Question

Titration of 100.00mL of a 0.1000M solution of a weak acid HA (K=1.80x105) with 0.100M NaOH solution. Using our knowledge of how weak acids react with strong bases, determine the pH at the following points. a) before addition of NaOH c) after addition of 20.0 ml NaOH e) after addition of 50.Oml NaOH b) after addition of 100 ml NaOH b) after addition of 10.0ml NaOH d) after addition of 30.0 ml NaOH f) after addition of 80.Oml NaOH h) after addition of 110 ml NaOH Titration of 100 mls 0.1 M Acetic Acid 0 + 0 20 40 60 80 100 120 mis 0.1 M NaOH added

Answer 1

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Solution chiren that 100 ml of BiMaletic and solutions Loncentration of NODH = 0.100M since acetic acid is weak auid Therefore Hacoon Chacoo the initial conc. oil Conc. at egnist 0.1 0.12 DOO Where a-degole sotho n crgCODH ko = [CH₂ coo) [nt) – ox dx 0-1 of Сси, соон) since ka=1.8x10 109 Therefore in 1 ka = 0.10² (H+ = 0.12 = fo.lxka = Jox1.8 x 15t 1:34*10-2M moles of Ht - Molarity & volume ist - 1.34 x 10 3 x 100 = 1.3410 a) before addition of NaOH bu = -log Cost] = -log (1.34x103) = 3 - log (634) - 3 -0.127 by After addition of 10 mL Naon = 12.873 ud 1011=nna 0.1% 10 , 0.001 moles of NAOH left = 0.001 – 0.000134 = 0.000866 Mularity of Naon let Mon 0.000866 x 1000 = 0.0866 -0.0079 pon = -log(on) = -log (0.0073) = bH + pun = 14 2) ĐH - 14 - 20H - 14 - 2) - (11-9 c) After addition of 20mL NAOH Noon = 0.1 X2 = 0.002 moles of on left = 0.002-000134 -0.001866 1000 119

120 (1353) 21-01 Molanty of Naon left - 0.00 18 6 6 1000 = 0.01555 рои - - logon): - PH = 141.81 - 112.13] After addition of 30 mil. Hoon NADH = 0.1 x 30 = 0.003 d) Nauh left = 0.001 - .000124-0.002866 Molarity of Neon left = 0.00 26 66 x 1000 = 0.022 bona alig Con] - log(0.022) - 1.657 bH - 14 - 1.657 = 112.343) e) After Addition of so ml Neon Noon - 01x59 -0.005 LOV V TS MNADH left = 0.005-0.000134 = 0.004866 Molarity of Non left = 0.004866x1807 = 0.01244 Pon=- log [on] =.- bag (0.03249) - 1.49 DH = 14-1.49 - 12:41 f) After addition of somL Na, oh Naon 0.1 x 80 - 0 UDG Noon left = 0.008 0.000134 = 0.007866 Motarty of Nauh left = 0.00786 6 x 100 = 0.0437 bons-log[on] = -log(0.0437) = + 1.36 PH = 14-1.36 = 1+2 64 [12.645 9 After addition of 100 ml Naon Naon = 0.18 100 = 0.01 n Naon left = 0.01 0.000 13 4 = 0.005866 It - 0.00986 x twoo = 0.04933 Molanty of Naon left = 2000 poh - wy(0.04933) = 1.3 PH = 44 - 1.3 = 12.7 200

) After additions et 110 mL Noon Noon = 0.1 x 11g - 0011 1000 NaOH left = 0.011-0.000134 - 0.010866 Molarity of NADH left = 0.010866 x 100 = 0.052 pona-log (0.052) - 1.284 bu = 14-1-28 4 - 112.716] neclarity of Nadh left - 210
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