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)]`

Answer

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

### Step 1: Determine the volumes of acid and base used
We have:
- Volume of acetic acid (V₁) = 25 mL = 0.025 L
- Molarity of acetic acid (C₁) = 0.1 M
- Molarity of sodium hydroxide (C₂) = 0.05 M

At the equivalence point, the moles of acetic acid will equal the moles of sodium hydroxide.

Using the formula for moles:
\[ \text{Moles of CH}_3\text{COOH} = C₁ \times V₁ = 0.1 \, \text{mol/L} \times 0.025 \, \text{L} = 0.0025 \, \text{mol} \]

Using the formula for moles of NaOH:
\[ \text{Moles of NaOH} = C₂ \times V₂ \]

Setting the moles equal:
\[ 0.0025 \, \text{mol} = 0.05 \, \text{mol/L} \times V₂ \]

Solving for \( V₂ \):
\[ V₂ = \frac{0.0025 \, \text{mol}}{0.05 \, \text{mol/L}} = 0.050 \, \text{L} = 50 \, \text{mL} \]

### Step 2: Calculate the total volume at the equivalence point
The total volume (V_total) at the equivalence point is the sum of the volumes of the acid and the base:
\[ V_{\text{total}} = V₁ + V₂ = 25 \, \text{mL} + 50 \, \text{mL} = 75 \, \text{mL} = 0.075 \, \text{L} \]

### Step 3: Calculate the concentration of the salt formed (CH₃COONa)
At the equivalence point, all the acetic acid has been converted to sodium acetate (CH₃COONa). The concentration of CH₃COONa can be calculated as follows:
\[ \text{Concentration of CH}_3\text{COONa} = \frac{\text{Moles of CH}_3\text{COOH}}{V_{\text{total}}} = \frac{0.0025 \, \text{mol}}{0.075 \, \text{L}} = \frac{0.0025}{0.075} = \frac{1}{30} \, \text{mol/L} \approx 0.0333 \, \text{mol/L} \]

### Step 4: Calculate the concentration of H⁺ ions at the equivalence point
Sodium acetate (CH₃COONa) will hydrolyze in water to produce CH₃COO⁻ ions, which will react with water:
\[ \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^- \]

The equilibrium expression for this reaction can be written as:
\[ K_b = \frac{K_w}{K_a} \]
Where:
- \( K_w = 1.0 \times 10^{-14} \)
- \( K_a = 1.8 \times 10^{-5} \)

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

Using the hydrolysis equation, we can set up the expression for \( K_b \):
\[ K_b = \frac{[CH_3COOH][OH^-]}{[CH_3COO^-]} \]

Let \( x \) be the concentration of OH⁻ produced:
\[ K_b = \frac{x^2}{C - x} \approx \frac{x^2}{C} \quad \text{(since } x \text{ is small compared to } C\text{)} \]

Where \( C \) is the concentration of CH₃COO⁻:
\[ 5.56 \times 10^{-10} = \frac{x^2}{0.0333} \]

Solving for \( x \):
\[ x^2 = 5.56 \times 10^{-10} \times 0.0333 \]
\[ x^2 \approx 1.85 \times 10^{-11} \]
\[ x \approx 4.30 \times 10^{-6} \, \text{mol/L} \]

### Step 5: Calculate the pH
Since \( x \) represents the concentration of OH⁻, we can find the pOH:
\[ \text{pOH} = -\log[OH^-] = -\log(4.30 \times 10^{-6}) \approx 5.37 \]

Now, we can find the pH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 5.37 = 8.63 \]

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