The pH of a solution obtained by mixing 100 ml of 0.2 M `CH_3COOH` with 100 ml of 0.2 N NaOH will be
`(pK_a "for " CH_3COOH=4.74 and log 2 =0.301)`

Options

A. 4.74
B. 8.87
C. 9.1
D. 8.57

Answer

To find the pH of a solution obtained by mixing 100 mL of 0.2 M acetic acid (CH₃COOH) with 100 mL of 0.2 N sodium hydroxide (NaOH), we will follow these steps:

### Step 1: Calculate the moles of CH₃COOH and NaOH
- **Moles of CH₃COOH**:
\[ \text{Moles of CH}_3\text{COOH} = \text{Molarity} \times \text{Volume} = 0.2 \, \text{mol/L} \times 0.1 \, \text{L} = 0.02 \, \text{mol} \]

- **Moles of NaOH**:
\[ \text{Moles of NaOH} = \text{Normality} \times \text{Volume} = 0.2 \, \text{N} \times 0.1 \, \text{L} = 0.02 \, \text{mol} \]

### Step 2: Determine the reaction between CH₃COOH and NaOH
The reaction between acetic acid and sodium hydroxide is:
\[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]
Since both reactants are present in equal moles (0.02 mol), they will completely react with each other.

### Step 3: Calculate the concentration of the acetate ion (CH₃COO⁻)
After the reaction, we will have the acetate ion (CH₃COO⁻) in solution. The total volume of the solution after mixing is:
\[ \text{Total Volume} = 100 \, \text{mL} + 100 \, \text{mL} = 200 \, \text{mL} = 0.2 \, \text{L} \]
The concentration of CH₃COO⁻ will be:
\[ \text{Concentration of CH}_3\text{COO}^- = \frac{\text{Moles of CH}_3\text{COO}^-}{\text{Total Volume}} = \frac{0.02 \, \text{mol}}{0.2 \, \text{L}} = 0.1 \, \text{M} \]

### Step 4: Use the Henderson-Hasselbalch equation to find the pH
The pH can be calculated using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
In this case, since all the acetic acid has reacted, we treat the acetate ion as the base and the concentration of acetic acid (HA) is 0:
\[ \text{pH} = \text{pK}_a + \log \left( \frac{0.1}{0} \right) \]
However, since we have no acetic acid left, we can consider the hydrolysis of the acetate ion:
\[ \text{pH} = \frac{1}{2} \left( \text{pK}_w + \text{pK}_a + \log [\text{CH}_3\text{COO}^-] \right) \]
Where \( \text{pK}_w = 14 \) and \( \text{pK}_a = 4.74 \):
\[ \text{pH} = \frac{1}{2} \left( 14 + 4.74 + \log(0.1) \right) \]
Since \( \log(0.1) = -1 \):
\[ \text{pH} = \frac{1}{2} \left( 14 + 4.74 - 1 \right) = \frac{1}{2} \left( 17.74 \right) = 8.87 \]

### Final Answer
The pH of the solution is approximately **8.87**.

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