Problem 94 - Chemistry and Chemical Reactivity

Question

Nicotine, \(\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{N}_{2},\) has two basic nitrogen atoms (page \(795),\) and both can react with water. $$\mathrm{Nic}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftarrows \mathrm{NicH}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})$$ $$\mathrm{NicH}^{+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftarrows \mathrm{NicH}_{2}^{2+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})$$ \(K_{\mathrm{b} 1}\) 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.

Step-by-Step Solution

Answer

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

Step 1: Determine dominant reaction

Nicotine has two basic nitrogen atoms which can accept protons, leading to two stepwise equilibria. First, identify which equilibrium reaction contributes more significantly to the basicity and hence, affects the pH more. Compare the values of \( K_{\mathrm{b}1} \) and \( K_{\mathrm{b}2} \). Since \( K_{\mathrm{b}1} = 7.0 \times 10^{-7} \) is significantly larger than \( K_{\mathrm{b}2} = 1.1 \times 10^{-10} \), the first deprotonation is the dominant process affecting pH.

Step 2: Use dominant reaction to setup expression

Calculate the concentration of hydroxide ions \( [\mathrm{OH}^-] \) using the first equilibrium:\[ \mathrm{K}_{\mathrm{b}1} = \frac{[\mathrm{NicH}^+][\mathrm{OH}^-]}{[\mathrm{Nic}]} \]Assume \( x = [\mathrm{OH}^-] = [\mathrm{NicH}^+] \). Since the initial concentration of nicotine \( [\mathrm{Nic}] = 0.020 \mathrm{M} \) is much larger than \( x \), approximate \( [\mathrm{Nic}] \approx 0.020 \mathrm{M} \).

Step 3: Solve for [OH⁻] concentration

Substitute into the equilibrium expression:\[ \mathrm{K}_{\mathrm{b}1} = \frac{x^2}{0.020} \]Solve for \( x \):\[ 7.0 \times 10^{-7} = \frac{x^2}{0.020} \]\[ x^2 = 7.0 \times 10^{-7} \times 0.020 \]\[ x^2 = 1.4 \times 10^{-8} \]\[ x = \sqrt{1.4 \times 10^{-8}} \approx 1.18 \times 10^{-4} \text{ M} \]

Step 4: Convert to pH

Convert the hydroxide ion concentration to pOH:\[ \mathrm{pOH} = -\log[\mathrm{OH}^-] \approx -\log(1.18 \times 10^{-4}) \approx 3.93 \]Since \( \mathrm{pH} + \mathrm{pOH} = 14 \), we find:\[ \mathrm{pH} = 14 - 3.93 \approx 10.07 \]

Step 5: Verify the assumptions

Verify the assumption that \( [\mathrm{Nic}] \approx 0.020 \mathrm{M} \). The calculated \( [\mathrm{OH}^-] = 1.18 \times 10^{-4} \mathrm{M} \) is significantly smaller than 0.020 M, validating the approximation used in Step 2. The contribution from the second equilibrium is negligible compared to the first, confirming that the basicity and pH are primarily influenced by the first reaction.

Key Concepts

Acid-Base Reactions

In aqueous solutions, nicotine, like many other organic compounds, participates in acid-base reactions. These reactions showcase the ability of nicotine to behave as a base, which involves accepting protons (H⁺ ions) from water. The reaction takes place in a stepwise manner:

• First, nicotine (Nic) accepts a proton to form NicH⁺ and releases a hydroxide ion (OH⁻).
• Next, NicH⁺ can accept another proton to form NicH₂²⁺, releasing another OH⁻.

The ability to accept protons in two steps exemplifies the polybasic nature of nicotine. Such reactions highlight acidic and basic behaviors based on the transfer of protons. Additionally, each step is characterized by its own equilibrium and equilibrium constant, showing how these reactions progress over time under stable conditions.
Understanding this concept helps in grasping how substances interact in solution and the changes in pH that result. It underscores the delicate balance present in chemical systems.

pH Calculation

Calculating the pH of a solution involves understanding the concentration of hydrogen ions (H⁺) or hydroxide ions (OH⁻) present. Depending on whether the solution is acidic or basic, you'll calculate pH or pOH. For basic solutions, such as when nicotine is dissolved in water, you will calculate pOH first.To find the pOH:

• Use the concentration of OH⁻ ions to determine pOH using the formula: \[\mathrm{pOH} = -\log[\mathrm{OH}^-]\]
• With pOH known, calculate pH using the relationship: \[\mathrm{pH} + \mathrm{pOH} = 14\]

In this exercise, the calculated hydroxide concentration leads to a pOH of about 3.93, resulting in a pH of approximately 10.07. This calculation shows that the nicotine solution is basic. Recognizing pH calculations is crucial for understanding solution properties, as pH influences chemical reactions and biological processes significantly.

Equilibrium Constant

The equilibrium constant (K) provides insight into the balance between reactants and products in a chemical reaction upon reaching equilibrium. For nicotine in water, two different equilibrium constants, Kₐ or Kᵦ, indicate the position of equilibrium for each step where nicotine accepts protons.For the reaction where Nic becomes NicH⁺:

• The equilibrium constant (Kᵦ₁) given by the expression:\[\mathrm{K}_{\mathrm{b}1} = \frac{[\mathrm{NicH}^+][\mathrm{OH}^-]}{[\mathrm{Nic}]}\]

Kᵦ values allow us to understand the extent to which nicotine dissociates in water, based on the ratios of products and reactants. A larger Kᵦ signifies a stronger base, indicating more significant ion formation, contributing to the pH. In this case, Kᵦ₁ is much higher than Kᵦ₂, so more emphasis is given to the first equilibrium. Understanding equilibrium constants is vital as they tell us how reactions can be manipulated or how conditions affect reaction direction.

Protonation

Protonation is a chemical process in which a base, like nicotine, gains protons to form positively charged ions. In this exercise, nicotine undergoes protonation twice:

• The first protonation forms NicH⁺.
• The second protonation further forms NicH₂²⁺.

These reactions reveal the number of basic sites available to accept protons, a feature commonly observed in amines and other nitrogen-containing bases. In terms of real-world implications, these reactions of nicotine reflect its role in biological systems where protonation can affect hydrophilicity and the ionization state of molecules, thus influencing its behavior in biological membranes. Overall, recognizing protonation highlights the transformative impact on molecular charge and structure, critical in both chemical reactions and biological interactions.