Answer:
About 0.151 M.
Explanation:
Because 32.20 mL of 0.250 M NaOH was used, determine the number of moles of NaOH consumed:
[tex]\displaystyle 32.20\text{ mL} \cdot \frac{0.250\text{ mol NaOH}}{1\text{L}} \cdot \frac{1\text{ L}}{1000\text{ mL}} = 0.00805\text{ mol NaOH}[/tex]
Find the number of moles of sulfuric acid reacted with using reaction stoichiometry:
[tex]\displaystyle 0.00805\text{ mol NaOH} \cdot \frac{1\text{ mol H$_2$SO$_4$}}{2\text{ mol NaOH}} = 0.00402\text{ mol H$_2$SO$_4$}[/tex]
Therefore, the molarity of the original sulfuric acid solution is:
[tex]\displaystyle \ [\text{H$_2$SO$_4$}] = \frac{0.00402\text{ mol}}{26.60\text{ mL}} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 0.151\text{ M}[/tex]
In conclusion, the molarity of the original sulfuric acid solution was about 0.151 M.
