The pH of a solution at `25°C` containing 0.20 M sodium acetate and 0.06 M acetic acid is `(pK_a for CH_3COOH = 4.74 and log 3 = 0.477)`

Options

A. 4.36
B. 5.26
C. 5.84
D. 6.32

Answer

To find the pH of a solution containing 0.20 M sodium acetate and 0.06 M acetic acid, we can use the Henderson-Hasselbalch equation, which is applicable for buffer solutions. The equation is given by:

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

### Step 1: Identify the components
- **Salt (Sodium Acetate)**: 0.20 M
- **Acid (Acetic Acid)**: 0.06 M
- **pK_a for Acetic Acid (CH₃COOH)**: 4.74

### Step 2: Plug in the values into the Henderson-Hasselbalch equation
Using the equation:

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

Substituting the known values:

\[ \text{pH} = 4.74 + \log\left(\frac{0.20}{0.06}\right) \]

### Step 3: Calculate the ratio of concentrations
Calculate the ratio of the concentrations of salt to acid:

\[ \frac{0.20}{0.06} = \frac{20}{6} = \frac{10}{3} \]

### Step 4: Calculate the logarithm
Now, we need to find the logarithm of the ratio:

\[ \log\left(\frac{10}{3}\right) = \log(10) - \log(3) = 1 - 0.477 = 0.523 \]

### Step 5: Substitute back into the pH equation
Now substitute this value back into the pH equation:

\[ \text{pH} = 4.74 + 0.523 \]

### Step 6: Final calculation
Perform the final addition:

\[ \text{pH} = 4.74 + 0.523 = 5.263 \]

### Conclusion
Thus, the pH of the solution is approximately **5.26**.

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