The pH of a buffer solution containing 0.2 mole per litre `CH_(3)COONa` and 1.5 mole per litre `CH_(3)COOH` is (`K_(a)` for acetic acid is `1.8 xx 10^(-5)`)

Options

A. 4.87
B. 5.8
C. 2.4
D. 9.2

Explanation

To find the pH of the buffer solution containing sodium acetate (CH₃COONa) and acetic acid (CH₃COOH), we can use the Henderson-Hasselbalch equation:

\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]

### Step 1: Calculate pKa
The first step is to calculate the pKa from the given Ka value of acetic acid.

\[ \text{pKa} = -\log(K_a) \]

Given:
\[ K_a = 1.8 \times 10^{-5} \]

Calculating pKa:
\[ \text{pKa} = -\log(1.8 \times 10^{-5}) \approx 4.74 \]

### Step 2: Identify concentrations
Next, we identify the concentrations of the salt (sodium acetate) and the acid (acetic acid):

- Concentration of salt (CH₃COONa) = 0.2 mol/L
- Concentration of acid (CH₃COOH) = 1.5 mol/L

### Step 3: Substitute values into the Henderson-Hasselbalch equation
Now we can substitute the values into the Henderson-Hasselbalch equation:

\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]

Substituting the values:
\[ \text{pH} = 4.74 + \log\left(\frac{0.2}{1.5}\right) \]

### Step 4: Calculate the log term
Calculating the ratio:
\[ \frac{0.2}{1.5} \approx 0.1333 \]

Now, calculate the logarithm:
\[ \log(0.1333) \approx -0.877 \]

### Step 5: Final pH calculation
Now substitute this value back into the pH equation:
\[ \text{pH} = 4.74 - 0.877 \approx 3.863 \]

### Final Answer
The pH of the buffer solution is approximately **3.86**.

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