`100 ml` of `0.1 M CH_(3)COOH` are mixed with `100ml` of `0.1 M NaOH`, the `pH` of the resulting solution would be

Options

A. zero
B. `7`
C. `gt7`
D. `lt7`

Explanation

To solve the problem of finding the pH of the resulting solution when 100 mL of 0.1 M acetic acid (CH₃COOH) is mixed with 100 mL of 0.1 M sodium hydroxide (NaOH), we can follow these steps:

### Step 1: Determine the moles of acetic acid and sodium hydroxide

1. **Calculate the moles of CH₃COOH:**
\[ \text{Moles of CH₃COOH} = \text{Volume (L)} \times \text{Concentration (M)} = 0.1 \, \text{L} \times 0.1 \, \text{M} = 0.01 \, \text{moles} \]

2. **Calculate the moles of NaOH:**
\[ \text{Moles of NaOH} = \text{Volume (L)} \times \text{Concentration (M)} = 0.1 \, \text{L} \times 0.1 \, \text{M} = 0.01 \, \text{moles} \]

### Step 2: Determine the reaction between acetic acid and sodium hydroxide

The reaction between acetic acid (a weak acid) and sodium hydroxide (a strong base) can be represented as:
\[ \text{CH₃COOH} + \text{NaOH} \rightarrow \text{CH₃COONa} + \text{H₂O} \]

Since both reactants are present in equal moles (0.01 moles), they will completely react with each other.

### Step 3: Identify the resulting solution

After the reaction, we will have:
- 0 moles of CH₃COOH (since it is completely consumed)
- 0 moles of NaOH (since it is completely consumed)
- 0.01 moles of sodium acetate (CH₃COONa) formed.

### Step 4: Calculate the concentration of the resulting solution

The total volume of the solution after mixing is:
\[ \text{Total Volume} = 100 \, \text{mL} + 100 \, \text{mL} = 200 \, \text{mL} = 0.2 \, \text{L} \]

Now, calculate the concentration of sodium acetate:
\[ \text{Concentration of CH₃COONa} = \frac{\text{Moles}}{\text{Volume (L)}} = \frac{0.01 \, \text{moles}}{0.2 \, \text{L}} = 0.05 \, \text{M} \]

### Step 5: Determine the pH of the resulting solution

Sodium acetate is the salt of a weak acid (acetic acid) and a strong base (sodium hydroxide). In solution, it will hydrolyze to produce hydroxide ions (OH⁻), making the solution basic.

The hydrolysis reaction is:
\[ \text{CH₃COO}^- + \text{H₂O} \rightleftharpoons \text{CH₃COOH} + \text{OH}^- \]

To find the pH, we can use the Kb of the acetate ion. The Kb can be calculated from the Ka of acetic acid:
\[ K_a \cdot K_b = K_w \quad \text{(where } K_w = 1.0 \times 10^{-14}\text{ at 25°C)} \]
For acetic acid, \( K_a \approx 1.8 \times 10^{-5} \):
\[ K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.56 \times 10^{-10} \]

Using the Kb to find the concentration of OH⁻ produced:
\[ K_b = \frac{[OH^-]^2}{[CH₃COO^-]} \approx \frac{x^2}{0.05} \]
Setting \( K_b = 5.56 \times 10^{-10} \):
\[ 5.56 \times 10^{-10} = \frac{x^2}{0.05} \Rightarrow x^2 = 5.56 \times 10^{-10} \times 0.05 \Rightarrow x^2 = 2.78 \times 10^{-11} \]
\[ x = \sqrt{2.78 \times 10^{-11}} \approx 5.27 \times 10^{-6} \, \text{M} \]

Now, calculate the pOH:
\[ \text{pOH} = -\log[OH^-] = -\log(5.27 \times 10^{-6}) \approx 5.28 \]

Finally, calculate the pH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 5.28 \approx 8.72 \]

### Final Answer
The pH of the resulting solution is approximately **8.72**.