The pH of a solution obtained by mixing 100 mL of 0.2 M `CH_3COOH` with 100 mL of 0.2 M NaOH would be :(`pK_a` for `CH_3COOH`=4.74)

Options

A. 4.74
B. 8.87
C. `9.10`
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 M sodium hydroxide (NaOH), we can follow these steps:

### Step 1: Calculate moles of CH₃COOH and NaOH
- Volume of CH₃COOH = 100 mL = 0.1 L
- Concentration of CH₃COOH = 0.2 M
- Moles of CH₃COOH = Volume × Concentration = 0.1 L × 0.2 mol/L = 0.02 moles

- Volume of NaOH = 100 mL = 0.1 L
- Concentration of NaOH = 0.2 M
- Moles of NaOH = Volume × Concentration = 0.1 L × 0.2 mol/L = 0.02 moles

### Step 2: Determine the reaction
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 moles), they will completely react with each other.

### Step 3: Calculate the concentration of the acetate ion (CH₃COO⁻)
After the reaction, we are left with a solution of sodium acetate (CH₃COONa). 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 acetate ions (CH₃COO⁻) in the solution is:
\[ \text{Concentration of CH}_3\text{COO}^- = \frac{\text{Moles of CH}_3\text{COO}^-}{\text{Total Volume}} = \frac{0.02 \text{ moles}}{0.2 \text{ L}} = 0.1 \text{ M} \]

### Step 4: Calculate the pH using the Henderson-Hasselbalch equation
Since we have a weak acid (acetic acid) and its salt (sodium acetate), we can use the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]

Where:
- \(\text{pK}_a\) for acetic acid = 4.74
- \([\text{A}^-]\) = concentration of acetate ion = 0.1 M
- \([\text{HA}]\) = concentration of acetic acid = 0 M (since it has completely reacted)

Since the concentration of acetic acid is 0, we cannot directly use the equation. However, we can consider that the acetate ion will hydrolyze in water to produce some acetic acid and hydroxide ions.

### Step 5: Calculate the pH using the hydrolysis of acetate
The hydrolysis reaction of acetate is:
\[ \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^- \]

Using the relation:
\[ K_b = \frac{K_w}{K_a} \]
Where:
- \(K_w = 1.0 \times 10^{-14}\)
- \(K_a = 1.8 \times 10^{-5}\) (for acetic acid)

Calculating \(K_b\):
\[ K_b = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10} \]

### Step 6: Set up the equilibrium expression
Let \(x\) be the concentration of OH⁻ produced:
\[ K_b = \frac{x^2}{0.1 - x} \approx \frac{x^2}{0.1} \]
\[ 5.56 \times 10^{-10} = \frac{x^2}{0.1} \]
\[ x^2 = 5.56 \times 10^{-11} \]
\[ x = \sqrt{5.56 \times 10^{-11}} \approx 7.45 \times 10^{-6} \]

### Step 7: Calculate pOH and then pH
\[ \text{pOH} = -\log(7.45 \times 10^{-6}) \approx 5.13 \]
\[ \text{pH} = 14 - \text{pOH} = 14 - 5.13 = 8.87 \]

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