Question

CHE 172 Acid-Base Equilibrium Worksheet 1. Identify the conjugate pairs in the following reaction: HC2H302(aq) + H2O + H20*24) + CH3O2 (aq) 2. Based on the Kb of the following weak bases, which is the strongest base? C6H5NH2 HONH2 H2NNH2 C2H5NH2 Kb = 4.3x10-10 Ko = 1.1x10-8 Kb = 1.3x106 Kb = 6.4x104 Which has the strongest conjugate acid? 3. Calculate the pH of the following solutions: a. 0.25M HBO b. 0.25M Ba(OH)2 C. 0.25M HCN (K. = 5.00x10-10)

Answer 1

HC₂ H₂O₂ (aq) + H₂011) - H₂ otcaa) + Calz oglan Here H62 H₂O donating 4t so acting as acid whereas H₂0 accepting it (proton) acting as base and after accepting foom H₂ ot that means Hot is conjugate acid. HQ H₂O₂ (aq) + H₂O e Acid t H₂ot cas) conjugate Base + C₂ H₂ O₂ (aq) conjugate base a id. (2 kb values Greater the Ky value basic character. stronger the kg = 4.3 8100 Co Hg NH₂ OHNH₂ H₂ N NH₂ GHS NH₂ kb : K6² 113610-6 6.4 x 10-4 ko value ie CHg NH₂ has highest Scanned with est base. CamScanner

having weakest base here lowest Ko value strong est conjugate strongest base is GHS NH₂ So form the acid. ? C2H5NH₂ strongest and Strong est conjugate acid : C6H5NH₂. pH of following solution 0.25M HBr Hydro bronic acid CH Bed is a strong acid and will dissociate completely in water to Ht and Bre for every mole of HB4, there will be I mole of Ht, so concentration of u will be same as concentration of Br ... [+] = 0.25M pH = -log [ht] pH = -log (0.25) pH = - €0. 602) pH = 0.602 of 0.25 m solution of HBr = 0.602 The pH Scanned with CamScanner

0.25M Ba(OH)2 Ba(OH) - Bar + 200 - since ratio of Ba 6H), and on is 1.2. the concentration of Balon), is 0.25M so the concentration of on would be 2x0.25 = 0.5% So, for 0.50M as Buloh basic рон - log lon) - = - log [015 ] -- 10:301) pon = 0.301 we know pH + pon = 14 ph = 14- pon pH = 14- 0.301 pH = 13.69 ph of 0.25 Babuz is 13.7 0.25M HIN C Ra= 5.00 x 1010) - HCN caq) + H₂O cas) = CN caq) + H₂ot aq) - o +x -Initial conch 0.25 concentration at home - Equilibrium 10.25-x] cs Scanned with CamScanner oko = 500810 + x x m xm

ka = [H₂ot] [w] [HIN] 5x10-10 = x² 0.25-x 5 810-10 4 x² 0.25 X y small X = 11.1 x 10-6 so at Equilibrium Hoot] = [EN] = lll xloom [HCN] = 0.25m. pH = -log (11.1x106) =-(-4.95 pn = 4.95 H. ph of 0.25m hen at Ka = 5:00 81010 ú 495 0.25 m Ca HSN (R = 1.6810-9) - CGHS N + H₂O - (g Hs Nut tou weak base in water , weake bases like Pyridine is dissociate a base B + Scanned with CamScanner Ho - But you

According to Handerson Hasselbalch equation, for weak bases pon = pkot dog (B4) [3] К. : Ви*] CouT [3) [ои г (s) К, Би:) : [s) К, (o-25) (р. 6x1019) [он'] - 2x to" Сс Сон:) : Lan') for pH, Therefore, pon - log cons ри: 14- роу Ри: 1ч + lea (2 x 105) 1ч - ч6 84 2 ч. 301 ри 4 б. 25 m Cs Иs (k= 1.6 x 70 °)= -30/ Scanned with CamScanner

(1) Ка? Бо 65 м) соlution o, C, Нs си, tоо и , н, ң, соби — си, си, соо + не Ка: (Сн, и, соо] Си*) [с, и, и, соти) 9ім сен, си, со и) : osof5 м) 2lum ри - 2 - 68 ри: - оз Си*] o [mt] = antilog GPH) 2 am o4 (-2 - 69) - (2-оя х 10-3) м K, 3 (204 x 103) * бовь ke : у 37 xp - 6 б: 085 5, 13 5 x 10 Scanned with CamScanner

Audic, basic or Neutral Naf Basic ( it is salt of strong base Naon and weak and HF) salt of weak base NH3 strong and hBr. N Hy Br ir a and Audie . Audic Su 1.2 strong acid and strong salt of base solution ( ... is Neutral a stronger aid. (a) H₂ H₂ Te Ho Te than higher Te lesser H₂S less bond bond dissociation .. acidity of enugy requuid and releasing I enthalpy. th te is to break (4+] easier Overall order to < (H₂S) <H₂ Se <Hz Te Scanned with CamScanner

HBr is Bei is more is the stronger a cid because dectronegative than se more stable than the Br. so HBrO₂ Mbro HBro Рококо асоц 4o-Br Bee -04 Halous halous O= Br -OH sulong est achd Chalic acid ) (Bromic acid) Increasing Increasing increasing (Hypo acid) acid) Bromous aud) Снуроbломоил аса]. number of oxygen attached and strength is due to electronegatively Scanned-with- CamScanner