Calculate the pH at the equivalence point when a solution of 0.1 M acetic acid is titrated with a solution of 0.1 M NaOH. `K_(a)` for acid `=1.9xx10^(-5)`.

Explanation

To calculate the pH at the equivalence point when titrating 0.1 M acetic acid with 0.1 M NaOH, we follow these steps:

### Step 1: Identify the Reaction
When acetic acid (CH₃COOH) is titrated with sodium hydroxide (NaOH), they react to form sodium acetate (CH₃COONa) and water. The reaction can be represented as:
\[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]

### Step 2: Determine the Concentration of the Salt
At the equivalence point, all the acetic acid will be converted to sodium acetate. Since both solutions are 0.1 M and equal volumes are used, the concentration of sodium acetate formed will be:
\[ \text{Concentration of CH}_3\text{COONa} = \frac{0.1 \, \text{mol/L} \times V}{V + V} = \frac{0.1}{2} = 0.05 \, \text{mol/L} \]

### Step 3: Hydrolysis of the Salt
Sodium acetate is a salt of a weak acid (acetic acid) and a strong base (sodium hydroxide). It will undergo hydrolysis in water:
\[ \text{CH}_3\text{COONa} \, \text{(aq)} \rightarrow \text{CH}_3\text{COO}^- \, \text{(aq)} + \text{Na}^+ \, \text{(aq)} \]

The acetate ion (CH₃COO⁻) will react with water to produce hydroxide ions (OH⁻):
\[ \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^- \]

### Step 4: Calculate the Hydrolysis Constant
The hydrolysis constant (K_h) can be calculated using the formula:
\[ K_h = \frac{K_w}{K_a} \]
Where:
- \( K_w = 1.0 \times 10^{-14} \) (ion product of water)
- \( K_a = 1.9 \times 10^{-5} \) (dissociation constant of acetic acid)

Calculating K_h:
\[ K_h = \frac{1.0 \times 10^{-14}}{1.9 \times 10^{-5}} = 5.26 \times 10^{-10} \]

### Step 5: Set Up the Hydrolysis Equation
Let \( x \) be the concentration of OH⁻ produced from the hydrolysis of sodium acetate. The equilibrium expression for the hydrolysis reaction is:
\[ K_h = \frac{x^2}{C - x} \]
Where \( C \) is the concentration of the salt (0.05 M).

Assuming \( x \) is small compared to \( C \):
\[ K_h \approx \frac{x^2}{C} \]
Thus:
\[ x^2 = K_h \cdot C \]
\[ x^2 = (5.26 \times 10^{-10}) \cdot (0.05) = 2.63 \times 10^{-11} \]

### Step 6: Solve for x
Taking the square root:
\[ x = \sqrt{2.63 \times 10^{-11}} \approx 5.12 \times 10^{-6} \, \text{mol/L} \]

### Step 7: Calculate pOH and pH
Now, we can calculate pOH:
\[ \text{pOH} = -\log[OH^-] = -\log(5.12 \times 10^{-6}) \approx 5.29 \]

Using the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]
Thus:
\[ \text{pH} = 14 - 5.29 = 8.71 \]

### Final Answer
The pH at the equivalence point is approximately **8.71**.

---