20.0 L of 0.2 M weak acid `(pK_(a)=5.0)` is titrated against 0.2 M strong base. What is the pH at the equivalence point ?

Options

A. `5.0`
B. `7.0`
C. `9.0`
D. `11.0`

Explanation

To find the pH at the equivalence point of the titration of a weak acid with a strong base, we can follow these steps:

### Step 1: Identify the components
We have:
- Volume of weak acid (V_a) = 20.0 L
- Concentration of weak acid (C_a) = 0.2 M
- pK_a of the weak acid = 5.0
- Concentration of strong base (C_b) = 0.2 M

### Step 2: Calculate the moles of the weak acid
Moles of weak acid (n_a) can be calculated using the formula:
\[ n_a = C_a \times V_a \]
\[ n_a = 0.2 \, \text{mol/L} \times 20.0 \, \text{L} = 4.0 \, \text{mol} \]

### Step 3: Determine the moles of strong base needed
At the equivalence point, the moles of strong base will equal the moles of weak acid:
\[ n_b = n_a = 4.0 \, \text{mol} \]

### Step 4: Calculate the concentration of the salt formed
The salt formed will be from the weak acid and strong base. The total volume at the equivalence point will be the sum of the volumes of the weak acid and the strong base. The volume of strong base needed can be calculated as:
\[ V_b = \frac{n_b}{C_b} = \frac{4.0 \, \text{mol}}{0.2 \, \text{mol/L}} = 20.0 \, \text{L} \]

Total volume at equivalence point:
\[ V_{total} = V_a + V_b = 20.0 \, \text{L} + 20.0 \, \text{L} = 40.0 \, \text{L} \]

The concentration of the salt (C_s) formed will be:
\[ C_s = \frac{n_a}{V_{total}} = \frac{4.0 \, \text{mol}}{40.0 \, \text{L}} = 0.1 \, \text{M} \]

### Step 5: Calculate the pH at the equivalence point
At the equivalence point, the pH can be calculated using the formula:
\[ \text{pH} = 7 + \frac{1}{2} \text{pK}_a + \frac{1}{2} \log C_s \]

Substituting the values:
- pK_a = 5.0
- C_s = 0.1 M

Calculating:
\[ \text{pH} = 7 + \frac{1}{2} \times 5 + \frac{1}{2} \log(0.1) \]
\[ \text{pH} = 7 + 2.5 + \frac{1}{2} \times (-1) \]
\[ \text{pH} = 7 + 2.5 - 0.5 \]
\[ \text{pH} = 9.0 \]

### Final Answer:
The pH at the equivalence point is **9.0**.