Question

Consider the titration of 40.0 mL of 0.0600 M C2H5NH2 (a weak base; Kb = 0.000640) with 0.100 M HCl. 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) 26.4 mL pH =

Answer 1

And given: volume Molarity of (245NH2 = 40mL of C245 MH2 = 0.06oom =0.0 Goomall Molarity of HCl = 0.100M = 0 loomalle (a) volume of Hcl = ome Constructing ICE table for the reaction CoMқ км, + K20 = СКss и + он — Inittal 0.0600 Change -x Equilibaum Co.0600-2) Kb = [C2H5NH37] [or] [C246 MHz = ? 0.000640 (0.0600-x) m' 0.000640 0.0600-1) = 3.84x135 - 0.000640x m +0.000 G 40% – 3.84 X10% =0

Solving for a a= 0.00588503 Thus, (on] - 0-00588503 pok= -log Core] = -log (0.00588503) = 2.230 25132 pH = 14-pon =14 – 2230 25132 = 11.7697487 = 11.8 (6) giveni volume g Hecl = .0mL 6.0x1632 Moles og Hela rolume of Hclx Molarity of nch = 6.0 xiok x .loomad = 0.0006 mol volume of C2H5 NH2 = 40mL = 40x103 L Moles og Calg Nike a rolume of GHG NH2X Molasity of Ca HG NH2 = 4oxiok x 0.0600 mol

= 0.0024 mol to the ² C2H5NH3 reaction + on- Constructing ICE table C2H5NH2 +420 Initial 0.0024 (moles) Change - 0.0006 Equilibalum (0.0024 -0.0006 +0.0006 to 0006 0.0006 0.0006 Moles of Calgal Hgt = 0.0006 Total volume = volume g C2 MyNWz+ Volume of - Kel 40mL + 6mL = 46mL = 46x103 L [{245 Nhat] = Moles of C2 HG NHgt Total rolume 0.0006 mol = 0.0130 434783M 46x10² L [Calg hH2] = Moles og C2H5NH2 = (0.0024 Total volume 46X10 -oooal)

-0.0018 468103 -0.0391 304348 Poll= Pkb Applying Henderson- Hasselbalch equation: Pol= pkb + log [C2kg MH37] [.45 мм) I K6= 0.000640 Pk b= -log (Kb) = -log(0-000640) = 3.19382003) Thus, pOH = 3.1938 2003 10.013043 4783 O 10.0391304348 = 2.71669878 PH= 14-pon = 14 - 2.71669878 = 1.2833012 = 11.3 volume of Hel = 12.0mL=12 x1c3L Moles of Hele 12x1őkx 0.100 mal = 0.0012 mal (6)

Thus, Moles ginclo / Ricles of Chanta) This is midpoint Thus, pola pkb = 3.1938 2003 PH= 14 - pon = 14- 3.1938 2003 = 10.80618 = 10.8 (d) rolume og kcl = 18.0mL = 18x103 L Moles og Hal= 18x10²4x o looma - 0.0018 mol Constructing ICE table for the reaction Cath NH2 + H2O È C2MG NH3 + out Inital 0.00 24 to.o0l8 Change -0.0018 +0.0018 Equilibrium Co-0024 -0.0018) 0.0018 0.0018 Total volume = volume of teelt volume of (moles) 2 Hg NH2 = 18mL + tom L- 58mL = 58x10L

(C215042] = (0.0024 -0.0018) mol 58x1034 = 0.0006 M20.0103 44 8276 M 58x1ő3 [Ca H5NH3+] = 0.0018 Mc 0.0310 344828M 58x103 Applying Henderson- Hasselbalch equation: pola pkbt log [C2 HGN HET] СК и) = 3.1938 2003+ log (0.0310344828 0.0103448276 = 3.67 099128 pH = 14-pon = 14 - 3.67094128 = 10.3290587 = 10.3 - @ volume of Hal = 24 mL = 24x10² L Moleg g Mal = 24 xTokx 0.loomallie = 0.0024 mol

This is equivalence point Constructing ICE table C2H5NH2 +420 Chittal for the reaction: Ą C2H5rikat ton (moles) 0.0024 Change -0.0024 Equilinsium o +0.0024 0.0024 +0.0024 0.0024 [c2hg hkg t] = Moles of Calg Nhat - Total volume Total volume = volume of C245NH2+ Volume of Hal = 40mL & 24mL = 64 mL = 64x103) = 0.0024 mol = 0.0375 M 64*103L table for the reaction ² (245NH2 + H₂ot Constructing ice C245 Nugt +420 Initfal 0.0375 -Mi Equilibrium 0.0.375-8 change th tu

Кь - t-00xo' aliooxic14 о: ooo 640 11. 5625 xia" Ка: Сери, чи,) Cugo) Со Mқми,t) 2 (-5325х1 = О-о за5- - (3 — Б. 85a315xo 20 т? (oo375 -x) (+5625xio" - 5-85 4395 xio3 – 1:56 25xio' - +/- 525xTo'x - 5-85азақ хто? -0 Solving for , 1. 1. 4545 8 x 1 Thus, Cugo) - -65+58 хію? Рие — lea Cugat) = -los (а- G568 xio?) 6. (16o 1863 — С-12

(m volume of Hal = 26.4mL = 26.4x103 Moles of real = 26.4x10²4 x o. looma) = 0.00264 mol Excers Moles of rel= Moles of real- Moles of Calg NH2 = (0.00264 - 0.0024) moll = 0.000 24 mol (4301) - Excess moles opel Total volume = 0.000 24 mol (26.4+ 40)x103 L = 0.00024 -0.003614457831 66.4x103 pke -log (1201) = -log (0.0036 14 45783) = 2.44195684 = 2.44