Final answer:
The volume of 0.108 M H2SO4 required to neutralize 25.0 mL of 0.145 M KOH is approximately 16.78 mL, based on the stoichiometry of the balanced chemical equation for their reaction.
Step-by-step explanation:
To calculate the volume of 0.108 M H2SO4 required to neutralize 25.0 mL of 0.145 M KOH, we must first write a balanced chemical equation:
H2SO4 + 2KOH → K2SO4 + 2H2O
This shows that one mole of sulfuric acid reacts with two moles of potassium hydroxide. Now we can use the stoichiometry of the reaction and the given molarities to find the volume needed for neutralization:
- Calculate the moles of KOH:
Moles of KOH = volume (L) × molarity (mol/L) = 0.025 L × 0.145 mol/L = 0.003625 mol
- Using the stoichiometry of the reaction, calculate the moles of H2SO4 needed:
Moles of H2SO4 = 0.003625 mol KOH × (1 mol H2SO4 / 2 mol KOH) = 0.0018125 mol H2SO4
- Calculate the volume of H2SO4 solution required:
Volume of H2SO4 = moles H2SO4 / molarity H2SO4 = 0.0018125 mol / 0.108 mol/L = 0.0167824 L
Therefore, the volume of H2SO4 required is 16.78 mL.