A buffer solution contains 100mL of 0.01 M `CH_(3)COOH` and 200mL of 0.02 M `CH_(3)COONa`. 700mL of water is added, pH before and after dilution are `(pK_(a)=4.74)`

Options

A. 5.04,5.04
B. 5.04,0.504
C. 5.04,1.54
D. 5.34,5.34

Explanation

To solve the problem, we need to calculate the pH of the buffer solution before and after dilution. We will use the Henderson-Hasselbalch equation:

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

### Step 1: Calculate the initial concentrations of acetic acid and sodium acetate.

1. **Volume of Acetic Acid (CH₃COOH)**: 100 mL of 0.01 M
- Moles of CH₃COOH = Concentration × Volume = 0.01 M × 0.1 L = 0.001 moles

2. **Volume of Sodium Acetate (CH₃COONa)**: 200 mL of 0.02 M
- Moles of CH₃COONa = Concentration × Volume = 0.02 M × 0.2 L = 0.004 moles

### Step 2: Calculate the concentrations before dilution.

- Total volume before dilution = 100 mL + 200 mL = 300 mL = 0.3 L

1. **Concentration of Acetic Acid (CH₃COOH)**:
\[ [\text{CH}_3\text{COOH}] = \frac{0.001 \text{ moles}}{0.3 \text{ L}} = 0.00333 \text{ M} \]

2. **Concentration of Sodium Acetate (CH₃COONa)**:
\[ [\text{CH}_3\text{COONa}] = \frac{0.004 \text{ moles}}{0.3 \text{ L}} = 0.01333 \text{ M} \]

### Step 3: Calculate the pH before dilution.

Using the Henderson-Hasselbalch equation:
\[ \text{pH} = 4.74 + \log\left(\frac{0.01333}{0.00333}\right) \]

Calculating the log term:
\[ \frac{0.01333}{0.00333} = 4 \]
\[ \log(4) \approx 0.602 \]

So the pH before dilution is:
\[ \text{pH} = 4.74 + 0.602 = 5.342 \]

### Step 4: Calculate the concentrations after dilution.

When 700 mL of water is added, the total volume becomes:
\[ 300 \text{ mL} + 700 \text{ mL} = 1000 \text{ mL} = 1 \text{ L} \]

1. **New concentration of Acetic Acid (CH₃COOH)**:
\[ [\text{CH}_3\text{COOH}] = \frac{0.001 \text{ moles}}{1 \text{ L}} = 0.001 \text{ M} \]

2. **New concentration of Sodium Acetate (CH₃COONa)**:
\[ [\text{CH}_3\text{COONa}] = \frac{0.004 \text{ moles}}{1 \text{ L}} = 0.004 \text{ M} \]

### Step 5: Calculate the pH after dilution.

Using the Henderson-Hasselbalch equation again:
\[ \text{pH} = 4.74 + \log\left(\frac{0.004}{0.001}\right) \]

Calculating the log term:
\[ \frac{0.004}{0.001} = 4 \]
\[ \log(4) \approx 0.602 \]

So the pH after dilution is:
\[ \text{pH} = 4.74 + 0.602 = 5.342 \]

### Conclusion

The pH before and after dilution remains approximately the same at 5.342, demonstrating the buffer's ability to resist changes in pH upon dilution.

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