Let's look at this problem step by step:
1. Identify the reactants and their volumes and molarities:
- Volume of [tex]\( H_2SO_4 \)[/tex] solution: 0.25 L
- Molarity of [tex]\( H_2SO_4 \)[/tex] solution: 2.00 M
- Volume of [tex]\( NaOH \)[/tex] solution: 2.00 L
2. Determine the number of moles of [tex]\( H_2SO_4 \)[/tex] used:
- Moles of [tex]\( H_2SO_4 \)[/tex] = Volume of [tex]\( H_2SO_4 \)[/tex] × Molarity of [tex]\( H_2SO_4 \)[/tex]
- Moles of [tex]\( H_2SO_4 \)[/tex] = 0.25 L × 2.00 M = 0.5 moles
3. Find the number of moles of [tex]\( NaOH \)[/tex] required for neutralization:
- According to the balanced chemical equation [tex]\( 2 NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2 H_2O \)[/tex], 1 mole of [tex]\( H_2SO_4 \)[/tex] reacts with 2 moles of [tex]\( NaOH \)[/tex].
- Therefore, moles of [tex]\( NaOH \)[/tex] required = 2 × moles of [tex]\( H_2SO_4 \)[/tex]
- Moles of [tex]\( NaOH \)[/tex] required = 2 × 0.5 moles = 1.0 mole
4. Calculate the molarity of the [tex]\( NaOH \)[/tex] solution:
- Molarity (M) = Moles of solute / Volume of solution in liters
- Molarity of [tex]\( NaOH \)[/tex] = Moles of [tex]\( NaOH \)[/tex] / Volume of [tex]\( NaOH \)[/tex]
- Molarity of [tex]\( NaOH \)[/tex] = 1.0 mole / 2.00 L = 0.5 M
Therefore, the concentration of [tex]\( NaOH \)[/tex] that is used is [tex]\( 0.50 M \)[/tex].
Mark the answer: [tex]\( 0.50 M \)[/tex]
