Question

Consider the titration of 80.0 mL of 0.0200 M NH₃ (a weak base; K_b = 1.80e-05) with 0.100 M HI. Calculate the pH after the following volumes of titrant have been added:

(a) 0.0 mL pH =	(b) 4.0 mL pH =	(c) 8.0 mL pH =
(d) 12.0 mL pH =	(e) 16.0 mL pH =	(f) 20.8 mL pH =

Answer 1

pkb_4.74 Run NH₂ + HI NH₂, I & w. base S.Acid salt - Kb NH2 = 1.8x10s @ NHy is a weak base so fol weak bases [ON] = J K6 TC = Jl. 8x 105 10.02 = 6.0X1b4 .: pot fog rout] -log 6 x 109 - polt= 3-22 pit + port = 14.00 - pH = 14 polt= 14.00-3.22 plt = 10.78 3 Vq = 400ml [NH3 lett = M,V, M₂ 2 0.02880-001x4 utu 80+4 = 0.0143M .: salt = [NHI] = MaY2 = 0.144 [sadf}= 4.760103 - 04 Kb Joolsalt] " [Base] = 4.74 tlog 4.76 x 12 DO .. ipt = 9.74 oH = 4.26

6 [atty Just U2 = 8 me is half equivalence point At this point pit=pka =4574 .../pH=9.26) ② 80 +16 @ = 16 me Equivalence point .: [NH I J = 0.100 X 16 = 0.01667 pH = + [prw_pkb Hoge] +4214.00 -4.74 Hoglo-omenat [ p* = 7.48] CNH J = 0.02480 – 01x12 J0+12 = 4.35x103 [With I] = 0.013 polt=8-22 ./pt-8.78 € 20.8m as [HIJ exrer = 0.1820-8-002x80 20-8180 = 444x183, pH = -log 4:44X103 q: HI is .. = 2.35] ( Stringaid 1. MAR