Question

A 3.9-mL sample of a 0.18 M HCl(aq) solution was mixed with a 5.7-mL sample of a 0.71 M HCl(aq) solution. Calculate the pH of the mixture. Provide your answer to the correct number of decimal places.

Answer 1

n(HCl) = n1(HCl) + n2(HCl) = C1(HCl) × V1(HCl) + C2(HCl) × V2(HCl) = 0.18M × 0.0039L + 0.71M × 0.0057L = 0.004749mol.

V(HCl) = V1(HCl) + V2(HCl) = 0.0039L + 0.0057L = 0.0096L.

HCl = H⁺ + Cl^-

C(H⁺) = C(HCl) = n(HCl)/V(HCl) = 0.004749mol/0.0096L = 0.4947M.

pH = -lg[H⁺] = -lg0.4947 = 0.3057 ≈ 0.31.