A solution of koh required 31.3 ml of 0.118 m hcl to reach neutralization. what mass of koh was in the titrated sample?

Neutralization reaction is a reaction that involves reaction of a base/alkali and an acid forming salts and water as the only products. In this case;

KOH (aq) + 2HCl (aq) = H2O (l) + 2KCl (aq)

Moles of HCl used = (31.3/1000) * 0.118 = 0.00369 moles

Using mole ratio of KOH : HCl = 1:2, the mole of KOH will be;

0.00369 : 2 = 0.001845 Moles

But, 1 mole of KOH = 56 g

Therefore; 0.001845 moles will have; 0.001845 * 56 = 0.10332g

Hence the mass of KOH titrated was 0.10332g