What is the pH of a 0.240 M solution of aniline (C6HsNH2, Kb = 4.3 × 10-¹⁰)?

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**Question:**

What is the pH of a 0.240 M solution of aniline (C₆H₅NH₂, Kb = 4.3 × 10⁻¹⁰)?

**Explanation:**

To find the pH of aniline, which is a weak base, you can use the formula for the base dissociation constant (Kb) to determine the concentration of hydroxide ions (OH⁻). Then, convert the OH⁻ concentration to pOH, and finally to pH.

1. **Calculate OH⁻ Concentration:**

   For a weak base like aniline (C₆H₅NH₂), use the equation:

   \[
   Kb = \frac{[C₆H₅NH₃⁺][OH⁻]}{[C₆H₅NH₂]}
   \]

   Assume that the change in concentration \([OH⁻] = x\), then:

   \[
   Kb = \frac{x^2}{[0.240 - x]} \approx \frac{x^2}{0.240}
   \]

   Solve for \( x \) knowing \( Kb = 4.3 \times 10^{-10} \).

2. **Calculate pOH:**

   \[
   pOH = -\log[OH⁻]
   \]

3. **Calculate pH:**

   \[
   pH = 14 - pOH
   \]

Use these steps to find the pH of the solution.
Transcribed Image Text:**Question:** What is the pH of a 0.240 M solution of aniline (C₆H₅NH₂, Kb = 4.3 × 10⁻¹⁰)? **Explanation:** To find the pH of aniline, which is a weak base, you can use the formula for the base dissociation constant (Kb) to determine the concentration of hydroxide ions (OH⁻). Then, convert the OH⁻ concentration to pOH, and finally to pH. 1. **Calculate OH⁻ Concentration:** For a weak base like aniline (C₆H₅NH₂), use the equation: \[ Kb = \frac{[C₆H₅NH₃⁺][OH⁻]}{[C₆H₅NH₂]} \] Assume that the change in concentration \([OH⁻] = x\), then: \[ Kb = \frac{x^2}{[0.240 - x]} \approx \frac{x^2}{0.240} \] Solve for \( x \) knowing \( Kb = 4.3 \times 10^{-10} \). 2. **Calculate pOH:** \[ pOH = -\log[OH⁻] \] 3. **Calculate pH:** \[ pH = 14 - pOH \] Use these steps to find the pH of the solution.
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