The pH of an aqueous solution of 0.01 M `CH_(3)COONH_(4)" at "25^@C` is `(K_(a) (CH_(3)COOH)=K_(b) (NH_(4)OH)=1.8 xx 10^(-5))`

Options

A. `gt 7`
B. `lt 7`
C. 7
D. 8

Answer

To find the pH of a 0.01 M solution of CH₃COONH₄ (ammonium acetate) at 25°C, we can use the relationship between the pKa and pKb of the weak acid and weak base involved in this solution. Here’s how to solve it step by step:

### Step 1: Identify the components
Ammonium acetate (CH₃COONH₄) is a salt formed from a weak acid (acetic acid, CH₃COOH) and a weak base (ammonium hydroxide, NH₄OH).

### Step 2: Write the dissociation equations
1. **Acetic acid dissociation:**
\[ CH₃COOH \rightleftharpoons H^+ + CH₃COO^- \]
The dissociation constant \( K_a \) is given as \( 1.8 \times 10^{-5} \).

2. **Ammonium hydroxide dissociation:**
\[ NH₄OH \rightleftharpoons NH₄^+ + OH^- \]
The dissociation constant \( K_b \) is also given as \( 1.8 \times 10^{-5} \).

### Step 3: Calculate pKa and pKb
To find pKa and pKb:
\[ pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5}) \approx 4.74 \]
\[ pK_b = -\log(K_b) = -\log(1.8 \times 10^{-5}) \approx 4.74 \]

### Step 4: Use the pH formula for weak acid and weak base
For a solution containing a weak acid and its conjugate base (or a weak base and its conjugate acid), the pH can be calculated using the formula:
\[ pH = 7 + \frac{1}{2}(pK_a - pK_b) \]

### Step 5: Substitute the values
Since \( pK_a \) and \( pK_b \) are equal:
\[ pH = 7 + \frac{1}{2}(4.74 - 4.74) = 7 + 0 = 7 \]

### Conclusion
The pH of the 0.01 M CH₃COONH₄ solution is **7**.

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