How many milliliters of 0.20 molar koh solution are needed to exactly neutralize 20. milliliters of 0.50 molar hcl?

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