A `0.25M` solution of pyridinium chloride `(C_(5)H_(5)overset(o+)NHCl^(Theta))` has `pH` of `2.89`. Calculate `pK_(b)` for pyridine `(C_(5)H_(5)N)`.

Explanation

To calculate the \( pK_b \) for pyridine from the given data about pyridinium chloride, we can follow these steps:

### Step 1: Understand the relationship between \( pH \), \( pK_w \), \( pK_b \), and concentration.
Pyridinium chloride is a salt formed from a weak base (pyridine) and a strong acid (HCl). The \( pH \) of the solution can be related to \( pK_b \) using the formula:

\[ pH = \frac{1}{2} pK_w - \frac{1}{2} pK_b - \frac{1}{2} \log C \]

Where:
- \( pK_w \) is the ion product of water, typically \( 14 \) at \( 25^\circ C \).
- \( C \) is the concentration of the salt solution.

### Step 2: Calculate \( pK_w \) and \( \log C \).
Given:
- \( pH = 2.89 \)
- \( C = 0.25 \, M \)

First, calculate \( \log C \):

\[ C = 0.25 \implies \log C = \log(0.25) = \log\left(\frac{25}{100}\right) = \log 25 - \log 100 = \log 25 - 2 \]

Using \( \log 25 \approx 1.39794 \):

\[ \log C = 1.39794 - 2 = -0.60206 \]

### Step 3: Substitute values into the equation.
Now, substitute \( pH \), \( pK_w \), and \( \log C \) into the equation:

\[ 2.89 = \frac{1}{2} \times 14 - \frac{1}{2} pK_b - \frac{1}{2} (-0.60206) \]

This simplifies to:

\[ 2.89 = 7 - \frac{1}{2} pK_b + 0.30103 \]

### Step 4: Rearrange the equation to solve for \( pK_b \).
Combine the constants on the right side:

\[ 2.89 = 7.30103 - \frac{1}{2} pK_b \]

Rearranging gives:

\[ \frac{1}{2} pK_b = 7.30103 - 2.89 \]

Calculating the right side:

\[ \frac{1}{2} pK_b = 4.41103 \]

Thus,

\[ pK_b = 2 \times 4.41103 = 8.82206 \]

### Step 5: Round to appropriate significant figures.
Rounding gives:

\[ pK_b \approx 8.82 \]

### Final Answer:
The \( pK_b \) for pyridine is approximately **8.82**.

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