moles of H+ = 42.0 x 0.120 millimoles = 5.04 millimoles final volume = 1.55 L [H+] = (5.04 x 10^-3)/1.55 = 3.25 x 10^-3M pH = -log[H+] = 2.49 Solution moles of H+ = 42.0 x 0.120 millimoles = 5.04 millimoles final volume = 1.55 L [H+] = (5.04 x 10^-3)/1.55 = 3.25 x 10^-3M pH = -log[H+] = 2.49.