Question

= 3 of 3 Questions Assignment Score: 33.3% Resources CX Give Up? Hint Check Answer 50% < Question 3 of 3 > Attempt 2 1 Question 1 of 0 Attempts In Progress 0% 2 Question O of Attempts A buffered solution containing dissolved aniline, CH NH,, and aniline hydrochloride, CH_NH, CI, has a pH of 5.74. A. Determine the concentration of CH NHin the solution if the concentration of CH_NH, is 0.230 M. The pKb of aniline is 9.13 50% 3 Question 1 of 0 Attempts In Progress [C,H,NH] = .03102614 M B. Calculate the change in pH of the solution, ApH, if 0.350 g NaOH is added to the buffer for a final volume of 1.85 L. Assume that any contribution of NaOH to the volume is negligible. ApH =

Answer 1

C6H5NH2 С, и на Aniline hydrochloride Aniline C6H5N143 C6H5NH2 Conqчаи базе Weakacid ph of soj = 5-7ч pkg of aniline = 9.13 pкat pk = 19 Ч87 E рка = 14-РkЬ 14-4-13 = pкa 1 C4 ч. мн. - 48а pH of 64 Her calculated using Hendersons Hasselbach equation pH = pka + log [onjugate base] Lesk aid) 5:14 Ч-3+ + lCC, 9; ми.) (Ens Nuz] 0.82 log Co-230 ] С че бя) antilog 10.87) Co-236] [Constans Со 13° ) (44; мм.) [Consts 1:41 1.ЧІ 2 0.23о = 0:03 о

molar as of Naot = 40.0d mol mol og Naok= given mas molan man 0 350g = 0.008750 40.og mott Calculating initial mol of C6 4g MHz B (645 Milya molf 645NH2 = Molarity x volume 0.2 30 menit mol x 18SK = 0.4255 mol mol of Cousins 0.0310 mos *1.85% = 0.05735 mol When NaOH added to butler it will reat weak acid present in butter and amount weak quid decreases and conjugate base increases. mol 0.00875 J Moon = After addition 0.4255+ 0.00875 = 0.434 mol - mol CongNha g of Cong Ng = 0:05735-0-00815 = 0.0486 mol mol New pay butter pH = 4.87 + log [0.434] [0.0486] ри 5.82 - SpH = ( new ph- old pu) unit (5.82 - 5.74) = 0.08 -