Question

help! i got it wrong

Question 14 of 62 Submit What is the pH of a 0.0380 M solution of HONH3CI (Kb of HONH2 is 1.1 x 10-8)? 3.6 1 2 3 7 +/- 8 : 9 0 +/- x 100 Tap here or pull up for additional resources

Answer 1

HON H3CI = 0.0380 M HONG + 480 s Hong Ky = 51810-8 K = 1.1x10- → HO NHS CL HON HO + cie 0.0380 0.0380 0.0380 HONH₂O + HO HO NHgOH + H I Kn= kw 0.0 380 to OH = 10-14 l1x10-8 -Y +y < Kn Rn = 10-6 0.0380-y - Kh = (Y) CY) (0.8380-Y) Kn=ke is very very small, to reaction will not go much in forward direction. » value of y is very small, Can be meglect in addition or substraction sº • o 3 8 –y = 0.0 3 8 ) 10-6 _ To you 0.038 10-6x368x10–? yol = 3.45 x10-8 u: y = 1o8 586 *10-4 = [H +) pH= -log Cht) = -log C1.8586x10-4) = 3.73)