How much 0.100 m hcl is required to completely neutralize 20.0 ml of 0.250 m naoh?

The balanced equation for the reaction between NaOH and HCl is as follows

NaOH + HCl - - - > NaCl + H₂O

stoichiometry of NaOH to HCl is 1:1

number of NaOH moles reacted - 0.250 mol/L x 0.0200 L = 0.005 mol

therefore number of HCl moles reacted - 0.005 mol

molarity of given HCl solution - 0.100 M

number of HCl moles in 1 L - 0.100 mol

then 0.005 mol of HCl in - 0.005 mol / 0.100 mol/L = 50.0 mL

volume of HCl required is 50.0 mL