Question

A 35.0-mL sample of 0.150 M acetic acid CH,CO,H is titrated with 0.150 M NaOH soldtion. Calcua pH after the following volumes of base have been added: (a) 0 mL, (b) 17.5 mL, (c) 34.5 mL, (d) 35.0 ml (e) 35.5 mL, (f) 50.0 mL

Answer 1

- a one ka of acetic acid a 1-88105 the chcool 0.150m CM₂ coo tiho I O * oxx * 0-150 - ka Ceh.coo) [Hot] [CH₂COOH) 1-88105 x2 0-150 - x² +1-8x10x - 2.7x10 zo. using quadratic equation solver we get X2 0.0016m a Chot) 5 p^ + clo, Ch) 280 17-5mLt calculate ealovalence point volume of wooH first moles of acetic acid & moles of Naoy 0.150 m x 35-0mL : 0150m xve ve 2 35-0mL so 185 mL as eo hak - eacralence point volume. so [ch cooH) 2 [crfcos) and solution acts as buffer use henderson Cavation. plos + top [сH с соб) (cthcool) pH = plato pika : tog Ka 2 -log(1. 805) 2 4.74 - pH 2 4.74

Base) c) 34 5 MLT moles of Chcool E 0.150 mx 35-0 mL z 5-25 mmoles moles of Naon added 20150m x 34-5mL 25-175 mmiles Chcoon & NaoH c h.cooNa the I 5-25 mmoles s-125 mmoles o e -5-175 -5.175 S-175 E o.075 mnoles o s.175 mmolkes as ratio d [6 10 , the solution not at as both for but we can use henderson Cavation, considering x Echange) is very small on a pla + log (6) - 4.74+ log [5175) To-075] ? 6-68. 350ml equivalence point. So all ch, coon is converted to choo Cathroo) a moles od z coat Cintat Total volume 0-180m x 35-0ml Joomt 2 0-0750 m. chycoo + the 7 (H₂coot ton 0-0750 C & G cotão reglect x as & value is very leads - K 2 roxloy 256 1010 1.8x100

دعوة ع ko 2 (CH₂ cooh) [one) (Chacool 5-6x16 10 2x² 0-0750 can = x 2 6.5x10 m. pon 2 5-19 - PH2 14-00-5-19 18-81 2 ier 35-5mL7 excess Naot Con a Total moles of Nach - moles realed! - Total volume 0-150M (35.5 - 35.0mL 35-5 +35-0mL 2 0.001 m. Con from CH₂ coő is very less and neglect = por a log [o-ool 3.0 | = e^ = 14-b 3-5 = t1-0 so-0mL 7 - Corect otrom (50.0-35-0 mL 35-0 +50-oml = o-0265M pon 2 1.577 2 14 - 1577 2 12-423. le