If 24.83 cm³ of 0.15 M NaOH is titrated to its end point with 39.45 cm³ of HCl, what is the molarity of the HCl?

NaOH + HCl → NaCl + H₂O. One mole of acid reacts with one mole of base.

Moles of NaOH = 0.15 × 24.83 ÷ 1000 = 0.003725 mol. So moles of HCl = 0.003725 mol.

Molarity of HCl = 0.003725 ÷ 0.03945 dm³ ≈ 0.094 M. Equivalently, M₁V₁ = M₂V₂ gives 0.15 × 24.83 ÷ 39.45 ≈ 0.094 M.