Problem 77 - Chemistry

Question

Write the equation relating \(K_{\mathrm{a}}\) for a weak acid and \(K_{\mathrm{b}}\) for its conjugate base. Use \(\mathrm{NH}_{3}\) and its conjugate acid \(\mathrm{NH}_{4}^{+}\) to derive the relationship between \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\)

Step-by-Step Solution

Answer

The equation is \( K_{\mathrm{a}}(\mathrm{NH}_{4}^{+}) \times K_{\mathrm{b}}(\mathrm{NH}_{3}) = 1.0 \times 10^{-14} \).

Step 1: Write the Equation for NH4+ Dissociation

Start with the dissociation reaction of the conjugate acid, \( \mathrm{NH}_{4}^{+} \):\[ \mathrm{NH}_{4}^{+} \rightarrow \mathrm{NH}_{3} + \mathrm{H}^{+} \]The dissociation constant for this reaction is the acid dissociation constant, \( K_{\mathrm{a}} \).

Step 2: Write the Equation for NH3 Dissociation

Next, consider the reaction for the base, \( \mathrm{NH}_{3} \), accepting a proton to form its conjugate acid, \( \mathrm{NH}_{4}^{+} \):\[ \mathrm{NH}_{3} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{NH}_{4}^{+} + \mathrm{OH}^{-} \]The equilibrium constant for this reaction is the base dissociation constant, \( K_{\mathrm{b}} \).

Step 3: Relate Ka and Kb to the Ion Product of Water

Recall that for a weak acid and its conjugate base, the product of \( K_{\mathrm{a}} \) and \( K_{\mathrm{b}} \) is equal to the ion product of water, \( K_{\mathrm{w}} \):\[ K_{\mathrm{a}} \times K_{\mathrm{b}} = K_{\mathrm{w}} \]For water, at 25°C, \( K_{\mathrm{w}} = 1.0 \times 10^{-14} \).

Step 4: Deriving the Relationship for NH4+ and NH3

Using the reactions for \( \mathrm{NH}_{3} \) and \( \mathrm{NH}_{4}^{+} \): 1. For \( \mathrm{NH}_{4}^{+} \), \( K_{\mathrm{a}} \) describes the dissociation of \( \mathrm{NH}_{4}^{+} \) into \( \mathrm{NH}_{3} \) and \( \mathrm{H}^{+} \).2. For \( \mathrm{NH}_{3} \), \( K_{\mathrm{b}} \) describes the formation of \( \mathrm{OH}^{-} \) from \( \mathrm{NH}_{3} \) and water.Substituting these values into the relation:\[ K_{\mathrm{a}}(\mathrm{NH}_{4}^{+}) \times K_{\mathrm{b}}(\mathrm{NH}_{3}) = 1.0 \times 10^{-14} \]

Step 5: Conclusion

The equation relating the dissociation constants \( K_{\mathrm{a}} \) for \( \mathrm{NH}_{4}^{+} \) and \( K_{\mathrm{b}} \) for \( \mathrm{NH}_{3} \) is:\[ K_{\mathrm{a}} \times K_{\mathrm{b}} = K_{\mathrm{w}} \]Specifically, this is \[ K_{\mathrm{a}}(\mathrm{NH}_{4}^{+}) \times K_{\mathrm{b}}(\mathrm{NH}_{3}) = 1.0 \times 10^{-14} \] at 25°C.

Key Concepts

Weak Acid Dissociation

Understanding weak acid dissociation is key to grasping acid-base equilibria, as it involves the partial release of hydrogen ions. Unlike strong acids that completely dissociate in a solution, weak acids only partially dissociate, establishing an equilibrium. This state of balance is represented by the acid dissociation constant, denoted as \( K_a \). The value of \( K_a \) describes the degree to which an acid releases protons (\( H^+ \)) into the solution.
In general, the dissociation of a weak acid \( ext{HA} \) can be expressed as:

• \( ext{HA} ightleftharpoons ext{H}^+ + ext{A}^- \)

Where \( ext{HA} \) is the weak acid, and \( ext{A}^- \) is its conjugate base. The equilibrium constant for this reaction is given by:

• \( K_a = \frac{ [ ext{H}^+][ ext{A}^-] }{ [ ext{HA}] } \)

Understanding this equilibrium allows chemists to predict the behavior of weak acids in various solutions. The smaller the \( K_a \), the weaker the acid, meaning it dissociates less in solution.

Conjugate Acid-Base Pairs

Conjugate acid-base pairs are integral in understanding the reversible nature of acid-base reactions. When an acid donates a proton (\( ext{H}^+ \)), it becomes a base capable of accepting a proton back — this is its conjugate base. Conversely, when a base accepts a proton, it forms an acid — the conjugate acid.
This concept is evident in the reaction of ammonia (\( ext{NH}_3 \)) and its conjugate acid, ammonium (\( ext{NH}_4^+ \)). In solution:

• \( ext{NH}_4^+ ightleftharpoons ext{NH}_3 + ext{H}^+ \)
• \( ext{NH}_3 + ext{H}_2 ext{O} ightleftharpoons ext{NH}_4^+ + ext{OH}^- \)

These reactions demonstrate the dynamic equilibrium between ammonia and ammonium, showcasing their roles in proton exchange. Recognizing these pairs helps in predicting the direction of proton transfer in chemical reactions. They establish a balance in the solution, crucial for understanding buffer systems and calculating the resultant pH.

Ion Product of Water

The ion product of water, \( K_w \), is foundational in studying the equilibrium in aqueous solutions, as it represents the constant product of hydrogen and hydroxide ions concentrations. In pure water or any aqueous solution, this equilibrium is described by the self-ionization of water:

• \( ext{H}_2 ext{O} ightleftharpoons ext{H}^+ + ext{OH}^- \)

The equilibrium constant for this self-ionization reaction is \( K_w \), which at 25°C has a value of \( 1.0 \times 10^{-14} \). This tiny value indicates that water is mostly un-ionized and that only a minimal amount of ions form naturally in pure water.
This constant becomes crucial when relating the dissociation constants of weak acids and bases. For a conjugate acid-base pair, such as ammonium and ammonia, the relationship is defined by:

• \( K_a \times K_b = K_w \)

where \( K_a \) and \( K_b \) are the dissociation constants of the acid and base, respectively. This relationship emphasizes the interdependence of the strength of acids and bases in water, illustrating how the balance of \( H^+ \) and \( OH^- \) governs the acidity and basicity in aqueous systems.