Problem 58 - Chemistry and Chemical Reactivity

Question

Calculate the \(\mathrm{pH}\) of a \(0.12 \mathrm{M}\) 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

Answer

The pH of the solution is 8.84.

Step 1: Write the Reaction Equation

Begin by writing the dissociation reaction for aniline in water:\[ \text{C}_6\text{H}_5\text{NH}_2 + \text{H}_2\text{O} \rightleftarrows \text{C}_6\text{H}_5\text{NH}_3^+ + \text{OH}^- \]

Step 2: Express the Equilibrium Expression

For the dissociation of aniline, use the base equilibrium constant, \( K_b \), to write the expression:\[ K_b = \frac{[\text{C}_6\text{H}_5\text{NH}_3^+][\text{OH}^-]}{[\text{C}_6\text{H}_5\text{NH}_2]} \]With \( K_b = 4.0 \times 10^{-10} \).

Step 3: Define Initial Concentrations

Initially, the concentration of aniline \([\text{C}_6\text{H}_5\text{NH}_2]\) is 0.12 M, and both \([\text{C}_6\text{H}_5\text{NH}_3^+]\) and \([\text{OH}^-]\) are 0 M.

Step 4: Set up the ICE Table

Create an ICE (Initial, Change, Equilibrium) table:- **Initial:** \([\text{C}_6\text{H}_5\text{NH}_2] = 0.12\), \([\text{C}_6\text{H}_5\text{NH}_3^+] = 0\), \([\text{OH}^-] = 0\)- **Change:** \([-x]\) for \([\text{C}_6\text{H}_5\text{NH}_2]\), \([+x]\) for \([\text{C}_6\text{H}_5\text{NH}_3^+]\) and \([\text{OH}^-]\)- **Equilibrium:** \([\text{C}_6\text{H}_5\text{NH}_2] = 0.12 - x\), \([\text{C}_6\text{H}_5\text{NH}_3^+] = x\), \([\text{OH}^-] = x\)

Step 5: Substitute Into the Equilibrium Expression

Plug the equilibrium concentrations into the \( K_b \) expression:\[ K_b = \frac{x^2}{0.12 - x} = 4.0 \times 10^{-10} \]

Step 6: Simplify Assumption and Solve for x

Assume that \( x \ll 0.12 \), so \( 0.12 - x \approx 0.12 \). This makes the equation:\[ \frac{x^2}{0.12} = 4.0 \times 10^{-10} \]Solving for \( x \):\[ x^2 = 4.8 \times 10^{-11} \]\[ x = \sqrt{4.8 \times 10^{-11}} = 6.9 \times 10^{-6} \]

Step 7: Calculate pOH and pH

The concentration of \( \text{OH}^- \) is \( 6.9 \times 10^{-6} \) M. Calculate \( \text{pOH} \):\[ \text{pOH} = -\log(6.9 \times 10^{-6}) = 5.16 \]Then, use the relation between \( \text{pH} \) and \( \text{pOH} \):\[ \text{pH} + \text{pOH} = 14 \]\[ \text{pH} = 14 - 5.16 = 8.84 \]

Key Concepts

Aniline Base

Aniline is a fascinating base commonly encountered in chemistry. Its molecular formula is \(\text{C}_6\text{H}_5\text{NH}_2\). Aniline is an aromatic amine, which means it has a nitrogen atom connected to a benzene ring. The presence of the nitrogen group gives it basic properties.
This makes aniline capable of accepting hydrogen ions from water, resulting in the formation of hydroxide ions \(\text{OH}^-\).
The aromatic ring influences its basicity, making it weaker than other amines. When calculating the \(\text{pH}\) of an aniline solution, understanding its chemical structure is essential.

ICE Table

An ICE table is a systematic tool used to keep track of concentrations throughout a chemical reaction. It stands for Initial, Change, and Equilibrium.

- **Initial**: Begin by noting the initial concentrations of all reactants and products. For our aniline example, the initial concentration of \(\text{C}_6\text{H}_5\text{NH}_2\) is 0.12 M, while \(\text{C}_6\text{H}_5\text{NH}_3^+\) and \(\text{OH}^-\) start at 0 M.- **Change**: Indicate how the concentrations change as the system moves towards equilibrium, represented as \([-x]\) for reactants and \([+x]\) for products.- **Equilibrium**: Use algebraic expressions to show the change applied to initial concentrations and find their values at equilibrium.

This helps us predict the concentrations of involved species before solving numerical values.

Equilibrium Expression

In an equilibrium reaction, you can express the balance through an equilibrium expression. This comes in handy for reactions involving weak bases like aniline.

Given the dissociation reaction:\[ \text{C}_6\text{H}_5\text{NH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{C}_6\text{H}_5\text{NH}_3^+ + \text{OH}^- \] The base equilibrium constant expression \(K_b\) is articulated as:\[ K_b = \frac{[\text{C}_6\text{H}_5\text{NH}_3^+][\text{OH}^-]}{[\text{C}_6\text{H}_5\text{NH}_2]} \] This equation relates the concentrations of all species at equilibrium. By substituting given or calculated values, you can determine unknown concentrations or constants.
This expression is key to analyzing the reaction's behavior in solution.

Base Dissociation Constant

The Base Dissociation Constant, \(K_b\), provides a measure of the base strength in aqueous solutions. For weak bases like aniline, \(K_b\) helps to quantify how readily the base ionizes in water.

It is defined by the equilibrium concentrations of the ions formed when the base reacts with water:\[ K_b = \frac{[\text{products}]}{[\text{reactants}]} \] In our aniline case, \(K_b = 4.0 \times 10^{-10}\) signifies a small amount of ionization, characteristic of weak bases. The smaller the \(K_b\), the less the base dissociates, and the weaker it is. Understanding \(K_b\) allows chemists to predict pH and other properties of the solution, making it a crucial parameter in chemistry studies.
You will often use \(K_b\) alongside expressions to calculate system variables like \(\text{pH}\) or concentrations.