pH of a solution of 0.1 M `[CH_3COONH_4(aq)]` is `[given: K_a(CH_3COOH) = K_b(NH_4OH) = 1.8 x 10^-5)]`

Options

A. 1
B. 9
C. 7
D. 13

Answer

To find the pH of a 0.1 M solution of ammonium acetate \([CH_3COONH_4(aq)]\), we can use the relationship between \(K_a\) and \(K_b\) for the weak acid and weak base that make up the salt. The formula we will use is:

\[ \text{pH} = 7 + \frac{1}{2}(\text{p}K_a - \text{p}K_b) \]

### Step 1: Calculate \(pK_a\)
Given \(K_a(CH_3COOH) = 1.8 \times 10^{-5}\):

\[ pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5}) \]

Using the properties of logarithms:

\[ pK_a = -\log(1.8) - \log(10^{-5}) = -\log(1.8) + 5 \]

Calculating \(-\log(1.8)\) (approximately):

\[ \log(1.8) \approx 0.2553 \quad \Rightarrow \quad -\log(1.8) \approx -0.2553 \]

Thus,

\[ pK_a \approx 5 - 0.2553 \approx 4.7447 \]

### Step 2: Calculate \(pK_b\)
Given \(K_b(NH_4OH) = 1.8 \times 10^{-5}\):

\[ pK_b = -\log(K_b) = -\log(1.8 \times 10^{-5}) = -\log(1.8) + 5 \]

Using the same calculation as for \(pK_a\):

\[ pK_b \approx 4.7447 \]

### Step 3: Substitute \(pK_a\) and \(pK_b\) into the pH formula
Now we can substitute \(pK_a\) and \(pK_b\) into the pH formula:

\[ \text{pH} = 7 + \frac{1}{2}(pK_a - pK_b) \]

Since \(pK_a \approx pK_b\):

\[ \text{pH} = 7 + \frac{1}{2}(4.7447 - 4.7447) = 7 + \frac{1}{2}(0) = 7 \]

### Final Answer
The pH of the solution is:

\[ \text{pH} = 7 \]