the balanced equation for the neutralisation reaction between KOH and HCl is as follows
KOH + HCl --> KCl + H₂O
stoichiometry of KOH to HCl is 1:1
the number of HCl moles reacted = HCl concentration x volume
number of HCl moles = 0.50 mol/L x 0.020 L = 0.01 mol
according to molar ratio of 1:1
number of KOH moles required to neutralise 0.01 mol of HCl = 0.01 mol of KOH
molarity of KOH = 0.20 M
there are 0.20 mol in 1 L Then 0.01 mol are - 0.01 mol / 0.20 mol/L = 50 mL
50 mL of 0.20 M KOH required