Question

MANCED CELL BLOY L.. Rowan University-CHEM09240 - Fall 19 - BANDEGE Chec + Ch Caroline Alemany BANDEGI > Activities and Due Dates Ch 6c nment Score: 681/3200 Resources Hint Check Answer estion 8 of 32 > Calculate the pH of a 0.219 M solution of ethylenediamine (H,NCH.CH, NH). The pk, values for the acidic form of ethylenediamine (HNCH,CH, NH) are 6.848 (pK.,) and 9.928 (pK2). pH- Calculate the concentration of each form of ethylenediamine in this solution at equilibrium. [H,NCH,CHÁNH,]= [H,NCH,CH, NH}]= [HỆ NCH,CH,NH]= butus Driverse came
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Answer 1

со мед, оң, уа, +50 — це,04м-той е седем, то е целом. В., roыг * first reaction & #Ney of Not too - NCANA FONT to O. 2 19 м भ 2 x Excitibrium 0.219M-2 БҚа, 2 - 89 8 Рка + РК, 2 рко 6. 39 + РК, 2 fЧ Рk, з 19– С. 48 2 п. 1 2 РКь, - — од Кь k, = 10-PRь Кь - to-7. 152 27.oxto-8 н.

Қ, = самоң, оң, ми. ) [он] 1ң, оң Сң, Nң. ] 1x 10-*- 2.x О. 29 м-х for weak acid det 0.219M-2 ~0.2299 Пх10-8 м = х2 0-29 м Пхь kasb- 29 У 2 x2 1- 53 alo8 *- 22 х = ((-S3 х188 v2. = 1.2.3 x 6-9 м. - Corl=[7 M Ch C NH) or theJ=1238104 Remaining ANCH, CH2 N = (0.219 m-a) = [б. 219 ч — 2-3 x) *м) = 0:24 ям — о- оо (23 ч - О, 28% м.

I Part -йон, чи, ти, о 9 недеяч, оғ 1.2.3 хоч (423xto ел. - ко. - 149ся, ст. 9) (o=") E e Car Cry Note) Кь, = х2 again ignowring & in (1.23x10 4x) 1:23 х10 Ч -х Кь, = 2 23 х10 ч. Рка трка = Іч

9-928 грқ, 2 (9 P% - 19- 9928 Рк» = чо+2 Кь, – L0-Ч-o+2 Қ, з 3-4 х (o-se From Equation 3-4ко би= 0 2* 9, 1:23 х х- - 8-ч хtorSM к , 2.3 хочу х1 lo- 33 х10-9 м- 2 -o3 x-8 н. х, - •o3 x1-94- - 1-о) хогч и — Г cң си, 4ч. ) - Гонг)

Lor = 2, +22 z 1.23 x 10-44 + 1.01 X10404 = 2:24 X10-4 M pon = -log[2.24x604) = -log 2.24 +E logole = -0.3511 tu . 3.64 pH + port = pkw 14. boy + 3-64 = 14 Pop a 14-3.64 py = 10.35

pa = 10.35 HM CH₂ Cola No₂ = 0.2188 M q Cola CH₂ Nota = 1.23 810-4 - X2 - 1.23 X164 - 101210-4ry 2 0-2 2 x 104 M. = 1:01 810-44 gh Ch Can f