The `pK_b` value of ammonium hydroxide is 4.75. An aqueous solution of ammonium hydroxide is titrated with HCl. The pH of the ammonium hydroxide has been neutralized will be

Options

A. 9.25
B. 8.25
C. `7.50`
D. 4.75

Answer

To solve the problem, we need to determine the pH of an aqueous solution of ammonium hydroxide that has been neutralized by hydrochloric acid (HCl). Here are the steps to arrive at the solution:

### Step-by-Step Solution:

1. **Identify the Reaction**:
Ammonium hydroxide (NH4OH) is a weak base, and hydrochloric acid (HCl) is a strong acid. When they react, they form ammonium chloride (NH4Cl) and water (H2O):
\[ \text{NH}_4\text{OH} + \text{HCl} \rightarrow \text{NH}_4\text{Cl} + \text{H}_2\text{O} \]

2. **Understanding the Neutralization Point**:
At the neutralization point, the amount of ammonium hydroxide that has reacted with hydrochloric acid will equal the amount of ammonium chloride formed. Therefore, the concentrations of NH4+ (from NH4Cl) and OH- (from NH4OH) will be equal.

3. **Using the pK_b Value**:
The given pK_b value of ammonium hydroxide is 4.75. To find the pOH at the neutralization point, we can use the Henderson-Hasselbalch equation:
\[ \text{pOH} = \text{pK}_b + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
At the neutralization point, the concentration of salt (NH4Cl) is equal to the concentration of the acid (HCl), which means:
\[ \frac{[\text{Salt}]}{[\text{Acid}]} = 1 \]
Thus, the logarithm term becomes:
\[ \log(1) = 0 \]
Therefore:
\[ \text{pOH} = \text{pK}_b = 4.75 \]

4. **Calculating pH**:
We know that:
\[ \text{pH} + \text{pOH} = 14 \]
Substituting the value of pOH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 4.75 = 9.25 \]

5. **Final Answer**:
The pH of the ammonium hydroxide solution after neutralization with hydrochloric acid is:
\[ \text{pH} = 9.25 \]