To solve the problem, we will follow these steps:
### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction between sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH) is:
\[
H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O
\]
### Step 2: Calculate the moles of reactants
We need to find the number of moles of sulfuric acid and sodium hydroxide present in the solution.
- For sulfuric acid (H₂SO₄):
\[
\text{Moles of } H_2SO_4 = \text{Molarity} \times \text{Volume} = 0.1 \, \text{M} \times 1 \, \text{L} = 0.1 \, \text{moles}
\]
- For sodium hydroxide (NaOH):
\[
\text{Moles of } NaOH = \text{Molarity} \times \text{Volume} = 0.1 \, \text{M} \times 1 \, \text{L} = 0.1 \, \text{moles}
\]
### Step 3: Determine the limiting reagent
From the balanced equation, we see that 1 mole of H₂SO₄ reacts with 2 moles of NaOH. Therefore, for 0.1 moles of H₂SO₄, we need:
\[
0.1 \, \text{moles of } H_2SO_4 \times 2 = 0.2 \, \text{moles of } NaOH
\]
However, we only have 0.1 moles of NaOH available. Thus, NaOH is the limiting reagent.
### Step 4: Calculate the moles of sodium sulfate formed
Since NaOH is the limiting reagent, we can calculate the moles of sodium sulfate (Na₂SO₄) produced. According to the stoichiometry of the reaction:
\[
2 \, \text{moles of } NaOH \rightarrow 1 \, \text{mole of } Na_2SO_4
\]
Thus, from 0.1 moles of NaOH, the moles of Na₂SO₄ formed will be:
\[
\text{Moles of } Na_2SO_4 = \frac{0.1 \, \text{moles of } NaOH}{2} = 0.05 \, \text{moles}
\]
### Step 5: Calculate the total volume of the solution
The total volume of the solution after mixing is:
\[
1 \, \text{L (H₂SO₄)} + 1 \, \text{L (NaOH)} = 2 \, \text{L}
\]
### Step 6: Calculate the molarity of sodium sulfate
Molarity (M) is defined as the number of moles of solute divided by the volume of solution in liters:
\[
\text{Molarity of } Na_2SO_4 = \frac{\text{Moles of } Na_2SO_4}{\text{Total Volume}} = \frac{0.05 \, \text{moles}}{2 \, \text{L}} = 0.025 \, \text{M}
\]
### Final Answer
The molarity of sodium sulfate formed is **0.025 M**.
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