If 30mL of 0.5M KOH is needed to neutralize 2M HCl, what was the volume of the acid?

The equation for the reaction between KOH and HCl is as follows

KOH + HCl - - - > KCl + H2O

the stoichiometry of KOH to HCl = 1:1

the number of KOH moles reacted = 0.5 mol / 1000 cm³ * 30 cm³

= 0.015 mol

the number of HCl moles reacted = number of KOH moles reacted

therefore HCl moles reacted = 0.015 mol

the molarity of HCl is 0.2 mol/dm³

0.2 mol of HCl in - 1000 cm³

Therefore volume required for 0.015 mol = 1000 cm³ / 0.2 mol * 0.015 mol

= 75 cm³

Therefore 75 cm³ of HCl is required