What is the pH of a 0.660 M solution of C3H;NHB (Kb of C5H5N is 1.7 x 10-9)?

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**Problem Statement:**
What is the pH of a 0.660 M solution of C₅H₅NHBr? (Kₙ of C₅H₅N is 1.7 × 10⁻⁹)

**Explanation:**
To find the pH of the solution, we first need to consider the nature of C₅H₅NHBr. 

- **C₅H₅NHBr** is the salt formed from the weak base **C₅H₅N (pyridine)** and a strong acid **HBr**. 
- In solution, C₅H₅NHBr dissociates into C₅H₅NH⁺ and Br⁻. The bromide ion (Br⁻) does not affect pH as it is the conjugate base of a strong acid. Therefore, the focus is on the acidic nature due to C₅H₅NH⁺.

**Steps for Calculation:**

1. **Calculate the Ka of C₅H₅NH⁺:**
   \[
   K_w = 1.0 \times 10^{-14}
   \]
   \[
   K_a \times K_b = K_w \Rightarrow K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.7 \times 10^{-9}} 
   \]

2. **Determine the pH using Ka:**
   Set up the equilibrium expression for dissociation of C₅H₅NH⁺ and use the concentration to find the pH of the solution.

These calculations involve substituting values in the formulas above, approximating where necessary (for concentrations much greater than x), and finally computing the pH using \(-\log[H⁺]\).

This process allows students to understand applications of equilibrium constants in real-world scenarios involving solution chemistry.
Transcribed Image Text:**Problem Statement:** What is the pH of a 0.660 M solution of C₅H₅NHBr? (Kₙ of C₅H₅N is 1.7 × 10⁻⁹) **Explanation:** To find the pH of the solution, we first need to consider the nature of C₅H₅NHBr. - **C₅H₅NHBr** is the salt formed from the weak base **C₅H₅N (pyridine)** and a strong acid **HBr**. - In solution, C₅H₅NHBr dissociates into C₅H₅NH⁺ and Br⁻. The bromide ion (Br⁻) does not affect pH as it is the conjugate base of a strong acid. Therefore, the focus is on the acidic nature due to C₅H₅NH⁺. **Steps for Calculation:** 1. **Calculate the Ka of C₅H₅NH⁺:** \[ K_w = 1.0 \times 10^{-14} \] \[ K_a \times K_b = K_w \Rightarrow K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.7 \times 10^{-9}} \] 2. **Determine the pH using Ka:** Set up the equilibrium expression for dissociation of C₅H₅NH⁺ and use the concentration to find the pH of the solution. These calculations involve substituting values in the formulas above, approximating where necessary (for concentrations much greater than x), and finally computing the pH using \(-\log[H⁺]\). This process allows students to understand applications of equilibrium constants in real-world scenarios involving solution chemistry.
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