Question

2. A 30.00 ml solution of 0.235 M CH2NH2 was titrated with 0.150 M H2S0O.. a) Calculate the pH of the 30.00 ml solution of 0.235 M CH2NH2 The Kb for CH2NH2 is 4.4x 104 b) Calculate the volume of 0.150 M H2S04 required to completely neutralize the 30.00 mL solution of 0.235 M CH2NH2. c)Calculate the volume of 0.150 M H2SO, required to neutralize half the mole of CH2NHz. present 30.00 mL solution of 0.235 M CH2NH2 d) Calculate the pH of the solution in (c). e) Calculate the pH of the resulting solution after the addition of 100.00 ml of 0.150 M H250.. f) Sketch the pH titration curve by graphing volume of 0.150 M H250. (x axis) vs pH for: ) pH before the start of titration, at V-0.0 mL of 0.150 M H SO (ii) pH at the halfway point, (c) (ii) pH at the equivalence point (complete neutralization) (iv) pH at (e) the pH of the resulting solution after the addition of 100.0 mL of 0.150 M H2504

Answer 1

iven Com centsati Base dissoriation O. 23 S Degree or dissociat on fc H13トーリ 4.4XIo o-235 or toリー OH он 14-1.99

0.20上4 tien vn Complete Neu alizathen Neutsalization o o 3o Sm.mo み。Sn.mau Om.mde osー3.osjm.mole

case Hal m molea -mdes 子. DS Bagic This ie a burten saution Hande Son-Hass Ca CH3 NH -4 ) -3-3s ー0572 m-mole 엉( V Total volum 14-3-35 PH - 10-65 wen