Question

Phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}, K_{\mathrm{a}}=1.3 \times 10^{-10}\right)\) is a weak acid used in mouthwashes, and pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}, K_{\mathrm{b}}=1.8 \times 10^{-9}\right)\) is a weak base used as a solvent. Calculate the value of \(K_{\mathrm{n}}\) for the neutralization of phenol by pyridine. Does the neutralization reaction proceed very far toward completion?

Short answer

\(K_{\mathrm{n}} = 4.27 \times 10^{4}\); the reaction proceeds far toward completion.

Step-by-step solution

Write the Acid and Base Equilibria Equations

Phenol (C6H5OH) dissociates in water according to the equation: \[ \text{C}_6\text{H}_5\text{OH (aq)} \rightleftharpoons \text{C}_6\text{H}_5\text{O}^- \text{(aq)} + \text{H}^+ \text{(aq)}\] Pyridine (C5H5N) accepts a proton in water according to the equation: \[ \text{C}_5\text{H}_5\text{N (aq)} + \text{H}_2\text{O (l)} \rightleftharpoons \text{C}_5\text{H}_5\text{NH}^+ + \text{OH}^-\]

Calculate the Neutralization Constant (K_n)

The neutralization reaction between phenol and pyridine can be represented as:\[ \text{C}_6\text{H}_5\text{OH (aq)} + \text{C}_5\text{H}_5\text{N (aq)} \rightleftharpoons \text{C}_6\text{H}_5\text{O}^- \text{(aq)} + \text{C}_5\text{H}_5\text{NH}^+ \text{(aq)}\]\(K_{\mathrm{n}}\) can be calculated using the formula:\[K_{\mathrm{n}} = \frac{K_{\mathrm{w}}}{K_{\mathrm{a}} \times K_{\mathrm{b}}}\]where \(K_{\mathrm{w}}\) is the ion product of water at 25°C, which is \(1.0 \times 10^{-14}\). Using given \(K_{\mathrm{a}} = 1.3 \times 10^{-10}\) and \(K_{\mathrm{b}} = 1.8 \times 10^{-9}\), substitute into the formula:\[ K_{\mathrm{n}} = \frac{1.0 \times 10^{-14}}{1.3 \times 10^{-10} \times 1.8 \times 10^{-9}}\]

Calculate K_{n}

Perform the calculation for the neutralization constant:\[K_{\mathrm{n}} = \frac{1.0 \times 10^{-14}}{2.34 \times 10^{-19}}\]Simplifying gives:\[K_{\mathrm{n}} = 4.27 \times 10^{4}\]

Determine the Extent of Reaction Completion

Since \(K_{\mathrm{n}} = 4.27 \times 10^{4}\) is much greater than 1, the neutralization reaction proceeds far toward completion, indicating that the reaction between phenol and pyridine proceeds significantly.

Key concepts

Weak Acids

Weak acids are those acids that do not completely dissociate in water. They only partially release hydrogen ions into the solution. A common characteristic of weak acids is their low acid dissociation constant, known as \(K_{a}\).
\(K_{a}\) is crucial because it quantifies the strength of an acid in solution. For example, phenol (\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\)) is a weak acid with an equilibrium constant \(K_{a} = 1.3 \times 10^{-10}\). This value tells us that phenol barely separates into ions in water. The equilibrium equation for a weak acid like phenol is shown below:
\[ \mathrm{C}_6\mathrm{H}_5\mathrm{OH (aq)} \rightleftharpoons \mathrm{C}_6\mathrm{H}_5\mathrm{O}^- \mathrm{(aq)} + \mathrm{H}^+ \mathrm{(aq)} \]Due to this partial dissociation, weak acids have a higher pH compared to strong acids. This makes them significantly important in reactions where selective ionization is needed, like the neutralization process with bases such as pyridine.

Weak Bases

Weak bases are like the reverse of weak acids; they do not fully accept hydrogen ions in water. Instead, they create an equilibrium where only some molecules convert to hydroxide ions. The base dissociation constant \(K_{b}\) measures the strength of a weak base.
Pyridine (\(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\)) is a well-known weak base with \(K_{b} = 1.8 \times 10^{-9}\), indicating only a partial reaction occurs with water. The equilibrium can be written as:\[ \mathrm{C}_5\mathrm{H}_5\mathrm{N (aq)} + \mathrm{H}_2\mathrm{O (l)} \rightleftharpoons \mathrm{C}_5\mathrm{H}_5\mathrm{NH}^+ + \mathrm{OH}^- \]Weak bases like pyridine are important in chemistry due to their structural properties that allow them to act in specific chemical environments without causing a large pH change. In addition, because pyridine is a weak base, it is frequently used as a solvent and in the neutralization reactions, providing controlled environments for chemical reactions.

Equilibrium Constants

Equilibrium constants are vital in chemistry for understanding how far a reaction proceeds. For acid-base reactions, two key constants are \(K_{a}\) and \(K_{b}\) for acids and bases, respectively. The relationship between these constants and the ion product of water \(K_{w}\) helps determine outcomes in neutralization reactions.
In a neutralization reaction, weak acids react with weak bases to form water and other products. The equilibrium constant for such reactions is \(K_{n}\), which can be calculated using the formula:\[ K_{n} = \frac{K_{w}}{K_{a} \times K_{b}} \]For example, in the neutralization of phenol by pyridine, we use \(K_{w} = 1.0 \times 10^{-14}\), \(K_{a} = 1.3 \times 10^{-10}\), and \(K_{b} = 1.8 \times 10^{-9}\). This results in:\[ K_{n} = \frac{1.0 \times 10^{-14}}{1.3 \times 10^{-10} \times 1.8 \times 10^{-9}} = 4.27 \times 10^{4} \]A large \(K_{n}\) value, like \(4.27 \times 10^{4}\), indicates the reaction proceeds significantly toward completion. This is a key concept, as it helps predict the behavior of reactions in laboratory and industrial settings.