Problem 54
Question
Calculate the pH of a 0.12 M aqueous solution of the base aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=4.0 \times 10^{-10}\right)\) $$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftarrows \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})$$
Step-by-Step Solution
Verified Answer
The pH of the 0.12 M aniline solution is approximately 8.84.
1Step 1: Understanding the Reaction
The reaction of the base aniline (\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\)) with water is represented by: \[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\] Aniline accepts a proton from water, forming aniline hydrochloride and hydroxide ions (\(\mathrm{OH}^-\)).
2Step 2: Setting up the Expression for \(K_b\)
Use the base dissociation constant expression to relate concentrations: \[K_b = \frac{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}]}\] Substituting given values: initial concentration \([\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}] = 0.12 \text{ M}\).
3Step 3: Simplifying the \(K_b\) Expression
Assume \([\mathrm{OH}^-] = x\). Due to the low \(K_b\) value, the concentration change will be small: \([\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}] = x\) and \([\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}] = 0.12 - x \approx 0.12\). Then, \[\frac{x^2}{0.12} = 4.0 \times 10^{-10}\]
4Step 4: Solving for \(x\)
From the simplified expression: \[x^2 = 4.0 \times 10^{-10} \times 0.12\] Calculate \[x = \sqrt{4.8 \times 10^{-11}} = 6.93 \times 10^{-6} \]. \([\mathrm{OH}^-] = x = 6.93 \times 10^{-6} \text{ M}\).
5Step 5: Calculating \(pOH\)
Determine \(pOH\) from \([\mathrm{OH}^-]\): \[pOH = -\log([\mathrm{OH}^-]) = -\log(6.93 \times 10^{-6}) \approx 5.16\].
6Step 6: Calculating \(pH\) from \(pOH\)
Since \(pH + pOH = 14\), calculate \(pH\): \[pH = 14 - pOH = 14 - 5.16 \approx 8.84\].
Key Concepts
Base Dissociation ConstantAnilineSolution ChemistrypOH Calculation
Base Dissociation Constant
The base dissociation constant, denoted as \( K_b \), measures the strength of a base in solution. It indicates how well a base dissociates to form its conjugate acid and hydroxide ions. In the scenario involving aniline, with the formula \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} \), the dissociation occurs when aniline reacts with water to form \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\) and \(\mathrm{OH}^{-}\).
\[ K_b = \frac{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}]} \]
The small value of \( K_b = 4.0 \times 10^{-10} \) suggests that aniline is a weak base. In weak bases, the extent of dissociation is minimal, resulting in low concentrations of \(\mathrm{OH}^{-}\) ions. This property is crucial when calculating the pH of the solution, as it implies that the concentration of the base remains largely undisturbed.
Knowing \( K_b \) helps us predict the position of the equilibrium and allows us to calculate the pH of the solution given initial concentrations.
\[ K_b = \frac{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}]} \]
The small value of \( K_b = 4.0 \times 10^{-10} \) suggests that aniline is a weak base. In weak bases, the extent of dissociation is minimal, resulting in low concentrations of \(\mathrm{OH}^{-}\) ions. This property is crucial when calculating the pH of the solution, as it implies that the concentration of the base remains largely undisturbed.
Knowing \( K_b \) helps us predict the position of the equilibrium and allows us to calculate the pH of the solution given initial concentrations.
Aniline
Aniline, known for its chemical formula \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} \), is an organic compound classified as an aromatic amine. The presence of the amino group bonded to the benzene ring makes it a weaker base compared to aliphatic amines because the lone pair on nitrogen is delocalized into the aromatic ring, making it less available to accept protons.
Aniline's reactivity with water in solution results in the transfer of a proton from water to aniline. This process forms aniline's conjugate acid (\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\)) and hydroxide ions \(\mathrm{OH}^{-}\).
While aniline's base characteristics are weak, it plays a significant role in solution chemistry because it often serves as a reference point to understand the behavior of other similar compounds. Understanding the dissociation behavior of aniline can provide insights into the properties of other, more complex amines in aqueous solutions.
Aniline's reactivity with water in solution results in the transfer of a proton from water to aniline. This process forms aniline's conjugate acid (\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\)) and hydroxide ions \(\mathrm{OH}^{-}\).
While aniline's base characteristics are weak, it plays a significant role in solution chemistry because it often serves as a reference point to understand the behavior of other similar compounds. Understanding the dissociation behavior of aniline can provide insights into the properties of other, more complex amines in aqueous solutions.
Solution Chemistry
Solution chemistry involves the study of chemical reactions that occur within aqueous solutions. When analyzing solutions, we consider the molecular interactions and effects of solute on solvent, typically water. In the context of aniline's behavior in water, solution chemistry helps us appreciate the subtle changes and equilibrium states within the system.
In a typical solution with aniline, the initial concentration of aniline is significant to predict how much it dissociates to release \(\mathrm{OH}^{-}\). As stated, aniline slightly alters the water's concentration due to its weak base behavior. The assumptions made in solution chemistry such as the negligible concentration change (\(x\)) due to low \( K_b \), highlight the fact that base dissociation in weak solutions often results in negligible changes to the solute concentration that's still largely intact.
Solution chemistry is a broad field, but, in this case, understanding the behavior of weak bases like aniline reaffirms the complexities and interactions an aqueous system can exhibit.
In a typical solution with aniline, the initial concentration of aniline is significant to predict how much it dissociates to release \(\mathrm{OH}^{-}\). As stated, aniline slightly alters the water's concentration due to its weak base behavior. The assumptions made in solution chemistry such as the negligible concentration change (\(x\)) due to low \( K_b \), highlight the fact that base dissociation in weak solutions often results in negligible changes to the solute concentration that's still largely intact.
Solution chemistry is a broad field, but, in this case, understanding the behavior of weak bases like aniline reaffirms the complexities and interactions an aqueous system can exhibit.
pOH Calculation
Calculating pOH from the hydroxide ion concentration is essential for finding the acidity or basicity of solutions. pOH is defined as the negative logarithm of the hydroxide ion concentration:
\[\text{pOH} = -\log([\mathrm{OH}^{-}])\]
In this example, after calculating the concentration of \(\mathrm{OH}^{-}\) ions following aniline's dissociation, we determined \([\mathrm{OH}^{-}] = 6.93 \times 10^{-6} \text{ M}\). This leads to:
\[\text{pOH} = -\log(6.93 \times 10^{-6}) \approx 5.16\]
Understanding the pOH is crucial because it allows us to calculate the pH of the solution through the relationship \(\text{pH} + \text{pOH} = 14\). From here, we found the pH as:
\[\text{pH} = 14 - 5.16 = 8.84\]
Being adept at converting between pH and pOH is important in solution chemistry, as it provides a complete picture of the acid-base balance within a solution.
\[\text{pOH} = -\log([\mathrm{OH}^{-}])\]
In this example, after calculating the concentration of \(\mathrm{OH}^{-}\) ions following aniline's dissociation, we determined \([\mathrm{OH}^{-}] = 6.93 \times 10^{-6} \text{ M}\). This leads to:
\[\text{pOH} = -\log(6.93 \times 10^{-6}) \approx 5.16\]
Understanding the pOH is crucial because it allows us to calculate the pH of the solution through the relationship \(\text{pH} + \text{pOH} = 14\). From here, we found the pH as:
\[\text{pH} = 14 - 5.16 = 8.84\]
Being adept at converting between pH and pOH is important in solution chemistry, as it provides a complete picture of the acid-base balance within a solution.
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