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 determine the volume ratio of `NH4OH` and `HCl` when mixed to create a total volume of 300 mL, given that the pH of the mixture is 9.26 and the `pKa` of `NH4+` is also 9.26.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction between `NH4OH` (ammonium hydroxide) and `HCl` (hydrochloric acid) can be represented as:
\[ NH4OH + HCl \rightarrow NH4Cl + H2O \]
Here, `NH4OH` is a weak base, and `HCl` is a strong acid.

2. **Setting Up Variables**:
Let \( V \) be the volume of `HCl` in mL. Therefore, the volume of `NH4OH` will be \( 300 - V \) mL since the total volume is 300 mL.

3. **Calculating Moles**:
Since both solutions are 1 M, the number of moles can be calculated as:
- Moles of `NH4OH` = \( 1 \, \text{mol/L} \times (300 - V) \, \text{L} = 300 - V \) moles
- Moles of `HCl` = \( 1 \, \text{mol/L} \times V \, \text{L} = V \) moles

4. **Determining Limiting Reactant**:
Given that the pH is 9.26, which is above 7, it indicates that `HCl` is the limiting reactant because if `HCl` were in excess, the pH would be below 7. Therefore, all of `HCl` will react, and some `NH4OH` will remain.

5. **Using the Henderson-Hasselbalch Equation**:
Since we have a buffer solution of `NH4OH` and `NH4Cl`, we can use the Henderson-Hasselbalch equation:
\[ pH = pKa + \log\left(\frac{[\text{base}]}{[\text{acid}]}\right) \]
Here, the base is `NH4OH` and the acid is `NH4Cl`.

6. **Substituting Known Values**:
Given that \( pH = 9.26 \) and \( pKa = 9.26 \):
\[ 9.26 = 9.26 + \log\left(\frac{300 - V}{V}\right) \]
This simplifies to:
\[ \log\left(\frac{300 - V}{V}\right) = 0 \]
Therefore, we have:
\[ \frac{300 - V}{V} = 1 \]

7. **Solving for V**:
From the equation \( 300 - V = V \):
\[ 300 = 2V \implies V = 150 \, \text{mL} \]
Thus, the volume of `HCl` is 150 mL and the volume of `NH4OH` is:
\[ 300 - V = 300 - 150 = 150 \, \text{mL} \]

8. **Calculating the Volume Ratio**:
The volume ratio of `NH4OH` to `HCl` is:
\[ \text{Volume Ratio} = \frac{150}{150} = 1:1 \]

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