Question

A chemist titrates 140.0 ml of a 0.8665 Methylamine (C,H,NH,) solution with 0.5484 M HBr solution at 25 °C. Calculate the pH at equivalence. The pky of ethylamine is 3.19. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HBr solution added.

Answer 1

ANSWER:: CHENH2 + HBr = CH NH₃ + Br weak base stidy weak Now, we have for GH5NH [C, H, NH₂] = 0.8665M __ Volume =__40.0 mban for HBr HBr] 0.5484M volume of HBr = ? mL. At Equivalence, molis of HBr = moles of GHqNH. [482] xvol_of Hb!= [GHS Miten] x 101:08 @43. Gus). Vol of HBr = 0.86658 140 mL = 221.207 mL. 0.5484 Total volume = 140.0+ 221.207 ml Total volume = 561-207 ml At squirolence point; Making I ĆE table а „Не мно) Initial 0.216 change - x Equilevalence 0:216-22 С.И; ми, +и+

No. of moles of NH₂ Cat's produced = 0. 12131 moles. (mºsqivalence) (0.8665x 1401) mol/L 1000 Initially conc. of Tuhalal = 0.1213) = 0.2161591 M. given pk of [C2H5NH] Hencepka for CCM EN Hz] pka = 14-pkb = 10.80999 =) Ka a épre ka = 1.5488 X 10" from ICE , we have; Ka = (x)(w) - 1.548 X 10 = x2 70246-) solving n, we get n = 1.8568791x10-5 ора W (0.216-2)

lat)= x= 1.86x105 pH = -log[+] = -log [ 1.868 105 py = 4.73. de