Question

Nicotine, \(\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{N}_{2}\), has two basic nitrogen atoms (Figure \(16.12),\) and both can react with water. $$\begin{array}{c} \text { Nic(aq) }+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{NicH}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \\ \mathrm{NicH}^{+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{NicH}_{2^{2+}}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \end{array}$$ \(K_{\mathrm{b} 1}\) is \(7.0 \times 10^{-7}\) and \(K_{\mathrm{b} 2}\) is \(1.1 \times 10^{-10} .\) Calculate the approximate \(\mathrm{pH}\) of a \(0.020 \mathrm{M}\) solution.

Short answer

The pH of the 0.020 M nicotine solution is approximately 10.07.

Step-by-step solution

Determine Dominant Reaction

Nicotine has two nitrogen atoms that can be protonated, represented by the equilibrium reactions with their corresponding base dissociation constants: \( K_{\mathrm{b} 1} \) for the first reaction and \( K_{\mathrm{b} 2} \) for the second. Since \( K_{\mathrm{b} 1} \) is much larger than \( K_{\mathrm{b} 2} \), the first reaction \( \text{Nic} + \mathrm{H}_2\mathrm{O} \leftrightarrow \mathrm{NicH}^+ + \mathrm{OH}^- \) is the dominant reaction for determining the pH.

Calculate Hydroxide Ion Concentration for Dominant Reaction

Using the reaction \( \text{Nic} + \mathrm{H}_2\mathrm{O} \leftrightarrow \mathrm{NicH}^+ + \mathrm{OH}^- \) with initial concentration \( [\text{Nic}]_0 = 0.020 \text{ M} \), set up the ice table and use \( K_{\mathrm{b} 1} = 7.0 \times 10^{-7} \) to calculate \([\mathrm{OH}^-]\). The equilibrium expression is: \[ K_{\mathrm{b} 1} = \frac{[\mathrm{NicH}^+][\mathrm{OH}^-]}{[\text{Nic}]} \approx \frac{x^2}{0.020 - x} \approx \frac{x^2}{0.020}.\]Solving, \( x^2 = 7.0 \times 10^{-7} \times 0.020 \). Find \( x \), which corresponds to \([\mathrm{OH}^-]\).

Calculate \([\mathrm{OH}^-]\) Approximation

\[ x = \sqrt{7.0 \times 10^{-7} \times 0.020} = \sqrt{1.4 \times 10^{-8}} \approx 1.18 \times 10^{-4} \text{ M}. \]Thus, \([\mathrm{OH}^-] \approx 1.18 \times 10^{-4} \text{ M}.\)

Calculate \(\mathrm{pOH}\) and \(\mathrm{pH}\)

Calculate \( \mathrm{pOH} \) using \( \mathrm{pOH} = -\log[\mathrm{OH}^-] \):\[ \mathrm{pOH} = -\log(1.18 \times 10^{-4}) \approx 3.93. \]Finally, calculate \( \mathrm{pH} \) using the relation \( \mathrm{pH} + \mathrm{pOH} = 14 \):\[ \mathrm{pH} = 14 - 3.93 = 10.07. \]

Final Calculation

With the calculated \( \mathrm{pH} \approx 10.07 \), our estimate for the 0.020 M nicotine solution aligns with the weakly basic nature due to the dominant reaction of the initial nitrogen atom.

Key concepts

Nicotine Chemistry

Nicotine is an organic compound classified as an alkaloid. It is composed of carbon, hydrogen, and nitrogen atoms, with the chemical formula \( \mathrm{C}_{10} \mathrm{H}_{14} \mathrm{N}_{2} \). Nicotine is found naturally in the nightshade family of plants, notably in tobacco. Its structure includes two nitrogen atoms that are capable of accepting protons, which is why it can participate in acid-base reactions.
In aqueous solutions, nicotine acts as a base. This is due to its ability to accept protons from water, forming ions. This acid-base behavior is central to understanding nicotine's chemical properties, especially in how it influences the pH level of a solution.- **Two Basic Sites**: Nicotine's two nitrogen atoms mean it has two potential sites for protonation, each contributing differently to the solution's behavior.- **Protonation Reactions**: These nitrogen atoms can react with water to form positively charged species like \( \mathrm{NicH}^+ \) and \( \mathrm{NicH}_{2}^{2+} \).
The chemical reactions are what lead to nicotine's ability to raise the pH of a solution, as they result in the formation of hydroxide ions \( \mathrm{OH}^- \). Understanding these properties is crucial for comprehending nicotine's role in acid-base equilibria.

Base Dissociation Constant

The base dissociation constant, \( K_{b} \), is a critical component in predicting the behavior of bases in solution. It quantifies the extent to which a base can dissociate in water to accept a proton and form hydroxide ions. Larger values of \( K_{b} \) indicate stronger bases, as they dissociate more completely in solution.
Nicotine, with its two nitrogen atoms, has two respective base dissociation constants, \( K_{b1} \) and \( K_{b2} \):- **Primary \( K_{b1} \)**: \( 7.0 \times 10^{-7} \). This is the larger constant and signifies the initial nitrogen site's ability to dissociate and form \( \mathrm{NicH}^+ \) and \( \mathrm{OH}^- \).- **Secondary \( K_{b2} \)**: \( 1.1 \times 10^{-10} \). This is much smaller, indicating the secondary protonation is less favorable.
When assessing which reaction dominates in a given scenario, comparing these constants is vital. In nicotine’s case, \( K_{b1} \) is significantly larger than \( K_{b2} \), making it the predominant reaction that influences the pH of the solution.

pH Calculation

Calculating the pH of a solution involves understanding the relationship between hydrogen ions \( (\mathrm{H}^+) \) and hydroxide ions \( (\mathrm{OH}^-) \). The pH is defined as the negative base-10 logarithm of the hydrogen ion concentration.For basic solutions, however, it is often easier to calculate the \( \mathrm{pOH} \), which is the negative logarithm of the \( \mathrm{OH}^- \) concentration. Nicotine’s base reaction produces \( \mathrm{OH}^- \), allowing us to first find \( \mathrm{pOH} \) and then pH:- **Hydroxide Concentration**: From the dominant reaction, calculate \( [\mathrm{OH}^-] \), which is approximately \( 1.18 \times 10^{-4} \) M for a \( 0.020 \) M nicotine solution.- **pOH Calculation**: \( \mathrm{pOH} = -\log([\mathrm{OH}^-]) \), leading to a \( \mathrm{pOH} \approx 3.93 \).- **Final pH**: Use the relation \( \mathrm{pH} + \mathrm{pOH} = 14 \) to find the pH: - \( \mathrm{pH} = 14 - 3.93 \approx 10.07 \).
Thus, the solution's pH is consistent with the weakly basic properties of nicotine due to the hydroxide production from the dominant nitrogen site.

Equilibrium Expression

Understanding the equilibrium expression is key to solving for unknown concentrations in chemical reactions involving acids and bases. In nicotine's case, the primary reaction that determines the pH involves forming an equilibrium with \( \mathrm{NicH}^+ \) and \( \mathrm{OH}^- \).The expression for the base dissociation constant \( K_{b1} \) is: \[ K_{b1} = \frac{[\mathrm{NicH}^{+}][\mathrm{OH}^{-}]}{[\mathrm{Nic}]} \]To solve equilibrium problems, we often use an ICE (Initial, Change, Equilibrium) table to organize initial concentrations, the change that occurs as the system reaches equilibrium, and the final equilibrium concentrations:- **Initial Concentrations**: Start with \( [\mathrm{Nic}]_0 = 0.020 \text{ M} \) and assume \( \mathrm{NicH}^+ \) and \( \mathrm{OH}^- \) start at zero.- **Changes**: As nicotine dissociates, \( x \) amount of \( \mathrm{NicH}^+ \) and \( \mathrm{OH}^- \) is produced.- **Equilibrium Concentrations**: These are \( 0.020 - x \) for \( \mathrm{Nic} \) and \( x \) for \( \mathrm{NicH}^+ \) and \( \mathrm{OH}^- \).
By substituting these into the equilibrium expression, you derive an equation in terms of \( x \). Solving this gives the concentration of \( \mathrm{OH}^- \), which is crucial for determining the pH of the solution.