Question

Consider the titration of 40.0 mL of 0.0600 M HONH2 (a weak base; Kb = 1.10e-08) with 0.100 M HBr. Calculate the pH after the following volumes of titrant have been added:

(a) 0.0 mL pH =

(b) 6.0 mL pH =

(c) 12.0 mL pH =

(d) 18.0 mL pH =

(e) 24.0 mL pH =

(f) 43.2 mL pH =

Answer 1

here. The reaction is n ot NH₂OH + HBr - whoth + Br Now port of the 40.0 ml 0.0600M HONH, solution is to be determined ka [NHOFH ₂ ] [OH) 2o1 [NH2OH] Llet [WA₂0H2] = [or] =] So koz [NH, OH] 2) Sz 11x103) * 0106 a) 32 2057 X 80 M 82

5.7 So [ot] : 2.57×105 m. pon. -S. (2.57.2165). 2-4.59 pt 2 14-polt gul 2 so come pra 9.4)

5,06636X105 - [HT 0.52 Схо. - tion in 6 ml HBr -4 CX10 holes Numb Number of moles of oth im innom NHOH 40x2.57110 miles 1000 21.028x10 moles As HBr is strong acid all the 1.028x10 moles will react, so remaining moles of Hta (6x184 1.028x106) moles 25.98972x 10-4 moles So [H+). 5.98972x1074 400 (0.50 +0.06) Ma 1.302x10-3M PH₂ 20885

mole Numbers of moles of Ht in lame of Abra 0.100212 1000 2 12x10-4 moles Number of moles of NH₃OH in nome a 1.028 X10 e moles Moles of At inexeers (12x1074_1.028x106) 21:10897 x10-3 moles [ht] . 1019897 X10-3 (0.4+0:12 th 2 2.3058x100 PH 2. 2.637 i mitt toute d) Number of moles of itt in 18md HBr 0.1% 18 moles 2 18x10-hmotes Number of moles of NH oH innomla 1.028x10t moles Moles of excers Ht₂ (18x109100288106) moles 2 1.798 x 103 moles [At] 1.7988103 Couto.18 M2 3.1810-3m. PH₂ 2.508

e) Number of moles of tt in 24.0 ml HBr 0.1 X 24 24 x 100 de 0.1824 1000 24 KID Mes Number of moles of NH oH in nomla 1.028 x 10 emoles moles of exeers Hta (24x10 6 1,028x102) moka - 2.398 810-3 moles THE 2.398x10 m 2 3.7468x10m (0.4+0.24) PH2 2.426 b to Number of moles of itt in 43.2 ml- Orlx 4302 Toso Moles = 40320109 mores Number of moles of NH₃OH in nome 21.028 x 10 cmoles Moles of exeurs Ht, (4.32x10 3 _ 1.028x156) moles 2 4.3189x10 3 moles [HTZ 4.3189x103 (o.u+ 0.432 Mo 5.190x103m. PH 2.2848