In what volume ratio should you mix 1.0 M solution of `NH_(4)CI` and `NH_(3)` to produce a buffer solution of pH 9.80 `[pK_(b) (NH_(3)) = 4.74]`

Options

A. `1 :3.5`
B. `3.5 : 1`
C. `2 : 1`
D. `1 : 2`

Answer

To solve the problem of mixing 1.0 M solutions of NH4Cl and NH3 to create a buffer solution with a pH of 9.80, we can use the Henderson-Hasselbalch equation. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Components
- **NH4Cl** is the conjugate acid of the weak base NH3.
- **NH3** is a weak base.
- We are creating a basic buffer solution.

### Step 2: Determine the pOH
Given that the pH of the buffer solution is 9.80, we can calculate the pOH:
\[ pOH = 14 - pH = 14 - 9.80 = 4.20 \]

### Step 3: Use the Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation for a basic buffer is given by:
\[ pOH = pK_b + \log \left( \frac{[NH4^+]}{[NH3]} \right) \]
Where:
- \( pK_b \) for NH3 is given as 4.74.
- Let \( V_{NH4Cl} \) be the volume of NH4Cl and \( V_{NH3} \) be the volume of NH3.

### Step 4: Substitute Known Values
Substituting the known values into the equation:
\[ 4.20 = 4.74 + \log \left( \frac{[NH4^+]}{[NH3]} \right) \]

### Step 5: Rearranging the Equation
Rearranging the equation to isolate the logarithmic term:
\[ \log \left( \frac{[NH4^+]}{[NH3]} \right) = 4.20 - 4.74 \]
\[ \log \left( \frac{[NH4^+]}{[NH3]} \right) = -0.54 \]

### Step 6: Convert Logarithm to Exponential Form
To eliminate the logarithm, we convert it to exponential form:
\[ \frac{[NH4^+]}{[NH3]} = 10^{-0.54} \]
Calculating \( 10^{-0.54} \):
\[ 10^{-0.54} \approx 0.291 \]

### Step 7: Express Concentrations in Terms of Volumes
Since both solutions are 1.0 M, the concentrations can be expressed in terms of volumes:
\[ \frac{V_{NH4Cl}}{V_{NH3}} = 0.291 \]

### Step 8: Find the Volume Ratio
This can be rearranged to find the volume ratio:
\[ \frac{V_{NH4Cl}}{V_{NH3}} = 0.291 \implies V_{NH4Cl} = 0.291 \cdot V_{NH3} \]
Thus, the volume ratio of \( V_{NH4Cl} : V_{NH3} \) is:
\[ 1 : 3.43 \quad (\text{approximately } 1 : 3.5) \]

### Conclusion
The final volume ratio in which to mix the solutions is approximately:
**1 : 3.5**