1 M `NH_(4)OH and 1 M HCl` are mixed to make a total volume of 300 mL. If pH of the mixture is 9.26 and `pK_(A)(NH_(4)^(+))=9.26` then what would be the volume ratio of `NH_(4)OH and HCl`

Options

A. `2:1`
B. `1:2`
C. `2:3`
D. `3:2`

Answer

To solve the problem, we need to find the volume ratio of `NH₄OH` and `HCl` when mixed to form a solution with a specific pH. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Reaction
When `NH₄OH` (a weak base) is mixed with `HCl` (a strong acid), they react to form `NH₄Cl` (a salt) and water. The reaction can be represented as:
\[ \text{NH}_4\text{OH} + \text{HCl} \rightarrow \text{NH}_4\text{Cl} + \text{H}_2\text{O} \]

### Step 2: Define Variables
Let:
- \( V \) = volume of `HCl` in mL
- Therefore, the volume of `NH₄OH` will be \( 300 - V \) mL (since the total volume is 300 mL).

### Step 3: Calculate pOH and pKb
Given that the pH of the mixture is 9.26, we can calculate the pOH:
\[ \text{pOH} = 14 - \text{pH} = 14 - 9.26 = 4.74 \]

Since \( pK_a(NH_4^+) = 9.26 \), we can find \( pK_b(NH_4OH) \) using the relationship:
\[ pK_w = pK_a + pK_b \]
Where \( pK_w = 14 \):
\[ pK_b = 14 - pK_a = 14 - 9.26 = 4.74 \]

### Step 4: Use the Henderson-Hasselbalch Equation
For a basic buffer solution, the Henderson-Hasselbalch equation is:
\[ \text{pOH} = pK_b + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) \]
Where:
- \([\text{Salt}] = \text{NH}_4\text{Cl}\) (formed from `HCl`)
- \([\text{Base}] = \text{NH}_4\text{OH}\)

### Step 5: Substitute Values
From the reaction, we know:
- Moles of `NH₄Cl` formed = \( V \) (since 1 M solution)
- Moles of `NH₄OH` remaining = \( 1 \times (300 - V) \)

Substituting into the equation:
\[ 4.74 = 4.74 + \log\left(\frac{V}{300 - V}\right) \]

### Step 6: Simplify the Equation
Since \( pOH \) and \( pK_b \) are equal, we can simplify:
\[ 0 = \log\left(\frac{V}{300 - V}\right) \]
Taking antilog:
\[ 1 = \frac{V}{300 - V} \]

### Step 7: Solve for V
Cross-multiplying gives:
\[ 300 - V = V \]
\[ 300 = 2V \]
\[ V = 150 \, \text{mL} \]

### Step 8: Find Volumes of Each Component
- Volume of `HCl` = \( V = 150 \, \text{mL} \)
- Volume of `NH₄OH` = \( 300 - V = 300 - 150 = 150 \, \text{mL} \)

### Step 9: Calculate the Volume Ratio
The volume ratio of `NH₄OH` to `HCl` is:
\[ \text{Volume Ratio} = \frac{150}{150} = 1:1 \]

### Conclusion
The volume ratio of `NH₄OH` to `HCl` is **1:1**.