Question

1.Exactly 12cm3 of H2SO4 were enough to neutralize 36cm3 of a solution made by dissolving 7grams of KOH in 250ml of water.
a) What is the molarity of the base used
b)what is the molarity of the acid used

Answer 1

a) C(KOH) = (7 / (56)) / (250 / 1000) = 0.5 (mol / L)

b) C(H₂SO₄) = 0.5 × (36 / 12) / 2 = 0.75 (mol / L)