Answer :
To determine the concentration of NaOH used in this neutralization reaction, we'll follow a systematic, step-by-step approach:
1. Identify the relevant information and balanced equation:
- Volume of \( \text{H}_2\text{SO}_4 \): \( 0.25 \, \text{L} \)
- Concentration of \( \text{H}_2\text{SO}_4 \): \( 2.00 \, \text{M} \)
- Volume of \( \text{NaOH} \): \( 2.00 \, \text{L} \)
- Balanced chemical equation: \( 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \)
2. Calculate the moles of \( \text{H}_2\text{SO}_4 \) used:
The number of moles of \( \text{H}_2\text{SO}_4 \) can be calculated using the formula:
[tex]\[ \text{moles of } \text{H}_2\text{SO}_4 = \text{volume} \times \text{concentration} \][/tex]
Substituting the given values:
[tex]\[ \text{moles of } \text{H}_2\text{SO}_4 = 0.25 \, \text{L} \times 2.00 \, \text{M} = 0.50 \, \text{moles} \][/tex]
3. Determine the moles of \( \text{NaOH} \) required:
According to the balanced chemical equation, 2 moles of NaOH react with 1 mole of \( \text{H}_2\text{SO}_4 \). Therefore, the moles of \( \text{NaOH} \) needed are:
[tex]\[ \text{moles of } \text{NaOH} = 2 \times \text{moles of } \text{H}_2\text{SO}_4 \][/tex]
Substituting the calculated moles of \( \text{H}_2\text{SO}_4 \):
[tex]\[ \text{moles of } \text{NaOH} = 2 \times 0.50 \, \text{moles} = 1.00 \, \text{moles} \][/tex]
4. Calculate the concentration of \( \text{NaOH} \):
The concentration of \( \text{NaOH} \) can be found using the formula:
[tex]\[ \text{concentration} = \frac{\text{moles}}{\text{volume}} \][/tex]
Substituting the values we found:
[tex]\[ \text{concentration of } \text{NaOH} = \frac{1.00 \, \text{moles}}{2.00 \, \text{L}} = 0.50 \, \text{M} \][/tex]
Therefore, the concentration of NaOH used in the reaction is [tex]\( 0.50 \, M \)[/tex]. The correct answer is [tex]\( 0.50 \, M \)[/tex].
1. Identify the relevant information and balanced equation:
- Volume of \( \text{H}_2\text{SO}_4 \): \( 0.25 \, \text{L} \)
- Concentration of \( \text{H}_2\text{SO}_4 \): \( 2.00 \, \text{M} \)
- Volume of \( \text{NaOH} \): \( 2.00 \, \text{L} \)
- Balanced chemical equation: \( 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \)
2. Calculate the moles of \( \text{H}_2\text{SO}_4 \) used:
The number of moles of \( \text{H}_2\text{SO}_4 \) can be calculated using the formula:
[tex]\[ \text{moles of } \text{H}_2\text{SO}_4 = \text{volume} \times \text{concentration} \][/tex]
Substituting the given values:
[tex]\[ \text{moles of } \text{H}_2\text{SO}_4 = 0.25 \, \text{L} \times 2.00 \, \text{M} = 0.50 \, \text{moles} \][/tex]
3. Determine the moles of \( \text{NaOH} \) required:
According to the balanced chemical equation, 2 moles of NaOH react with 1 mole of \( \text{H}_2\text{SO}_4 \). Therefore, the moles of \( \text{NaOH} \) needed are:
[tex]\[ \text{moles of } \text{NaOH} = 2 \times \text{moles of } \text{H}_2\text{SO}_4 \][/tex]
Substituting the calculated moles of \( \text{H}_2\text{SO}_4 \):
[tex]\[ \text{moles of } \text{NaOH} = 2 \times 0.50 \, \text{moles} = 1.00 \, \text{moles} \][/tex]
4. Calculate the concentration of \( \text{NaOH} \):
The concentration of \( \text{NaOH} \) can be found using the formula:
[tex]\[ \text{concentration} = \frac{\text{moles}}{\text{volume}} \][/tex]
Substituting the values we found:
[tex]\[ \text{concentration of } \text{NaOH} = \frac{1.00 \, \text{moles}}{2.00 \, \text{L}} = 0.50 \, \text{M} \][/tex]
Therefore, the concentration of NaOH used in the reaction is [tex]\( 0.50 \, M \)[/tex]. The correct answer is [tex]\( 0.50 \, M \)[/tex].
