Question

In a titration of sulfuric acid against sodium hydroxide, 32.20 mL of 0.250 M NaOH is required to neutralize 26.60 mL of H2so

Answer 1

About 0.151 M.

Step-by-step explanation:

Because 32.20 mL of 0.250 M NaOH was used, determine the number of moles of NaOH consumed:

`\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}`

Find the number of moles of sulfuric acid reacted with using reaction stoichiometry:

`\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$_4$}`

Therefore, the molarity of the original sulfuric acid solution is:

`\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}`

In conclusion, the molarity of the original sulfuric acid solution was about 0.151 M.