Question

Consider the titration of 60.0 mL of 0.0400 M HONH₂ (a weak base; K_b = 1.10e-08) with 0.100 M HIO₄. 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) 40.8 mL pH =

Answer 1

Aug) 3 оfurn * * ginen: Molarity of 1104 = o. 100M volume 9 N4201 - 6000m L=60.0x16L Molasity g MH204 = 0.0400M (a) volume of H104 = omL ta Constucting ice table for the reactions NM2 OH Cag] + H20 (Nyomt (aq) + or caq) Inpttal 0.0400 ta change an Equilibrium (6.0400-1) kb = 1.10x108 = [MH2O Hat] [on] - [Nhaon] (6.0400-2) mc (0.0400-x) HH10x108 = 4.4xiolo - 1.loxio ge? +1.60 XBox – 4.4x100 =0 solving for ny = 2.09707 X105

Thus, Cou')-*-.2.09707x105 polo -log Con') = -log(2.09707x165) = 4.67838707 PH = 14 - pon = 14 - 4.67838707 = 9.32161293 = 9.32 (b) volume of 4104 = 6.0mL = 6.0x10²L Moles of HiOA = Molaapty of H10 4x volume of Hoa = 0.100mol * 6.0 xidy - 0.0006 mol Moles g 14204 = Molasity ay NM1201 Volumeg NH2o4 = 0.0400 mal x 60 xToy = 0.0024 mol Constructing Ice table for the reaction: W 120H caq) +1200) = Nha ong cag] + or (09) Inittal 0.0024 Change -0.0006 +0.0006 +0.0006 Equilibrium (0.0024 -0.000b) 0.0006 0.0006

Applying the Henderson - Hasselbalch equallon pol = pkb + log (But to [B] whose [but] - [MHz out [B]= [NHon] Pk6= -log(Kb) = -log (1.10x108) = 7.95860731 (But). Moles of NH30ng7 Volumeg Nhy ont volume g 1104 = 0.0006 mol (bomLt 6mL) = 0.0006 mol 666103L = 0.00909090909 [B]= [NH2 onl= Moles of NH2O volume o NH 2 ontvolune grica = (0.0024 -0.0006) md = 0.0272727273 66x103L

Substituting the values in ean POH = 7.95860731 +log (0.0090909099 0.0272727273 = 7.481 48605 PH= 14- poH = 14 - 7.48148605 = 6.51851395 = 6:52 (c) volume of 410 4 = 12.0mL = 12.0x10²L Moles y H104 = 0.100 mel x12.0x1o'y = 0.0012 mol Thus, Moles og nooq = 1/2 Moles og NH 2 OH This is midpoint. At this point, pon = pkb = -log(Kb) E-log (1.10x108) = 7.9586 0731 PU= 14-pol = 14 - 7.95860731 -6.64139269 = 6.04

(d) Volume g 1104 - 18.0ml = 18.0x103L Moles of Hoq = 0.100 mal x 18.8x10²4 = 0.0018 mol Constructing ICE table for the reaction NH2OH can + H20 (1) NHL Chat (aq) + on (aq) Initial 0.0024 Cmoles) Change -0.0018 Equilibrium 0.0024-0.0018 + 0.0018 +0.0018 0.0018 0.0018 [NM2 Onzt) - 0.0018_mol = 0.0018 mol (60+18)xio' 78x10²L =0.0230769231M [urbon] -(0.0024 -0.0018) mol (60t18) xio3L = 0.0006 mol = 0.00769230769M 78x103 Substituting the values in ean.(1), pOH = 7.96860731 t log (0.023076923 0.00769230769

Pon= 8.4357 2857 PHI 11 - POW - 14 - . 435+18 51 = 5.56427143 = 5.56 (e) volume of 1104 = 24 mL = 24x16L Moles g 1104 = 0.100 mol x 24x1osy = 0.0024 mol Thus, Moles of Hloq = Moles of Whoon This i equiralence pant.

Constructing ICE table for the reaction: SW2OH (aq) + Học (49) » N, OH + H20(4) 0.0024 Inittal 0.0024 Cmoles) Change -0.0024 Eçmillbrim o -0.0024 +0-0024 0-0024

[amongst = Moles of NH 2 O Hat volume g 11047 Volume oj NIKOM = 0.0024 mol = 0.0024 moll 24+60) xio3L 84x18L = 0.0285714286M Constructing Ice table NH2 Chat +420 to the reaction: Ô NH2 OH Cag) +H30tcaa) (aq) Inital 0.0285714286 Change -x Equilpbrium (0.0285714286-20 tu X ta X ka- l.oxió! 1 = loxio't -9.09090909xio Tlox108 Kb ka= [MH2on] [r3ot] = n2 CAM2 OH21] (0.0285714286-~) =9.09090909x10

M (0.02.85714286-w) 9.09090909 x107 = 2.9994026 X108 - 9.09090909xione my?+ 9.09090909 xiołne - 2.5974026x108-0 Solveig for a ME 0.000160711 'Thee, (1301) - ~= 0.000160711 PHE-log (1201) = -log Co. 000160 711) = 3. 7939544 = 3.79 (f) volume of H104 = 40.8ml -40.8*10'L Meles og Heads 0.loomolx 40-8x Toy = 0.00 408 mol Excops moles g 1104 - Moleg g MICA - Moles of Noon -0.00408 - 0.0024) mol = 0.00168 mol

Tatal volume Volume g 1104+ Volume oj Nyon (40.8+60) mL = 100.8x10?L (M304] - Excasse moles g 4104 Total volume 0.00 168 mol = 0.0166666667 100.8x10%L PHE -loa (kgo) = -log (0.01666 66667) = 1.77815125 = 1.78