Question

Be sure to answer all parts. Find the pH of the two equivalence points and the volume (mL) of 0.427 M HNO3 needed to reach them in the titration of 0.188 L of 0.250 M ethylenediamine (H2NCH2CH2NH2). First equivalent point: mL HNO3 pH- Second equivalent point: mL HNO3 pH

Answer 1

classmate Date Page Etrylene di amine has two different -Nho group. Thus it is diprotic base. for the first equivarance point. V = volume of HON CHO CHaNto a 0.1882 - 188 ml. Sastrength of HẠN CHOCHO NH2 = 0.250m. S, 2 strength of ANO₂ = 0.427M. Now, Vis, = V₂52 = 1880.250 = V2 X 0.427 => V2 z 1884 0.25o = 110.07 . 0.427 Thus, 110.07 ml HNO3 required Now, PkbU = 4.072 pkb/2) = 7.152. Af equivalance point is bicidic since strong acid and weak base have been used. 1... pH= 7 - 1 pk6) - the loge [salt] = 0.25x188 - 7- / x 4.072 Ixloge 188+ 110-07 - 0.1577 T = 7 1 2 4:072 - - 2 x 10g (0.157) = 460 5.368 Me volume of HNO₃ = 110.07ml. pH = 450 5. 366 Ics Scannes-....... CamScanner

classmate Date Page Now, for and equivalance pourt Siz concentration of e oor5 0.25 188 188 + 110.07 0.157 Va 188 + 10.07= 898:07 Sa = strength of HNO₂ = 0.487 M. I Now, vius, = Vaso = 298.07 x 0.157 - 82 x 0.427 7 V = 298.07X0.157 0427 - = 109.59. then after second lausvalance I sall/= 0.15 7 x 298.07 298.07 + 109.59 This plz 7 - 3 pk 6-12 log @ ta =7-3x7.152 - 3xloq (0.115) = logo 3.89. Scanned with CamScanner