Equilibrium constant for the following reaction is `1xx10^(-9)` :
`C_(5)H_(5)N(aq.)+H_(2)O(l)hArrC_(5)H_(5)NH^(+)(aq.)+OH^(-)(aq.)`
Determine the moles of pyridinium chloride `(C_(5)H_(5)N.HCL)` is added to 500ml of 0.4M of pyridine`(C_(5)H_(5)N)` obtain a buffer solution of pH=5 :

Options

A. `0.1` mole
B. `0.2` mole
C. `0.3` mole
D. `0.4` mole

Answer

To solve the problem, we need to determine the moles of pyridinium chloride (C₅H₅NHCl) that must be added to 500 mL of 0.4 M pyridine (C₅H₅N) to create a buffer solution with a pH of 5.

### Step-by-Step Solution:

1. **Identify the Reaction and Constants**:
The equilibrium reaction is:
\[ C_5H_5N(aq) + H_2O(l) \rightleftharpoons C_5H_5NH^+(aq) + OH^-(aq) \]
The equilibrium constant \( K_b \) for this reaction is given as \( 1 \times 10^{-9} \).

2. **Calculate \( pK_b \)**:
The relationship between \( K_b \) and \( pK_b \) is given by:
\[ pK_b = -\log K_b \]
Substituting the value of \( K_b \):
\[ pK_b = -\log(1 \times 10^{-9}) = 9 \]

3. **Determine \( pH \) and \( pOH \)**:
Since the pH of the buffer solution is given as 5, we can calculate \( pOH \) using the relationship:
\[ pH + pOH = 14 \]
Therefore,
\[ pOH = 14 - pH = 14 - 5 = 9 \]

4. **Use the Buffer Equation**:
For a basic buffer, the equation is:
\[ pOH = pK_b + \log \left( \frac{[\text{Salt}]}{[\text{Base}]} \right) \]
Substituting the known values:
\[ 9 = 9 + \log \left( \frac{[\text{Salt}]}{[\text{Base}]} \right) \]
This simplifies to:
\[ \log \left( \frac{[\text{Salt}]}{[\text{Base}]} \right) = 0 \]
Therefore,
\[ \frac{[\text{Salt}]}{[\text{Base}]} = 1 \]
This means that the concentration of salt (C₅H₅NHCl) is equal to the concentration of base (C₅H₅N).

5. **Calculate Moles of Pyridine**:
The volume of pyridine solution is 500 mL (0.5 L) with a concentration of 0.4 M:
\[ \text{Moles of pyridine} = \text{Volume} \times \text{Concentration} = 0.5 \, \text{L} \times 0.4 \, \text{mol/L} = 0.2 \, \text{mol} \]

6. **Determine Moles of Pyridinium Chloride**:
Since the concentration of the salt must equal the concentration of the base, the moles of pyridinium chloride required is also 0.2 mol.

### Final Answer:
The moles of pyridinium chloride (C₅H₅NHCl) that must be added to the solution is **0.2 moles**.