Calculate the pH at the equivalence point during the titration of `0.1M, 25 mL CH_(3)COOH` with `0.05M NaOH` solution. `[K_(a)(CH_(3)COOH) = 1.8 xx 10^(-5)]`

Options

A. `9.63`
B. `8.63`
C. `10.63`
D. `11.63`

Explanation

To calculate the pH at the equivalence point during the titration of 0.1 M acetic acid (CH₃COOH) with 0.05 M NaOH, we can follow these steps:

### Step 1: Calculate the moles of acetic acid
First, we need to find the number of moles of acetic acid present in the solution.

\[ \text{Moles of CH}_3\text{COOH} = \text{Concentration} \times \text{Volume} \]
Given:
- Concentration of CH₃COOH = 0.1 M
- Volume of CH₃COOH = 25 mL = 0.025 L

\[ \text{Moles of CH}_3\text{COOH} = 0.1 \, \text{mol/L} \times 0.025 \, \text{L} = 0.0025 \, \text{mol} \]

### Step 2: Calculate the volume of NaOH required to reach the equivalence point
At the equivalence point, the moles of acetic acid will equal the moles of NaOH.

\[ \text{Moles of NaOH} = \text{Moles of CH}_3\text{COOH} = 0.0025 \, \text{mol} \]

Now, we can find the volume of NaOH needed:

\[ \text{Volume of NaOH} = \frac{\text{Moles of NaOH}}{\text{Concentration of NaOH}} = \frac{0.0025 \, \text{mol}}{0.05 \, \text{mol/L}} = 0.050 \, \text{L} = 50 \, \text{mL} \]

### Step 3: Calculate the total volume at the equivalence point
The total volume at the equivalence point is the sum of the volumes of acetic acid and NaOH:

\[ \text{Total Volume} = 25 \, \text{mL} + 50 \, \text{mL} = 75 \, \text{mL} = 0.075 \, \text{L} \]

### Step 4: Calculate the concentration of the acetate ion (CH₃COO⁻)
At the equivalence point, all the acetic acid has been converted to its conjugate base, acetate ion (CH₃COO⁻). The concentration of acetate ion can be calculated as follows:

\[ \text{Concentration of CH}_3\text{COO}^- = \frac{\text{Moles of CH}_3\text{COOH}}{\text{Total Volume}} = \frac{0.0025 \, \text{mol}}{0.075 \, \text{L}} = 0.0333 \, \text{M} \]

### Step 5: Calculate the pH using the Kb of acetate ion
To find the pH, we need to calculate the concentration of hydroxide ions (OH⁻) produced by the hydrolysis of acetate ion. The equilibrium expression for the hydrolysis of acetate ion is:

\[ \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^- \]

Using the relation between Ka and Kb:

\[ K_w = K_a \times K_b \implies K_b = \frac{K_w}{K_a} \]
Where:
- \(K_w = 1.0 \times 10^{-14}\)
- \(K_a = 1.8 \times 10^{-5}\)

Calculating Kb:

\[ K_b = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.56 \times 10^{-10} \]

Now, we can set up the equilibrium expression for the hydrolysis:

\[ K_b = \frac{[CH_3COOH][OH^-]}{[CH_3COO^-]} \approx \frac{x^2}{0.0333} \]

Assuming \(x\) is small compared to 0.0333 M, we can simplify:

\[ 5.56 \times 10^{-10} = \frac{x^2}{0.0333} \]

Solving for \(x\):

\[ x^2 = 5.56 \times 10^{-10} \times 0.0333 \approx 1.85 \times 10^{-11} \]
\[ x \approx \sqrt{1.85 \times 10^{-11}} \approx 4.30 \times 10^{-6} \, \text{M} \quad (\text{This is } [OH^-]) \]

### Step 6: Calculate pOH and then pH
Now we can calculate pOH:

\[ pOH = -\log[OH^-] = -\log(4.30 \times 10^{-6}) \approx 5.37 \]

Finally, we find the pH:

\[ pH = 14 - pOH = 14 - 5.37 \approx 8.63 \]

### Final Answer
The pH at the equivalence point during the titration is approximately **8.63**.

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