I need help with this please. Show all work. Thank you. 8. Calculate a pH of a solution in which 20 ml of 0.10 M NaOH has been added to 25 mL of 0.10 M H2SO4 Solution reaction is H2SO4 + 2 NaOH ...........> Na2SO4 + 2 H2O one mole H2So4 is neutralized by two mole NaOh. 20 ml of 0.10 M NaOH = 0.020 * 0.10 = 0.0020 mole. 25 ml of 0.10 M H2So4 = 0.025 * 0.10 = 0.0025 mole. NaOH is the limiting reactant. excess H2SO4 = 0.0025 - 0.0020 / 2 = 0.0015 mole. assuming all the protons in H2SO4 are completely ionized. so moles of H+ = 0.0015 * 2 = 0.0030 mole. concentration of H+ = 0.0030 * 1000 / (20 + 25) = 0.06667 (M) PH = -log [H+] = - log (0.06667) = 1.18 PH = 1.18 .