Sure, let's solve this problem step-by-step to find the concentration of the sulfuric acid solution.
### Given Data:
- Volume of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] (Sulfuric Acid) solution: [tex]\( 50.0 \, \text{mL} \)[/tex]
- Volume of [tex]\( \text{NaOH} \)[/tex] (Sodium Hydroxide) solution: [tex]\( 62.5 \, \text{mL} \)[/tex]
- Concentration of [tex]\( \text{NaOH} \)[/tex] solution: [tex]\( 0.375 \, \text{M} \)[/tex]
### Step-by-Step Solution:
1. Convert volumes to liters:
- Volume of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] in liters:
[tex]\[
50.0 \, \text{mL} = 50.0 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.050 \, \text{L}
\][/tex]
- Volume of [tex]\( \text{NaOH} \)[/tex] in liters:
[tex]\[
62.5 \, \text{mL} = 62.5 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.0625 \, \text{L}
\][/tex]
2. Calculate moles of [tex]\( \text{NaOH} \)[/tex]:
- Moles of [tex]\( \text{NaOH} = \text{Concentration} \, (\text{M}) \times \text{Volume} \, (\text{L}) \)[/tex]
[tex]\[
\text{Moles of NaOH} = 0.375 \, \text{M} \times 0.0625 \, \text{L} = 0.0234375 \, \text{moles}
\][/tex]
3. Determine moles of [tex]\( \text{H}_2\text{SO}_4 \)[/tex]:
- The reaction between [tex]\( \text{NaOH} \)[/tex] and [tex]\( \text{H}_2\text{SO}_4 \)[/tex] is as follows:
[tex]\[
\text{H}_2\text{SO}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O}
\][/tex]
- This indicates that 1 mole of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] reacts with 2 moles of [tex]\( \text{NaOH} \)[/tex]. Therefore, we need to divide the moles of [tex]\( \text{NaOH} \)[/tex] by 2 to find the moles of [tex]\( \text{H}_2\text{SO}_4 \)[/tex]:
[tex]\[
\text{Moles of H}_2\text{SO}_4 = \frac{\text{Moles of NaOH}}{2} = \frac{0.0234375 \, \text{moles}}{2} = 0.01171875 \, \text{moles}
\][/tex]
4. Calculate the concentration of [tex]\( \text{H}_2\text{SO}_4 \)[/tex]:
- Concentration ([tex]\( \text{M} \)[/tex]) is defined as moles of solute divided by volume of solution in liters:
[tex]\[
\text{Concentration of H}_2\text{SO}_4 = \frac{\text{Moles of H}_2\text{SO}_4}{\text{Volume of H}_2\text{SO}_4 (\text{L})} = \frac{0.01171875 \, \text{moles}}{0.050 \, \text{L}} = 0.234375 \, \text{M}
\][/tex]
Hence, the concentration of the sulfuric acid solution is [tex]\( 0.234375 \, \text{M} \)[/tex].
