The pH of an acetic acid ` + ` sodium acetate buffer is given by `pH = pK_(a) + log . "[Salt]"/"[Acid]" " where " K_(a)` of acetic acid ` = 1.8 xx 10^(-5)`
If [Salt] = [Acid] ` = 0.1` M, then the pH of the solution would be about

Options

A. 7
B. `4.7`
C. `5.3`
D. `1.4`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Henderson-Hasselbalch Equation
The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]

### Step 2: Identify Given Values
From the problem, we know:
- \( K_a \) of acetic acid = \( 1.8 \times 10^{-5} \)
- Concentration of Salt = Concentration of Acid = \( 0.1 \, M \)

### Step 3: Calculate \( pK_a \)
To find \( pK_a \), we use the formula:
\[ \text{pK}_a = -\log(K_a) \]
Substituting the value of \( K_a \):
\[ \text{pK}_a = -\log(1.8 \times 10^{-5}) \]

### Step 4: Break Down the Logarithm
Using properties of logarithms:
\[ \text{pK}_a = -\log(1.8) - \log(10^{-5}) \]
\[ \text{pK}_a = -\log(1.8) + 5 \]

### Step 5: Calculate \( -\log(1.8) \)
Using a calculator or logarithm table:
\[ -\log(1.8) \approx -0.255 \]
Thus,
\[ \text{pK}_a \approx -0.255 + 5 = 4.745 \]

### Step 6: Substitute Values into the Henderson-Hasselbalch Equation
Since the concentrations of salt and acid are equal, we have:
\[ \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) = \log(1) = 0 \]
Therefore, the pH can be simplified to:
\[ \text{pH} = \text{pK}_a + 0 = \text{pK}_a \]

### Step 7: Final Calculation
Thus, the pH of the solution is approximately:
\[ \text{pH} \approx 4.745 \approx 4.74 \]

### Conclusion
The pH of the acetic acid and sodium acetate buffer solution is approximately **4.74**.

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