Question

Derive an equation analogous to the Henderson-Hasselbalch equation but relating \(\mathrm{pOH}\) and \(\mathrm{p} K_{\mathrm{b}}\) of a buffered solution composed of a weak base and its conjugate acid, such as \(\mathrm{NH}_{3}\) and \(\mathrm{NH}_{4}^{+} .\)

Short answer

The analogous equation to the Henderson-Hasselbalch equation relating pOH and pKb for a buffered solution composed of a weak base and its conjugate acid is given by: \[pOH = pK_b - \log_{10}\frac{[Conjugate\: Acid]}{[Base]}\]

Step-by-step solution

Write the dissociation reaction of a weak base and its conjugate acid

For a weak base and its conjugate acid, we can use NH₃ as a base and NH₄⁺ as the conjugate acid. The reaction will be: \[NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-\]

Write the equilibrium expression for the reaction

The equilibrium expression for the above reaction can be written using the equilibrium constant, Kb. \[K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}\]

Isolate the Kb for the reaction

We can rewrite the equilibrium expression to isolate Kb. \[K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}\]

Convert the expression to involve pOH, pKb, and the base-to-acid concentration ratio

Now, let's introduce the following logarithmic expressions: \[pOH = -\log_{10}([OH^-])\] \[pK_b = -\log_{10}(K_b)\] \[\frac{[NH_4^+]}{[NH_3]} = \frac{[Conjugate\: Acid]}{[Base]}\] Taking the logarithmic form of the Kb expression, we get: \[\log_{10}(K_b) = \log_{10}\frac{[NH_4^+][OH^-]}{[NH_3]}\] \[-pK_b = \log_{10}\frac{[NH_4^+][OH^-]}{[NH_3]}\]

Rearrange the expression to form the desired equation

Now, let's rearrange the previous equation to isolate the pOH. \[-pK_b = \log_{10}\frac{[NH_4^+][OH^-]}{[NH_3]}\] \[pOH = pK_b - \log_{10}\frac{[NH_4^+]}{[NH_3]}\] This equation relates pOH, pKb, and the concentration ratio of the conjugate acid to the weak base. It is analogous to the Henderson-Hasselbalch equation for pOH and pKb in a buffered solution.

Key concepts

pOH

The concept of pOH is closely related to pH, which most students are familiar with. While pH is a measure of the hydrogen ion concentration in a solution, pOH provides the measure of the hydroxide ion concentration. Mathematically, it is defined as the negative logarithm of the hydroxide ion concentration, represented as

• \( pOH = -\log_{10}([OH^-])\).

A key understanding here is that the pH and the pOH are connected through the ion-product constant of water at 25°C,

• \( Kw = 1.0 \times 10^{-14}\).

The relationship is expressed as

• \( pH + pOH = 14\).

This means that if you know either the pH or pOH of a solution, you can easily find the other. When considering buffer solutions involving weak bases, understanding pOH is essential. It helps in calculating the basic component's contribution to the overall chemical behavior of the buffer.

pKb

The strength of a base is indicated by its base dissociation constant, commonly denoted as Kb. Think of it as a measure of a base's ability to accept protons. For practical purposes, we often use pKb, which is the negative logarithm of the base dissociation constant:

• \(pK_b = -\log_{10}(K_b) \).

This helps simplify calculations, especially when working with very small numbers like Kb values typically are.

Smaller pKb values mean that the base is strong, just as smaller pKa values point to a stronger acid. It's important to understand that pKb and pKa are related through the equation:

• \(pK_b + pK_a = 14\),

at 25°C, which provides a direct connection between the strength of a base and its conjugate acid. By understanding pKb, students can effectively gauge the positions of equilibrium in reactions involving bases.

Buffer Solution

A buffer solution is a special type of solution that resists drastic changes in pH. This is particularly crucial when small amounts of acids or bases are added. A typical buffer is composed of a weak acid and its conjugate base or, as described in the exercise, a weak base and its conjugate acid.

For instance, a buffer can consist of ammonia ( NH₃ ) and ammonium ions ( NH₄^+ ).

The magic of buffer solutions arises from their equilibrium directly involving both components that can react with added acids or bases to neutralize them:

• When an acid ( H^+ ) is added, the weak base component neutralizes it.
• When a base ( OH^- ) is added, the conjugate acid neutralizes it.

This balance prevents significant shifts in pH, making buffer solutions incredibly useful in both laboratory and biological settings. Understanding the Henderson-Hasselbalch equation for buffers helps predict the pH (or pOH) given the concentrations of components in the buffer. This is crucial in maintaining pH stability in many chemical applications.