Question

When the conjugate acid of aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+}\), reacts with the acetate ion, the following reaction takes place: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+}(a q)+\mathrm{CH}_{3} \mathrm{COO}^{-}(a q) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}(a q)+\mathrm{CH}_{3} \mathrm{COOH}(a q)\) If \(K_{\mathrm{a}}\) for \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+}\) is \(1.35 \times 10^{-5}\) and \(K_{\mathrm{a}}\) for \(\mathrm{CH}_{3} \mathrm{COOH}\) is \(1.86 \times 10^{-5}\) what is \(K\) for the reaction?

Short answer

Answer: The equilibrium constant (K) for the reaction between aniline and acetic acid is approximately 1.38.

Step-by-step solution

Identify the Ka values provided

We are given that: \(K_{a}\)(C6H5NH3+) = \(1.35 \times 10^{-5}\), this is for the conjugate acid of aniline \(K_{a}\)(CH3COOH) = \(1.86 \times 10^{-5}\), this is for the acetic acid

Apply the relationship to find the equilibrium constant of the reaction

We know that \(K_{reaction} = \frac{K_{a \space acid}}{K_{a \space conjugate \space acid}}\). In this case, our reaction involves the acid CH3COOH and the conjugate acid C6H5NH3+. So, we can plug in the given Ka values to find the equilibrium constant: \(K = \frac{K_{a \space (CH3COOH)}}{K_{a \space (C6H5NH3+)}}\) \(K = \frac{1.86 \times 10^{-5}}{1.35 \times 10^{-5}}\)

Calculate the equilibrium constant K

Now divide the given values to find the equilibrium constant K: \(K \approx \frac{1.86}{1.35}\) \(K \approx 1.38\) Thus, the equilibrium constant K for the reaction is approximately 1.38.

Key concepts

Conjugate Acids

In an acid-base reaction, conjugate acids and bases are pivotal concepts. The term "conjugate acid" refers to the species that is formed when a base gains a proton (H\(^+\)). In our exercise,
the conjugate acid is the phenylammonium ion, represented by C\(_6\)H\(_5\)NH\(_3^+\).
When aniline (C\(_6\)H\(_5\)NH\(_2\)) accepts a proton, it forms this conjugate acid.

• A conjugate acid has one more proton than its associated base.
• The strength of a conjugate acid is inversely related to the strength of its base—
  meaning if the base is weak, the conjugate acid is stronger.
• Understanding the role of conjugate acids helps in predicting the direction of equilibria in acid-base reactions.

The relationship between an acid and its conjugate base is crucial in maintaining equilibrium in chemical reactions. This balance affects the overall reaction rate and
clarifies how shifts in concentration could impact the system.

Equilibrium Constant

The equilibrium constant, often designated as \(K\), is a fundamental measure of the position of equilibrium in a chemical reaction.
It quantifies how far a reaction proceeds before reaching equilibrium.
In this exercise, it helps us understand how the reaction between conjugate acid of aniline and acetate ion reaches its equilibrium state.

• The expression for the equilibrium constant depends on the concentration of the products divided by the concentration of reactants.
• For a general reaction: \(aA + bB \rightleftharpoons cC + dD\),
  the equilibrium constant \(K\) is denoted by \(K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\).
• This value does not depend on the concentrations but is
  influenced by temperature changes.

With the provided \(K_{a}\) values for acetic acid and the conjugate acid of aniline,
we can calculate the equilibrium constant for their reaction
by applying the relationship: \(K = \frac{K_{a (\text{acid})}}{K_{a (\text{conjugate acid})}}\). This helps clearly in determining how the equilibrium is affected between these two reactive species.

Chemical Reactions

Chemical reactions are processes in which substances transform into other substances. In acid-base equilibria, this involves the transfer of protons.
Our specific focus here is on reactions involving conjugate acids and bases. In this exercise,
we observed the reaction of phenylammonium ion with acetate ion
to produce aniline and acetic acid.

• Chemical reactions can be reversible, as denoted by the double arrow (\(\rightleftharpoons\)) indicating a dynamic equilibrium.
• Reversible reactions continue to proceed in both directions,
  due to ongoing formation and conversion of reactants \(\rightleftharpoons\) products.
• The concept of dynamic equilibrium means that while reactants and products are formed,
  their concentrations become constant.
• The role of Ka values in reactions helps in predicting the extent to which the products are formed.

Furthermore, reactions can shift depending upon changes in conditions such as concentration, temperature, and pressure.
The understanding of the chemical equilibrium and these shifts are essential for grasping the impact and importance of acid-base reactions in real-world applications.