`0.15` mole of pyridinium chloride has been added into `500 cm^(3)` of `0.2M` pyridine solution. Calculate pH and hydroxyl ion contration in the resulting solution, assuming no change in volume. `(K_(b)"for pyridine" = 1.5xx10^(-9)M)`

Options

A. `5,10^(-8)` mol/litre
B. `5,10^(-9)` mol/litre
C. `6,10^(-9)` mol/litre
D. `7,10^(-9)` mol/litre

Answer

To solve the problem, we need to calculate the pH and hydroxyl ion concentration in a solution containing pyridinium chloride and pyridine. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the concentration of pyridinium chloride
Given:
- Moles of pyridinium chloride = 0.15 moles
- Volume of solution = 500 cm³ = 0.5 L

Using the formula for molarity:
\[ \text{Molarity} = \frac{\text{Number of moles}}{\text{Volume in liters}} \]
\[ \text{Molarity of pyridinium chloride} = \frac{0.15 \text{ moles}}{0.5 \text{ L}} = 0.3 \text{ M} \]

### Step 2: Identify the concentration of pyridine
Given:
- Concentration of pyridine = 0.2 M

### Step 3: Use the Henderson-Hasselbalch equation to find pOH
The Henderson-Hasselbalch equation for a basic buffer is:
\[ \text{pOH} = \text{pK}_b + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) \]
Where:
- \(K_b\) for pyridine = \(1.5 \times 10^{-9}\)

First, calculate \(pK_b\):
\[ pK_b = -\log(K_b) = -\log(1.5 \times 10^{-9}) \approx 8.82 \]

Now, substitute the values into the equation:
\[ \text{pOH} = 8.82 + \log\left(\frac{0.3}{0.2}\right) \]
Calculating the logarithm:
\[ \log\left(\frac{0.3}{0.2}\right) = \log(1.5) \approx 0.176 \]
Now substitute this back into the pOH equation:
\[ \text{pOH} = 8.82 + 0.176 \approx 9.00 \]

### Step 4: Calculate hydroxyl ion concentration
Using the relationship between pOH and hydroxyl ion concentration:
\[ \text{pOH} = -\log[\text{OH}^-] \]
From the calculated pOH:
\[ 9 = -\log[\text{OH}^-] \]
This implies:
\[ [\text{OH}^-] = 10^{-9} \text{ M} \]

### Step 5: Calculate pH
Using the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]
Substituting the value of pOH:
\[ \text{pH} + 9 = 14 \]
Thus:
\[ \text{pH} = 14 - 9 = 5 \]

### Final Results
- pH = 5
- Hydroxyl ion concentration = \(10^{-9} \text{ M}\)