Question

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

Answer 1

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