Question

A 10.00 mL sample of 0.300 M NH3 is titrated with 0.100 M HCl (aq). what is the initial pH? Calculate the pH after the addition of 10.0, 20.0, 30.0 and 40.0 mL of HCl. The Kb for NH3 is 1.8 x 10^-5

Answer 1

Initial pH Ntg cast Hoca, Nytorica) too 0.3M 0.3nx teg. kh= x.x & no 0.3-1 x = rostla v kq xo:3 - J 1.8x105x013 [CH] = 2.324x10²M polt=2-6336 pH = pkw-polt- 14-2-6338 = 11.3662 Now Nho can & Hll cap ? Nhy cay, & cu cap t=0 mmoles toxo.3 0.010 (MxVmi) t-op. 2 - poll= pkb + log (Nhy) [Hendersonneq. for Basic pot = 4.74 + lag F Pola 4.74 - 0.30 = 4.44 PH= 14-4.44 = 9.56

# On addition of 2002 NHg capt till cae & NHght cu can 10x03 20x0 l mmolen Basic Buffer polt= pkb + log (NH) = 4.74 +log 2 = 5.04 pH = 14-5.04 - 8.96 # On addition of 30mL NH Caps & Helcap - NMy capt cu cap loxo.3 30X0l t=0 mmolen tal (NAUT J = 3 note = RM Due to catainoic hydrolysis pH = 7 - 1 pky - log c = 4-1 (4:44) - 1 dag (3) 7 -2.34 +0.562 = 5.192 # t o tee on addition of youL NHg cay + HClip 2 Nhy latet cap mimolen 3 4 - - - 1 3 3 [HT J = [hu] - Immote = 16M pHa 1.7 50 mL ch,