`20 mol` of `M//10 CH_(3)COOH` solution is titrated with `M//10 NaOH` solution. After addition of `16 mL` solution of `NaOH`. What is the `pH` of the solution `(pK_(a) = 4.74)`

Options

A. `5.05`
B. `4.15`
C. `4.75`
D. `5.35`

Explanation

To find the pH of the solution after titrating 20 mL of 0.1 M acetic acid (CH₃COOH) with 0.1 M NaOH, follow these steps:

### Step 1: Calculate the number of millimoles of acetic acid (CH₃COOH)
- **Formula**: Millimoles = Molarity × Volume (in mL)
- For CH₃COOH:
\[ \text{Millimoles of CH₃COOH} = 0.1 \, \text{mol/L} \times 20 \, \text{mL} = 2 \, \text{mmol} \]

### Step 2: Calculate the number of millimoles of NaOH added
- For NaOH:
\[ \text{Millimoles of NaOH} = 0.1 \, \text{mol/L} \times 16 \, \text{mL} = 1.6 \, \text{mmol} \]

### Step 3: Determine the remaining acetic acid and formed salt after neutralization
- Since NaOH is a strong base, it will completely neutralize the acetic acid:
- Millimoles of CH₃COOH remaining after reaction:
\[ \text{Remaining CH₃COOH} = 2 \, \text{mmol} - 1.6 \, \text{mmol} = 0.4 \, \text{mmol} \]
- Millimoles of sodium acetate (CH₃COONa) formed:
\[ \text{Millimoles of CH₃COONa} = 1.6 \, \text{mmol} \]

### Step 4: Use the Henderson-Hasselbalch equation to calculate pH
- The Henderson-Hasselbalch equation is:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
- Here, \(\text{pK}_a = 4.74\), \([\text{Salt}] = 1.6 \, \text{mmol}\), and \([\text{Acid}] = 0.4 \, \text{mmol}\).
- Substitute the values into the equation:
\[ \text{pH} = 4.74 + \log\left(\frac{1.6}{0.4}\right) \]
- Calculate the ratio:
\[ \frac{1.6}{0.4} = 4 \]
- Now substitute back into the equation:
\[ \text{pH} = 4.74 + \log(4) \]
- Since \(\log(4) \approx 0.6\):
\[ \text{pH} = 4.74 + 0.6 = 5.34 \]

### Final Answer:
The pH of the solution after adding 16 mL of NaOH is approximately **5.34**.

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