The pH of basic buffer mixtures is given by : `pH=pK_(a)+log((["Base"])/(["Salt"]))` , whereas pH of acidic buffer mixtures is given by: `pH= pK_(a)+log((["Salt"])/(["Acid"]))`. Addition of little acid or base although shows no appreciable change for all practical purpose, but since the ratio `(["Base"])/(["Salt"])` or `(["Salt"])/(["Acid"])` change, a slight decrease or increase in pH results in.
Mole of HCI required to prepare a buffer solution of `pH=8.5` with `0.1` mole of NaCN in one litre solution is: `(pK_(a) for CN^(-)=4.61)`

Options

A. `8.85xx10^(-2)`
B. `7.85xx10^(-2)`
C. `9.85xx10^(-2)`
D. `6.85xx10^(-2)`

Explanation

To solve the problem of determining the moles of HCl required to prepare a buffer solution with a pH of 8.5 using 0.1 moles of NaCN, we will follow these steps:

### Step 1: Identify the type of buffer
Since the pH (8.5) is greater than 7, we are dealing with a basic buffer solution. The relevant formula for a basic buffer is:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Base}]}{[\text{Salt}]}\right) \]

### Step 2: Gather the known values
- pH = 8.5
- pK_a for CN^- = 4.61
- Initial concentration of NaCN (acting as the base) = 0.1 moles in 1 liter

### Step 3: Calculate pK_b
To find the pK_b for CN^-, we use the relationship:
\[ \text{pK}_b = 14 - \text{pK}_a \]
Calculating this gives:
\[ \text{pK}_b = 14 - 4.61 = 9.39 \]

### Step 4: Set up the equation
Using the buffer equation, we can substitute the known values:
\[ 8.5 = 4.61 + \log\left(\frac{0.1 - A}{A}\right) \]
Where \( A \) is the moles of HCl added, which will also be the concentration of the acid formed.

### Step 5: Rearranging the equation
Subtract pK_a from both sides:
\[ 8.5 - 4.61 = \log\left(\frac{0.1 - A}{A}\right) \]
This simplifies to:
\[ 3.89 = \log\left(\frac{0.1 - A}{A}\right) \]

### Step 6: Convert logarithmic form to exponential form
Convert the logarithmic equation to its exponential form:
\[ \frac{0.1 - A}{A} = 10^{3.89} \]
Calculating \( 10^{3.89} \) gives approximately 7757.43.

### Step 7: Solve for A
Now we can set up the equation:
\[ 0.1 - A = 7757.43A \]
Combine like terms:
\[ 0.1 = 7758.43A \]
Now solve for \( A \):
\[ A = \frac{0.1}{7758.43} \approx 0.00001287 \text{ moles} \]

### Step 8: Final answer
Thus, the moles of HCl required to prepare the buffer solution of pH 8.5 is approximately:
\[ A \approx 0.00001287 \text{ moles} \]