Question

Calculate the pH of a 0.292 M solution of ethylenediamine (H2NCH2CH, NH,). The pK values for the acidic form of ethylenediamine (H NCH,CH2NH) are 6.848 (pKal) and 9.928 (pKa2). pH 10.16 Calculate the concentration of each form of ethylenediamine in this solution at equilibrium 0.294 H2NCH2CH2NH, ] = М 1.44 x10-4 [H2NCH2CH2NH = -5 [HNCH2CH2NH 3.51 x10 MA M

Answer 1

calculate plaz - - plol + pkb plaz and pk br Flust. 2 14.00 z 14.000-9.928 2 4.072 104-072 28-47x10.5 2 14.00 pkb 2+ pllal 2 pkb 2 +6848 2 14.00 Let =) pkb 2 a 7-152 2) Kb a 107.152 m CH₂ CH2 NH ₂ 2 3 2054 10 8 ANCH, CH NH, the hit Chau Nr to 0.992m -X x x 0-942 x - kb, 2 Chtch, CH, NH 2) [at] Croch, RAM) x² 2) 8-47x10 6992-X 2) x² + 8.42 6105x - 2 21 34 165 20. Usingquadvancequohon Solver a get X: 0.003 20-00493 m Con) 20-00493M Cho BH) - 0.00493

second lonisation of ħ by t2 ton Now consider But the 20-00493m cx G o-00493-X o-00493m. X 0-00493+x neglect x as - Kb as low less Koz - = ) (BH + Cot Cont). Kb a 7-05x105 x 2 (BH +2 m Cond o-00493m =) 2 pon a 2.307 in 14.000 .-2-307 1-683) [HO CHO NH) 0.292 - x at equilibrum & 0.292 - 0.00493m. - 20287 m Cont) a [Brht) E ANCHCH N 20 00493 m. Ehet CICH NAT] 2 17:05 0 m