Question

Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.190 M pyridine, C5H5N(aq) with 0.190 M HBr(aq):

a.) Before addition of HBr?

b.) After addition of 12.5 mL of HBr?

c.) After addition of 15.0 mL of HBr?

d.) After addition of 25.0 mL of HBr?

e.) After addition of 34.0 mL of Hbr?

Answer 1

pH formula's for salt's weak acid + strong baise salt tho pH at { pk w tp ka +Log e] strong acid + weak base salt tho pH = [ pk wapkb_loge]

I < Ky & SHANE 17X10 9 altsN+ HO H H + oH' 0.19 A O -* 0.19 * [est Nit] Lott] [5H₃N] +X 1-2810a 0-19- 1-28109 (0.19 -x) = x² 0-323 x109 - 1:721096x)2 x? x?+1.7x109 CX) - 0.323 x109 20 + bx + c = 0 -bt b²-4ac x= 2a - 1-28109 11C1-7X007) 4(1)(-0.323 x109) 261) Į - 0000 360555 -1-78109 -1-28109 +0.0000 360555 20.0000 180 269M Х. 204] = 0.0000 180 2.69M pot-tog 2014] 2 -log (0.0000 180269) 24-7441 pH alu-p Ott 214-4.744129-2559

® stan molarity 2 0.19M volume 225 ml=0.0252 HBr molarity a o.19M volume 12.5ml=0.01252 no of moles o.19 X 0.0 125 -0.002375 20.19 x 0-025 -0.00475 HNT HBr SH, NH Br 0.00475 1-0.002375 0.002375 -0.002375 l o +0.002375 a-002375 10.002375 total volume = 0.025+ 0.0125= 0.03752 -0.0633M -0.002375 EGGH,N] = 2 2 = 0.0315 0.002375 IGH NHB) - 0.0375 - 30-0633M k of Catena 12X109 pkb= log kb= log (1-7 X10 9) = 8.27 pota pkb tloq [sths NABY] ng [satsn] 0.0633 PCH 8-17 +10g .0022 a 8.77 - 14-d17a5-23 pt = 14-Pon

HBr molarity = o- LAM GHN Molarity oLaM volume = 15ml = 0.0152 volume - 25 ml=0.0252 no- of moles= 0.19 x 0.015 no ob moles= o.19 X0-025 =0.00285 20.00475 C.H. NH BY HN + HBr - Co-00475 0.00285 To e-0-00285 1-0.00285 T +0.00285 Elo.ool9 I o 0-00285 Tofal volume 20.025+ 0.01570-042 0.0019 - 0.0475M 25tJ² 0.04 0.00285 -0.07125M L 9 N8 33] 0 0 4 {H NH Br] polf- pkb tog [ çats N3 0-07125 a 8-946 pOH = 8.77 + log 0.0425 pH = 14- pot = 14- 8.94625.054

H₂N Molarity - O-lam po HBr molarity 0.19M volume = 25ml = 0.0252 volume = 25ml=0.0252 no ob moleso. L9X 0.025 no- of moles-0-19X0-025 -0.00475 -0.00475 GHANA HBr - GH_NHBY weakhase strong acid salt Io.00475 0.00475 o cho.00475 -0.00475 +0.00475 Elo Lo 0.00415 Total volumeao-025 +0.025= 6.052 0.00455 & CH, NH Br] = -=0.095M=[E] > 0-05 pkW 214 okb=8.77 For this salf pH formula pH = + 1 pkw _pkb log e pH + [14-8.17-109 (0.095)] pH = 3.126

GHSN Molarity o-LaM HBr Molarity = 0.19M volume = 25m1 = 0.0252 volume = 34m120.0342 no of molesa 8. L9X0.025 no of moles - 0.19 Xo-034 = 0.00425 -0-00 646 çf Nt HBr galt NHIBA I I0.00475 0.00 646 I clo.00475 -0.00425 El 1o.oo17 Total volume = 0.025+ 0.034= 0.0592 o-ool1 0-059 ElfBr] a [Ht72 [47] = 0.0289839 plfa -log[wt} = -log (0-0 28 983) pH = 1.53786