millimoles of HCl = 15*0.6 = 9 KOH = 25*0.35 = 8.75 so millimoles of H+ ions (in form of HCl that remain un-neutralized) = 9-8.75 = 0.25 so [H+] = 0.25 millimoles or 2.5e- 4M pH = -log(2.5e-4) = 3.6 Solution millimoles of HCl = 15*0.6 = 9 KOH = 25*0.35 = 8.75 so millimoles of H+ ions (in form of HCl that remain un-neutralized) = 9-8.75 = 0.25 so [H+] = 0.25 millimoles or 2.5e- 4M pH = -log(2.5e-4) = 3.6.