Question

11. How many mL of a 0.525 M NaOH solution is needed to neutralize 15.00 mL of a 1.25 M H2SO4 solution? 12. What is the molarity of an HCl solution if 25.0 mL of 0.212 M NaOH is used to neutralize 13.6 mL of the HCI solution? 13. If 10.0 mL of 0.121 M H2SO4 is neutralized by 17.1 mL of KOH solution, what is the molarity of the KOH solution?

Answer 1

H_2SO_4 + 2NaOH = Na_2SO_4 + 2H_2O According to the principle of volumetric Analysis, gm equivalents of Acid will react with gm equivalents of base. V in ml times strength in (M) = moles of substance. Let V ml of .525 M NaOH will neutralize 15 ml 1.25 M H_2SO_4 milli moles of NaOH = v times 5.25 milli moles of H_2SO_4 = 15 times 1.25 1 milli mole H_2SO_4 = 2 m moles of NaOH Therefore 15 times 1.25 H_2SO_4 = 2 times 15 times 1.25 m moles of NaOH Therefore V times .525 = 2 times 15 times 1.25 Therefore V = 71.428 ml NaOH + HCl = NaCl + H_2O 1 milli mole NaOH reacts with 1 milli mole HCl 25 ml 0.212 M NaOH = 25 times .212 milli moles of NaOH let molarity of HCl = x (M) Therefore 13.6 ml x(M) HCl = 13.6 times x milli moles of HCl