To solve the problem step by step, we will follow the titration process of the weak diprotic acid \( H_2A \) with \( NaOH \) and calculate the concentration at the second equivalent point.
### Step 1: Determine the moles of the weak diprotic acid \( H_2A \)
Given:
- Volume of \( H_2A \) solution = 20.0 mL = 0.020 L
- Molarity of \( H_2A \) = 0.20 M
Using the formula for moles:
\[
\text{Moles of } H_2A = \text{Molarity} \times \text{Volume} = 0.20 \, \text{mol/L} \times 0.020 \, \text{L} = 0.004 \, \text{mol}
\]
### Step 2: Determine the moles of \( NaOH \) required for complete neutralization at the second equivalent point
Since \( H_2A \) is a diprotic acid, it can donate two protons. Therefore, the reaction can be represented as:
\[
H_2A + 2 NaOH \rightarrow Na_2A + 2 H_2O
\]
From the stoichiometry of the reaction, 1 mole of \( H_2A \) reacts with 2 moles of \( NaOH \). Thus, for 0.004 moles of \( H_2A \):
\[
\text{Moles of } NaOH = 2 \times \text{Moles of } H_2A = 2 \times 0.004 \, \text{mol} = 0.008 \, \text{mol}
\]
### Step 3: Calculate the volume of \( NaOH \) solution needed
Given:
- Molarity of \( NaOH \) = 0.250 M
Using the formula for moles:
\[
\text{Volume of } NaOH = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.008 \, \text{mol}}{0.250 \, \text{mol/L}} = 0.032 \, \text{L} = 32.0 \, \text{mL}
\]
### Step 4: Calculate the total volume of the solution after titration
Total volume after adding \( NaOH \):
\[
\text{Total Volume} = \text{Volume of } H_2A + \text{Volume of } NaOH = 20.0 \, \text{mL} + 32.0 \, \text{mL} = 52.0 \, \text{mL}
\]
### Step 5: Calculate the concentration of the resulting solution at the second equivalent point
At the second equivalent point, all \( H_2A \) has been converted to \( Na_2A \). The moles of \( Na_2A \) formed is equal to the moles of \( H_2A \) initially present, which is 0.004 mol.
Now, we calculate the molarity of \( Na_2A \):
\[
\text{Molarity of } Na_2A = \frac{\text{Moles of } Na_2A}{\text{Total Volume in L}} = \frac{0.004 \, \text{mol}}{0.052 \, \text{L}} = \frac{0.004}{0.052} \approx 0.0769 \, \text{M}
\]
### Final Answer
The concentration of the solution at the second equivalent point is approximately **0.0769 M**.
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