one litre of an aqueous solution contains 0.15 mole of `CH_(3)COOH(pK_(a=4.8))` and 0.15 mole of `CH_(3)` COONa. After the addition of 0.05 mole of solid NaOH to this solution, the pH will be :

Options

A. 4.5
B. 4.8
C. 5.1
D. 5.4

Answer

To solve the problem, we need to calculate the pH of a buffer solution after the addition of NaOH. Here's a step-by-step breakdown of the solution:

### Step 1: Identify the components of the buffer solution
We have:
- Acetic acid (CH₃COOH) = 0.15 moles
- Sodium acetate (CH₃COONa) = 0.15 moles

### Step 2: Understand the effect of adding NaOH
When we add 0.05 moles of NaOH to the solution:
- NaOH dissociates into Na⁺ and OH⁻ ions.
- The OH⁻ ions will react with the H⁺ ions from the acetic acid, reducing the concentration of H⁺ ions.

### Step 3: Calculate the change in concentrations
1. **Decrease in acetic acid (CH₃COOH)**:
- The H⁺ ions from acetic acid will react with OH⁻ ions to form water.
- Therefore, the concentration of acetic acid decreases by 0.05 moles:
\[ \text{New concentration of CH₃COOH} = 0.15 - 0.05 = 0.10 \text{ moles} \]

2. **Increase in sodium acetate (CH₃COONa)**:
- The OH⁻ ions do not affect the sodium acetate directly, but since NaOH provides Na⁺ ions, it can be considered that the concentration of acetate ions (from sodium acetate) effectively increases by the amount of NaOH added:
\[ \text{New concentration of CH₃COONa} = 0.15 + 0.05 = 0.20 \text{ moles} \]

### Step 4: Use the Henderson-Hasselbalch equation
The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]
Where:
- \(\text{pK}_a = 4.8\)
- \([\text{Base}] = [\text{CH₃COO}^-] = 0.20 \text{ moles}\)
- \([\text{Acid}] = [\text{CH₃COOH}] = 0.10 \text{ moles}\)

### Step 5: Substitute values into the equation
\[ \text{pH} = 4.8 + \log\left(\frac{0.20}{0.10}\right) \]
\[ \text{pH} = 4.8 + \log(2) \]
Using the approximate value \(\log(2) \approx 0.301\):
\[ \text{pH} = 4.8 + 0.301 = 5.101 \]

### Step 6: Round the pH value
The pH can be rounded to one decimal place:
\[ \text{pH} \approx 5.1 \]

### Final Answer:
The pH of the resulting solution will be **5.1**.

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