under given conditions.

Acid and Base Reactants

Acid-base reactions involve reactants that either donate or accept protons. In these reactions, acids are substances that donate protons (H\(^+\)), while bases are those that accept protons. By identifying these characteristics, we can predict the products of the reaction.
In the example of \(\mathrm{NH}_{4}{\underline{\phantom{xx}}}^{+}(a q)+\mathrm{OH}^{-}(a q)\), \(\mathrm{NH}_{4}^{+}\) acts as an acid because it donates a proton, turning into \(\mathrm{NH}_3\), while \(\mathrm{OH}^{-}\) acts as a base by accepting the proton to form water.
Identifying the acid and base reactants is the first step in predicting the reaction's products, as it allows us to see which substances are donating and accepting protons.

Conjugate Acid-Base Pairs

Conjugate acid-base pairs are vital in understanding acid-base reactions. When an acid loses a proton, it forms its conjugate base, and when a base gains a proton, it forms its conjugate acid.
For instance, in \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)\), the acetate ion \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) is a base that forms acetic acid \(\mathrm{CH}_3 \mathrm{COOH}\), its conjugate acid. The hydronium ion \(\mathrm{H}_{3} \mathrm{O}^{+}\) acts as an acid, forming water, its conjugate base.
Understanding conjugate pairs is crucial for predicting equilibrium positions, as the strengths of these conjugates help determine where the balance between the reactants and products will settle.