Sure, let's go through the problem step by step to find the concentration of NaOH used in the reaction.
1. Identify the Given Data:
- Volume of [tex]\( H_2SO_4 \)[/tex]: [tex]\(0.25 \, L \)[/tex]
- Concentration of [tex]\( H_2SO_4 \)[/tex]: [tex]\(2.00 \, M \)[/tex]
- Volume of [tex]\( NaOH \)[/tex]: [tex]\(2.00 \, L \)[/tex]
2. Chemical Reaction:
According to the balanced chemical equation:
[tex]\[
2 \, NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2 \, H_2O
\][/tex]
This shows that 1 mole of [tex]\( H_2SO_4 \)[/tex] neutralizes 2 moles of [tex]\( NaOH \)[/tex].
3. Calculate Moles of [tex]\( H_2SO_4 \)[/tex]:
Using the formula [tex]\( \text{moles} = \text{volume} \times \text{concentration} \)[/tex]:
[tex]\[
\text{Moles of } H_2SO_4 = 0.25 \, L \times 2.00 \, M = 0.5 \, \text{moles}
\][/tex]
4. Determine Moles of [tex]\( NaOH \)[/tex] Required for Neutralization:
From the balanced equation, with a mole ratio of [tex]\( 2 \, NaOH : 1 \, H_2SO_4 \)[/tex]:
[tex]\[
\text{Moles of } NaOH = 2 \times \text{Moles of } H_2SO_4 = 2 \times 0.5 = 1.0 \, \text{moles}
\][/tex]
5. Calculate Concentration of [tex]\( NaOH \)[/tex]:
Using the formula [tex]\( \text{concentration} = \frac{\text{moles}}{\text{volume}} \)[/tex]:
[tex]\[
\text{Concentration of } NaOH = \frac{1.0 \, \text{moles}}{2.0 \, L} = 0.5 \, M
\][/tex]
Thus, the concentration of [tex]\( NaOH \)[/tex] used is [tex]\( 0.50 \, M \)[/tex].
So, the correct answer is:
[tex]\[ 0.50 \, M \][/tex]
