Question

(8. Calculate the pH for the following cases in the titration of 25.00 ml 01 0.200-M acetic acid, CH3COOH(aq), with 0.200 M NaOH(aq): (a) before addition of any NaOH(aq) (b) after addition of 5.00 mL of NaOH(aq) (c) after addition of 12.50 mL of NaOH(aq) (d) after addition of 25.00 mL of NaOH(aq) (e) after addition of 26.00 mL of NaOH(aq) QlCalculate the pH for each of the following cases in the titration of 35.0 mL of 0.200-M methylamine, CH3NH2(aq), with 0.200 M HCl(aq): (a) before addition of any HCl(aq) (b) after addition of 17.5 mL of HCl(aq) (C) after addition of 34.9 mL of HCl(aq) (d) after addition of 35.0 mL of HCl(aq) (e) after addition of 35.1 mL of HCl(aq) 10. Sketch the titration curves for Problem 8 and 9. Label the initial point, midpoint, buffer region, and equivalence point on your curve.
read #10

Answer 1

Solution solution: A @ . volume do acehi acid = 25ml. Molasity of acetic acid= 0.200 M 1. Ka of chocoon = 1.8xions before addition of any Naun .e pune acetic acid 8o acelic and is a weak acid which dissociate . as follows - CH3 coon 0.2M 0.2-8 ka = CH₂ 600 +ht 0 x x 0. 2k [in coo*] [*] [chocoon = xxx 10.2- since a is small 1.8x10" honce 0.2-x=012 18x105 = x2 . x = 0,2X1,8 X105 x = 5 1.897 xio? m. :(1+] = 1.899 X10'3 m pha - Pog(n) = - Boy (1697x103) (pH = 2.72 | Scanned with CamScanner

LLLLL 1 6 5.ome Naon addition . Molusity of NaOH = 0.2m moles of Naon added = molasity & Volume = 0.2 x 5 ml ! = 10 mmole initial moles of acetic and = molarity & volume = 0.2 x 25 mL 5.0 mmole remaining miley y acid = initial mole – reached mole 15.04 10) = 4 mmole moles of (Hycoo (salt) formed = 1.0 on mole It will form a buffer solution ph is giren as pH = pka + log [cm3 coo] [chocoon ] pH = -log ka + log (1 mmolle ph = 4.74 + dog (4) pH = 4.74 -0.6020 lipn = 4.14 ICS Scanned with CamScanner

12. 5 mL of addition Naon This is the case of half equivalence point I ph='pka = 4.74 Calculation: moles of Nadh added a medasidy X Volume = 0.2 x.: 12.5 = 2.5 mmole mole of Chacoo formed = 2.5 m male mole of chacool remaining = 5-2.5 0 = 2.5 mmole [ chacron] pm = pkn + gry [cenzer I .pH = 4.74+, Jay 2:55 [ph = 474 | 25 ml Naon added. at the equivalence point at this stage all chacool will convey into (H₂ coo- moles of cha cdo fermed = 25 ml X0.2 = 5 m male Total volume = 25 + 25 = 50ml. concentration of CH360o = mole CamScanner Scanned with Volume

Concentration of Cllg Coo son - 5 m mode 50 mL ря , с ов" 4. Да kеm а При = 1 + + Рka + ; 2 4 ) wrLлло (= conce thahon 4 «Изоо" При = + 2x4 эч + + Зеј (61) р. a+ 2 - 3) + ; x (-4) | Рут 4-%7) following equilibrium exist in solution of chzcio- смусов" + Мъв е си; соли + игр — } И 0:1 . . o o Or) - . К ? 0.1 - Х • L since nis small. . k4 04 ДҮР ka Lonj= ✓ o! X kW al X kw E4 1. OX09 from this equation, you can find the value of ut, and hence can easily calculate the value of ph, Answer will be same. Scanned with CamScanner cs

Titration Curre: - - a J ..... - 4.744 4 + w - 2.72 ! !26ml 12.5ml 25mL o . Wolume of Naon added 26.0 ml addition moles of Na on added = 26m. Xo.2=5.2 mmole extramale of. Naon added = 5.2.-5.0 P = 0.2 nmode tohel volume =:26 +25= 51 me. concentration of [or ] = 0.2 m male - 51 mL = 0.003921 m (n+] = loxioly Cou] - 1o X10''y 0.003921 rs Scanned with (94+] = 254.99 X10-14 = 2.549 X10 -CamScanner -12

ph= -log(nu) Finlay ( 2.54981012) PT= 11.59 1 S Scanned with CamScanner