In which volume ratio `NH_(4)Cl` and `NH_(4)OH` solutions (each 1 M ) should be mixed to get a buffer solution of pH 9.80 ? `(pK_(b) "of" NH_(4)OH=4.74)`

Options

A. `1:2.5`
B. `2.5:1`
C. `1:3.5`
D. `3.5:1`

Explanation

To solve the problem of determining the volume ratio of \( NH_4Cl \) and \( NH_4OH \) solutions (both 1 M) needed to create a buffer solution with a pH of 9.80, we can follow these steps:

### Step 1: Determine the pOH
Since we are given the pH, we first need to calculate the pOH:
\[ \text{pOH} = 14 - \text{pH} = 14 - 9.80 = 4.20 \]

### Step 2: Use the Henderson-Hasselbalch Equation
We can use the Henderson-Hasselbalch equation for basic buffers, which is given by:
\[ \text{pOH} = pK_b + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
Here, \( [\text{Salt}] \) corresponds to the concentration of \( NH_4Cl \) and \( [\text{Acid}] \) corresponds to the concentration of \( NH_4OH \).

### Step 3: Substitute Known Values
We know:
- \( pK_b = 4.74 \)
- \( \text{pOH} = 4.20 \)

Substituting these values into the equation:
\[ 4.20 = 4.74 + \log\left(\frac{[NH_4Cl]}{[NH_4OH]}\right) \]

### Step 4: Rearranging the Equation
Rearranging the equation to isolate the logarithmic term:
\[ \log\left(\frac{[NH_4Cl]}{[NH_4OH]}\right) = 4.20 - 4.74 = -0.54 \]

### Step 5: Convert Logarithm to Ratio
To eliminate the logarithm, we can exponentiate both sides:
\[ \frac{[NH_4Cl]}{[NH_4OH]} = 10^{-0.54} \]

Calculating \( 10^{-0.54} \):
\[ \frac{[NH_4Cl]}{[NH_4OH]} \approx 0.28 \]

### Step 6: Determine the Volume Ratio
The ratio \( \frac{[NH_4Cl]}{[NH_4OH]} = 0.28 \) implies that for every 1 part of \( NH_4OH \), we need 0.28 parts of \( NH_4Cl \). Therefore, the volume ratio of \( NH_4Cl \) to \( NH_4OH \) is:
\[ \text{Volume ratio} = \frac{0.28}{1} = 0.28:1 \]

### Conclusion
To achieve a buffer solution with a pH of 9.80, mix \( NH_4Cl \) and \( NH_4OH \) in a volume ratio of approximately \( 0.28:1 \).

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