How many moles of HCl will be required to preapare one litre of a buffer solution (containing NaCN and HCN) of pH 8.5 using 0.01 gram formula mass of NaCN .` K_a` for HCN `= 4.1 xx 10^(-10)? `

Explanation

To solve the problem of how many moles of HCl are required to prepare a 1-liter buffer solution containing NaCN and HCN with a pH of 8.5, we can follow these steps:

### Step 1: Understand the Buffer System
A buffer solution consists of a weak acid and its conjugate base. In this case, HCN is the weak acid and NaCN is the conjugate base. The presence of HCl will convert some of the NaCN into HCN.

### Step 2: Use the Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is given by:
\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \]

### Step 3: Calculate pKa
Given \( K_a \) for HCN is \( 4.1 \times 10^{-10} \):
\[ \text{pKa} = -\log(K_a) = -\log(4.1 \times 10^{-10}) \]

Calculating this:
\[ \text{pKa} \approx 9.39 \]

### Step 4: Set Up the Equation
Substituting the known values into the Henderson-Hasselbalch equation:
\[ 8.5 = 9.39 + \log \left( \frac{[NaCN] - \alpha}{\alpha} \right) \]
Where \( \alpha \) is the moles of HCl added.

### Step 5: Rearranging the Equation
Rearranging gives:
\[ 8.5 - 9.39 = \log \left( \frac{0.01 - \alpha}{\alpha} \right) \]
\[ -0.89 = \log \left( \frac{0.01 - \alpha}{\alpha} \right) \]

### Step 6: Convert Logarithm to Exponential Form
Converting the logarithmic equation to its exponential form:
\[ \frac{0.01 - \alpha}{\alpha} = 10^{-0.89} \]
Calculating \( 10^{-0.89} \):
\[ 10^{-0.89} \approx 0.129 \]

### Step 7: Solve for α
Now we can set up the equation:
\[ 0.01 - \alpha = 0.129\alpha \]
\[ 0.01 = 0.129\alpha + \alpha \]
\[ 0.01 = 1.129\alpha \]
\[ \alpha = \frac{0.01}{1.129} \approx 0.00885 \text{ moles} \]

### Step 8: Conclusion
Thus, the number of moles of HCl required to prepare the buffer solution is approximately \( 0.00885 \) moles.