A `0.02 M` solution of pyridinium hydrochloride has `pH=3.44`. Calculate the ionization constant of pyridine.

Explanation

To calculate the ionization constant of pyridine from the given data about pyridinium hydrochloride, follow these steps:

### Step 1: Understand the given information
We have a 0.02 M solution of pyridinium hydrochloride with a pH of 3.44. Pyridinium hydrochloride is the salt of pyridine and hydrochloric acid, and it undergoes hydrolysis in solution.

### Step 2: Calculate the concentration of hydrogen ions \([H^+]\)
Using the pH value, we can find the concentration of hydrogen ions:
\[ \text{pH} = -\log[H^+] \]
Rearranging this gives:
\[ [H^+] = 10^{-\text{pH}} = 10^{-3.44} \]
Calculating this:
\[ [H^+] \approx 3.63 \times 10^{-4} \, \text{M} \]

### Step 3: Relate the ionization constant of the salt to the concentration of hydrogen ions
The hydrolysis constant \(K_H\) can be expressed as:
\[ K_H = C \alpha^2 \]
Where:
- \(C\) is the concentration of the salt (0.02 M)
- \(\alpha\) is the degree of ionization

Since \([H^+] = C \alpha\), we can express \(\alpha\) as:
\[ \alpha = \frac{[H^+]}{C} \]
Substituting the values:
\[ \alpha = \frac{3.63 \times 10^{-4}}{0.02} = 0.01815 \]

### Step 4: Calculate \(K_H\)
Now substituting \(\alpha\) back into the equation for \(K_H\):
\[ K_H = C \alpha^2 = 0.02 \times (0.01815)^2 \]
Calculating this gives:
\[ K_H \approx 6.6 \times 10^{-6} \]

### Step 5: Calculate the ionization constant \(K_a\) of pyridine
The relationship between \(K_H\) and \(K_a\) is given by:
\[ K_H = \frac{K_w}{K_a} \]
Where \(K_w\) (the ion-product constant of water) is \(1.0 \times 10^{-14}\) at 25°C. Rearranging gives:
\[ K_a = \frac{K_w}{K_H} \]
Substituting the values:
\[ K_a = \frac{1.0 \times 10^{-14}}{6.6 \times 10^{-6}} \approx 1.51 \times 10^{-9} \]

### Final Result
The ionization constant \(K_a\) of pyridine is approximately:
\[ K_a \approx 1.51 \times 10^{-9} \]

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