Determine the pH of a 0.21 M solution of pyridinium nitrate (C5H5NHNO3) at 25°C. [Pyridinium nitrate dissociates in water to give pyridinium ions (C5H5NH*), the conjugate acid of pyridine (K = 1.7 × 10-9), and nitrate ions (NO3).] pH =
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![**Problem: Determining the pH of Pyridinium Nitrate Solution**
**Objective:**
Calculate the pH of a 0.21 M solution of pyridinium nitrate (C₅H₅NHNO₃) at 25°C.
**Background Information:**
- Pyridinium nitrate dissociates in water to form pyridinium ions (C₅H₅NH⁺) and nitrate ions (NO₃⁻).
- The pyridinium ion is the conjugate acid of pyridine.
- The base dissociation constant, \(K_b\), for pyridine is \(1.7 \times 10^{-9}\).
**Task:**
Calculate the pH of the solution using the given data.
**Calculation:**
\[ \text{pH} = \_\_\_\_ \]
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**Notes for Students:**
- Use the given \(K_b\) value to find \(K_a\) for the pyridinium ion, noting that \(K_w = 1.0 \times 10^{-14}\).
- Apply the formula \(K_a = \frac{K_w}{K_b}\).
- Set up the equilibrium expression for the dissociation of pyridinium ions and solve for \(H^+\) concentration.
- Lastly, use \(\text{pH} = -\log[H^+]\) to find the pH.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2ceb2d65-29b4-49d4-9830-4ba06589987b%2F27675e7b-d631-48b6-b402-495cd84ab75c%2Fcqb2lpt_processed.jpeg&w=3840&q=75)
To calculate pH , we need to calculate the concentration of hydronium ion given by pyridinium ion on dissociation in water.
So first we need to calculate Ka of pyridinium ion , then using its Ka and concentration we can calculate [H3O+] . From this we can calculate pH.
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